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1, then ω(x)1−p ∈ Ap0 by (ii). Using Theorem 1.3.3, it is clear that there exists an ε > 0 such that 1+ε Z Z 1 1 1+ε ω(x) dx ≤ C1 ω(x)dx |Q| Q |Q| Q and 1 |Q|
Z
Q
ω(x)
(1−p0 )(1+ε)
dx ≤ C2
1 |Q|
Z
Q
ω(x)
1−p0
dx
1+ε
.
hold at the same time. Thus, p−1 Z Z 1 1 1+ε 1+ε 1−p0 ω(x) dx [ω(x) ] dx |Q| Q |Q| Q p−1 )1+ε Z Z 1 1 0 ≤C ω(x)dx ω(x)1−p dx ≤ C. |Q| Q |Q| Q
28
Chapter 1. Hardy-Littlewood Maximal Operator
This shows that ω 1+ε ∈ Ap . (xi) Since ω(x) ∈ Ap (1 ≤ p < ∞), using H¨older’s inequality for 1 + ε and (1 + ε)/ε, where ε is fixed by (1.4.6), we have 1/(1+ε) Z Z 1+ε · |E|ε/(1+ε) ω(x) dx ω(x)dx ≤ E
E
1/(1+ε) Z 1 1+ε = |Q|1/(1+ε) |E|ε/(1+ε) ω(x) dx |Q| E Z C ≤ ω(x)dx|Q|1/(1+ε) |E|ε/(1+ε) |Q| Q |E| ε/(1+ε) = Cω(Q) . |Q|
Thus we have showed that ω satisfies (1.4.7) if taking δ = ε/(1 + ε). Remark 1.4.2 Let ω be a nonnegative locally integrable function on R n . We S say that ω ∈ A∞ if ω satisfies (1.4.7). The property (xi) shows that 1≤p<∞ Ap ⊂ A∞ . However, it can be proved that the S above containing relationship may be reversed. So, we have indeed A ∞ = 1≤p<∞ Ap . In this section, we shall see that Ap weights give a characterization of weighted weak Lp and strong Lp boundedness for the Hardy-Littlewood maximal operator M . Theorem 1.4.2 (Characterization of the weighted weak type (p, p)) Suppose that 1 ≤ p < ∞. Then the Hardy-Littlewood maximal operator M is of weak type (Lp (ωdx), Lp (ωdx)) if and only if ω ∈ Ap . That is, there exists a constant C > 0 such that for any λ > 0 and f (x) ∈ Lp (ωdx) (1 ≤ p < ∞) Z Z C |f (x)|p ω(x)dx (1.4.8) ω(x)dx ≤ p λ n n R {x∈R : M f (x)>λ} if and only if ω ∈ Ap . Proof. Let us first prove that ω ∈ Ap is the necessary condition of (1.4.8). For p = 1, let Q be any cube and Q1 ⊂ Q. If denote f = χQ1 , then for any 0 < λ < |Q1 |/|Q|, we have Q ⊂ {x ∈ Rn : M f (x) > λ}. By (1.4.8) Z Z Z λ ω(x)dx ≤ λ ω(x)dx ≤ C ω(x)dx. Q
{x∈Rn : M f (x)>λ}
Q1
29
1.4 Weighted norm inequalities By the arbitrariness of λ < |Q1 |/|Q|, we get Z Z C 1 ω(x)dx ≤ ω(x)dx. |Q| Q |Q1 | Q1
Applying Lebesgue differentiation theorem (Theorem 1.1.2), we have that Z 1 ω(y)dy ≤ Cω(x) a.e. x ∈ Q. |Q| Q By the arbitrariness of cube Q again, we have M ω(x) ≤ Cω(x)
a.e. x ∈ Rn .
Thus ω ∈ A1 . 0 For p > 1, let Q be any cube. If we take f = ω 1−p χQ , then for any 0 0 < λ < ω 1−p (Q)/|Q|, Q ⊂ {x ∈ Rn : M f (x) > λ}. By (1.4.8) Z Z λp λp ω(x)dx ≤ ω(x)dx |Q| Q |Q| {x∈Rn : M f (x)>λ} Z C 0 ≤ [ω(x)1−p χQ (x)]p ω(x)dx |Q| Rn Z C 0 ω(x)1−p dx. = |Q| Q Hence by arbitrariness of λ, we get 1−p Z Z 1 1 1−p0 . ω(x)dx ≤ C ω(x) dx |Q| Q |Q| Q So ω ∈ Ap . Below we will prove that ω ∈ Ap is also a sufficient condition of (1.4.8). When p = 1, (1.4.8) is just a direct corollary of (1.3.3). Now let us consider the case p > 1. By H¨older’s inequality and the condition ω ∈ A p , we have p Z Z 1 1 ω(x)dx |f (x)|dx |Q| Q |Q| Q p Z Z 1 1 ω(x)dx |f (x)| ω(x)1/p ω(x)−1/p dx = |Q| Q |Q| Q p−1 Z Z Z 1 1 1 p 1−p0 ω(x)dx |f (x)| ω(x)dx ω(x) dx ≤ |Q| Q |Q| Q |Q| Q Z 1 ≤C |f (x)|p ω(x)dx. |Q| Q
30
Chapter 1. Hardy-Littlewood Maximal Operator
Thus, this shows that if f ∈ Lp (ωdx) then f ∈ L1loc (Rn ), and
Z
1 ω(x)dx ≤ C |Q| Q
Z
Q
|f (x)|dx
−p Z
p
Q
|f (x)| ω(x)dx .
(1.4.9)
So without loss of generality, we may assume that f ≥ 0 and f ∈ L 1 (Rn ). Applying the Calder´on-Zygmund decomposition (Theorem 1.2.1) for f at height 7−n λ, we get a cube sequence {Qk } such that 7−n λ <
1 |Qk |
Z
f (x)dx
for all
Qk .
(1.4.10)
Qk
By Lemma 1.3.1, we have that {x ∈ Rn : M f (x) > λ} ⊂
[
Q∗k ,
k
where Q∗k = 2Qk . Thus by the property (vi) of Ap weights and (1.4.9) and (1.4.10), we conclude that Z ω(x)dx {x∈Rn : M f (x)>λ} XZ XZ ω(x)dx ≤ C2np ω(x)dx ≤ 2Qk
k
≤ C2np ≤C
X
14np λp
k
Z
Rn
1 |Qk |
Z
k
Qk
|f (x)|dx
Qk
−p Z
Qk
|f (x)|p ω(x)dx
|f (x)|p ω(x)dx.
This shows that the Hardy-Littlewood maximal operator M is of weak type (Lp (ωdx), Lp (ωdx)). Hence we finish the proof of Theorem 1.4.2. Theorem 1.4.3 (Characterization of the weighted strong type (p, p)) Suppose that 1 < p < ∞. Then the Hardy-Littlewood maximal operator M is of strong type (Lp (ωdx), Lp (ωdx)) if and only if ω ∈ Ap . Proof. The necessity is a corollary of Theorem 1.4.2. In fact, since M is of strong type (Lp (ωdx), Lp (ωdx)), and is also weak type (Lp (ωdx), Lp (ωdx)). Then by Theorem 1.4.2, ω ∈ Ap .
31
1.4 Weighted norm inequalities
On the other hand, if ω ∈ Ap for 1 < p < ∞, then by the property (viii) there exists 1 < q < p such that ω ∈ Aq . By Theorem 1.4.2, M is of weak type (Lp (ωdx), Lp (ωdx)). That is, Z Z C |f (x)|q ω(x)dx. (1.4.11) ω(x)dx ≤ q λ n n R {x∈R : M f (x)>λ} Notice that if ω ∈ Ap , then for any measurable set E ∈ Rn ω(E) = 0 if and only if |E| = 0. So, L∞ (ωdx) = L∞ in the sense of equality of norms. Hence by the L∞ (Rn )-boundedness of M (Theorem 1.1.1), we have kM f k∞,ω ≤ kf k∞,ω . Using the Marcinkiewicz interpolation theorem (Theorem 1.3.1) between this and (1.4.11) we obtain Z Z p |f (x)|p ω(x)dx. M f (x) ω(x)dx ≤ C Rn
Rn
In other words, M is of strong type (L p (ωdx), Lp (ωdx)). In this section, we shall discuss further some important properties of A 1 weights, such as the construction of A 1 weights, the relationship between A 1 weights and Ap weights and its some application, etc. Theorem 1.4.4 (Constructive characterization of A 1 weights) (a) If f (x) ∈ L1loc (Rn ) and M f (x) < ∞, a.e. x ∈ Rn , where M is the Hardy-Littlewood maximal operator. Then for any 0 < ε < 1, ω(x) = M f (x)
ε
∈ A1 and whose A1 constant depends only on ε.
(b) If ω(x) ∈ A1 , then there exists f (x) ∈ L1loc (Rn ), 0 < ε < 1 and a function b(x) such that
(i) 0 < C1 ≤ b(x) ≤ C2 , a.e. x ∈ Rn ; ε (ii) M f (x) < ∞, a.e. x ∈ Rn ;
ε (iii) ω(x) = b(x) M f (x) .
32
Chapter 1. Hardy-Littlewood Maximal Operator
Proof of (a). By Remark 1.4.1, we need to show that for any 0 < ε < 1, there exists a constant C such that for any cube Q in R n Z ε ε 1 for a.e. x ∈ Q. M f (y) dy ≤ C M f (x) |Q| Q Fix Q and 0 ≤ ε < 1, we denote f = f χ2Q + f χRn \2Q := f1 + f2 . Then ε ε ε M f (x) ≤ M f1 (x) + M f2 (x) .
By the weak (1, 1) boundedness of M and Kolmogorov’s inequality (Theorem 1.3.3) Z 1 C M f1 (x)ε dx ≤ |Q|1−ε kf1 kε1 |Q| Q |Q| ε Z 1 ≤C |f1 (y)|dy |Q| 2Q ε ≤ C2nε M f (x)
for any x ∈ Q. ε Now we give the estimate of M f2 (x) . Clearly, if y ∈ Q and a cube
Q0 3 y such that Q0 ∩ (Rn \ 2Q) 6= ∅, then 4Q0 ⊃ Q. Hence if x ∈ Q Z Z 1 4n |f2 (y)|dy ≤ |f2 (y)|dy ≤ 4n M f (x). |Q0 | Q0 |4Q0 | 4Q0 This shows that M f2 (y) ≤ 4n M f (x) for any y ∈ Q, and so is Z ε ε 1 M f2 (y) dy ≤ 4nε M f (x) . |Q| Q
Proof of (b). Since ω(x) ∈ A1 , then by the Reverse H¨older inequality (Theorem 1.4.1) there exists η > 0 such that for any cube Q 1/(1+η) Z Z 1 1 1+η ≤ ω(x) dx ω(x)dx ≤ Cω(x). |Q| Q |Q| Q This implies that M (ω 1+η )(x)1/(1+η) ≤ Cω(x) for a.e.
x ∈ Rn .
On the other hand, by the Lebesgue differentiation theorem (Theorem 1.1.2) ω(x)1+η ≤ M (ω 1+η )(x).
33
1.4 Weighted norm inequalities
ε Therefore, if let ε = 1/(1 + η), f (x) = ω(x) 1+η and b(x) = ω(x)/ M f (x) , then we obtain the conclusions of (b). Thus we finish the proof of Theorem 1.4.4. Below we shall turn to discuss the relationship between A 1 weights and Ap weights. By Remark 1.4.1, if ω ∈ A1 , then we have the following equivalent form of (1.4.2) Z 1 ω(x)dx · sup(ω(x)−1 ) ≤ C, |Q| Q x∈Q where and below, sup is the essential supremun. On the other hand, it is easy to see that when p → 1 p−1 Z 1 1−p0 ω(x) dx → kω −1 k∞,Q = sup(ω(x)−1 ). |Q| Q x∈Q
Therefore, the A1 weights can be seen as the limit of Ap weights when p → 1. By the property (iii) of Ap weights, we have known that if ω1 , ω2 ∈ A1 then ω1 · ω21−p ∈ Ap . A very deep result is that the converse of above conclusion is also true. Theorem 1.4.5 (Jones decomposition of A p weights) Let ω be nonnegative locally integrable function. Then for 1 < p < ∞, ω ∈ A p if and only if there are ω1 , ω2 ∈ A1 such that ω(x) = ω1 (x) · ω2 (x)1−p . Proof. we only consider the necessity. Since ω ∈ A p , by the property (ii) 0 ω(x)1−p ∈ Ap0 . First let 1 < p ≤ 2, then s = p0 − 1 > 1. Take a nonnegative function 0 u0 (x) ∈ Lp (Rn , ωdx), let us construct a function sequence {u j (x)} by the Hardy-Littlewood maximal operator M as follows. uj+1 (x) = [M (usj )(x)]1/s + ω(x)−1 M (uj ω)(x), for j = 0, 1, · · · . Then we have the following two conclusions. (a) There exists a constant C such that ku j+1 kp0 , j = 0, 1, · · · .
ωdx
≤ Ckuj kp0 ,
(b) For the constant C appearing in (a), take δ > C and denote U (x) =
∞ X j=0
then U ω ∈ A1 and U s ∈ A1 .
δ −j uj (x),
ωdx ,
for
34
Chapter 1. Hardy-Littlewood Maximal Operator
In fact, by Theorem 1.4.3, we have that 0
kuj+1 kpp0 , ωdx Z Z 0 0 ≤C [M (usj )(x)]p /s ω(x)dx + [ω(x)−1 M (uj ω)(x)]p ω(x)dx n n ZR Z R s p p0 1−p0 [M (uj )(x)] ω(x)dx + =C [M (uj ω)(x)] ω(x) dx Rn Rn Z Z p0 p0 1−p0 [uj (x)ω(x)] ω(x) dx uj (x)ω(x)dx + ≤ C1 n Rn Z R 0 ≤C upj (x)ω(x)dx. Rn
So the conclusion (a) holds and it is easy to check that the constant C is independent of uj for all j ≥ 0. Now let us verify (b). By the definition of uj+1 , we have that M (U ω)(x) ≤ ≤
∞ X
j=0 ∞ X
δ −j M (uj ω)(x) δ −j uj+1 (x)ω(x)
j=0 ∞ X
=δ
δ −j uj (x)ω(x)
j=1
≤ δ(U (x)ω(x)). Hence U ω ∈ A1 . On the other hand, we have that s
[M (U )(x)]
1/s
≤
∞ X j=0
δ
−j
[M (usj )(x)]1/s
≤
∞ X j=0
δ −j uj+1 (x) ≤ δU (x).
So U s ∈ A1 too. Thus, if denote ω1 = U ω and ω2 = U s , then ω1 , ω2 ∈ A1 and ω = (U ω) · (U s )−1/s = (U ω) · (U s )1−p = ω1 · ω21−p .
Hence the necessity of theorem holds for 1 < p ≤ 2. 0 If 2 < p < ∞ and ω ∈ Ap , then ω 1−p ∈ Ap0 and 1 < p0 ≤ 2. Thus by 0 0 the above proof process, there are ν 1 , ν2 ∈ A1 such that ω 1−p = ν1 · ν21−p . Equivalently, ω = ν2 · ν11−p . Thus we complete the proof of Theorem 1.4.5. Below we give a sharp result about A1 weights.
35
1.5 Notes and references
Proposition 1.4.3 Let x ∈ Rn . Then |x|α ∈ A1 if and only if −n < α ≤ 0. Proof. If |x|α ∈ A1 . Then |x|α ∈ L1loc (Rn ), hence the condition α > −n is necessary. On the other hand, if α > 0 and |x| α ∈ A1 , we take 1 < p ≤ 1 + α/n, then |x|α ∈ Ap by the property (i) of Ap weights. However, 0 |x|α(1−p ) ∈ / L1loc (Rn ) for the choice of p. Conversely, suppose that −n < α ≤ 0, we will show that |x| α ∈ A1 . Indeed, for any fixed cube Q, we denote Q 0 is the translation of Q with the center at origin. The case (i): 2Q0 ∩ Q 6= ∅. In this case, we have 4Q0 ⊃ Q, and Z Z 1 1 α |x|α dx ≤ C|Q|α/n ≤ C inf |x|α . |x| dx ≤ x∈Q |Q| Q |Q| 4Q0 The case (ii): 2Q0 ∩ Q = ∅. Notice that if x, y ∈ Q, then |x| ≤ |x − y| + |y| ≤ C|Q|1/n + |y| ≤ (C + 1)|y|, Thus
1 |Q|
Z
Q
|x|α dx ≤ C inf |x|α . x∈Q
It is easy to see that the constants C in the cases both (i) and (ii) depend only on n. Therefore, we have proved that |x| α ∈ A1 . By the property (iii) of Ap weights, we get immediately the following sharp result for Ap weights.
Proposition 1.4.4 Let x ∈ Rn . Then for 1 < p < ∞, |x|α ∈ Ap if and only if −n < α < n(p − 1).
1.5
Notes and references
Theorem 1.1.1 was first proved by Hardy and Littlewood [HaL] for n = 1 and then by Wiener [Wi] for n > 1. The idea of proof given here was taken from Stein [St4] which is one of the most important monograph in harmonic analysis. The Calder´on-Zygmund decomposition (Theorems 1.2.1 and 1.2.3) first appeared in Calder´on and Zygmund [CaZ1] which is regarded as the foundation of several variables singular integrals theory.
36
Chapter 1. Hardy-Littlewood Maximal Operator
Theorem 1.2.2 is due to Stein [St2]. The idea of proof given here was taken from Garcia-Cuerva and Rubia de Francia [GaR], which is a nice monograph on the topic of weighted norm inequalities. The Marcinkiewicz interpolation theorem was first announced by Marcinkiewicz [Ma]. Its complete proof can be found in Zygmund [Zy]. Theorem 1.3.1 in this chapter is general form of the Marcinkiewicz interpolation theorem, whose proof was taken from Duoandikoetxea [Du2]. Theorem 1.3.2 is due to Fefferman and Stein [FeS1]. The Reverse H¨older inequality (Theorem 1.4.1) first was proved by Coifman and Fefferman [CoiF]. Theorem 1.4.2 and Theorem 1.4.3 were first proved by Muckenhoupt [Mu] for n = 1. So, the Ap weights are also called as Muckenhoupt A p weights. Precisely, a function ω is called to satisfy Muckenhoupt A p condition if ω ∈ Ap . The idea of proving Theorems 1.4.2 and 1.4.3 given here was taken from Journ´e [Jou]. Theorem 1.4.4 is due to Coifman and Rochberg [CoiR]. Theorem 1.4.5 was first proved by Jones [Jon]. The proof of Theorems 1.4.5 given here was taken from Coifman, Jones and Rubia de Francia [CoiJR].
Chapter 2
Singular Integral Operators Calder´on-Zygmund singular integral operator is a direct generalization of the Hilbert transform and the Riesz transform. The former is originated from researches of boundary value of conjugate harmonic functions on the upper half-plane, and the latter is tightly associated to the regularity of solution of second order elliptic equation. Now we will introduce their backgrounds briefly. Suppose f ∈ Lp (R) (1 ≤ p < ∞). Consider the Cauchy integral on R: F (z) =
1 2πi
Z
R
f (t) dt, t−z
where z = x + iy, y > 0. It is clear to see that F (z) is analytic on R 2+ . Note that Z Z 1 i y x−t F (z) = f (t)dt + f (t)dt 2π R (x − t)2 + y 2 2π R (x − t)2 + y 2 i 1h := (Py ∗ f )(x) + i(Qy ∗ f )(x) , 2
where
1 y 2 π t + y2
Py (t) = is called the Poisson kernel, and Qy (t) =
t 1 2 π t + y2
37
38
Chapter 2. Singular Integral Operators
is called the conjugate Poisson kernel. Correspondingly, P y ∗ f is called the Poisson integral of f , and Qy ∗ f is called the conjugate Poisson integral of f . From the property of boundary values of harmonic functions, it follows that Py ∗ f → f , a.e. as y → 0 and Qy ∗ f → Hf , a.e. as y → 0. Hf is called the Hilbert transform of f . Z f (t) 1 dt. (2.0.1) Hf (x) = p.v. π R x−t It can be proved that if f ∈ L2 (R), then d ˆ Hf(ξ) = −isgnξ f(ξ).
(2.0.2)
1 Let K(x) = p.v. , then Hf = (K ∗ f ). x Now suppose that f ∈ L2 (Rn ) and give the Poisson equation 4u = f , where n X ∂2 4= ∂x2j j=1
is the Laplacian operator on Rn . By taking the Fourier transform on both sides of the equation we have that fˆ(ξ) = −4π 2 |ξ|2 u ˆ(ξ). This is equivalent to u ˆ(ξ) = − Thus, for 1 ≤ j, k ≤ n,
1 4π 2 |ξ|2
ˆ f(ξ).
\ ∂2u ξj ξk ˆ f(ξ). (ξ) = −4π 2 ξj ξk u ˆ(ξ) = ∂xj ∂xk |ξ|2
If we define the operator
ξj ˆ d R f (ξ), j f(ξ) = −i |ξ|
j = 1, 2, · · · , n,
then it implies that
\ ∂2u (ξ) = −R\ j Rk f(ξ). ∂xj ∂xk
(2.0.3)
Chapter 2. Singular Integral Operators
39
The operator Rj defined by (2.0.3) is called the Riesz transform. Thus the problem of regularity of solution for the Poisson equation is transfered to that of boundedness of the Riesz transform. By comparing (2.0.2) with (2.0.3), it is evident to see that the Riesz transform is a generalization of the Hilbert transform from one dimension to n dimension. If f ∈ L p (Rn ) (1 ≤ p < ∞), then the Riesz transform of f has the following form: Z xj − y j f (y)dy, 1 ≤ j ≤ n. (2.0.4) Rj f (x) = p.v.Cn n+1 Rn |x − y| If we let Kj (x) = p.v.
xj (j = 1, 2, · · · , n), then |x|n+1 Rj f = Kj ∗ f.
Let us now see another background of singular integral operator. We know, when n ≥ 3, the basic solution of the Laplacian operator 4 is Γ(x) =
1 1 . (2 − n)ωn−1 |x|n−2
Thus when f has good properties, for example f ∈ S (R n ), Γ∗f is a solution of the Poisson equation 4u = f , that is Z f (y) u(x) = Γ ∗ f (x) = Cn dy. n−2 Rn |x − y| Formally, by taking partial derivatives of second order of u, we obtain that Z Z Ωj (x − y) Ωj (x − y) ∂ 2 u(x) = f (y)dy := lim f (y)dy, 2 n n + ∂xj ε→0 Rn |x − y| |x−y|>ε |x − y| where Ωj (y) = Cn (1 − n|y|−2 yj2 ). It is not difficult to prove that Ωj satisfies the following properties: (a) Ωj (λy) = Ωj (y), ∀λ > 0; Z (b) Ωj (y 0 )dσ(y 0 ) = 0; Sn−1
(c) Ωj ∈ L1 (Sn−1 ).
Z
Ωj (x − y) f (y)dy, then Lp regularity of soε→0+ |x−y|>ε |x − y|n lution of equation 4u = f is converted to the L p boundedness of operator Tj . Put Tj f (x) = lim
40
Chapter 2. Singular Integral Operators
Suppose that a function Ω satisfies above three conditions (a), (b) and (c), then for f ∈ Lp (Rn )(1 ≤ p < ∞), define an integral operator T Ω by Z Ω(x − y) TΩ f (x) = p.v. f (y)dy. (2.0.5) n Rn |x − y| xj , then TΩ becomes the Riesz transform Rj (j = |x| 1, 2, · · · , n). If we take n = 1 and Ω(x) = sgnx, then T Ω is just the Hilbert transform H. The operator TΩ defined by (2.0.5) is the object of study in this chapter.
If we take Ω(x) =
2.1
Calder´ on-Zygmund singular integral operators
Definition 2.1.1 Suppose that K(x) ∈ L 1loc (Rn \{0}) and satisfies the following conditions: |K(x)| ≤ B|x|−n , ∀ x 6= 0; (2.1.1) Z K(x)dx = 0, ∀ 0 < r < R < ∞; (2.1.2) Z
r≤|x|≤R
|x|≥2|y|
|K(x − y) − K(x)|dx ≤ B,
∀ y 6= 0.
(2.1.3)
Then K is called the Calder´ on-Zygmund kernel, where B is a constant independent of x and y. And the condition (2.1.3) is called as H¨ omander’s condition. Theorem 2.1.1 Suppose that K is the Calder´ on-Zygmund kernel. For ε > 0 and f ∈ Lp (Rn )(1 < p < ∞), let Z Tε f (x) = f (x − y)K(y)dy, |y|≥ε
then the following statements hold. (i) kTε f kp ≤ Ap kf kp , where Ap is independent of ε and f . (ii) For any f ∈ Lp (Rn ), lim Tε f exists in the sense of Lp norm. That is, these exists a T such that
ε→0
T f (x) = p.v.
Z
Rn
(iii) kT f kp ≤ Ap kf kp .
f (x − y)K(y)dy.
41
2.1 Calder´on-Zygmund singular integral operators
Remark 2.1.1 The linear operator T defined by Theorem 2.1.1 (ii) is called the Calder´ on-Zygmund singular integral operator. T ε is also called the truncated operator of T . Proof. For ε > 0, let Kε (x) = K(x)χ{|x|≥ε} (x), then Tε f (x) = Kε ∗ f (x). First we will prove that Tε is of type (2, 2), and {Tε } is uniformly bounded on L2 (Rn ). It is clear to check that Kε satisfies (2.1.1) and (2.1.2) uniformly in ε. Next we will show that Kε also satisfies (2.1.3) uniformly. Actually, for any x, y ∈ Rn with y 6= 0 and |x| ≥ 2|y|, if both x and x − y are in B(0, ε), then Kε (x) = Kε (x − y) = 0; If both x and x − y are in (B(0, ε)) c , then Kε (x) = K(x), Kε (x − y) = K(x − y). In this case, Kε satisfies (2.1.3). If |x| > ε and |x − y| < ε, then we have |x|/2 ≤ |x − y| < ε and ε < |x| < 2ε. Therefore, we have that Z Z |Kε (x)|dx ≤ CB, |Kε (x − y) − Kε (x)|dx ≤ |x|≥2|y|
ε≤|x|≤2ε
where C is independent of ε. By similar way we can prove, when |x| < ε and |x − y| > ε, Kε satisfies (2.1.3) uniformly in ε. Next we will show that {Tε } is uniformly bounded in ε on L2 (Rn ). Actually, for any ε > 0, Kε ∈ L2 (Rn ), so by the Plancherel Theorem, it suffices to prove that there exists a constant C > 0 such that, for any ε > 0, sup |Kˆε (ξ)| ≤ CB.
(2.1.4)
ξ∈Rn
In fact, for ξ ∈ Rn , Z ˆ Kε (ξ) = lim
R→∞ |x|≤R
= lim
R→∞
Z
e−2πix·ξ Kε (x)dx
a |x|≤ |ξ|
e
−2πix·ξ
Kε (x)dx +
Z
a <|x|≤R |ξ|
e
:= lim (I1 + I2 ). R→∞
From condition (2.1.1) and (2.1.2), it follows that Z −2πix·ξ |I1 | = (e − 1)Kε (x)dx |x|≤ a |ξ| Z ≤ C|ξ| |x||Kε (x)|dx a |x|≤ |ξ|
≤ CaB.
−2πix·ξ
Kε (x)dx
!
42
Chapter 2. Singular Integral Operators
Now consider I2 . Take y = Z I2 =
ξ , 2|ξ|2
then e2πiy·ξ = −1. Thus
a <|x−y|≤R |ξ|
=− =− where
Z
J= Hence we have I2 = As |y| =
1 2|ξ|
1 2
Z Z
a <|x−y|≤R |ξ|
a <|x|≤R |ξ|
a <|x|≤R |ξ|
Z
e−2πi(x−y)·ξ Kε (x − y)dx
−
a <|x|≤R |ξ|
Z
e−2πix·ξ Kε (x − y)dx
e−2πix·ξ Kε (x − y)dx + J,
a <|x−y|≤R |ξ|
!
e−2πix·ξ Kε (x − y)dx.
[Kε (x) − Kε (x − y)]e−2πix·ξ dx +
J . 2
and a > 1, we have that Z [Kε (x) − Kε (x − y)]e−2πix·ξ dx a <|x|≤R |ξ| Z ≤ |Kε (x) − Kε (x − y)|dx ≤ CB, |x|≥2|y|
where C, B are independent of ε, ξ. On the other hand, if we let E be the symmetric difference of both sets a a {x : |ξ| < |x| ≤ R} and {x : |ξ| < |x − y| ≤ R}, then Z |Kε (x − y)|dx. |J| ≤ E
1 and a > 1, it follows that Noting that |y| = 2|ξ| a R 2a [ E⊂ x: x: ≤ |x| ≤ ≤ |x| ≤ 2R . 2|ξ| |ξ| 2
Thus by (2.1.1), it implies that Z Z |Kε (x − y)|dx + |J| ≤ 2a a ≤|x|≤ |ξ| 2|ξ|
≤ CB,
R ≤|x|≤2R 2
|Kε (x − y)|dx
43
2.1 Calder´on-Zygmund singular integral operators
where C, B are independent of ξ, ε. Sum up all above we get (2.1.4), therefore {Tε } is uniformly bounded in ε on L2 (Rn ). Now we show that Tε is of weak type (1, 1), and the bound is independent of ε. For every f ∈ L1 (Rn ) and λ > 0, by the Calder´on-Zygmund decompositin of a function (Theorem 1.2.3) we can get a series of non-overlapping cubes {Qj } and two functions g, b, such that f = g + b which satisfy the following properties: (a) kgk22 ≤ Cλkf k1 , |g(x)| ≤ 2n λ, a.e. x ∈ Rn ; Z 1 (b) λ ≤ |f (x)|dx ≤ 2n λ, for every Qj ; |Qj | Qj X 1 (c) |Qj | ≤ kf k1 ; λ j Z Z X bj (x)dx = 0, suppbj ⊂ Qj and kbj k1 ≤ 2 (d) b(x) = bj (x), Qj
j
Qj
|f (x)|dx.
Since Tε f (x) = Tε g(x) + Tε b(x), we have that
|{x ∈ Rn : |Tε f (x)| > λ}| λ λ n n ≤ x ∈ R : |Tε g(x)| > + x ∈ R : |Tε b(x)| > 2 2 := I1 + I2 .
The first step and (a) imply that 2 Z Z 2 4C 4C 2 I1 ≤ |Tε g(x)| dx ≤ 2 |g(x)|2 dx ≤ kf k1 . λ λ Rn λ Rn √ To estimate I2 , let Q∗j = 2 nQj be a cube whose S ∗ center is the same as Q j √ ∗ and side is 2 n times that of Qj . Let E = j Qj , then (c) implies that |E ∗ | ≤
Thus
X j
|Q∗j | ≤
Cn kf k1 . λ
λ ∗ I2 ≤ |E | + x ∈ / E : |Tε b(x)| > 2 Z Cn 2 ≤ |Tε b(x)|dx. kf k1 + λ λ Rn \E ∗ ∗
44
Chapter 2. Singular Integral Operators
Notice that |Tε b(x)| ≤
X
|Tε bj (x)|, it suffices to prove that
j
Rn \E ∗
j
XZ
|Tε bj (x)|dx ≤ Ckf k1 .
(2.1.5)
Denote the center of Qj by yj , then by (2.1.3) we have that Z Z Z |Tε bj (x)|dx ≤ |Kε (x − y) − Kε (x − yj )||bj (y)|dydx Rn \E ∗
≤
Z
Rn \Q∗j Qj
≤ CB
Qj
|bj (y)| Z
Qj
Z
Rn \Q∗j
|Kε (x − y) − Kε (x − yj )|dxdy
|bj (y)|dy ≤ 2CB
Z
Qj
|f (y)|dy,
which, together with (b) and (c), yields (2.1.5). It follows that T ε is of weak type (1, 1) and its bound is independent of ε or f . Now we will show that Tε is of type (p, p) (1 < p < ∞). Applying the Marcinkiewiz Interpolation Theorem (Theorem 1.3.1) we know that T ε is of type (p, p) (1 < p < 2), and its bound is independent of ε or f . Now suppose 2 < p < ∞, and p1 + p10 = 1, then 1 < p0 < 2. If we let Teε be fε ∗ f (x), where K fε (x) = Kε (−x). the dual operator of Tε , then Teε f (x) = K fε satisfies all the conditions of Kε . Thus Teε is of type (p0 , p0 ). Clearly, K Thus for any f ∈ Lp (Rn ), we have that Z kTε f kp = sup Tε f (x)g(x)dx kgkp0 ≤1 Rn Z e = sup f (x)Tε g(x)dx kgkp0 ≤1
Rn
≤ kf kp · sup Teε g kgkp0 ≤1
p0
≤ Ap kf kp .
Here Ap is independent of ε or f . This finishes the proof of (i). Next we will illustrate that for any f ∈ L p (Rn ) (1 < p < ∞), T f exists and is the limit of {Tε f } in Lp . First suppose that f ∈ C0∞ (Rn ). For any y (y 6= 0) we wish to obtain 1 Z p p ≤ C|y|. (2.1.6) |f (x − y) − f (x)| dx Rn
45
2.1 Calder´on-Zygmund singular integral operators Actually, from
d f (x − ty) = h∇f, −yi(x − ty), dt
it follows that f (x − y) − f (x) = =
Z
1
0
Z
1
h∇f, −yi(x − ty)dt
0
=
Z
|y|
0
where y 0 = Z
y |y| .
h∇f, −y 0 i(x − sy 0 )ds,
Thus p
Rn
d f (x − ty)dt dt
|f (x − y) − f (x)| dx
1
p
p ! 1 Z p |y| 0 0 = h∇f, −y i(x − sy )ds dx Rn 0 1 Z |y| Z p 0 0 p |h∇f, −y i(x − sy )| dx ≤ ds Z
Rn
0
n X
∂f
≤ |y| ·
∂xj . p
j=1
Now assume 0 < η < ε, then (2.1.6) and (2.1.1) implies that kTη f − Tε f kp ≤
Z
η<|y|≤ε
≤C
Z
|K(y)|
η<|y|≤ε
Z
Rn
p
|f (x − y) − f (x)| dx
1
p
dy
|y| · |K(y)|dy
≤ CBε −→ 0
(as η, ε −→ 0).
This shows that for every function f in C 0∞ (Rn ), {Tε f } is a Cauchy sequence in Lp (Rn ), therefore there exists T f ∈ Lp such that lim kTε f − T f kp = 0.
ε→0
It immediately follows that kT f kp ≤ kT f − Tε f kp + kTε f kp ≤ kT f − Tε f kp + Ap kf kp , so kT f kp ≤ Ap kf kp .
46
Chapter 2. Singular Integral Operators
For any f ∈ Lp (Rn ) and δ > 0, there exists g ∈ C0∞ (Rn ) such that f = g + h and khkp < δ. Thus, for 0 < η < ε, kTη f − Tε f kp ≤ kTη (f − g)kp + kTη g − Tε gkp + kTε (g − f )kp ≤ Ap kf − gkp + kTη g − Tε gkp + Ap kg − f kp −→ 2Ap δ, as η, ε → 0.
Since δ is arbitrary, we conclude that {T ε f } is still a Cauchy sequence in Lp (Rn ) for any f ∈ Lp (Rn ). Thus there exists T f ∈ Lp , such that lim kT f − Tε f kp = 0
ε→0
and kT f kp ≤ Ap kf kp .
This completes the proof of Theorem 2.1.1.
Let us now consider a dilation δε in Rn . Define δε f (x) = f (εx) for ε > 0 and x ∈ Rn . Suppose that T f = K ∗ f , and T commute with dilation, i.e., T δε = δε T . Then the kernel K(x) of T satisfies: K(εx) = ε−n K(x).
(2.1.7)
The formula (2.1.7) shows that K is homogeneous of degree −n. Thus we can rewrite K(x) as Ω(x) |x|n , where Ω satisfies the homogeneous condition of degree zero, i.e., Ω(λx) = Ω(x) for every λ > 0 and x 6= 0. In this case, due to the conditions (2.1.1) and (2.1.3), Ω(x 0 ) should satisfy: B 0 0 n−1 ; (a) |K(x)| ≤ |x| n ⇐⇒ |Ω(x )| ≤ B, for every x ∈ S Z Z Ω(x0 )dσ(x0 ) = 0; where σ is a meaK(x)dx = 0 ⇐⇒ (b) Sn−1
r<|x|≤R
sure on Sn−1 induced by the Lebesgue measure. (c) The condition (2.1.3) will be changed to a stronger Dini’s condition: Z 1 ω∞ (δ) dδ < ∞, δ 0
where
ω∞ (δ) =
sup x0 ,y 0 ∈S n−1 ,|x0 −y 0 |<δ
|Ω(x0 ) − Ω(y 0 )|.
The condition (b) is derived from the following equality. Z ∞Z Z R dr K(x)dx = Ω(x0 )dσ(x0 ) = log Ω(x0 )dσ(x0 ). r r n−1 n−1 r S S r<|x|≤R
Z
47
2.1 Calder´on-Zygmund singular integral operators
Remark 2.1.2 The condition mentioned above is called L ∞ -Dini condition. Theorem 2.1.2 Suppose that Ω(x) is a homogeneous bounded function of degree 0 on Rn , with Z Ω(x0 )dσ(x0 ) = 0 (2.1.8) Sn−1
and
Z
Let
1 0
Tε f (x) =
ω∞ (δ) dδ < ∞. δ Z
|y|≥ε
(2.1.9)
Ω(y) f (x − y)dy |y|n
for f ∈ Lp (Rn ), 1 < p < ∞. Then the following three statements hold: (i) there exists a constant Ap independent of f, ε such that kTε f kp ≤ Ap kf kp ; (ii) there exists T f such that lim Tε f (x) = T f (x) in Lp norm; ε→0
(iii) kT f kp ≤ Ap kf kp . Proof. By Theorem 2.1.1, we merely need to show that K(x) = Ω(x) |x|n satisfies the condition (2.1.3). Ω(x − y) − Ω(x) 1 1 K(x − y) − K(x) = + Ω(x) − := I1 + I2 . |x − y|n |x − y|n |x|n when |x| ≥ 2|y| with 0 < θ < 1, we have that |x − θy| ≤ |x| + |y| ≤
3 |x| 2
|x − y| ≥ |x| − |y| ≥
1 |x|. 2
and
Thus
1 |y||x − θy|n−1 |y| 1 − ≤ C ≤ C n+1 . |x − y|n |x|n n n |x − y| |x| |x|
On the other hand, when |x| ≥ 2|y|, x−y x |y| |x − y| − |x| ≤ 2 |x| .
(2.1.10)
(2.1.11)
48
Chapter 2. Singular Integral Operators
Therefore from (2.1.10) and (2.1.11) it follows that Z Z Z |K(x − y) − K(x)|dx ≤ |I1 |dx + |I2 |dx |x|≥2|y| |x|≥2|y| |x|≥2|y| Z x dx x−y ≤ Ω |x − y| − Ω |x| |x − y|n |x|≥2|y| Z dx + CkΩk∞ |y| n+1 |x|≥2|y| |x| Z |y| dx + C 0 kΩk∞ ≤C ω∞ 2 n |x| |x| |x|≥2|y| Z ∞Z dr |y| dσ(x0 ) + C 0 kΩk∞ =C ω∞ 2 r r n−1 2|y| S Z 1 ω∞ (δ) dδ + CkΩk∞ ≤ C0 δ 0 ≤ B.
Remark 2.1.3 If K(x) = then TΩ defined by
Ω(x) |x|n ,
TΩ f (x) = p.v.
where Ω is homogeneous of degree zero, Z
Rn
Ω(y) f (x − y)dy |y|n
(2.1.12)
is also called a singular integral operator with homogeneous kernel. Both Theorem 2.1.1 and Theorem 2.1.2 show that the L p -norm limit of Calder´on-Zygmund singular integral operator and singular integral operator with homogeneous kernel exist while considering them as the truncated operator family, and they are both operators of type (p, p)(1 < p < ∞). A natural question is whether the limit of {T ε f (x)} in pointwise sense exist for any f ∈ Lp (Rn )(1 ≤ p < ∞). In the following we will give an affirmative answer by introducing the maximal operator of singular integral operator, meanwhile we will show the weak (1, 1) boundedness of T Ω . Suppose that TΩ is a singular integral operator defined by (2.1.12). Let Ω be a homogeneous of degree 0 on Rn and satisfy (2.1.8) and (2.1.9). For f ∈ Lp (Rn )(1 ≤ p < ∞), TΩ∗ f (x) = sup |TΩ,ε f (x)| ε>0
2.1 Calder´on-Zygmund singular integral operators
49
is called the maximal singular integral operator, where T Ω,ε is the truncated operator of TΩ defined by Z Ω(y) f (x − y)dy, ε > 0. TΩ,ε f (x) = n |y|≥ε |y| Next we will formulate a pointwise inequality of T Ω∗ . Lemma 2.1.1 (Cotlar inequality) There exist two constant C 1 , c2 > 0 such that TΩ∗ f (x) ≤ C1 M (TΩ f )(x) + C2 M f (x), (2.1.13) for every f ∈ Lp (Rn ) (1 < p < ∞) and x ∈ Rn , where M is the HardyLittlewood maximal operator. Proof. Let Kε (x) = |x|−n Ω(x)χ{|x|≥ε} (x). Choose nonnegative, compact supported, radial function ϕ ∈ S (Rn ) such that supp(ϕ) ⊂ {x : |x| ≤ 1} and Z ϕ(x)dx = 1. Rn
Without loss of generality, we may assume that ϕ(|x|) decrease in |x|. Thus lim Kε ∗ ϕ = K ∗ ϕ
ε→0
holds pointwisely. Now denote Φ(x) = K ∗ϕ(x)−K1 (x). Since K is homogeneous of degree −n, we have for ε > 0 that Φε (x) = ϕε ∗ K(x) − Kε (x),
(2.1.14)
where
1 x 1 x Φ ϕ and ϕ (x) = . ε εn ε εn ε By (2.1.14), for any f ∈ Lp (Rn ), we have Φε (x) =
TΩ,ε f (x) = Kε ∗ f (x) = (ϕε ∗ K) ∗ f (x) − Φε ∗ f (x).
(2.1.15)
On the other hand, for every η > 0 and x ∈ R n , it follows that (ϕε ∗ Kη ) ∗ f (x) = ϕε ∗ (Kη ∗ f )(x) = ϕε ∗ (TΩ,η f )(x). 0
Assume ϕε ∈ Lp . Then ϕε ∗ Kη converges to ϕε ∗ K in Lp as η → 0, (see the proof of Theorem 2.1.1), meanwhile T Ω,η f converges to TΩ f in L∞ . Thus (ϕε ∗ K) ∗ f (x) = ϕε ∗ (TΩ f )(x).
50
Chapter 2. Singular Integral Operators
This formula together with (2.1.15) implies that TΩ,ε f (x) = ϕε ∗ (TΩ f )(x) − Φε ∗ f (x).
(2.1.16)
Next we will prove that Φ can be dominated by a radial integrable function. When |x| < 1, Z [ϕ(x − y) − ϕ(x)]K(y)dy. Φ(x) = ϕ ∗ K(x) = Rn
Note that K(y) = |y|−n Ω(y), ϕ ∈ C0∞ (Rn ) and suppϕ ⊂ {x : |x| ≤ 1}. Clearly Φ is bounded when |x| < 1. Also, when 1 ≤ |x| ≤ 2, Φ is still bounded. When |x| > 2, we have that Z Φ(x) = K(x − y)ϕ(y)dy − K(x) n ZR = [K(x − y) − K(x)]ϕ(y)dy. |y|≤1
Since 1 1 |Ω(x − y) − Ω(x)| + |Ω(x)| − , |K(x − y) − K(x)| ≤ |x − y|n |x − y|n |x|n
The proof of Theorem 2.1.2 implies that
0
|K(x − y) − K(x)| ≤ C |x|
−n
· ω∞
2 |x|
.
2 Thus when |x| > 2, |Φ(x)| ≤ C 0 |x|−n ω∞ |x| . Since Ω satisfies (2.1.10), the minimum radical dominated function of Φ, denoted by Ψ(x) = sup |Φ(y)|, |y|≥|x|
must be integrable. Thus by (2.1.16) and the properties of the HardyLittlewood maximal operator we obtain that there exist C 1 , C2 > 0, such that TΩ∗ f (x) = sup |TΩ,ε f (x)| ε>0
≤ sup |ϕε ∗ (TΩ f )(x)| + sup |Φε ∗ f (x)| ε>0
ε>0
≤ C1 M (TΩ f )(x) + C2 M f (x).
2.1 Calder´on-Zygmund singular integral operators
51
This implies (2.1.13). Since M and TΩ are both operators of (p, p) type (see Theorem 1.1.1 and Theorem 2.1.2), it immediately follows that T Ω∗ is of type (p, p). Theorem 2.1.3 Suppose that Ω is a homogeneous bounded function of degree 0 and satisfies (2.1.8) and (2.1.9). Then T Ω∗ is of type (p, p) (1 < p < ∞) and weak type (1, 1). Proof. It suffices to show that TΩ∗ is of weak type (1, 1). The idea in the proof is the same as that of Theorem 2.1.1. For any f ∈ L 1 (Rn ) and λ > 0, using the Calder´on-Zygmund decomposition, we have f = g + b and a sequence of non-overlapping cubes {Q j }. Thus λ λ ∗ ∗ ∗ + x : T b(x) > . (2.1.17) |{x : TΩ f (x) > λ}| ≤ x : TΩ g(x) > Ω 2 2 Since kgk22 ≤ Cλkf k1 and TΩ∗ is of type (2, 2), we have x : TΩ∗ g(x) > λ ≤ C 0 λ−2 kTΩ∗ gk22 ≤ C 00 1 kf k1 . 2 λ
Now denote the center S of Qj by yj and the side-length of Qj by dj . Let √ Sj = nQj and E = j Sj . Thus |E| ≤
X j
|Sj | =
X j
Cn |Qj | ≤
Cn kf k1 . λ
It follows that λ λ ∗ c ∗ x : TΩ b(x) > ≤ |E| + x ∈ E : TΩ b(x) > . 2 2 Fix x ∈ E c and ε > 0, then
TΩ,ε b(x) =
XZ j
Qj
Kε (x − y)b(y)dy.
We consider the following three cases of Q j : (i) for any y ∈ Qj , |x − y| < ε; (ii) for any y ∈ Qj , |x − y| > ε;
(2.1.18)
52
Chapter 2. Singular Integral Operators (iii) there exists y ∈ Qj , such that |x − y| = ε.
For the first case, Kε (x − y) = 0, thus TΩ,ε b(x) = 0. For the second case, Kε (x − y) = K(x − y), thus Z Z [K(x − y) − K(x − y )]b (y)dy = K (x − y)b (y)dy j j j Qj Qj ε Z ≤ |K(x − y) − K(x − yj )||bj (y)|dy. Qj
As to the third case, notice that x ∈ E c ⊂ Sjc , there exist two constants Cn and Cn0 only depending on n such that Qj ⊂ S(x, r), where S(x, r) is a closed ball with the center at x and radius r = C n ε. If y ∈ Qj , then |x − y| ≥ Cn0 ε. Thus, for y ∈ Qj , |Kε (x − y)| ≤
|Ω(x − y)| ≤ kΩk∞ (Cn0 ε)−n . |x − y|n
Therefore Z Z Kε (x − y)bj (y)dy ≤ |Kε (x − y)||b(y)|dy Qj Qj ∩S(x,r) Z 0 −n ≤ C kΩk∞ ε |b(y)|dy S(x,r) Z 1 00 ≤C |b(y)|dy. |S(x, r)| S(x,r) Taking the sum of all cubes yields that |TΩ,ε b(x)| ≤
XZ j
Qj
|K(x−y)−K(x−yj )||bj (y)|dy+
C 00 |S(x, r)|
Z
S(x,r)
|b(y)|dy.
Thus TΩ∗ b(x) ≤
XZ j
Qj
|K(x − y) − K(x − yj )||b(y)|dy + CM b(x).
2.1 Calder´on-Zygmund singular integral operators
53
Hence x ∈ E c : TΩ∗ b(x) > λ 2 Z X λ c ≤ x∈E : |k(x − y) − K(x − yj )||b(y)|dy > 4 Qj j λ c . + x ∈ E : CM b(x) > 4
By (2.1.5) and the weak (1, 1) boundedness of the Hardy-Littlewood maximal operator we have C0 λ c ∗ ≤ x ∈ E : TΩ b(x) > kf k1 . 2 λ
This inequality together with (2.1.17) and (2.1.18) shows that T Ω∗ is operator of weak type (1, 1).
Corollary 2.1.1 Suppose that Ω satisfies the conditions in Theorem 2.1.3. Then, for f ∈ Lp (Rn ) (1 ≤ p < ∞), we have that lim TΩ,ε f (x) = TΩ f (x),
ε→0
a.e. x ∈ Rn .
Proof. For f ∈ Lp (Rn ) (1 ≤ p < ∞), let Λ(f )(x) = lim sup TΩ,ε f (x) − lim inf TΩ,ε f (x) , x ∈ Rn , ε→0
ε→0
2TΩ∗ f (x).
then Λ(f )(x) ≤ For any δ > 0, let f = g + h such that g ∈ ∞ n C0 (R ) and khkp < δ. Since Ω satisfies (2.1.8) and g is a smooth function with compact support, TΩ,ε g converges to TΩ g uniformly as ε → 0. So Λ(g)(x) = 0. Then, for 1 < p < ∞, kΛ(f )kp ≤ kΛ(h)kp ≤ 2Ap khkp ≤ 2Ap δ. Since δ is arbitrary, it follows that Λ(f )(x) = 0 a.e. x ∈ R n for 1 < p < ∞. Thus the limit of TΩ,ε f (x) exists for a.e. x ∈ Rn . When p = 1, for any λ > 0, we also have |{x : Λ(f )(x) > λ}| ≤
2Aδ 2A khk1 ≤ . λ λ
Therefore we still have Λ(f )(x) = 0 a.e. x ∈ R n . Thus the limit of TΩ,ε f (x) exists for a.e. x ∈ Rn with f ∈ L1 (Rn ).
54
Chapter 2. Singular Integral Operators
Corollary 2.1.2 Suppose that Ω satisfies the conditions in Theorem 2.1.3. Then TΩ is of weak type (1, 1). Remark 2.1.4 The Riesz transforms R j (j = 1, 2, · · · , n) are defined by Z xj − y j Rj f (x) = p.v.Cn f (y)dy, n+1 Rn |x − y| where Cn = Γ
n+1 2
/π
n+1 2
. x
It is clear that the kernels Ωj (x) = Cn |x|j (j = 1, 2, · · · , n) satisfies the conditions in Theorem 2.1.2. So we have Theorem 2.1.4 The Riesz transforms R j (j = 1, 2, · · · , n) are of type (p, p) (1 < p < ∞) and of weak type (1, 1). Since the Hilbert transform H is the Riesz transform of dimension 1, i.e., Z 1 ∞ f (y) dy, Hf (x) = p.v. π −∞ x − y of course we have Theorem 2.1.5 The Hilbert transform H is operator of type (p, p) (1 < p < ∞) and of weak type (1, 1). Remark 2.1.5 Similarly, we can also define the maximal Hilbert transform H ∗ and the maximal Riesz transforms Rj∗ , where Z f (x − y) H ∗ f (x) = sup Cn dy , y ε>0 |y|≥ε Z y j f (x − y)dy Rj∗ f (x) = sup Cn , n+1 |y| ε>0 |y|≥ε
(j = 1, 2, · · · , n).
Theorem 2.1.3 implies H ∗ and Rj∗ (j = 1, 2, · · · , n) are all of type (p, p) (1 < p < ∞) and of weak type (1, 1). In the following we will give the weighed boundedness of the Calder´onZygmund singular integral operator and its maximal operator. First we will formulate the definition of sharp maximal function.
2.1 Calder´on-Zygmund singular integral operators
55
Suppose f ∈ L1loc (Rn ). The sharp maximal function M ] f (x) of f is defined by Z 1 |f (y) − fQ |dy, x ∈ Rn . M ] f (x) = sup |Q| Q3x Q Here the supremum is taken over all the cubes Q in R n which contain x, and Z 1 f (t)dt fQ = |Q| Q
is the average of f on Q. An obvious fact is that, for every a ∈ C and Q ⊂ Rn , Z Z Z 1 1 2 |f (x) − fQ|dx ≤ |f (x) − a|dx + |a − fQ| ≤ |f (x) − a|dx. |Q| Q |Q| Q |Q| Q (2.1.19)
Lemma 2.1.2 Suppose that Ω is a homogeneous bounded function of degree 0 in Rn and satisfies (2.1.8) and (2.1.9). Then M ] (TΩ f )(x) ≤ C(n, s) (M (|f |s ) (x)) , x ∈ Rn holds for every s > 1, where M is the Hardy-Littlewood maximal operator. Proof. For any fixed s > 1 and x ∈ Rn , suppose Q is cube containing x, and let B be a ball of the same center and radius as Q. Let f 1 = f χ16B and f2 = f χ(16B)c , then f = f1 + f2 . By (2.1.19) we only need to show that Z 1 1 |TΩ f (y) − TΩ f2 (x)|dy ≤ C (M (|f |s ) (x)) s . (2.1.20) |Q| Q Since 1 |Q|
Z
|TΩ f (y) − TΩ f2 (x)|dy Z Z 1 1 |TΩ f1 (y)|dy + |TΩ f2 (y) − TΩ f2 (x)|dy ≤ |Q| Q |Q| Q Q
and s > 1, applying Theorem 2.1.2 we have 1 Z Z s 1 1 s |TΩ f1 (y)|dy ≤ |TΩ f1 (y)| dy |Q| Q |Q| Q 1 Z s 1 s |f (y)| dy ≤C |Q| 16B 1 s ≤ C(n, s) M (|f |s )(x) .
(2.1.21)
56
Chapter 2. Singular Integral Operators
On the other hand, let d be the radius of B, then Z 1 |TΩ f2 (y) − TΩ f2 (x)|dy |Q| Q Z Z 1 Ω(y − z) Ω(x − z) = f (z)dz − dy n n |Q| Q Rn \16B |y − z| |x − z| Z X ∞ Z Ω(y − z) Ω(x − z) 1 ≤ |y − z|n − |x − z|n |f (z)|dzdy. |Q| k k+1 Q k=2
2 d<|x−z|≤2
d
(2.1.22) Note that |x − z| ∼ |y − z|, thus Ω(y − z) Ω(x − z) |x − y| |Ω(y − z) − Ω(x − z)| . |y − z|n − |x − z|n ≤ C |Ω(x − z)| |x − z|n+1 + |x − z|n And when 2k d < |x − z| ≤ 2k+1 d, |Ω(x − z)|
|x − y| |x − y| ≤ kΩk∞ k n+1 . n+1 |x − z| (2 d)
In addition, the homogeneity of degree 0 of Ω yields that y−z x − z |Ω(y − z) − Ω(x − z)| = Ω −Ω |y − z| |x − z| (x − z) − (x − y) x − z . = Ω −Ω |(x − z) − (x − y)| |x − z|
Note |x − z| ≥ 2|x − y|, so (2.1.11) implies that |Ω(y − z) − Ω(x − z)| ≤ ω∞
|x − y| 2 |x − z|
.
Since ω∞ (δ) is nondecreasing on δ, when 2k d < |x − z| ≤ 2k+1 d, we have that |Ω(y − z) − Ω(x − z)| 1 |x − y| ≤ k n ω∞ 2 |x − z|n (2 d) |x − z| d 1 ≤ k n ω∞ (2 d) 2k−1 d 1 1 . = k n ω∞ (2 d) 2k−1
2.1 Calder´on-Zygmund singular integral operators
57
Thus Ω(y − z) Ω(x − z) |y − z|n − |x − z|n |f (z)|dz k d<|x−z|≤2k+1 d 2 k=2 ∞ Z X |x − y| |Ω(x − z)| ≤C |f (z)|dz |x − z|n+1 k k+1 2 d<|x−z|≤2 d k=2 ∞ Z X |Ω(y − z) − Ω(x − z)| +C |f (z)|dz |x − z|n k k+1 d k=2 2 d<|x−z|≤2 Z ∞ X |x − y| |f (z)|dz ≤C kΩk∞ k n+1 (2 d) |x−z|≤2k+1 d k=2 Z ∞ X 1 1 +C |f (z)|dz ω∞ (2k d)n 2k−1 |x−z|≤2k+1 d k=2 ∞ X 1 0 ≤ C M f (x) + CM f (x) . ω∞ 2k−1
∞ Z X
k=2
However, ∞ X
k=2
ω∞
1 2k−1
Z 1 ∞ 1 1 X 2k−2 dδ = ω∞ 1 log 2 2k−1 δ k=2 2k−1 1 Z ∞ 1 X 2k−2 dδ ≤ ω∞ (δ) 1 log 2 δ k=2 2k−1 Z 1 1 dδ ≤ < ∞. ω∞ (δ) log 2 0 δ
All estimates above imply that Z 1 1 |TΩ f2 (y) − TΩ f2 (x)|dy ≤ CM f (x) ≤ C(M (|f |s )(x)) s , |Q| Q which combining with (2.1.21) leads to (2.1.20). In order to formulate the next lemma, we first give definitions of dyadic cubes, the dyadic Hardy-Littlewood maximal function, and the dyadic sharp maximal function. Let Q0 = [0, 1)n be the unit cube in Rn , and Q0 is the total of cubes which are translated from Q0 by one unit. That is, the length of the side of the cubes in Q0 are all 1 and they are nonoverlapping whose union is R n .
58
Chapter 2. Singular Integral Operators
Similarly, for k ∈ Z, let Qk = [0, 2k )n , and Qk are the total cubes which are translated from Qk by 2k units. That is, the length of the side of the cubes in Qk are all 2k and they are nonoverlapping whose union is R n . Thus we have the following properties. (i) For any fixed x ∈ Rn and k ∈ Z, there is only one cube Qk 3 x. (ii) For any given two dyadic cubes, either their intersection is empty, or one lies in the other. (iii) Any one cube in Qk must contain 2n cubes in Qk−1 . (iv) When k < j, any cube in Qk must lie in a unique cube in Qj . Lemma 2.1.3 Suppose ω ∈ A∞ and there exists 0 < p0 < ∞ such that M f ∈ Lp0 (ω). Then Z Z p M ] f (x) ω(x)dx (M f (x))p ω(x)dx ≤ C (2.1.23) Rn
Rn
holds for p0 ≤ p < ∞. Proof. Since M ] (|f |)(x) ≤ 2M ] f (x), without loss of generality, we may assume that f ≥ 0. For any λ > 0, we use the Calder´on-Zygmund decomposition. By the property (ii) of dyadic cubes, if there exists a sequence of dyadic cubes satisfying Q1 ⊂ Q2 ⊂ · · · and fQk > λ, then {ω(Qk )} is uniformly bounded about k. In fact, for any x ∈ Q k , we have Z 1 t< f (y)dy ≤ M f (x). |Qk | Qk Thus ω(Qk ) =
Z
Qk
Z 1 (M f (x))p0 ω(x)dx t p0 Q k 1 ≤ p0 kM f kpL0p0 (ω) . t
ω(x)dx ≤
Since ω(x)dx is a double measure and ω ∈ A ∞ , thus {|Qk |} is uniformly bounded. Therefore all the dyadic cubes Q satisfying f Q > t are definitely included in a dyadic cube, which is called the maximal dyadic cube. Suppose that {Qj } is the set of all maximal dyadic cubes, then any Q j must satisfy Z 1 t< f (y)dy ≤ 2n t. |Qj | Qj
59
2.1 Calder´on-Zygmund singular integral operators
Since {Qj } depend on λ, without loss of generality we denote it by {Q λ,j }. Note that when λ < µ, for any Qµ,j , there must be a k such that Qµ,j ⊂ Qλ,k . Now for any λ > 0, let Q0 = Q2−n−1 λ,j0 and A > 0. If Q0 ⊂ then
X
{j:Qλ,j ⊂Q0 }
λ x : M f (x) > A ]
ω(Qλ,j ) ≤ ω
If Q0 * x : M ] f (x) >
λ A
,
λ x : M f (x) > A ]
.
(2.1.24)
, then
1 |Q0 |
Z
Q0
and f Q0
1 = |Q0 |
λ 2
Z
|f (y) − fQ0 |dy ≤
Q0
λ A
λ . 2
f (t)dt ≤ 2n 2−n−1 λ =
Thus X
{j:Qλ,j ⊂Q0 }
λ−
|Qλ,j | = ≤ ≤
X
{j:Qλ,j ⊂Q0 } Qλ,j
X
{j:Qλ,j ⊂Q0 }
X
Z
{j:Qλ,j ⊂Q0 } Qλ,j
Z
Q0 −1
≤A which gives
Z
λ|Q0 |,
|Qλ,j | ≤ 2A−1 |Q0 |.
{j:Qλ,j ⊂Q0 }
λ−
Qλ,j ⊂ Q0
λ 2
dy
|f (y) − fQ0 |dy
|f (y) − fQ0 |dy
On the other hand, since [
60
Chapter 2. Singular Integral Operators
and ω ∈ A∞ , there exist δ > 0 and C > 0, such that δ [ X Qλ,j Qλ,j ω {j:Qλ,j ⊂Q0 } {j:Qλ,j ⊂Q0 } . ≤C ω(Q0 ) |Q0 | Thus we get
X
{j:Qλ,j ⊂Q0 }
ω(Qλ,j ) ≤ C 2A−1
δ
ω(Q0 ).
(2.1.25)
By (2.1.24) and (2.1.25) we have X δ λ ] + C 2A−1 ω(Q0 ). ω(Qλ,j ) ≤ ω x : M f (x) > A {j:Qλ,j ⊂Q0 }
Now taking over all the Q0 , this inequality yields X δ X λ ] ω(Qλ,j ) ≤ ω x : M f (x) > + C 2A−1 ω Q2−n−1 λ,k . A j
k
P S (2.1.26) Let α(λ) = j ω (Qλ,j ) and β(λ) = ω({x : M f (x) > λ}), then j Qλ,j ⊂ {x : M f (x) > λ} and the disjoint of {Qλ,j } imply α(λ) ≤ β(λ). Next we wish to obtain that X λ . β(λ) ≤ ω 3Q4−n λ,j ≤ C1 α C2
(2.1.27)
(2.1.28)
j
Put Eλ = {x : M f (x) > λ}, where 3Q is the cube whose side length is 3 times that of Q with the center at the same as that of Q. If we can show [ Eλ ⊂ 3Q4−n λ,j , (2.1.29) j
then the left inequality of (2.1.28) naturally holds. Therefore (2.1.28) holds, since the second one is clear. Now for any given x ∈ E λ , R is a cube which contains x and satisfies Z 1 f (y)dy > λ. |R| R
61
2.1 Calder´on-Zygmund singular integral operators
If we denote d as the side length of R, then there exists the unique integer k such that 2k−1 < d ≤ 2k . Thus in Qk there are at most 2n dyadic cubes that intersect with R, and there are at most one dyadic cube Q such that Z
|f (y)|dy >
R∩Q
λ|R| . 2n
Since |R| ≤ |Q| < 2n |R|, we have that Z
R∩Q
|f (y)|dy >
This is equivalent to 1 |Q|
Z
Q
λ|R| λ > n |Q|. n 2 4
|f (y)|dy >
λ . 4n
Consequently there exists j such that the maximal dyadic cube Q 4−n λ,j ⊃ Q. Since Q ∩ R 6= φ and |R| ≤ |Q|, then R ⊂ 3Q ⊂ 3Q 4−n λ,j , which establishes (2.1.29). It follows from (2.1.26) that α(λ) ≤ ω
λ x : M f (x) > A ]
+ C 2A−1
δ
α 2−n−1 λ .
For N > 0, by (2.1.27), applying the following equality Z
∞
p0 λ
p0 −1
β(λ)dλ =
Z
[M f (x)]p0 ω(x)dx,
Rn
0
we conclude that IN =
Z
N
pλp−1 α(λ)dλ
0
Z
N
pλp−1 β(λ)dλ Z N −1 p−p0 p0 λp0 −1 β(λ)dλ ≤ pp0 N 0 Z −1 p−p0 (M f (x))p0 ω(x)dx ≤ pp0 N
≤
0
Rn
< ∞.
(2.1.30)
62
Chapter 2. Singular Integral Operators On the other hand, (2.1.30) implies that Z N λ p−1 ] pλ ω x : M f (x) > dλ IN ≤ A 0 Z N −1 δ + C 2A pλp−1 α(2−n−1 λ)dλ 0 Z N λ p−1 ] pλ ω x : M f (x) > dλ = A 0 Z 2−n−1 N (n+1)p −1 δ + C2 2A pλp−1 α(λ)dλ 0 Z N δ λ p−1 ] ≤ pλ ω x : M f (x) > dλ + C2(n+1)p 2A−1 IN . A 0
Now we take A > 0 such that C2(n+1)p (2A−1 )δ = 21 , then Z N λ p−1 ] pλ ω x : M f (x) > dλ. IN ≤ 2 A 0
As N tends to ∞, we have Z ∞ Z pλp−1 α(λ)dλ ≤ 2 0
∞
pλ
p−1
0
ω
λ x : M f (x) > A ]
dλ.
This together with (2.1.28) implies that Z ∞ Z p [M f (x)] ω(x)dx = pλp−1 β(λ)dλ n R 0 Z ∞ λ p−1 pλ α ≤ C1 dλ C2 0 Z ∞ =C pλp−1 α(λ)dλ Z0 ∞ ≤C pλp−1 ω({x : M ] f (x) > λ})dλ Z0 p =C M ] f (x) ω(x)dx. Rn
Theorem 2.1.6 Suppose that Ω is a homogeneous bounded function of degree 0 and satisfies (2.1.8) and (2.1.9). Then there exists a constant C independent of f , for 1 < p < ∞ and ω ∈ Ap (Rn ), such that Z Z |TΩ f (x)|p ω(x)dx ≤ C |f (x)|p ω(x)dx. (2.1.31) Rn
Rn
63
2.1 Calder´on-Zygmund singular integral operators
Proof. Since ω ∈ Ap , there exists 1 < s < p such that ω ∈ A p . First we s will prove (2.1.31) under the condition that f is a bounded function with compact support. To the end, we will first illustrate M (T Ω f )(x) ∈ Lp (ω). It suffices to prove TΩ f ∈ Lp (ω). Without loss of generality we may assume that supp(f ) ⊂ {y : |y| < R}. If |x| > 2R, then |x − y| ≥ |x| − |y| >
|x| . 2
It follows that |TΩ f (x)| ≤ Set Z
Rn
p
Z
|y|
|TΩ f (x)| ω(x)dx =
Z
C |Ω(x − y)| |f (y)|dy ≤ kf k∞ . n |x − y| |x|n
p
|x|≤2R
|TΩ f (x)| ω(x)dx +
:= I1 + I2 .
Z
|x|>2R
(2.1.32)
|TΩ f (x)|p ω(x)dx
For some ε > 0, H¨older’s inequality implies that I1 ≤
Z
|x|≤2R
|TΩ f (x)|p(1+ε)/ε dx
!
ε 1+ε
Z
|x|≤2R
ω(x)1+ε dx
!
1 1+ε
.
Then the Lq −boundedness of TΩ (to see Theorem 2.1.2) implies that the first part of right side of the above inequality is bounded. Since ω satisfies the inverse H¨older inequality, as ε is small enough, we deduce that the second part of right side of above inequality is also bounded. In order to illustrate that I2 is bounded, we need the following property of A p weight. If ω ∈ Ap (1 ≤ p < ∞), then for any cube Q ⊂ Rn and a > 1 we have ω(aQ) ≤ Canp ω(Q),
(2.1.33)
where C only depends on ω but not on a. In fact, for any f ∈ L p (ω) and λ > 0, Theorem 1.4.2 implies that Z C |f (x)|p ω(x)dx. ω({x ∈ Rn : M f (x) > λ}) ≤ p λ Rn Now we take f ≥ 0 and cube Q such that Z f (y)dy > 0. Q
64
Chapter 2. Singular Integral Operators
Thus for all 0 < λ < fQ , we have Q ⊂ {x : M (f χQ )(x) > λ}, which implies that Z C ω(Q) ≤ p |f (x)|p ω(x)dx. λ Q So p
ω(Q)λ ≤ C
Z
|f (x)|p ω(x)dx.
Q
Now let λ → fQ , then p
ω(Q) (fQ ) ≤ C
Z
Q
|f (x)|p ω(x)dx.
If we take f = χS with S ⊂ Q being measurable set, then the above inequality becomes |S| p ≤ Cω(S). ω(Q) |Q| Replacing Q by aQ and S by Q in the above formula will yield (2.1.33). Clearly if we replace cube Q by a ball in (2.1.33), the inequality still holds. Now we turn our attention to the estimate of I 2 . For ω ∈ Ap there exists 1 < q < p such that ω ∈ Aq . From (2.1.32) and (2.1.33), it follows that I2 ≤ C ≤C ≤C
∞ Z X
k k+1 R k=1 2 R<|x|≤2 ∞ −np X k
ω B(0, 2k+1 R)
2 R
k=1
∞ X k=1
2k R
ω(x) dx |x|np
−np
≤ C(n, R, ω) < ∞.
nq C 2k+1 R ω(B(0, 1))
We have proved that TΩ f ∈ Lp (ω), therefore M (TΩ f ) ∈ Lp (ω). If f is a bounded function with compact support, then applying Lemma 2.1.2 and
65
2.1 Calder´on-Zygmund singular integral operators Lemma 2.1.3 we have Z Z p |TΩ f (x)| ω(x)dx ≤ Rn
Rn
≤C ≤C ≤C
Z
Z
(M (TΩ f )(x))p ω(x)dx
Rn
ZR
n
Rn
M ] (TΩ f )(x)
p
ω(x)dx
p
(M (|f |s )(x)) s ω(x)dx
|f (x)|p ω(x)dx.
Finally, since the set of bounded functions with compact support is dense in Lp (ω), TΩ can be extended to Lp (ω) continuously. Theorem 2.1.7 Suppose that Ω satisfies the conditions of Theorem 2.1.6 and ω ∈ A1 . Then there exists a constant C, for any λ > 0 and f ∈ L 1 (ω), such that Z C ω({x ∈ Rn : |TΩ f (x)| > λ}) ≤ |f (x)|ω(x)dx. λ Rn Proof. The proof is similar to that of truncated operator T ε being of weak type (1,1) in Theorem 2.1.1. Using the Calder´on-Zygmund decomposition of f and λ, we have f = g + b and a series of nonoverlapping cubes {Q k }. Clearly
λ ω({x : |TΩ f (x)| > λ}) ≤ ω x : |TΩ g(x)| > 2 := I1 + I2 .
+ω
λ x : |TΩ b(x)| > 2
We will give the estimate of I1 and I2 respectively. Since ω ∈ A1 ⊂ A2 and TΩ is bounded on L2 (ω), we have Z 4 I1 ≤ 2 |TΩ g(x)|2 ω(x)dx λ Rn Z 4C |g(x)|2 ω(x)dx ≤ 2 λ Rn Z 2n 4C ≤ |g(x)|ω(x)dx. λ Rn
66
Chapter 2. Singular Integral Operators
Noting the definition of g in the Calder´on-Zygmund decomposition, then C I1 ≤ λ ≤
C λ
≤
C λ
≤
C λ
Z
Z Z 1 CX |f (x)|ω(x)dx + |f (y)|dyω(x)dx S λ R n \ Qk Qk |Qk | Qk k Z Z 1X ω(Qk ) |f (x)|ω(x)dx + |f (y)| dy S λ |Qk | n R \ Qk Q k k Z Z CX |f (x)|ω(x)dx + |f (y)|ω(y)dy S λ R n \ Qk Qk k Z |f (x)|ω(x)dx. Rn
To estimate I2 , we will use some ideas from the proof of Lemma 2.1.2. For every Qk , let Bk∗ be the ball of which the center is the same as Q k and the diameter is 16 times of that of Qk . Then by (2.1.33), we obtain that
ω
[ k
Bk∗
!
≤
X k
≤C
≤C
ω(Bk∗ )
X
ω(Qk )
k
|Qk |
k
X ω(Qk )
|Qk |
X ω(Qk ) 1 Z
|f (y)|dy |Qk | λ Qk Z CX ≤ |f (y)|ω(y)dy λ Q k Zk C |f (y)|ω(y)dy. ≤ λ Rn ≤C
k
Finally, we will give the estimate of
ω
(
x ∈ Rn \
[ k
λ Bk∗ : |TΩ b(x)| > 2
Denote the center of Qk by ck . Since
R
Qk bk (x)dx
)!
.
= 0 for every k, we have
2.1 Calder´on-Zygmund singular integral operators that ω
(
n
x∈R \
[ k
Bk∗
λ : |TΩ b(x)| > 2
67
)!
Z CX ≤ |TΩ bk (x)|ω(x)dx λ Rn \Bk∗ k Z Z CX Ω(x − c ) Ω(x − y) k = − bk (y)dy ω(x)dx n n λ |x − y| |x − ck | Rn \Bk∗ Qk k ! Z Z Ω(x − y) Ω(x − ck ) CX ω(x)dx dy. |bk (y)| − ≤ n λ |x − ck |n Rn \Bk∗ |x − y| Qk k
Using the method in the proof of (2.1.22) and the property of A 1 . it is not hard to get that Z Ω(x − y) Ω(x − ck ) ω(x)dx ≤ CM ω(y) ≤ Cω(y) − (2.1.34) n |x − ck |n Rn \Bk∗ |x − y| for any y ∈ Qk . Thus by (2.1.34), we obtain that )! ( Z [ CX λ n ∗ ≤ |bk (y)|ω(y)dy ω x ∈ R \ Bk : |TΩ b(x)| > 2 λ Q k k Zk C (|f (y)| + |g(y)|)ω(y)dy ≤ λ Rn Z C |f (y)|ω(y)dy. ≤ λ Rn
Using the Cotlar inequality (Lemma 2.1.1), the weighed norm inequality of the Hardy-Littlewood maximal operator M (Theorem 1.4.2 and Theorem 1.4.3) as well as the weighed norm inequality of the Calder´on-Zygmund singular integral operator TΩ , we can immediately obtain the weighed norm inequality of the maximal singular integral operator. Theorem 2.1.8 Suppose that Ω is a homogeneous bounded function of degree 0 and satisfies (2.1.8) and (2.1.9). Then T Ω∗ is bounded on Lp (ω) for 1 < p < ∞ and ω ∈ Ap , and TΩ∗ is bounded from L1 (ω) to L1,∞ (ω) with ω ∈ A1 . Corollary 2.1.3 Let 1 ≤ p < ∞ and ω ∈ Ap (Rn ). Then (i) H and H ∗ are both bounded from Lp (ω) to Lp (ω) for 1 < p < ∞; (ii) H and H ∗ are both bounded from L1 (ω) to L1,∞ (ω).
68
Chapter 2. Singular Integral Operators
Corollary 2.1.4 Let 1 ≤ p < ∞ and ω ∈ Ap (Rn ), Rj and Rj∗ be the Riesz transform and the maximal Riesz transform respectively (j = 1, 2, · · · ). Then (i) Rj and Rj∗ are both bounded operator from Lp (ω) to Lp (ω) for 1 < p < ∞; (ii) Rj and Rj∗ are both bounded operators from L1 (ω) to L1,∞ (ω). Next we are going to investigate the vector-valued singular integral operator and its applications. Suppose that B is a separable Banach space and B ∗ is the dual space of B. The mapping f : R n → B is called a B-value function. A B-value function f is called measurable, if for any b 0 ∈ B ∗ , the mapping x → hf (x), b0 i is measurable. If a B-value function f is measurable, then the function x → kf (x)kB is also measurable. For 1 ≤ p ≤ ∞, define ( ) 1 Z Lp (B) =
f : kf kLp (B) =
Rn
kf (x)kpB dx
p
< ∞ , 1 ≤ p < ∞,
L∞ (B) = f : kf kL∞ (B) = sup {kf kB : x ∈ Rn } < ∞ .
Lp (B)(1 ≤ p ≤ ∞) are all Banach spaces.
Similarly, weak Lp (B) (1 ≤ p < ∞) spaces are defined by 1 p,∞ n p LB = f : kf kLp,∞ (B) = sup t |{x ∈ R : kf (x)kB > t}| < ∞ . t>0
We now formulate a dense subspace of L p (B). If f ∈ Lp (Rn ) is a realvalued function and b ∈ B, then the function (f · b) ∈ L p (B), where (f · b)(x) := f (x)b. The set of all the finite linear combination of all the vectorvalued functions of this form is a dense subspace of L p (B), and we denote N p it by L B. For X O F = fj bj ∈ L 1 B, j
its integral is defined by Z X Z F (x)dx = Rn
j
Rn
fj (x)dx bj ∈ B.
N Thus since L1 B is dense in L1 (B), the above definition can be continuously extended to L1 (B). And for F ∈ L1 (B), its integral on Rn is the unique member in B satisfying the following equation Z Z 0 F (x)dx, b = hF (x), b0 idx Rn
Rn
69
2.1 Calder´on-Zygmund singular integral operators 0
for every b0 ∈ B ∗ . Suppose that F (x) ∈ Lp (B), G ∈ Lp (B ∗ ), then hF (x), G(x)i is an integrable function on Rn and Z kGkLp0 (B ∗ ) = sup hF (x), G(x)idx : kF kLp (B) ≤ 1 . Rn
0
Especially, when 1 ≤ p < ∞ and B is a reflexible space, L p (B ∗ ) = (Lp (B))∗ .
Suppose that both A and B are two Banach spaces. Denote L (A, B) to be the set of all bounded linear operators from A to B. Suppose that K is a L (A, B)-valued function defined on R n \{0}. Let f ∈ L∞ (A) be A-valued function and be compactly supported. Then the operator T associated to K is defined by Z K(x − y)f (y)dy, x∈ / suppf. T f (x) = Rn
This operator T is called vector-valued singular integral operator. To obtain the Lp boundedness of vector-valued singular integral operator, we first introduce a lemma.
Lemma 2.1.4 Suppose that T is a linear operator on L ∞ (A) defined above with compact support. (i) There exists 1 < p0 < ∞ such that |Eλ (T f )| ≤
C kf kpL0p0 (A) , λ p0
where λ > 0 and Eλ (T f ) = {x ∈ Rn : kT f (x)kB > λ}; (ii) Let U = U (x0 , r) be the ball with the center at x0 and radius r and 2U be the twice expansion of U at Rthe same center. For all functions f in L1 (A) satisfying supp(f ) ⊂ U and Rn f (x)dx = 0, we have Z
Rn \2U
kT f (x)kB dx ≤ Ckf kL1 (A) .
(2.1.35)
If both (i) and (ii) hold, then kT f kLp (B) ≤ Ckf kLp (A) holds for 1 < p < p0 , where C is independent of f .
70
Chapter 2. Singular Integral Operators
Proof. We will first prove that, for any λ > 0, |Eλ (T f )| ≤
C kf kL1 (A) . λ
(2.1.36)
For λ > 0, using the Calder´on-Zygmund decomposition to real-valued function kf (x)kA , we get a seguence {Qj } with nonoverlapping cubes. Now we define A-value functions g and h such that f = g + h with [ f (x), x∈ / Qj ; j Z g(x) = 1 f (x)dx x ∈ Qj , |Qj | Qj Z h(x)dx = 0, for every j. Qj
Thus it follows that kgkA ≤ 2n λ, kgkL1 (A) ≤ kf kL1 (A) and khkL1 (A) ≤ 2kf kL1 (A) . Thus from the condition (i) it follows that 2 p0 C p (T g) E ≤ p0 kgkL0p0 (A) λ 2 λ Z 2 p0 C (2n λ)p0 −1 kg(x)kA dx ≤ p0 λ n R C n(p0 −1)+p0 ≤ 2 kgkL1 (A) λ C2n(p0 −1)+p0 ≤ kf kL1 (A) . λ Let Sj be a ball with the same center and radius as Q j , Sj∗ = 2Sj and N X S ∗ D1 = j Sj . Put hj = hχQj . Then hj converges to h in Lp0 (A). It j=1
follows from condition (i) that N X hj → T h, as N → ∞ T j=1
71
2.1 Calder´on-Zygmund singular integral operators in the sense of measure. Thus kT hkB ≤
∞ X j=1
kT hj kB .
Consequently, it follows from the condition (ii) that Z ∞ Z X kT hj kB dx kT hkB dx ≤ Rn \D1
≤
n j=1 R \D1 ∞ Z X ∗ n j=1 R \Sj ∞ X
≤C
j=1
kT hj kB dx
khj kL1 (A)
≤ C 0 kf kL1 (A) . On the other hand, we have X X X1Z Cn n ∗ kf kL1 (A) . kf kA dx ≤ |D1 | ≤ |Sj | ≤ C2 |Qj | ≤ Cn λ Qj λ j
Thus
j
j
Z C 2 kT hkB dx ≤ kf kL1 (A) . E λ (T h) ≤ |D1 | + 2 λ Rn \D1 λ
From above it follows (2.1.36). By (2.1.36) and the condition (i) as well as using the Marcinkiewicz interpolation theorem, it follows that kT f kLp (B) ≤ Ckf kLp (A) for 1 < p < p0 . Theorem 2.1.9 Let A and B be two reflexive Banach spaces. Suppose that there exists a p0 , 1 < p0 < ∞, such that T is a bounded operator from L p0 (A) to Lp0 (B) and K(x) satisfies Z kK(x − y) − K(x)kL (A,B) dx ≤ C, (2.1.37) |x|≥2|y|
Then T is bounded from Lp (A) to Lp (B) with 1 < p < ∞ and T is of weak type (1,1), i.e., there is C > 0 such that, for any λ > 0 and f ∈ L 1 (A), |Eλ (T f )| = |{x ∈ Rn : kT f (x)kB > λ}| ≤
C kf kL1 (A) . λ
72
Chapter 2. Singular Integral Operators
Proof. By the given condition we can easily deduce that T is of weak (p0 , p0 ) type, i.e., for any λ > 0, C kf kpL0p0 (A) . λ p0 Next we will show that T satisfies the condition (ii) in Lemma 2.1.4 by (2.1.37). Actually, suppose that a function f satisfies that supp(f ) ⊂ U (x0 , r) and its mean value is zero. From (2.1.37) it follows that
Z Z Z
K(x − y)f (y)dy kT f (x)kB dx =
dx
Rn \2U Rn Rn \2U B
Z Z
=
n K(x − y)f (y − x0 )dy dx R |x|>2r B
Z
Z
=
n [K(x − y) − K(x)]f (y − x0 )dy dx R |x|>2r B Z Z kf (y − x0 )kA ≤ kK(x − y) − K(x)kL (A,B) |Eλ (T f )| ≤
|y|≤r
|x|>2r
× dxdy
≤ Ckf kL1 (A) . Thus applying Lemma 2.1.4 we have kT f kLp (B) ≤ Ckf kLp (A)
(2.1.38)
for 1 < p < p0 . Now let K ∗ be the conjugate operator of K. K ∗ is a function from n R → L (B ∗ , A∗ ). For any function g in L∞ (B) with compact support, define linear operator Z T g(x) = K ∗ (y − x)g(y)dy, Rn
then T is the conjugate operator of T . It is easy to see that, for 1 < p < ∞, p1 + p10 = 1, kT gkLp0 (A∗ ) ≤ CkgkLp0 (B ∗ ) (2.1.39) if and only if
kT f kLp (B) ≤ Ckf kLp (A) . 0
0
It follows that T is a bounded operator from L p0 (B ∗ ) to Lp0 (A∗ ). Clearly (2.1.37) implies that K ∗ satisfies Z kK ∗ (y − x) − K ∗ (x)kL (B ∗ ,A∗ ) dx ≤ C. |x|≥2|y|
73
2.1 Calder´on-Zygmund singular integral operators Thus, for 1 < q < p00 , we have kT gkLq (A∗ ) ≤ CkgkLq (B ∗ ) . This, together with (2.1.39), yields that kT f kLp (B) ≤ Ckf kLp (A) ,
p0 < p < ∞.
Thus we have proved that T is a bounded linear operator from L p (A) to Lp (B) (1 < p < ∞). Thus Lemma 2.1.4 implies that T is of weak type (1,1). Remark 2.1.6 For simplicity of the proof, we add a condition on reflexivite in Theorem 2.1.9. However, it should be pointed out that the condition on the reflexivite of A and B in Theorem 2.1.9 can be removed (see [RuRT]). Next we will give some application of Theorem 2.1.9. Theorem 2.1.10 Suppose that K satisfies the H¨ ormander condition (2.1.3) and T is a convolution operator generated by K. If T is of type (2, 2), then for any r, p, 1 < p, r < ∞, we have
1 1
r r
X
X
r r |fj | |T fj |
≤ Cp,r
.
j
j
p
p
When p = 1, we have
1 1
r r C X
X
r |T fj (x)|r > λ ≤ |fj |r . x ∈ Rn :
λ j
j
1
Proof. It is clear that T is operator of type (p, p) (1 < p < ∞) under the given conditions. Take A = B = l r and let fe = (f1 , f2 , · · · ) ∈ Lp (lr ). Then Te is bounded from Lr (lr )N to Lr (lr ), where Tefe = (T f1 , T f2 , · · · ). The kernel corresponding to Te is K I, where I is the identity operator on l r . Thus, Z Z k(K(x − y) − K(x))Ik L (lr ,lr ) dx = |K(x − y) − K(x)|dx ≤ C. |x|>2|y|
|x|≥2|y|
Thus Theorem 2.1.10 holds. Another application of Theorem 2.1.9 is to show the boundedness of the Littlewood-Paley operator.
74
Chapter 2. Singular Integral Operators
Theorem 2.1.11 Suppose that ϕ(x) ∈ L 1 (Rn ) satisfies the following conditions: Z ϕ(x)dx = 0; (i) Rn
(ii) |ϕ(x)| ≤ C(1 + |x|)−n−α , x ∈ Rn ; Z (iii) |ϕ(x + h) − ϕ(x)|dx ≤ C|h|α , h ∈ Rn , Rn
for some C > 0 and α > 0. Then the Littlewood-Paley operators 1 2 ∞ X 2 |ϕ2j ∗ f (x)| Gf (x) = j=−∞
and
4f (x) =
Z
∞
0
dt |ϕt ∗ f (x)| t 2
1
2
are of type (p, p) (1 < p < ∞) and of weak type ( 1, 1), where 1 x ϕ2j (x) = jn ϕ j 2 2
and
ϕt (x) =
1 x ϕ . tn t
Proof. For the operator G, let A = C, B = l 2 and K(x) = {ϕ2j (x)}∞ j=−∞ , then by the Plancherel Theorem and Theorem 2.1.9, if we can prove , there exists C > 0 such that X |d ϕ2j (ξ)|2 ≤ C, ξ ∈ Rn (2.1.40) j
and Z
|x|≥2|y|
1 2 X 2 dx ≤ C, |ϕ2j (x − y) − ϕ2j (x)| j
y ∈ Rn ,
(2.1.41)
then the operator G is of type (p, p) and of weak type (1,1) (1 < p < ∞). Similarly, for the operator 4, take B = L 2 R+ , dtt , then if we can prove that there exists C > 0 such that Z ∞ 2 dt ≤ C, ξ ∈ Rn (2.1.42) |ϕ(tξ)| b t 0
75
2.1 Calder´on-Zygmund singular integral operators and Z
|x|≥2|y|
Z
∞ 0
|ϕt (x − y) − ϕt (x)|
2 dt
t
1
2
y ∈ Rn ,
dx ≤ C,
(2.1.43)
then 4 is operator of type (p, p) (1 < p < ∞) and of weak type (1,1). Since validations of the former and the latter are essentially the same, it suffices to prove (2.1.42) and (2.1.43). When 0 < |ξ| < 1, from the vanishing condition (i) of ϕ, it follows that Z −2πix·ξ |ϕ(ξ)| b = ϕ(x) e − 1 dx Rn Z Z −2πix·ξ |ϕ(x)| e |ϕ(x)|dx − 1 dx + C ≤ 1 1 |x|>|ξ|− 2 |x|≤|ξ|− 2 Z ∞ 1 dt 2 ≤ 2πkϕk1 |ξ| + C 1 1+α −2 t |ξ| ≤ C|ξ|β ,
where β = inf 12 , α2 . When |ξ| ≥ 1, note that
Taking h = that
ξ , 2|ξ|2
[ϕ(· + h) − ϕ(·)ˆ](ξ) = ϕ(ξ) b e2πih·ξ − 1 .
from the above equality and the condition (iii) it follows
2 |ϕ(ξ)| b ≤
Z
Rn
|ϕ(x + h) − ϕ(x)|dx ≤ C|h|α ≤ C|ξ|−α .
For any ξ 6= 0, denote ξ 0 = Z
∞ 0
ξ |ξ| .
Z
dt = |ϕ(tξ)| b t 2
0
Then
∞
0 2 dt ϕ(t|ξ|ξ b ) = t
The above estimates implies that Z
0
∞
2 dt ϕ(tξ b 0 ) ≤ t
Z
0
1
Ct
2β−1
dt +
Z
∞ 1
Z
0
∞
2 dt ϕ(tξ b 0 ) . t
Ct−1−α dt ≤ C1 < ∞.
Next we will validate (2.1.43). It is easy to see that the left side of (2.1.43) is dilation invariant with respect to y. Thus we merely need to prove (2.1.43)
76
Chapter 2. Singular Integral Operators
in the case |y| = 1. The Schwarz inequality implies that Z
|x|>2
≤
∞
0
Z
≤ Cα
Z
|x|>2
"Z
0
|ϕt (x − y) − ϕt (x)|
|x|
∞
t
−n− α 2
α 2
2
dx
Z
|x|> 2t
Now let I(t) = If |x| >
2 t
!1
Z
Z
2 dt
|x|>2
t Z
1
0
2
∞
dx
|ϕt (x − y) − ϕt (x)| |x|
2 α y − ϕ(x) |x|n+ 2 dx ϕ x − t
|x|> 2t
2
!
dt t
n+ α 2
#1
dt dx t
2
.
2 α y − ϕ(x) |x|n+ 2 dx. ϕ x − t
and |y| = 1, then we have y 1 |x| . x − ≥ |x| − ≥ t t 2
When 0 < t < 1, the condition (ii) yields Z 3α dx 2 . I(t) ≤ C 3α ≤ Ct n+ 2 2 |x|> t |x|
On the other hand, take 0 < ε < α2 , by (ii) we have y C C C ≤ ≤ n+ε . − φ(x) ≤ φ x − t (1 + |x|)n+α (1 + |x|)n+ε |x| When t ≥ 1, the condition (ii) gives, noting that |x| > 2t , α α α y − ϕ(x) · |x|n+ 2 ≤ C|x| 2 −ε ≤ Ctε− 2 . ϕ x − t
Therefore applying the condition (iii) we have Z α 3α y I(t) ≤ Ctε− 2 x − − ϕ(x) ϕ dx ≤ Ctε− 2 . 2 t |x|> t
Consequently Z ∞ Z ∞ Z 1 α α 3α α 3α dt + −1 t 2 I(t) ≤ Ct 2 − 2 +ε−1 dt < ∞, Ct 2 2 dt + t 0 1 0 which proves (2.1.43).
!1
2
77
2.1 Calder´on-Zygmund singular integral operators Remark 2.1.7
(1) Clearly, if ϕ ∈ S (Rn ), then ϕ satisfies (ii) and (iii).
(2) The conditions (ii) and (iii) can be weakened. Later we will see that there exists a class of functions in L 1 (Rn ), which only satisfies the vanishing condition (i) and weaker conditions on size, such that the Littlewood-Paley operators G and 4 are both of type (p, p)(1 < p < ∞) and weak type (1,1). The next consequence is a corollary of Theorem 2.1.9 and Theorem 2.1.11, and it is very useful in our following discussion.
Corollary 2.1.5 Suppose that ϕ satisfies the conditions of Theorem 2.1.11. → − If f = (· · · , f−1 , f0 , f1 , · · · ) ∈ Lp (l2 )(Rn ) for 1 < p < ∞, then we have that
1 1
2 2
X
XX
2 2 |f (·)| |ϕ ∗ f (·)| ≤ C
j j p 2k
j∈Z
j∈Z k∈Z
p
and
(2.1.44) p
1 1
Z
2 2
∞X
X
2 2 dt |f (·)| |ϕ ∗ f (·)| ≤ C
. j t j p
0
t j∈Z
j∈Z
p
Proof. assert
(2.1.45)
p
Let us now give the proof of (2.1.44). Let A = l 2 , B = l2 (l2 ). we 2 2 {ajk }∞ j,k=−∞ ∈ l (l ),
provided that
X
k∈Z
|ajk |2
!1 2
∈ l2 . j∈Z
→ − Now for f ∈ L2 (l2 )(Rn ), define an operator T as follows: → − T f (x) =
X
k∈Z
|ϕ2k ∗ fj (x)|2
!1 2
. j∈Z
78
Chapter 2. Singular Integral Operators
Then when p = 2, Theorem 2.1.11 shows that Z XX
→
− 2 |ϕ2k ∗ fj (x)|2 dx =
T f 2 L (B)
Rn
≤C
j
XZ j
k
Rn
|fj (x)|2 dx
→
− 2 = C f 2
L (A)
.
e e Let K(x) be the kernel of T . Then K(x) : Rn → L (l2 , l2 (l2 )). On the other hand, let K(x) = {ϕ2k (x)}k∈Z . Then for any x ∈ Rn , we have
e ≤ kK(x)kL (C,A) . Thus, we have that
K(x) L (A,B)
Z
|x|≥2|y|
≤ =
Z
Z
e e
K(x − y) − K(x)
|x|≥2|y|
|x|≥2|y|
≤ C,
L (A,B)
dx
kK(x − y) − K(x)kL (C,A) dx X k
!1 2
|ϕ2k (x − y) − ϕ2k (x)|
2
dx
for every y ∈ Rn . The last inequality is derived from Theorem 2.1.11. Theorem 2.1.9 tells us that T is bounded operator from L p (A) to Lp (B), i.e. (2.1.44) holds. Remark 2.1.8 (i) The conclusion of Theorem 2.1.9 shows that the corresponding inequality of weak type still holds under the condition of Corollary 2.1.5. (ii) The conclusion of Corollary 2.1.5 is still valid for ϕ ∈ S (R n ) and Z ϕ(x)dx = 0. Rn
2.2
Singular integral operators with homogeneous kernels
In the first section of this chapter, we have studied the L p boundedness (1 < p < ∞) and weak (1, 1) boundedness of the Calder´on-Zygmund singular integral operators. However, it is noticed that under the condition on
79
2.2 Singular integral operators with homogeneous kernels
the kernel K, we require that K satisfies the size-condition (2.1.1) and the smooth condition (2.1.3) (i.e., H¨ormander condition) besides the vanishing condition (2.1.2). As to singular integral operators with homogeneous kernel (for its definition, see (2.1.12)), besides vanishing condition (2.1.8), we still require that Ω satisfies L∞ -Dini condition (2.1.9). Then a natural question is raised up: can we weaken the condition of Ω but still ensure the L p boundedness (1 < p < ∞) and weak (1, 1) boundedness of the corresponding singular integral operator TΩ ? In this section, we will discuss this problem. Note that the operator TΩ commutes with dilation δε (ε > 0), so Ω should satisfy the homogeneous condition of degree 0, that is Ω(λx) = Ω(x)
(2.2.1)
holds for any λ > 0 and every x ∈ Rn . In addition, from the discussion before Theorem 2.1.2, we know that Ω(x0 ) must satisfy the vanishing condition on Sn−1 , i.e., Z Ω(x0 )dσ(x0 ) = 0.
(2.2.2)
Sn−1
Thus we can only weaken the size-condition and smooth condition on Ω. Let us now pay attention to investigating the L p boundedness and weak type (1,1). Since L∞ (Sn−1 ) & Lq (Sn−1 ) (1 ≤ q < ∞), we first consider Lq -Dini condition instead of L∞ -Dini condition. Now we give the following definition. Definition 2.2.1 (Lq -Dini condition) We say a function Ω(x 0 ) on Sn−1 satisfies Lq -Dini condition if (i) Ω ∈ Lq (Sn−1 )(1 ≤ q < ∞), Z 1 ωq (δ) (ii) dδ < ∞, δ 0 where ωq (δ) is called integral continuous modulus of Ω of degree q. Its definition is, for δ > 0, ωq (δ) = sup kρk<δ
Z
Sn−1
Ω(ρx0 ) − Ω(x0 ) q dσ(x0 )
1/q
,
where ρ is a rotation in Rn . ||ρ|| = sup |ρx0 − x0 | : x0 ∈ S n−1 .
80
Chapter 2. Singular Integral Operators
Remark 2.2.1 1. If Ω(x0 ) satisfies Lipschitz condition on S n−1 , that is, there exists a constant M > 0, such that ∀x0 , y 0 ∈ S n−1 , |Ω(x0 ) − Ω(y 0 )| ≤ M |x0 − y 0 |, then Ω must satisfy Lq -Dini condition (1 ≤ q ≤ ∞). 2. If 1 ≤ r < q ≤ ∞, then Ω satisfies Lr -Dini condition whenever it satisfies Lq -Dini condition. Suppose Ω is a homogeneous function of degree 0 on R n . Put K(x) = Ω(x)|x|−n , x ∈ Rn . The following theorem gives some important relations. Theorem 2.2.1 (a) If Ω satisfies L1 -Dini condition, then Ω ∈ L(log + L) (Sn−1 ) and K satisfies H¨ ormander condition, that is Z |K(x − y) − K(x)|dx ≤ C, ∀y ∈ Rn . (2.2.3) |x|>2|y|
(b) If K satisfies H¨ ormander condition (2.2.3), then Ω ∈ L(log + L)(Sn−1 ) 1 and Ω satisfies L -Dini condition. The proof of Theorem 2.2.1 can be found in Calder´on- Weiss- Zygmund [CaWZ] and Calder´on- Zygmund [CaZ4]. The following consequence improves Theorem 2.1.2. Theorem 2.2.2 Suppose that Ω satisfies (2.2.1), (2.2.2) and L 1 -Dini condition. Let Z Ω(x − y) f (y)dy TΩ f (x) = p.v. |x − y|n for f ∈ Lp (Rn ) (1 ≤ p < ∞). Then
(i) TΩ is of type (p, p) (1 < p < ∞); (ii) TΩ is of weak type (1,1). First we formulate a lemma which will be used in the proof of Theorem 2.2.2. Lemma 2.2.1 Suppose that Ω ∈ L∞ (Sn−1 ). Then, for any ε, 0 < ε < 1/2, there exists a constant C > 0 such that Z 2j+1 Z dr 0 ·y 0 0 −2πirx 0 ≤ CkΩkL∞ (Sn−1 ) 2−jε (2.2.4) Ω(y )e dσ(y ) r j n−1 2 S holds for every j ∈ N and x0 ∈ Sn−1 .
81
2.2 Singular integral operators with homogeneous kernels Proof. By H¨older’s inequality, the left side of (2.2.4) is dominated by
(log 2)
1 2
= (log 2)
Z
2j+1 2j
1 2
Z
0
Ω(y )e
−2πirx0 ·y 0
Sn−1
Z
Sn−1
Z
Ω(y Sn−1
Set I=
0
)Ω(u0 )
Z
2j+1
Z
2 ! 12 dr dσ(y ) r 0
2j+1
e
−2πirx0 ·(y 0 −u0 ) dr
r
2j
0
0
0
e−2πirx ·(y −u )
2j
!1 2
0
0
dσ(y )dσ(u )
.
dr . r
Clearly, we have |I| ≤ log 2. On the other hand, we obtain −1 |I| ≤ C 2j x0 · (y 0 − u0 ) ,
where C is independent of j, x0 , y 0 and u0 . Hence, for any α, 0 < α < 1, it follows that n −1 o −α |I| ≤ C min 1, 2j x0 · (y 0 − u0 ) ≤ C 2j x0 · (y 0 − u0 ) . (2.2.5) Since 0 < α < 1, there exists a constant C 1 independent of x0 , such that Z Z 1 dσ(y 0 )dσ(u0 ) ≤ C1 . 0 0 0 α Sn−1 Sn−1 |x · (y − u )|
Now let ε = (2.2.4).
α 2,
then the above inequality together with (2.2.5) implies
Next we turn to the proof of Theorem 2.2.2. First consider the case p = 2. Set K(x) = Ω(x)|x|−n . Then TΩ f = p.v. K ∗ f . We now give the ˆ estimate of K(ξ). Let ϕ ∈ S (Rn ). Then it follows that b ϕi = hK, ϕi hK, ˆ =
= =
lim
Z
ε→0,N →∞ ε≤|x|≤N
lim
Z
ε→0,N →∞ Rn
lim
Z
ε→0,N →∞ Rn
ϕ(y) ϕ(y)
K(x) Z Z
Z
e−2πix·y ϕ(y)dydx Rn
ε≤|x|≤N N/|y| Z ε/|y|
K(x)e−2πix·y dxdy 0
0
Ω(x0 )e−2πirx ·y dσ(x0 ) Sn−1
dr dy. r
82
Chapter 2. Singular Integral Operators
Clearly, if we can show that there exists a constant C independent of y 0 , such that Z ∞ Z 0 −2πirx0 ·y 0 0 dr (2.2.6) Ω(x )e dσ(x ) r ≤ C, n−1 0
S
b then the dominated convergence theorem implies | K(y)| ≤ C for every y ∈ n R ). Consequently TΩ is an operator of type (2, 2). Now set E0 = {x0 ∈ Sn−1 : |Ω(x0 )| < 2}
Ek = {x0 ∈ Sn−1 : 2k ≤ |Ω(x0 )| < 2k+1 },
k∈N
and Ωk (x0 ) = Ω(x0 )χEk (x0 )
(k ≥ 0).
Since Ω satisfies the vanishing condition, it implies that Z
0
While
2 Z
0
Ω(x ) e Sn−1
Z
2
∞ Z
−2πirx0 ·y 0
0
Ω(x )e
dr ≤ CkΩkL1 (Sn−1 ) . − 1 dσ(x ) r
−2πirx0 ·y 0
Sn−1 ∞ Z ∞ Z 2j+1 X X
0
dr dσ(x ) r 0
dr ≤ Ωk (x )e dσ(x ) r n−1 j j=1 2 k=0 S ∞ Z 2j+1 Z X dr 0 0 = Ω0 (x0 )e−2πirx ·y dσ(x0 ) r j Sn−1 j=1 2 ∞ Z 2j+1 Z ∞ X X 0 −2πirx0 ·y 0 0 dr + Ωk (x )e dσ(x ) r 2j Sn−1
0
−2πirx0 ·y 0
j=1 k=1
:= I1 + I2 .
Thus applying (2.2.4) leads to I1 ≤
∞ X j=1
C · 2−jε ≤ C.
0
83
2.2 Singular integral operators with homogeneous kernels Now take l ∈ N such that lε > 1. We have that ∞ X Z 2j+1 Z X 0 −2πirx0 ·y 0 0 dr I2 = Ω (x )e dσ(x ) k r j n−1 k=1 1≤j≤lk 2 ∞ X Z 2j+1 X
+
j k=1 j>lk 2
:=I21 + I22 .
S
Z
0
Sn−1
Ωk (x )e
−2πirx0 ·y 0
dr dσ(x ) r 0
Applying (2.2.4), we obtain I22 ≤ C
∞ X X
k=1 j>lk
kΩk kL∞ (Sn−1 ) 2−jε < ∞,
while I21 ≤
∞ X X
k=1 1≤j≤lk
2k+1 · |Ek | · log 2 ≤ CkΩkLlog+ L(Sn−1 ) < ∞.
Thus from the estimates of I1 , I21 and I22 , it follows (2.2.6). Therefore TΩ is an operator of type (2, 2). Now we illustrate that TΩ is of weak type (1, 1). In fact, by the condition (a) in Theorem 2.2.1, we know that K(x) = Ω(x)|x| −n satisfies H¨ormander condition (2.2.3). This, together with that T Ω is of type (2, 2), concludes that TΩ is of weak type (1, 1) (See the proof of Theorem 2.1.1). Finally, applying the Marcinkiewicz interpolation theorem and dual method yields that TΩ is of type (p, p) (1 < p < ∞). Remark 2.2.2 As stated above, when Ω ∈ Llog + L(Sn−1 ) and satisfies the vanishing condition, we can get L2 -boundedness of TΩ without requiring that Ω satisfies L1 -Dini condition. In fact, later we will obtain that T Ω is of type (p, p)(1 < p < ∞) and weak type(1, 1), provided that Ω ∈ Llog + L(Sn−1 ) and satisfies the vanishing condition (2.2.2). (See Theorem 2.3.4 and Remark 2.3.1). Now we start to discuss the weighted L p -boundedness of TΩ . By Theorem 2.1.6, if Ω satisfies the homogeneity of degree 0 as well as the vanishing condition, then L∞ -Dini condition (2.1.9) can ensure T Ω is bounded from Lp (ω) to Lp (ω) for 1 < p < ∞ and ω ∈ Ap . As to the conclusion of Theorem 2.2.2, a natural question is: if we weaken the L ∞ -Dini condition to L1 -Dini
84
Chapter 2. Singular Integral Operators
condition with 1 < p < ∞ and ω ∈ Ap , can we still ensure TΩ is bounded from Lp (ω) to Lp (ω)? However, there are counterexamples to illustrate that conclusion would not hold even ω ∈ A1 . So in the following we consider the Lq -Dini condition (1 < q < ∞) instead of the smooth condition of Ω. We will also discuss the weighted Lp -boundedness of the corresponding singular integral operator TΩ . Obviously we may predict that, in this situation, the weight function ω will depend not only on p, but also on q. We have the following statement.
Theorem 2.2.3 Suppose that Ω satisfies (2.2.1), (2.2.2) and L q -Dini condition (1 < q < ∞). When either of the following two conditions (i) or (ii) holds, TΩ is bounded from Lp (ω) to Lp (ω). (i) q 0 ≤ p < ∞ and ω ∈ Ap/q0 ; (ii) 1 < p ≤ q and ω −1/(p−1) ∈ Ap0 /q0 . 0
In addition, if ω q ∈ A1 , then TΩ is bounded from L1 (ω) to L1,∞ (ω), i.e., there exits a constant C > 0 such that ω ({x ∈ Rn : |TΩ f (x)| > λ}) ≤
C kf kL1 (ω) λ
for every λ > 0 and f ∈ L1 (ω). The proof of Theorem 2.2.3 will be obtained by applying the idea in the proof of Theorem 2.1.6. We first give some lemmas. Lemma 2.2.2 Suppose that Ω satisfies (2.2.1) and L q -Dini condition (1 ≤ q < ∞). Then for any R > 0 and x ∈ Rn , when |x| < R/2, there exists a constant C > 0 such that ) !1 ( Z 2|x| q Ω(y − x) Ω(y) q −n R ωq (δ) |x| dy − + dδ . ≤ CR q0 n |x| |y|n R δ R<|y|≤2R |x − y| R (2.2.7)
Z
Proof. Since |x − y| ∼ |y|, we have Ω(y − x) Ω(y) |Ω(y − x) − Ω(y)| |x| . |x − y|n − |y|n ≤ C |Ω(y)| |y|n+1 + |y|n
85
2.2 Singular integral operators with homogeneous kernels Hence
!1/q Ω(y − x) Ω(y) q dy − n |y|n R<|y|≤2R |x − y| !1/q Z q |x| ≤C |Ω(y)|q (n+1)q dy |y| R<|y|≤2R Z
Z
+C
R<|y|≤2R
|Ω(y − x) − Ω(y)|q dy |y|nq
!1/q
:= I1 + I2 . Since Ω ∈ Lq (Sn−1 ), it follows that 0
I1 ≤ CkΩkLq (Sn−1 ) |x| · R−(n+1) · Rn/q = CR−n/q ·
|x| . R
On the other hand, we have that Z 2R 1/q Z −nq+n−1 0 0 q 0 I2 ≤ C t |Ω(ty − x) − Ω(y )| dσ(y )dt R
≤ CR x t.
−n/q 0
Sn−1
Z
2R Z
R
S
0 q 1/q y − α 0 Ω dσ(y 0 ) dt − Ω(y ) , |y 0 − α| t n−1
where α = Applying a result of Calder`on- Weiss-Zygmund (see [CaWZ]) |x| to |α| = t < 21 and 1 ≤ q < ∞, we conclude that q 0 Z Z Ω y − α − Ω(y 0 ) dσ(y 0 ) ≤ C sup |Ω(ρx0 ) − Ω(x0 )|q dσ(x0 ) |y 0 − α| kρk<α Sn−1 Sn−1 q |x| ≤ Cωq t |x| ≤ Cωqq . R Here we use the property that ω(δ) is nondecreasing in δ. Thus, |x| 0 I2 ≤ CR−n/q · ωq · (log 2)1/q R Z 2|x|/R 0 |x| dδ ≤ C 0 R−n/q · ωq · R δ |x|/R Z 2|x|/R ωq (δ) 0 ≤ C 0 R−n/q dδ. δ |x|/R
86
Chapter 2. Singular Integral Operators
So (2.2.7) follows. Lemma 2.2.3 Suppose that Ω satisfies (2.2.1) and L q -Dini condition (1 < q < ∞). Then there exists a constant C > 0 such that Z i1/q0 h Ω(x − y) Ω(x − y) 0 q0 − (x ) f (y)dy ≤ C M |f | 0 |y−x0 |>4t |x − y|n |x0 − y|n (2.2.8) for any x0 ∈ Rn , t > 0 and every x with |x − x0 | < t, where M is the Hardy-Littlewood maximal function. Proof. From H¨older’s inequality and (2.2.7) it follows that Z Ω(x − y) Ω(x0 − y) − f (y)dy n n |y−x0 |>4t |x − y| |x0 − y| !1/q Z ∞ X Ω(x − y) Ω(x0 − y) q ≤ |x − y|n − |x0 − y|n dy j t<|y−x |≤2j+1 t 2 0 j=2 !1/q0 Z ×
0
|y−x0 |≤2j+1 t
|f (y)|q dy
! !1/q0 Z ωq (δ) 1 0 ≤C dδ |f (y)|q dy j+1 t)n δ (2 jt j+1 t |y−x |≤2 |x −x|/2 0 0 j=2 Z 0 1 1/q ωq (δ) 0 dδ ≤C M |f |q (x0 ) 1+ δ 0 1/q0 0 ≤C M |f |q (x0 ) . ∞ X
1 + 2j
Z
|x0 −x|/2j−1 t
Lemma 2.2.4 Suppose that Ω satisfies (2.2.1) and L q -Dini condition (1 < q < ∞). Then there exists a constant C = C(n, q) such that 1/q0 0 M ] (TΩ f ) (x) ≤ C(n, q) M |f |q (x) ,
x ∈ Rn ,
(2.2.9)
where M ] is the sharp maximal function defined in the section 2.1. Proof. The idea of proof is similar to that of Lemma 2.1.2. Choose n x ∈ R and a cube Q containing x. Let B be a ball with the same center
2.2 Singular integral operators with homogeneous kernels
87
and diameter as that of Q. Set f = f1 + f2 , where f1 = f · χ16B . Theorem 2.2.2 yields that 1 |Q|
Z
Q
|TΩ f1 (y)|dy ≤ C
1 |Q|
Z
16B
q0
|f (y)| dy
1/q0
1/q0 0 . ≤ C M |f |q (x)
On the other hand, denote the diameter of B by d. Then applying (2.2.8) implies that Z 1 |TΩ f2 (y) − TΩ f2 (x)|dy |Q| Q Z Z Ω(y − z) Ω(x − z) 1 − = f (z)dz dy n n |Q| Q Rn \16B |y − z| |x − z| 0 i h 1/q 0 . ≤C M |f |q (x) From (2.1.19), Lemma 2.2.4 follows. To prove Theorem 2.2.3, we still need a celebrated interpolation theorem. Lemma 2.2.5 (Stein-Weiss interpolation theorem with change of measure) Suppose that u0 , v0 , u1 , v1 are positive weight functions and 1 < p0 , p1 < ∞. Assume sublinear operator S satisfies : kSf kLp0 (u0 ) ≤ C0 kf kLp0 (v0 ) and kSf kLp1 (u1 ) ≤ C1 kf kLp1 (v1 ) . Then kSf kLp (u) ≤ Ckf kLp (v) holds for any 0 < θ < 1 and 1−θ θ 1 + , = p p0 p1 where pθ/p0 p(1−θ)/p1 , u1
u = u0 and C ≤ C0θ C11−θ .
pθ/p0 p(1−θ)/p1 v1
v = v0
88
Chapter 2. Singular Integral Operators
For the proof of Lemma 2.2.5, see Stein and Weiss [StW]. Proof of Theorem 2.2.3. First consider the case which the condition (i) holds. Analogous to the proof of Theorem 2.1.6, we merely need to discuss the bounded function with compact support. Assume f will be the case. We need to show M (TΩ f )(x) ∈ Lp (ω), provided that q 0 < p < ∞ and ω ∈ Ap/q0 . By the Lp -boundedness of M , it suffices to verify T Ω f ∈ Lp (ω). We may suppose supp(f ) ⊂ {x ∈ Rn : |x| < R} as well. Set Z Z Z p p |TΩ f (x)|p ω(x)dx |TΩ f (x)| ω(x)dx + |TΩ f (x)| ω(x)dx = Rn
|x|>2R
|x|≤2R
:= I1 + I2 .
By the Lr (1 < r < ∞)-boundedness of TΩ (Theorem 2.2.2) and the Reverse H¨older inequality, we see that I1 is finite. Thus we mainly consider I2 . Note that |y| < R and |x| > 2R. it implies that |x − y| ∼ |x|. Consequently, it follows that !p Z Z ω(x) dx I2 ≤ Ckf kpL∞ |Ω(x − y)|dy |x|np |x|>2R |y|
∞ X
0
(2j R)−np/q ω(B(0, 2j+1 R)).
j=1
Since ω ∈ Ap/q0 , there exists 1 < s < p/q 0 such that ω ∈ As . Hence applying (2.1.33) yields that ω(B(0, 2j+1 R)) ≤ C · (2j+1 R)ns ω(B(0, 1)), where C is independent of j. So we obtain that I2 ≤ C 00
∞ X j=1
0
(2j R)−n(p/q −s) ω(B(0, 1)) < ∞,
2.2 Singular integral operators with homogeneous kernels
89
which implies that M (TΩ f ) ∈ Lp (ω). Applying Lemma 2.1.3, Lemma 2.2.4 and the weighted boundedness of M (Theorem 1.4.3), we conclude that, for a bounded function f with compact support, Z Z p |TΩ f (x)| ω(x)dx ≤ [M (TΩ f )(x)]p ω(x)dx n Rn RZ ≤C [M ] (TΩ f )(x)]p ω(x)dx Rn Z h ip/q0 0 ≤C ω(x)dx M (|f |q )(x) n R Z ≤ C0 |f (x)|p ω(x)dx. Rn
When q 0 < p < ∞ and ω ∈ Ap/q0 , TΩ is bounded from Lp (ω) to Lp (ω). Consider the case p = q 0 , clearly ω ∈ A1 . Thus there exists p1 > p = q 0 such that ω p1 /p ∈ A1 ⊂ Ap1 /q0 . Thus there exists ε > 0 such that ω p1 (1+ε)/p ∈ Ap1 /q0 . Now set θ =
ε 1+ε
and choose p0 such that 1−θ 1 θ + . = p p0 p1
Clearly 1 < p0 < p. First we have kTΩ f kLp0 ≤ C0 kf kLp0 . On the other hand, the above conclusion implies that kTΩ f kLp1 (u1 ) ≤ C1 kf kLp1 (u1 ) , where u1 = ω p1 (1+ε)/p ∈ Ap1 /q0 .
It follows that TΩ is bounded from Lp (ω) to Lp (ω) by applying Lemma 2.2.5 to u0 = v0 = 1 and u1 = v1 = ω p1 (1+ε)/p . Up to now, we have shown that TΩ is bounded from Lp (ω) to Lp (ω) under the condition (i). Now we discuss the case which the condition (ii) holds. Let Tf Ω be the conjugate operator of TΩ . Then we have Tf Ω f (x) = TΩ e f (x),
90
Chapter 2. Singular Integral Operators
e e satisfies all the conditions on Ω. Consewith Ω(x) = Ω(−x). Clearly Ω p quently, for any f ∈ L (ω)(1 < p ≤ q), we have Z TΩ f (x) · g(x)dx , kTΩ f kLp (ω) = sup Rn
0
where the supremum is taken over all functions g ∈ L p (ω −1/(p−1) ) with kgkLp0 (ω−1/(p−1) ) ≤ 1. Clearly q 0 ≤ p0 < ∞ follows from 1 < p ≤ q. By ω −1/(p−1) ∈ Ap0 /q0 together with the above conclusion we know that TeΩ is 0
0
bounded from Lp (ω −1/(p−1) ) to Lp (ω −1/(p−1) ). Thus Z kTΩ f kLp (ω) = sup f (x)TeΩ g(x)dx Rn
≤ kf kLp (ω) sup kTeΩ gkLp0 (ω−1/(p−1) ) ≤ Ckf kLp (ω) .
Therefore this completes the proof of Theorem 2.2.3 under the condition (ii). Finally we wish to show that TΩ is bounded from L1 (ω) to L1,∞ (ω) as 0 q ω ∈ A1 . Since the idea is similar to that of Theorem 2.1.7, we only give the differentia. For any λ > 0, we make the Calder´on-Zygmund decomposition: 0 f = g +b, where g is the ”good” function and b is the ”bad” one. If ω q ∈ A1 , then ω ∈ A1 . The estimate of I1 follows from applying the boundedness of 0 TΩ on Lq (ω). The estimate of I2 can be obtained by applying Lemma 2.2.3 0 and ω q ∈ A1 . Here we omit the details. We have finished the proof of Theorem 2.2.3. 0 In the proof of Theorem 2.2.3 we see that ω q ∈ A1 implies that TΩ is of 0 weighted weak type (1,1). The following conclusion shows that ω q ∈ Ap (1 < p < ∞) is also a sufficient condition for T Ω to be of weighted type (p, p). Theorem 2.2.4 Suppose that Ω satisfies (2.2.1), (2.2.2) and L q -Dini con0 dition (1 < q < ∞). If ω q ∈ Ap , then TΩ is bounded from Lp (ω) to Lp (ω) for 1 < p < ∞. Proof. We will use Theorem 2.2.3 and Lemma 2.2.5 to prove Theorem 2.2.4. For the purpose, we have to verify that there exist θ (0 < θ < 1), p0 , p1 and weight function ω0 , ω1 satisfying the following conditions: (i)
1 p
=
θ p0
+
1−θ p1
pθ/p0
(ii) ω = ω0
and 1 < p0 < min{q, p}, max{q 0 , p} < p1 < ∞;
p(1−θ)/p1
ω1
1−p00
and ω0
∈ Ap00 /q0 , ω1 ∈ Ap1 /q0 .
2.2 Singular integral operators with homogeneous kernels
91
0
Since ω q ∈ Ap (1 < p < ∞), applying the Jones decomposition theorem of Ap weight (Theorem 1.4.5), there exist u, v ∈ A 1 such that 0
ω q = u · v 1−p ,
i.e.,
0
0
ω = u1/q v (1−p)/q .
Set ω = (uα v β )s (uγ v δ )1−s , then we have
1 q0
(2.2.10)
1−ρ . q0
(2.2.11)
αs + γ(1 − s) = and βs + δ(1 − s) =
Put ω0 = uα v β and ω1 = uγ v δ . First we illustrate that 1−p00
ω0
∈ Ap00 /q0
if 1 < p0 < min{q, p} and α = [(p00 /q 0 ) − 1]/(p00 − 1), β = 1/(1 − p00 ). In fact, since u, v ∈ A1 , we have that (p00 /q0 )−1 Z Z 1 1 1−p00 −(1−p00 )/[(p00 /q 0 )−1] dx dx ω0 (x) ω0 (x) |Q| Q |Q| Q Z 1 0 0 = u(x)α(1−p0 ) v(x)β(1−p0 ) dx |Q| Q (p00 /q0 )−1 Z 1 −β(1−p00 )/[(p00 /q 0 )−1] −α(1−p00 )/[(p00 /q 0 )−1] dx v(x) u(x) × |Q| Q α(1−p00 ) β(1−p00 ) Z Z 1 1 ≤C u(x)dx v(x)dx |Q| Q |Q| Q (p00 /q0 )−1 −β(1−p00 ) Z Z 1 1 −α(1−p00 )/[(p00 /q 0 )−1] × dx v(x)dx u(x) |Q| Q |Q| Q
≤ C.
Here C is independent of Q. In the same way we can prove ω1 ∈ Ap1 /q0 if max{q 0 , p} < p1 < ∞ and γ = 1, δ = 1 − p1 /q 0 . Thus the value of α and γ, together with the equality (2.2.10), imply s = 1/p 0 . Similarly, the value of β and δ, together with s = p10 , lead to p1 = p00 (p − 1) by invoking the
92
Chapter 2. Singular Integral Operators
equality (2.2.11). Set θ = p0 (p1 − p)/p(p1 − p0 ). Then we have 0 < θ < 1 and s = pθ/p0 , 1 − s = p(1 − θ)/p1 . Thus by our choice of p0 , p1 , θ, ω0 , ω1 we know both (i) and (ii) hold. Finally, we obtain Theorem 2.2.4. Remark 2.2.3 1. When q = ∞, Theorem 2.2.3 and Theorem 2.2.4 are in accordance with Theorem 2.1.6 and Theorem 2.1.7 respectively. 2. In Section 2.1 we call K is the Calder´ on-Zygmund kernel (see 2.1.1), if K satisfies H¨ ormander condition (2.1.3). Actually, we can also define L q H¨ ormander condition Hq (1 ≤ q ≤ ∞). If K ∈ L1loc (R\{0}) and there exist C, Cr > 0, such that for any y ∈ Rn and R > C|y|, the following condition (Hq )
∞ X
2k R
k=1
n/q0
Z
2k R<|x|≤2k+1 R
|K(x − y) − K(x)|q dx
!1/q
≤ Cr
and (H∞ )
∞ X k=1
2k R
n
sup 2k R<|x|≤2k+1 R
|K(x − y) − K(x)| ≤ C∞
hold. Sometimes (Hq ) and (H∞ ) are called Lq -H¨ ormander conditions (1 ≤ q ≤ ∞). Clearly, (2.1.3) is just condition (H 1 ). By the proof of Lemma 2.2.2 and (2.1.4) we know that if K(x) = Ω(x) |x|n q and Ω satisfies (2.2.1) as well as L -Dini condition (1 ≤ q ≤ ∞), then K must satisfy the condition (Hq ) (1 ≤ q ≤ ∞). In this sense, the condition (Hq ) (1 ≤ q ≤ ∞) is weaker than Lq -Dini condition.
2.3
Singular integral operators with rough kernels
In the last section we have seen, for the singular integral operator T Ω with homogeneous kernel, that Ω satisfies L 1 -Dini condition can ensure (p, p) boundedness and weak (1,1) boundedness of T Ω , whenever Ω has the homogeneity of degree 0 and satisfies the vanishing condition. The L 1 -Dini condition is much weaker than L∞ -Dini condition. However, the L1 -Dini condition still involves certain smoothness of Ω on the unit sphere. On the other hand, from the proof of Theorem 2.2.2 we note that L 1 -Dini condition was not used in the proof of the (2, 2) boundedness of T Ω (see Remark 2.2.2). This fact suggests us whether or not we can still keep the (p, p) boundedness
2.3 Singular integral operators with rough kernels
93
and weak (1,1) boundedness of TΩ without any smoothness of Ω on the unit sphere. It is just the question that we will discuss in this section. We begin with discussing the rotation method and singular integral operators with rough odd kernel. In 1956, Calderon and Zygmund [CaZ3] first considered the above problem, and introduced ”rotation method” to deal with the Lp -boundedness (1 < p < ∞) of a class of singular integral operators with rough odd kernels. In short, the rotation method, by sphere coordinate transform, is to reduce the study of convolution operator on R n to the study of an operator in one dimension in some direction. For any y ∈ Rn , y 6= 0, we write y = ry 0 , where 0 < r < ∞ and y 0 ∈ Sn−1 . Set T f (x) = K ∗ f (x), we have T f (x) =
Z
Rn
Z
K(y)f (x − y)dy =
Sn−1
Z
∞ 0
K(ry 0 )f (x − ry 0 )r n−1 drdσ(y 0 ).
Set K(ry 0 ) = ω(y 0 )h(r). We have T f (x) = :=
Z
0
ω(y )
n−1 ZS
Sn−1
Z
∞
h(r)f (x − ry 0 )r n−1 drdσ(y 0 )
0
ω(y 0 )Ty0 f (x)dσ(y 0 ).
Now for fixed y 0 ∈ Sn−1 , let Y be the hyperplane through the origin orthogonal to y 0 . Then for all x ∈ Rn , there exist s ∈ R and z ∈ Y , such that x = z + sy 0 . Hence Ty0 f (x) =
Z
∞ 0
h(r)f [z + (s − r)y 0 ]r n−1 dr := L(fz,y0 )(s).
If L is of type (p, p) (1 < p < ∞) on R1 , then it follows that Z
Rn
Ty0 f (x) p dx =
Z Z
∞
Z −∞ Z ∞
Y
≤C
=C
L(fz,y0 )(s) p dsdz
ZY
Rn
−∞
f (z + sy 0 ) p dsdz
|f (x)|p dx,
where C is independent of z, y 0 and f . Thus Minkowski’s inequality yields
94
Chapter 2. Singular Integral Operators
that
Z
0 0 0 ω(y )T f (·)dσ(y ) kT f kp = y
n−1 S p Z |ω(y 0 )| · kTy0 f kp dσ(y 0 ) ≤ n−1 S Z |ω(y 0 )|dσ(y 0 ). ≤ Ckf kp · Sn−1
Therefore when
Z
Sn−1
|ω(y 0 )|dσ(y 0 ) < ∞,
Lp -boundedness of T follows immediately. Now using the above idea, we will obtain the boundedness of a class of singular integral operators with homogeneous kernels. Theorem 2.3.1 Suppose that Ω satisfies (2.2.1) and Ω(−x 0 ) = −Ω(x0 ), for every x0 ∈ Sn−1 . If Ω ∈ L1 (Sn−1 ), then TΩ is of type (p, p) for 1 < p < ∞. Proof. Using the rotation method, we have that Z Ω(y) TΩ f (x) = p.v. f (x − y)dy n Rn |y| Z Z ∞ π 1 f (x − ry 0 ) 0 = dσ(y 0 ) Ω(y ) p.v. 2 Sn−1 π r −∞ Z π := Ω(y 0 )Hy0 f (x)dσ(y 0 ), 2 Sn−1 where Hy0 is the Hilbert transform along y 0 direction. With the above idea and (p, p) boundedness of the Hilbert transform (Theorem 2.1.5), and also noticing Ω ∈ L1 (Sn−1 ), we immediately deduce that TΩ is of type (p, p). Similarly, we can define the maximal Hilbert transform H y∗0 and the Hardy-Littlewood maximal operator M y0 along y 0 direction: Z 0) f (x − ty 1 Hy∗0 f (x) = sup dt , π ε>0 |t|>ε t Z r 1 My0 f (x) = sup |f (x − ty 0 )|dt. 2r r>0 −r Using the rotation method we get the following result.
2.3 Singular integral operators with rough kernels
95
Theorem 2.3.2 Suppose that Ω satisfies (2.2.1) and Ω(−x 0 ) = −Ω(x0 ), for every x0 ∈ Sn−1 . If Ω ∈ L1 (Sn−1 ), then TΩ∗ is of type (p, p) for 1 < p < ∞. It follows that TΩ∗ f (x)
Z Ω(y) = sup f (x − y)dy n |y| ε>0 |y|>ε Z π |Ω(y 0 )|Hy∗0 f (x)dσ(y 0 ). ≤ 2 Sn−1
By the rotation method and Remark 2.1.5, Theorem 2.3.2 is easily proved. Now define the maximal operator MΩ with rough kernel by Z 1 |Ω(y)||f (x − y)|dy. MΩ f (x) = sup n r>0 r |y|≤r
(2.3.1)
It is a very important tool in the study of singular integral operators. Since Z MΩ f (x) ≤ |Ω(y 0 )|My0 f (x)dσ(y 0 ), Sn−1
using the rotation method and (p, p)-boundedness of the Hardy-Littlewood maximal operator (Theorem 1.1.1 (c)) we easily obtain the following theorem. Theorem 2.3.3 Suppose that Ω satisfies (2.2.1) and Ω ∈ L 1 (Sn−1 ). Then MΩ is of type (p, p) for 1 < p ≤ ∞. We have got Lp -boundedness (1 < p < ∞) of a class of singular integral operators with odd rough kernel by the rotation method, where the kernel Ω is integrable on Sn−1 and is an odd function, but without any smoothness. We know that any function Ω on Sn−1 can always be decomposed into a sum of an odd function and an even one on S n−1 . Thus, to deal with the boundedness of the corresponding singular integral operator with general kernel, we just need to consider the case that Ω is an even function on Sn−1 . Early in 1956, Calder´on and Zygmund [CaZ3] considered the L p boundedness (1 < p < ∞) of the corresponding singular integral operator TΩ as Ω ∈ L log + L(Sn−1 ) is an even function. They obtained the following result. Theorem 2.3.4 Suppose that Ω satisfies (2.2.1), (2.2.2) and Ω(−x 0 ) = Ω(x0 ) for every x0 ∈ Sn−1 . If Ω ∈ Llog + L(Sn−1 ), then TΩ is of type (p, p) for 1 < p < ∞.
96
Chapter 2. Singular Integral Operators
The basic idea of the proof is to turn the even kernel into the odd kernel by the Riesz transform, then using Theorem 2.3.1 we can deduce the conclusion. However, here we do not give the detailed proof of Theorem 2.3.4. Instead, we will discuss this problem in more general situation, and obtain the same conclusion, where Theorem 2.3.4 is just a natural consequence. In detail, since Llog + L(Sn−1 ) is a proper subspace of H 1 (Sn−1 ) (under the condition (2.2.2)), where H 1 (Sn−1 ) is Hardy space on Sn−1 , we will show Theorem 2.3.4 holds whenever Ω ∈ H 1 (Sn−1 ) is an even function. First we give the definition of H 1 (Sn−1 ). Suppose that α(x0 ) ∈ L1 (Sn−1 ), x0 ∈ Sn−1 . Set Z + 0 0 0 0 P α(x ) = sup Ptx0 (y )α(y )dσ(y ) , 0
where
Ptx0 (y 0 ) =
Now define
Sn−1
1 − t2 , |y 0 − tx0 |n
y 0 ∈ Sn−1 .
H 1 (Sn−1 ) = {α ∈ L1 (Sn−1 ) : kP + αkL1 (Sn−1 ) < ∞}, and write kαkH 1 (Sn−1 ) = kP + αkL1 (Sn−1 ) .
An important property of H 1 (Sn−1 ) is its atom decomposition. We first give the definition of atom. We call a function a on S n−1 regular atom if n−1 and 0 < r ≤ 2 such that a ∈ L∞ (Sn−1 ) and there T exist ξ ∈ S n−1 (i) supp(a) ⊂ S B(ξ, r), where B(ξ, r) = {y ∈ Rn : |y − ξ| < r}; −n+1 (ii) kak ; Z L∞ (Sn−1 ) ≤ r a(x0 )dσ(x0 ) = 0.
(iii)
Sn−1
A function a on Sn−1 is called exceptional atom, if a ∈ L∞ (Sn−1 ) and kakL∞ (Sn−1 ) ≤ 1. Lemma 2.3.1 For any α ∈ H 1 (Sn−1 ), there exist P complex number λj and atom aj (regular or exceptional) such that α = j λj aj , and X kαkH 1 (Sn−1 ) ∼ |λj |. j
Now we formulate main result. Theorem 2.3.5 Suppose that Ω satisfies (2.2.1), (2.2.2), and Ω(−x 0 ) = Ω(x0 ), for every x0 ∈ Sn−1 . If Ω ∈ H 1 (Sn−1 ), then TΩ is of type (p, p) for 1 < p < ∞.
97
2.3 Singular integral operators with rough kernels We prove the following lemma first.
Lemma 2.3.2 Suppose that Ω satisfies (2.2.1), (2.2.2) and Ω ∈ H 1 (Sn−1 ). Then Ω(x) χ 1 (x) ∈ H 1 (Rn ) |x|s { 2 <|x|≤2}
for every s ∈ R. Here we denote by H 1 (Rn ) the real Hardy space on Rn . Proof. By Lemma 2.3.1 and Z
Ω(x0 )dσ(x0 ) = 0,
Sn−1
we have Ω(x0 ) =
X
λj aj (x0 ),
(2.3.2)
j
where {aj } is a regular atom sequence in H 1 (Sn−1 ). Assume, without loss of generality, that \ supp(aj ) ⊂ B(ξj0 , ρj ) Sn−1 ,
where ξj0 ∈ Sn−1 . First recall the definition of atom on real Hardy space H 1 (Rn ). The function b on Rn is called H 1 (Rn ) atom, if (i) there exist x0 ∈ Rn and r > 0 such that supp(b) ⊂ B(x0 , r); Z b(x)dx = 0;
(ii)
Rn
(iii) kbkL∞ ≤ r −n . Moreover,
1
n
f ∈ H (R ) ⇐⇒ f = where bj is an atom in H 1 (Rn ), and kf kH 1 ∼
∞ X j=1
∞ X
µj bj
(2.3.3)
j=1
|µj | < ∞.
For the above conclusion and other properties on real Hardy spaces, readers can refer to Lu [Lu1]. Now turn to the proof of Lemma 2.3.2. First extend a j to Rn \{0} along the radial direction, i.e., for x ∈ R n \ {0}, put x . aej (x) = aj |x|
98
Chapter 2. Singular Integral Operators
We split it into two cases. Case I ρj > 14 . Applying the assumption that a is a regular atom, we have e aj (x) (x) = Cbj (x), (2.3.4) χ 1 |x|s { 2 <|x|≤2}
where bj is an atom in H 1 (Rn ), and C is a constant depending only on n, s. Case II 0 < ρj < 41 . Set 3 1 n 1 Ek = x ∈ R : + 2(k − 1)ρj < |x| ≤ + 2kρj , k = 1, 2, · · · , 2 2 4ρj
and E0 =
3 1 2ρj < |x| ≤ 2 , x∈R : + 2 4ρj n
where [t] denotes the largest integer not exceeding t (E 0 may be an empty set.) In this way, we have [ 4ρ3 ]
j X e aj (x) e aj (x) χ{ 1 <|x|≤2} (x) = χEk (x). s 2 |x| |x|s
(2.3.5)
k=0
Applying the condition of aj being a regular atom together with the construction of Ek , it is easy to see that, there exists a constant C depending only on n and s, but independent of K and ρ j , such that e aj (x) 3 k = 0, 1, · · · , χEk (x) = Cρj bjk (x), , |x|s 4ρj where bjk is an atom in H 1 (Rn ). Thus, by (2.3.2), (2.3.4) and (2.3.5) we conclude that Ω(x) (x) χ 1 |x|s { 2 <|x|≤2}
can be expressed as the sum of a sequence of atoms on H 1 (Rn ), the sum of ∞ X |λj | < ∞. By (2.3.3) we have whose coefficients is dominated by C j=1
and
Ω(x) χ 1 (x) ∈ H 1 (Rn ) |x|s { 2 <|x|≤2}
Ω(x)
χ (x)
|x|s { 12 <|x|≤2} 1 n ≤ CkΩkH 1 (Sn−1 ) . H (R )
(2.3.6)
2.3 Singular integral operators with rough kernels
99
Lemma 2.3.3 Suppose that Ω satisfies the condition of Theorem 2.3.5 and Rj are the Riesz transform (j = 1, 2, · · · , n). Then Z Ω(·) xj − yj Ω(y) Kj (x) = Rj dy, (x) = p.v.C n n n+1 |y|n |·| Rn |x − y| with Cn = Γ( n+1 2 )/π
n+1 2
, which satisfies the following properties:
(i) Kj (x) is a homogeneous function of order −n on R n ; (ii) Kj (−x) = −Kj (x), x ∈ Rn ; Z Kj (x0 ) dσ(x0 ) ≤ CkΩkH 1 (Sn−1 ) . (iii) Sn−1
Proof. Properties (i) and (ii) are obvious, so it suffices to check (iii). For 3 3 4 < |x| < 2 , set Z Z xj − yj Ω(y) xj − yj Ω(y) dy + Cn dy Kj (x) = p.v.Cn n+1 n n+1 |y|n 1 |y| <|y|≤2 |x − y| |y|≤ 21 |x − y| 2 Z xj − yj Ω(y) + Cn dy n+1 |y|n |y|>2 |x − y| := Cn (I1 + I2 + I3 ).
Since Ω satisfies (2.2.2), it implies that Z x − y Ω(y) x j j j |I1 | = − dy |y|≤ 1 |x − y|n+1 |x|n+1 |y|n 2
Z 1 Z 2 x j 0 0 0 0 )Ω(y )dσ(y )dr y0 · ∇ (x − ry t = x,ry n+1 0 Sn−1 |x| xj 1 ≤ kΩkL1 (Sn−1 ) · max ∇ n+1 7 1 2 |x| ≤|x|≤ 4 4
= CkΩkL1 (Sn−1 ) ,
where tx,ry0 ∈ [0, 1]. Note that the Riesz transform is (H 1 , L1 ) bounded, therefore by Lemma 2.3.2 and (2.3.6) we have
Z
Ω(·)
|I2 |dx ≤ C n χ{ 1 <|·|≤2} (·)
1 n ≤ CkΩkH 1 (Sn−1 ) . 2 |·| 3/4<|x|<3/2 H (R )
100
Chapter 2. Singular Integral Operators
As to I3 , it follows that |I3 | ≤ ≤
Z
|y|>2
Z
|y|>2
|Ω(y)| 1 dy n |x − y| |y|n 4n |Ω(y)|dy |y|2n
= CkΩkL1 (Sn−1 ) . Thus, we have
Kj (x)χ{ 3 <|x|< 3 } (x) ∈ L1 (Rn ). 4
2
(x0 )
L1 (Sn−1 )
This fact is equivalent to Kj ∈ Z Z |Kj (x0 )|dσ(x0 ) = C Sn−1
and
3 <|x|< 23 4
|Kj (x)|dx
≤ CkΩkL1 (Sn−1 ) + CkΩkH 1 (Sn−1 ) ≤ CkΩkH 1 (Sn−1 ) .
The Proof of Theorem 2.3.5. Let us first illustrate the following fact that n X Rj2 = −I, (2.3.7) j=1
where I is an identity operator. In fact, for any Schwartz function f , from (2.0.3) it follows that \ n n X X ξj 2 ˆ 2 ˆ −i f(ξ) = −f(ξ). Rj f (ξ) = |ξ| j=1
j=1
Then (2.3.7) follows. By (2.3.7), we have
TΩ =
n X j=1
−Rj2 TΩ := −
n X
Rj Tj ,
j=1
where Tj = Rj TΩ . If we denote the kernel of Tj by Kj , then Ω(·) Kj (x) = Rj (x), j = 1, 2, · · · , n. | · |n
(2.3.8)
101
2.3 Singular integral operators with rough kernels
Vj (x) , where Vj satisfies (2.2.1) and is an odd |x|n integrable function on Sn−1 . Thus, by Theorem 2.3.1, Tj can be extended to be a linear bounded operator on Lp (Rn ) (1 < p < ∞). Combining it with Lp boundedness of Rj (Theorem 2.1.4) and (2.3.8), we deduce that T Ω is bounded on Lp (Rn )(1 < p < ∞). By Lemma 2.3.3, let Kj (x) =
Next we give the Lp boundedness (1 < p < ∞) of the corresponding maximal singular integral operator T Ω∗ for Ω ∈ H 1 (Sn−1 ). Theorem 2.3.6 Suppose that Ω satisfies the conditions in Theorem 2.3.5. Then the maximal operator TΩ∗ , TΩ∗ f (x)
Z Ω(y) f (x − y)dy = sup , n |y| ε>0 |y|≥ε
is of type (p, p) (1 < p < ∞).
Proof. Choose a radial function ϕ ∈ C ∞ such that 0 ≤ ϕ ≤ 1 and ϕ(x) = 0 when |x| ≤ 14 as well as ϕ(x) = 1 when |x| ≥ 21 . Then TΩε f (x)
Z
Ω(x − y) f (y)dy n |x−y|≥ε |x − y| Z Ω(x − y) x−y = ϕ f (y)dy n ε Rn |x − y| Z Ω(x − y) x−y − ϕ f (y)dy. n ε |x−y|<ε |x − y|
:=
It follows from (2.3.1) that Z Ω(x − y) x−y sup ϕ f (y)dy n |x − y| ε ε>0 |x−y|<ε Z |Ω(x − y)| ≤ sup |f (y)|dy ≤ MΩ f (x). n ε>0 ε/4<|x−y|<ε |x − y| Theorem 2.3.3 immediately implies that operator: Z
Ω(x − y) f→ ϕ n |x−y|<ε |x − y|
x−y ε
f (y)dy
102
Chapter 2. Singular Integral Operators
is bounded on Lp (1 < p ≤ ∞). Thus it suffices to show the smooth truncated maximal singular integral operator TeΩ∗ f (x)
Z = sup ε>0
Rn
Ω(x − y) ϕ |x − y|n
x−y ε
is bounded on Lp (1 < p < ∞). For 1 ≤ j ≤ n, set Kj (x) = Rj and Vej (x) = Rj
Put
TeΩε f (x) =
Then TeΩε f
=−
n X j=1
Rj
Z
Rn
Ω(·) | · |n
Ω(·)ϕ(·) | · |n
Ω(x − y) ϕ |x − y|n
f (y)dy
(x)
(x).
x−y ε
f (y)dy.
n X Ω(·) · 1 e · ϕ ∗ Rj (f ). (2.3.9) ∗ R (f ) = − Vj j | · |n ε εn ε j=1
We need the following Lemma 2.3.4 whose proof will be given after that of Theorem 2.3.6. Lemma 2.3.4 There exists a function ω j on Rn satisfying (2.2.1), and ωj (x0 ) ∈ L1 (Sn−1 ) such that (i) |Vej (x)| ≤ ωj (x),
|x| ≤ 1;
(ii) |Vej (x) − Kj (x)| ≤ CkωkL1 (Sn−1 ) |x|−n−1 ,
|x| > 1.
The above lemma and (2.3.9) implies that
Z n X x − y 1 eε e Vj (Rj f )(y)dy TΩ f (x) = n ε j=1 ε Rn ≤ I1 (f, ε) + I2 (f, ε) + I3 (f, ε),
2.3 Singular integral operators with rough kernels where n X 1 I1 (f, ε) = εn j=1
I2 (f, ε) =
n X 1 εn j=1
n X 1 I3 (f, ε) = εn j=1
Since
103
Z x−y K (R f )(y)dy , j |x−y|>ε j ε Z x − y x − y Vej − Kj (Rj f )(y)dy , |x−y|>ε ε ε Z x−y e (Rj f )(y)dy . Vj |x−y|<ε ε sup |I1 (f, ε)| ≤ ε>0
Sj∗
n X
Sj∗ (Rj f )(x),
j=1
where is the maximal singular integral operator with the kernel K j for 1 ≤ j ≤ n. By Lemma 2.3.3, Theorem 2.3.2 and L p -boundedness of the Riesz transform, we can easily deduce that sup I1 (f, ε) ε>0
is bounded on (ii), we have
Lp
(1 < p < ∞). On the other hand, applying Lemma 2.3.4
Z n X sup I2 (f, ε) ≤ sup C · ε ε>0
ε>0
≤C·
j=1
n X
|x−y|>ε
|Rj f (y)| dy |x − y|n+1
M (Rj f )(x),
j=1
where M is the Hardy-Littlewood maximal operator. The L p -boundedness of M (Theorem 1.1.1) and Theorem 2.1.4 imply that sup I2 (f, ε) ε>0
is also bounded on Lp (1 < p < ∞). Finally, it follows from Lemma 2.3.4 (i) that Z n X 1 sup I3 (f, ε) ≤ sup |ωj (x − y)||Rj f (y)|dy εn |x−y|<ε ε>0 ε>0 j=1
≤
n X j=1
Mωj (Rj f )(x).
104
Chapter 2. Singular Integral Operators
Thus we obtain that sup I3 (f, ε) ε>0
Lp
is also bounded on (1 < p < ∞) by applying Theorem 2.3.3 and Theorem 2.1.4. Therefore TeΩ∗ is bounded on Lp (1 < p < ∞). This completes the proof of Theorem 2.3.6. The proof of Lemma 2.3.4. Suppose |x| > 1. Since ϕ(y) = 1 when |y| ≥ 12 , it follows from the vanishing condition (2.2.2) of Ω that Z e Vj (x) − Kj (x) ≤
xj − y j Ω(y) (ϕ(y) − 1) n dy n+1 |y| Rn |x − y| Z xj |Ω(y)| xj − yj ≤C − n+1 dy n |x − y|n+1 |x| |y|< 21 |y| Z C |Ω(y)| ≤ dy n+1 n−1 |x| |y|< 1 |y| 2
≤ CkΩkL1 (Sn−1 ) |x|−n−1 .
Now let |x| ≤ 1. Note that |Vej (x)| ≤ CkΩkL1 when |x| ≤ 81 , since ϕ(y) ≡ 0 when |y| ≤ 14 . When 18 ≤ |x| ≤ 1, we have that e Vj (x) − ϕ(x)Kj (x) Z |xj − yj | |Ω(y)| |ϕ(y) − ϕ(x)| dy ≤ n |x − y|n+1 |y|>2 |y| Z |xj − yj | |Ω(y)| + |ϕ(y) − ϕ(x)| dy n 1 |x − y|n+1 <|y|≤2 |y| 16 Z xj − y j xj |Ω(y)| + |ϕ(y) − ϕ(x)| − dy |y|n |x − y|n+1 |x|n+1 0<|y|≤ 1 16
:= P1 (x) + P2 (x) + P3 (x).
We next calculate the estimate of P1 , P2 and P3 , respectively. Note that if 81 ≤ |x| ≤ 1 and |y| > 2, then |x − y| ≥ C|y|. Thus P1 (x) ≤ C
Z
|y|>2
|Ω(y)| dy ≤ CkΩkL1 (Sn−1 ) . |y|2n
The smoothness of ϕ implies |ϕ(x) − ϕ(y)| ≤ C|x − y|. So for
1 8
≤ |x| ≤ 1,
105
2.3 Singular integral operators with rough kernels we have that P2 (x) ≤ C
Z
1 <|y|≤2 16
≤ C|x| As to P3 , we have that P3 (x) ≤ C
Z
n− 32
Z
Rn
η(x) = |x|
n− 23
|y|
|Ω(y)|
n− 21
|x − y|n−1
dy.
|Ω(y)| dy ≤ CkΩkL1 (Sn−1 ) . |y|n−1
1 0<|y|≤ 16
Let
|Ω(y)| dy n |y| |x − y|n−1
Z
Rn
and
|y|
|Ω(y)|
n− 12
|x − y|n−1
dy
Vj (x) = |x|n Kj (x). Take ωj (x) = C |Vj (x)| + kΩkL1 (Sn−1 ) + η(x) , then clearly ω j is a homoge neous function of degree 0, i.e., satisfying (2.2.1). Moreover, Vej (x) ≤ ωj (x) 0 ) ∈ L1 (Sn−1 ). In fact, we only when |x| ≤ 1. We will next prove that nω j (x need to show η is integrable on x ∈ R : 21 < |x| < 2 . Obviously, if |y| ≤ 41 , then we have that Z 3 |x|n− 2 dx ≤ C0 ; n−1 1 <|x|<2 |x − y| 2
if |y| ≥ 4, then
and if
1 4
Z
3
1 <|x|<2 2
|x|n− 2 dx ≤ C1 |y|−n+1 ; |x − y|n−1
< |y| < 4, then Z 3 Z dx |x|n− 2 dx ≤ C2 ≤ C20 . n−1 n−1 1 |x−y|<6 |x − y| <|x|<2 |x − y| 2
It follows from the above estimates that Z η(x)dx ≤ CkΩkL1 (Sn−1 ) . 1 <|x|<2 2
This finishes the proof of Lemma 2.3.4. As a natural consequence of Theorem 2.3.6, we have
106
Chapter 2. Singular Integral Operators
Theorem 2.3.7 Suppose that Ω satisfies (2.2.1) and (2.2.2). If Ω ∈ L log + L (Sn−1 ), then the maximal operator TΩ∗ is of type (p, p) (1 < p < ∞). Next we will give the weighted boundedness of singular integral operator TΩ with rough kernel and the corresponding maximal operator T Ω∗ . In the proof of the weighted boundedness of the Calder´on-Zygmund singular integral operator (Theorem 2.1.6), we used the technique of good-λ inequality. For singular integral operators with homogeneous kernel, in the proof of their weighted boundedness (Theorem 2.2.3) we essentially still use this inequality. Since the kernel Ω of singular integral operator with rough kernel does not have any smoothness on the unit sphere, the good-λ inequality is not applicable. In order to deal with this problem, Duoandikoetxea and Rubio de Francia synthetically used the Fourier transform estimate, weighted Littlewood-Paley theory and Stein-Weiss interpolation method with change of measure, then obtained the weighted boundedness of T Ω and TΩ∗ . In their proof, the weighted boundedness of the maximal operator M Ω with rough kernel (for its definition, see (2.3.1)) is needed, while the latter itself is of great significance. Theorem 2.3.8 Suppose that Ω is a homogeneous function of degree 0 on Rn and Ω ∈ Lq (Sn−1 ), q > 1. If p, q and the weight function ω satisfy one of the following statements: (i) q 0 ≤ p < ∞, p 6= 1 and ω ∈ Ap/q0 ; 0
(ii) 1 < p ≤ q, p 6= ∞ and ω 1−p ∈ Ap0 /q0 ; 0
(iii) 1 < p < ∞, and ω q ∈ Ap , then MΩ is bounded on Lp (ω). Proof. It suffices to prove this result under the statements (i) and (ii). The statement (iii) will be easily deduced by applying a method similar to that in the proof of Theorem 2.2.4. The conclusion under the statement (i) is easy to prove. In fact, h i1/q0 0 MΩ f (x) ≤ kΩkLq (Sn−1 ) M |f |q (x) ,
where M is the Hardy-Littlewood maximal operator. If p/q 0 > 1, then the conclusion follows directly from Theorem 1.4.3. If p/q 0 = 1, the conclusion can be obtained by applying the method in the proof of Theorem 2.2.3.
107
2.3 Singular integral operators with rough kernels
Now consider the case of the statement (ii). Let us first introduce some notations. For j ∈ Z, set KΩ,j (x) = Ω(x)χ[2j ,2j+1 ) (x)|x|−n , TΩ,j f (x) = KΩ,j ∗ f (x) = and
gΩ (f )(x) =
Z
2j ≤|x−y|<2j+1
X
Ω(x − y) f (y)dy |x − y|n 1 2
|TΩ,j f (x)|
2
.
(2.3.10)
MΩ f (x) ≤ C · sup T|Ω|,j (|f |)(x).
(2.3.11)
j∈Z
It is clear that there exists C > 0, such that j
Let Ω0 (x) = |Ω(x)| −
kΩkL1 (Sn−1 ) , |Sn−1 |
then Ω0 ∈ Lq (Sn−1 ) satisfies (2.2.1) and (2.2.2). By (2.3.11) we have MΩ f (x) ≤ C sup j
Z
2j ≤|x−y|<2j+1
kΩkL1 + n−1 |S |
Z
Ω0 (x − y) f (y)dy |x − y|n
2j ≤|x−y|<2j+1
|f (y)| dy |x − y|n
(2.3.12)
≤ CgΩ0 (|f |)(x) + CM f (x). We split the proof of this theorem into that of several propositions by the value of q. Analogous to the statement (i), we merely need to consider the case 1 < p < q. Proposition 2.3.1 Suppose that Ω satisfies (2.2.1) and Ω ∈ L q (Sn−1 ), q > 0 max{p, 2}. If ω 1−p ∈ Ap0 /q0 , then there exists a constant C > 0 such that kMΩ f kLp (ω) ≤ Ckf kLp (ω) for every f ∈ Lp (ω).
108
Chapter 2. Singular Integral Operators 0
It follows from ω 1−p ∈ Ap0 /q0 ⊂ Ap0 and the property (I) (Proposition 1.4.1) of Ap that ω ∈ Ap . Hence there exists a constant C > 0 such that kM f kLp (ω) ≤ Ckf kLp (ω) by applying Theorem 1.4.3. Noting (2.3.12), it remains to check kgΩ0 (|f |)kLp (ω) ≤ Ckf kLp (ω)
(2.3.13)
under the hypothesis of Proposition 2.3.1, where
gΩ0 (|f |)(x) =
X j
1
2
|TΩ0 ,j (|f |)(x)|
2
.
For any sequence ε = {εj }, εj = ±1, define a linear operator X εj (KΩ0 ,j ∗ f )(x). Tε,Ω0 f (x) = j∈Z
Then by a result of Rademacher function (see Kurtz [Ku]), to prove (2.3.13), it suffices to show that there exist C, independent of f and {ε j }, such that kTε,Ω0 f kLp (ω) ≤ Ckf kLp (ω) .
(2.3.14)
Thus we reduce the proof of Proposition 2.3.1 to the validation of (2.3.14). Next we will show that (2.3.14) holds by two lemmas. Lemma 2.3.5 Suppose that Ω ∈ Lq (Sn−1 ) satisfies (2.2.1) and (2.2.2). If p > q 0 , q > 2 and ω ∈ Ap/q0 , then there exists a constant C > 0, independent of f and {εj }, such that kTε,Ω f kLp (ω) ≤ Ckf kLp (ω) . Proof. Take a radial function ψ ∈ C0∞ (Rn ) such that n 1 0 ≤ ψ ≤ 1, suppψ ⊂ x ∈ R : ≤ |x| ≤ 2 2 and
X
ψ 2 (2k x) = 1,
k∈Z
x 6= 0.
For any f ∈ S (Rn ), define a multiplier k \ ˆ Sk : (S k f )(ξ) = ψ(2 ξ)f(ξ).
109
2.3 Singular integral operators with rough kernels Then
X
Sk2 f (x) = f (x).
k∈Z
Consequently, it follows that X Tε,Ω f (x) = εj (KΩ,j ∗ f ) (x) j∈Z
= = :=
X
εj KΩ,j ∗
j∈Z X X
2 (Sj+k f
k∈Z
!
(x) (2.3.15)
εj Sj+k (KΩ,j ∗ Sj+k f ) (x)
k∈Z Xj∈Z k Tε,Ω f (x). k∈Z
By the Plancherel theorem, we have
XZ
k 2 ∞
Tε,Ω f 2 ≤ Ck{εj }kl L
≤C
X
XZ j∈Z
j∈Z
Rn
|Sj+k (KΩ,j ∗ Sj+k f )(x)|2 dx
2−j−k−1 ≤|ξ|≤2−j−k+1
2 2 b KΩ,j (ξ) fb(ξ) dξ.
Next we will show that there exist C > 0 and 0 < θ < 1, such that n o b Ω,j (ξ)| ≤ C min |2j ξ|θ , |2j ξ|−θ |K (2.3.16)
for every j ∈ Z and ξ ∈ Rn . Obviously, there exists C, depending only on Ω and n, such that b Ω,j (ξ)| ≤ C. |K (2.3.17)
Since Ω satisfies the vanishing condition (2.2.2), it follows that Z Ω(x) b e−2πix·ξ − 1 dx ≤ C|2j ξ|. KΩ,j (ξ) = 2j ≤|x|≤2j+1 |x|n
This formula and (2.3.17) imply b KΩ,j (ξ) ≤ C|2j ξ|θ
for any 0 < θ < 1, where C is independent of j and ξ.
110
Chapter 2. Singular Integral Operators On the other hand, H¨older’s inequality implies Z 2 b KΩ,j (ξ) ≤ log 2 ·
2j+1
2j
where
Ir (ξ) =
Z
|Ir (ξ)|2
dr , r
0
Ω(x0 )e−2πirx ·ξ dσ(x0 ).
Sn−1
Since Z j+1 2 0 0 dr e−2πirξ·(x −y ) ≤ C min 1, |2j ξ · (x0 − y 0 )|−1 2j r −α ≤ C|2j ξ|−α ξ 0 · (x0 − y 0 )
with 0 < α < 1 and aq 0 < 1, H¨older’s inequality again yields that b KΩ,j (ξ) ≤ C
Z
2j+1
2j
Z Z
Sn−1 ×Sn−1
−α/2 ≤ C 2j ξ kΩkLq (Sn−1 ) −α/2 ≤ C 2j ξ .
Ω(x0 )Ω(y 0 )e Z Z
!1/2 dr dσ(x0 )dσ(y 0 ) r 0 dσ(x0 )dσ(y 0 ) 1/2q |x0 − y 0 |αq0
−2πirξ·(x0 −y 0 )
Sn−1 ×Sn−1
Choose θ = α/2. We then get (2.3.16) which concludes that
k
Tε,Ω f
L2
≤ C2−θ|k| kf kL2
(2.3.18)
for every k ∈ Z, where C is independent of k, f, {ε j } and 0 < θ < 1. Let us now show that there exists a constant C > 0, independent of f, {ε j } and k ∈ Z, such that k kTε,Ω f kLp (ω) ≤ Ckf kLp (ω) .
(2.3.19)
Indeed, the conditions of Lemma 2.3.5 implies ω ∈ A p . By applying the weighted Littlewood-Paley theory (see Chapter 5, Theorem 5.2.3), there
111
2.3 Singular integral operators with rough kernels exists a constant C, independent of f and {ε j }, such that
k
Tε,Ω f
Lp (ω)
p 1/p X εj Sj+k (KΩ,j ∗ Sj+k f )(x) ω(x)dx = Rn j∈Z
1/2
X
≤ Ck{εj }kl∞ |Sj+k (KΩ,j ∗ Sj+k f )(·)|2
j∈Z
p L (ω)
1/2
X
≤ C |(KΩ,j ∗ Sj+k f )(·)|2 .
j∈Z
p
Z
L (ω)
For any fixed k, denote hj (x) = Sj+k f (x). Then |KΩ,j ∗ Sj+k f (x)| Z ≤
Thus
|Ω(x − y)|q dy n 2j ≤|x−y|≤2j+1 |x − y| h i1/q0 0 ≤ C M |hj |q (x) .
k
Tε,Ω f
Lp (ω)
!1/q
Z
0
2j ≤|x−y|≤2j+1
|hj (y)|q dy |x − y|n
1/q0 0
≤ C ·
M |hj |q (·) 2/q0 0 l
Lp/q (ω)
!1/q0
.
Since ω ∈ Ap/q0 , by the weighted norm inequality of vector-valued maximal operator (see Andersen and John [AnJ]) we obtain that
k T f
ε,Ω
Lp (ω)
1/q0
0
≤ C
hqj (·) 2/q0 0 l Lp/q (ω)
1/2
X
2 |Sj+k f (·)| =C
j∈Z
Lp (ω)
≤ Ckf kLp (ω) .
Here we use again the weighted Littlewood-Paley theory(Theorem 5.2.3). Thus we obtain (2.3.19). Next we shall use the Stein-Weiss interpolation theorem with change of measure (Lemma 2.2.5) to complete the proof of Lemma 2.3.5. We deal with it in three cases.
112
Chapter 2. Singular Integral Operators
Case 1: p > 2. Since ω ∈ Ap/q0 , Property (II) of Ap (Proposition 1.4.2) tells us that there exists σ > 0 such that ω 1+σ ∈ Ap/q0 . Take p1 such that (p1 − p) = σ, (p − 2) then p1 > p and ω 1+σ ∈ Ap1 /q0 . The proof of (2.3.19) implies that there exists a constant C1 , independent of f and k, such that
k (2.3.20)
Tε,Ω f p 1+σ ≤ C1 kf kLp1 (ω1+σ ) . L
1 (ω
Set
t= then 0 < t < 1 and
)
p1 , (1 + σ)p
1−t t 1 = + . p 2 p1
Applying Lemma 2.2.5 to (2.3.18) and (2.3.20) yields
k
Tε,Ω f p ≤ C1 2−θγ|k| kf kLp (ω) , L (ω)
(2.3.21)
where C1 , θ, γ > 0 are independent of f and k ∈ Z. Case 2: p < 2. Since ω ∈ Ap/q0 , there exists δ > 0 such that ω 1+δ ∈ Ap/q0 . Again Property (II) of Ap implies that there exists q 0 < l < p such that ω 1+δ ∈ Al/q0 . Thus we can choose σ, p0 such that 0 < σ ≤ δ and q 0 < l ≤ p0 < p satisfying σ=
(p − p0 ) , (2 − p)
ω 1+σ ∈ Ap0 /q0 .
Actually, if δ = (p − l)/(2 − p), then we can take σ = δ and p 0 = l. If δ < (p − l)/(2 − p), then we can take σ = δ and l < p 0 < p such that σ = (p − p0 )/(2 − p). If δ > (p − l)/(2 − p), then we can take 0 < σ < δ and p0 = l such that σ = (p − p0 )/(2 − p). We still have ω 1+σ ∈ Ap0 /q0 . Therefore it follows from (2.3.19) that
k (2.3.22)
Tε,Ω f p 1+σ ≤ C2 kf kLp0 (ω1+σ ) . L
0 (ω
)
Analogously, set t = p0 /(1 + σ)p, then applying Lemma 2.2.5 to (2.3.18) and (2.3.22) yields
0
k (2.3.23)
Tε,Ω f p ≤ C2 2−θγ |k| kf kLp (ω) , L (ω)
2.3 Singular integral operators with rough kernels
113
where C2 , θ, γ 0 > 0 are independent of f and k ∈ Z. Case 3: p = 2. Since ω ∈ A2/q0 , there exists σ > 0 such that ω 1+δ ∈ A2/q0 . Consequently, we have
k (2.3.24)
Tε,Ω f 2 1+σ ≤ C3 kf kL2 (ω1+σ ) . L (ω
)
Set t = 1/(1 + σ), then applying Lemma 2.2.5 to (2.3.18) and (2.3.24) leads to
00
k (2.3.25)
Tε,Ω f 2 ≤ C3 2−θγ |k| kf kL2 (ω) , L (ω)
, θ, γ 00
where C3 > 0 are independent of f and k ∈ Z. Now put C = max{C1 , C2 , C3 }, η = min{γ, γ 0 , γ 00 }, then for p > q 0 and q > 2, by (2.3.21), (2.3.23) and (2.3.25) we obtain
k
Tε,Ω f p ≤ C2−θη|k| kf kLp (ω) . L (ω)
This formula together with (2.3.15) leads to Lemma 2.3.5.
Lemma 2.3.6 Suppose that Ω(x0 ) ∈ Lq (Sn−1 ) satisfies (2.2.1) and (2.2.2). 0 If q > max{p, 2} and ω 1−p ∈ Ap0 /q0 , then there exists a constant C > 0, independent of f and {εj }, such that kTε,Ω f kLp (ω) ≤ Ckf kLp (ω) . Using Lemma 2.3.5 and the dual method (see the proof of Theorem 2.2.3) we immediate obtain Lemma 2.3.6. Thus the inequality (2.3.14) can be deduced from Lemma 2.3.6 under the condition of Proposition 2.3.1. Hence we complete the proof of Proposition 2.3.1. Proposition 2.3.2 Suppose that Ω ∈ L q (Sn−1 ) satisfies (2.2.1). If q > 0 max{p, 43 } and ω 1−p ∈ Ap0 /q0 , then there exists a constant C > 0, independent of f , such that kMΩ f kLp (ω) ≤ Ckf kLp (ω) . Clearly, it is enough to consider the case 4 < q ≤ 2. max p, 3 Also from the proof of Proposition 2.3.1 we observe that the proof of Proposition 2.3.2 lies in the following lemma.
114
Chapter 2. Singular Integral Operators
Lemma 2.3.7 Suppose that Ω(x0 ) ∈ Lq (Sn−1 ) satisfies (2.2.1) and (2.2.2). If q 0 < p, 4/3 < q ≤ 2 and ω ∈ Ap/q0 , then there exists a constant C > 0, independent of f and {εj }, such that kTε,Ω f kLp (ω) ≤ Ckf kLp (ω) . Proof. From the proof of Lemma 2.3.5, it is easy to see that if we can show, under the assumption of Lemma 2.3.7, there exists a constant C > 0 independent of f and {εj } such that
k
Tε,Ω f
Lp (ω)
≤ Ckf kLp (ω) ,
(2.3.26)
then we can obtain the conclusion of Lemma 2.3.7 by applying the SteinWeiss interpolation theorem with change of measure (Lemma 2.2.5) to (2.3.18) and (2.3.26). 4 3
First suppose that
e Ω,j (x) = |KΩ,j (x)|2−q . K
Then it follows that 2
|KΩ,j ∗ g(x)| ≤ ≤ and
< q < 2. Then 2 < q 0 < p. Set
Z
q
Z
|KΩ,j (x − y)| dy Rn Rn j(n−nq) e 2 C2 KΩ,j ∗ |g| (x)
e Ω,j ∗ (|h|)(x) ≤ K
Z
2j ≤|x−y|<2j+1 Z
≤ C2−jn(2−q)
|KΩ,j (x − y)|
|Ω(x − y)| |x − y|n
|x−y|<2j+1
2−q
2−q
2
|g(y)| dy
|h(y)|dy
|Ω(x − y)|2−q |h(y)|dy
(2.3.27)
≤ C2−j(n−nq) MΩ2−q (h)(x). Since ω ∈ Ap , by the weighted Littlewood Paley theory (see Theorem 5.2.3)
115
2.3 Singular integral operators with rough kernels we conclude that
k 2 f
Tε,Ω p
L (ω)
p 2/p X εj Sj+k (KΩ,j ∗ Sj+k f )(x) ω(x)dx = Rn j∈Z
1/2
2
X
2 ≤C k{εj }kl∞ |Sj+k (KΩ,j ∗ Sj+k f )(·)|
j∈Z
p L (ω)
1/2
2
X
≤C |(KΩ,j ∗ Sj+k f )(·)|2
p
j∈Z L (ω)
1/2
2
X
e Ω,j ∗ |Sj+k f |2 (·) ≤C 2j(n−nq) K
j∈Z
p L (ω) Z X j(n−nq) e 2 2 KΩ,j ∗ |Sj+k f | (x) h(x)dx , =C sup n Rn j∈Z
Z
where the supremum is taken over all functions h satisfying khkL(p/2)0 (ω1−(p/2)0 ) ≤ 1. By (2.3.27) we have that Z X e Ω,j ∗ |Sj+k f |2 (x) h(x)dx 2j(n−nq) K Rn
=
Z
j∈Z
X
Rn j∈Z
≤C
Z
Rn
≤C ×
e Ω,j ∗ h (x)dx 2j(n−nq) |Sj+k f (x)|2 K
X j∈Z
Z
Rn
Z
|Sj+k f (x)|2 MΩ2−q (h)(x)dx
2/p p/2 X |Sj+k f (x)|2 ω(x)dx
Rn
j
[MΩ2−q (h)(x)]
(p/2)0
ω(x)
1−(p/2)0
dx
1/(p/2)0
.
116
Chapter 2. Singular Integral Operators
Since 4/3 < q < 2, if we set r = n q/(2 − q)o (correspondingly, r 0 = q 0 /2), then 0 Ω2−q ∈ Lr (Sn−1 ) and r > max 2p , 2 as well as (p/2)/r 0 = p/q 0 . Thus by Proposition 2.3.1 we obtain kMΩ2−q (h)kL(p/2)0 (ω1−(p/2)0 ) ≤ CkhkL(p/2)0 (ω1−(p/2)0 ) .
(2.3.28)
Using again the weighted Littlewood-Paley theory together with (2.3.28) yields that 2 p p 2 Z
2 X
k |Sj+k f (x)|2 ω(x)dx
Tε,Ω f p ≤ C sup khk ( p )0 1−( p )0 L (ω)
h
L
2
ω
2
Rn
j
1/2
2
X
≤ C |Sj+k f (·)|2
j
p
L (ω)
≤
Ckf k2Lp (ω) .
Thus we obtain (2.3.26) for 34 < q < 2. Now consider q = 2, then we have 2 = q 0 < p and ω ∈ Ap/2 . By the property (i) and (ii) of Ap (Proposition 1.4.1 and Proposition 1.4.2), we can choose σ > 0 such that (i) (2 − σ)0 < p, (ii) ω ∈ Ap/(2−σ)0 , (iii)
4 3
< (2 − σ) < 2,
(iv) Ω ∈ L2 (Sn−1 ) ⊂ L2−σ (Sn−1 ).
Consequently by the conclusion for 34 < q < 2, (2.3.26) still holds for q = 2. Therefore we have proved Lemma 2.3.7. Applying Lemma 2.3.7 and the dual method, we can obtain the conclusion of Proposition 2.3.2. By induction, if Proposition 2.3.2 holds for 2m−1 q > max p, m−1 , m ≥ 2, 2 −1
then it still holds for q > max{p, 2m /(2m −1)}. That is, we have the following result.
2.3 Singular integral operators with rough kernels
117
Proposition 2.3.3 Suppose that Ω ∈ L q (Sn−1 ) satisfies (2.2.1). If m ∈ 0 N, m ≥ 2, q > max{p, 2m /(2m − 1)} and ω 1−p ∈ Ap0 /q0 , then there exists a constant C > 0, independent of f , such that kMΩ f kLp (ω) ≤ Ckf kLp (ω) . Now we turn to the proof of Theorem 2.3.8. If p ≥ 2, then Theorem 2.3.8 follows from Proposition 2.3.1. If p < 2, then there exists m ∈ N, m ≥ 2 such that 2m−1 2m ≤ p < . 2m − 1 2m−1 − 1
So q > p is equivalent to q > max{p, 2m /(2m − 1)}. Thus by Proposition 2.3.3 we prove that Theorem 2.3.8 holds under the condition (ii).
Theorem 2.3.9 Suppose that Ω ∈ Lq (Sn−1 ) (q > 1) satisfies (2.2.1) and (2.2.2). If p, q and weight function ω satisfy one of the following conditions: (i) q 0 ≤ p < ∞, p 6= 1
and
ω ∈ Ap/q0 ;
(ii) 1 < p ≤ q, p 6= ∞
and
ω 1−p ∈ Ap0 /q0 ;
(iii) 1 < p < ∞
and
0
0
ω q ∈ Ap ,
then TΩ is bounded on Lp (ω). Proof. In fact, the proof of Theorem 2.3.8 under the condition (ii) has implied that of Theorem 2.3.9. First we give the proof of Theorem 2.3.9 under the condition (i). It suffices to consider the case p > q 0 . We use the notations as in Theorem 2.3.8. It follows that ! X X 2 Sj+k f (x) TΩ f (x) = KΩ,j ∗ j
=
XX j
k
:=
X
k
Sj+k (KΩ,j ∗ Sj+k f )(x)
TΩk f (x).
k
k by T k , then (2.3.18) still holds. By the Plancherel theorem, if we replace T ε,Ω Ω That is, there exist constants C, θ > 0 independent of f and k ∈ Z, such that
k (2.3.29)
TΩ f 2 ≤ C2−θ|k| kf kL2 . L
118
Chapter 2. Singular Integral Operators
If we can show, under the condition (i), there exists a constant C > 0 , independent of f and k ∈ Z, such that
k T f (2.3.30)
Ω p ≤ Ckf kLp (ω) , L (ω)
then we can get the conclusion under the condition (i) by applying the Stein-Weiss interpolation theorem with change of measure (Lemma (2.2.5)) to (2.3.29) and (2.3.30). If q > 2, then (2.3.30) follows from (2.3.19). If q < 2 (then p > q 0 > 2), analogous to the proof of Lemma 2.3.7, we have that Z
2 X
k j(n−nq) e 2 2 KΩ,j ∗ |Sj+k f | (x) h(x)dx
TΩ f p ≤C sup L (ω) h Rn j∈Z
1/2 2
X
≤C |Sj+k f (·)|2
j
p × sup h
Z
Rn
L (ω)
[MΩ2−q h(x) ]
( p2 )0
ω(x)
1−(p/2)0
dx
1/(p/2)0
,
where the supremum is taken over all functions h satisfying khkL(p/2)0 (ω1−(p/2)0 ) ≤ 1. Since q < 2, it follows that r = q/(2 − q) > (p/2) 0 , (p/2)/r 0 = p/q 0 and 0 Ω2−q ∈ Lr (Sn−1 ). It is clear that r, (p/2)0 and ω 1−(p/2) satisfy the condition (ii) of Theorem 2.3.8 under the condition (i). Thus (2.3.28) holds. In this way, for q < 2, the formula (2.3.30) can be deduced from (2.3.28) and the weighted Littlewood-Paley theory. The way of dealing with the case q = 2 is completely the same as Lemma 2.3.7. Here we will not give details. Thus we have proved Theorem 2.3.9 under the condition (i). Applying the conclusion under the condition (i) and dual method (see the proof of Theorem 2.2.3), we can obtain Theorem 2.3.9 under the condition (ii). Finally, by the conclusion under the condition (i) and (ii), together with the method in the proof of Theorem 2.2.4 we can show that Theorem 2.3.9 still holds under the condition (iii). Thus we complete the proof of Theorem 2.3.9.
2.4 Commutators of singular integral operators
119
Remark 2.3.1 As space is limited, all results on boundedness of T Ω and MΩ are only involved in Lp -boundedness. In 1967, Calder´ on, Weiss and Zygmund [CaWZ] showed that if Ω ∈ L(log + L)1−ε (Sn−1 ), 0 < ε < 1 and Ω satisfies (2.2.2), then there is an L 2 (Rn ) function f such that TΩ f ∈ / L2 (Rn ). For the weak type (1,1) boundedness of T Ω and MΩ , we will mentioned several results here. In 1988, Christ [Chr] proved that M Ω is of weak type (1,1) when Ω ∈ Lq (S 1 )(q > 1). Hofmann [Ho] showed that TΩ is of weak type (1,1) if Ω ∈ Lq (S1 )(q > 1) and satisfies (2.2.2). And Christ and Rubio de Francia [ChrR] improved the result of Christ above and proved that M Ω is of weak type (1,1) if Ω ∈ Llog + L(Sn−1 ) for all n ≥ 2. In the same paper, Christ and Rubio de Francia [ChrR] proved that T Ω is of weak type (1,1) if Ω ∈ Llog + L(Sn−1 ) for n ≤ 5 and satisfies (2.2.2). In 1996, Seeger [Se] proved that TΩ is of weak type (1,1) when Ω ∈ Llog + L(Sn−1 ) for all n ≥ 2 and satisfies (2.2.2).
Remark 2.3.2 It is still an open problem whether T Ω is of weak type (1,1) when Ω ∈ H 1 (Sn−1 ) with the condition (2.2.2). It is also an open problem whether MΩ is of weak type (1,1) if Ω ∈ L1 (Sn−1 ).
2.4
Commutators of singular integral operators
In 1965, Calder´on defined the Calder´on commutator in studying the boundedness of the Cauchy integral on Lipschitz curves, and its definition is Z ∞ ϕ(x) − ϕ(y) f (y) dy, Ch,ϕ (f )(x) = p.v. h x−y x−y −∞
where h ∈ C ∞ (R), ϕ is a Lipschitz function on R. It is clear that, if h(t) = (1 + it)−1 , then Ch,ϕ (f ) is the Cauchy integral along the curve y = ϕ(x); if h = 1, then Ch,ϕ (f ) is Hilbert transform; if h(t) = t k (k is a natural number), then Ch,ϕ (f ) is commutator of degree k of the Hilbert transform about ϕ. In 1976, Coifman, Rochberg and Weiss studied the L p boundedness (1 < p < ∞) of commutator [b, TΩ ] generated by the Calder´on-Zygmund singular integral operator TΩ and a function b, where [b, TΩ ] is defined by [b, TΩ ](f )(x) = p.v.
Z
Rn
Ω(x − y) [b(x) − b(y)]f (y)dy, |x − y|n
(2.4.1)
120
Chapter 2. Singular Integral Operators
where Ω satisfies the condition of homogeneity of degree zero (2.2.1) and the vanishing condition (2.2.2), moreover, Ω ∈ Lip1(S n−1 ) and b is a BMO(Rn ) function. Using the Lp boundedness of commutator [b, TΩ ], Coifman, Rochberg and Weiss successfully gave a decomposition of Hardy space H 1 (Rn ). The commutator defined by (2.4.1) is called CRW-type commutator, and CRW-type commutator plays an important role in studying the regularity of solutions of elliptic partial differential equations of second order. In this section we will introduce some results on L p boundedness and weighted Lp boundedness (1 < p < ∞) of CRW-type commutator. First we introduce the definition of the space of BMO (Bounded Mean Oscillation) functions on Rn . Suppose that f is a locally integrable function on Rn , let Z 1 |f (x) − fQ |dx, kf kBMO = sup Q |Q| Q where the supreme is taken over all cubes Q ⊂ R n , and Z 1 fQ = f (y)dy. |Q| Q Define BMO(Rn ) = {f : kf kBMO < ∞}. If one regards two functions whose difference is a constant as one, then space BMO is a Banach space with respect to norm k · k BMO . We next formulate some remarks about BMO(R n ) (we refer to Lu[Lu1]). Remark 2.4.1 (1) The John-Nirenberg inequality: there are constants C 1 , C2 > 0, such that for all f ∈ BM O(Rn ) and α > 0 |{x ∈ Q : |f (x) − fQ | > α}| ≤ C1 |Q|e−C2 α/kf kBM O , ∀Q ⊂ Rn . (2) The John-Nirenberg inequality implies that kf kBMO ∼ sup Q
1 |Q|
Z
Q
p
|f (x) − fQ | dx
1
p
(2.4.2)
for 1 < p < ∞. (3) By the definition of BMO and the Sharp maximal function (see Section 2.1), if f ∈ Lloc (Rn ), then f ∈ BMO(Rn ) ⇐⇒ M ] f ∈ L∞ (Rn ).
(2.4.3)
2.4 Commutators of singular integral operators
121
(4) If f ∈ BMO(Rn ) and h ∈ Rn , then f (· − h), the translation of f , satisfies that f (· − h) ∈ BMO(Rn ), and kf (· − h)kBMO = kf kBMO .
(2.4.4)
(5) If f ∈ BMO(Rn ) and h ∈ Rn , and λ > 0, then f (λx) ∈ BMO(Rn ), and kf (λ·)kBMO = kf kBMO . (2.4.5) (6) If f ∈ BMO(Rn ), then kf kBMO
1 ∼ sup inf Q c∈R |Q|
Z
Q
|f (x) − c|dx.
(2.4.6)
Theorem 2.4.1 Suppose that Ω ∈ Lip1(S n−1 ) satisfies (2.2.1) and (2.2.2). TΩ is a singular integral operator with the kernel Ω. Then the following two statements hold. (i) If b ∈ BMO(Rn ), then [b, TΩ ] is bounded on Lp (Rn ) (1 < p < ∞); (ii) Suppose 1 < p0 < ∞ and b ∈ ∪q>1 Lqloc (Rn ). If [b, TΩ ] is bounded on Lp0 (Rn ), then b ∈ BMO(Rn ). Let us first establish the following lemma. Lemma 2.4.1 Suppose b ∈ BMO(Rn ) and 1 < p < ∞. Then for any 1 < s < p, there exists a constant C, independent of b and f , such that i h 1 1 M ] ([b, TΩ ]f ) (x) ≤ CkbkBMO (M (|TΩ f |s )(x)) s + (M (|f |s )(x)) s , x ∈ Rn , (2.4.7) where M is the Hardy-Littlewood maximal operator. Proof. For x ∈ Rn , let Q be any cube containing x. For y ∈ Q, set [b, TΩ ](f )(y) = (b(y) − bQ )TΩ f (y) − TΩ (b − bQ )f χ4√nQ (y) − TΩ (b − bQ )f χ(4√nQ)c (y) := a1 (y) − a2 (y) − a3 (y).
It follows from H¨older’s inequality and (2.4.2) that 1 10 Z Z Z s s 1 1 1 s s0 |a1 (y)|dy ≤ |b(y) − bQ | dy |TΩ f (y)| dy |Q| Q |Q| Q |Q| Q h i1 s ≤ CkbkBMO M (|TΩ f |s ) (x) . (2.4.8)
122
Chapter 2. Singular Integral Operators
Take 1 < u, q < ∞ such that uq = s. Thus by the L q boundedness of TΩ (see Theorem 2.1.2), we conclude that
1 |Q|
Z
Q
|a2 (y)|dy ≤
1 |Q|
Z q 1q TΩ (b − bQ )f χ4√nQ (y) dy Q
≤C
1 |Q|
Z
≤C
1 |Q|
Z
×
q
√ 4 nQ
√ 4 nQ
1 |Q|
Z h
q
|b(y) − bQ | |f (y)| dy |b(y) − bQ | qu
√ 4 nQ
qu0
|f (y)| dy s
≤ CkbkBMO M (|f | )(x)
i1
s
dy
1 q
1 qu0
(2.4.9)
1 qu
.
Now denote the center and side length of Q by x 0 and d, respectively. If y ∈ √ Q, z ∈ (4 nQ)c , then |z − y| ∼ |z − x0 | ∼ |z − x|. Since Ω(x0 ) ∈ Lip1(Sn−1 ), it follows from (2.1.2) that Ω(y − z) Ω(x0 − z) 1 1 |y − z|n − |x0 − z|n ≤ |Ω(y − z)| |y − z|n − |x0 − z|n 1 + |x0 − z|n
≤ kΩkL∞ (Sn−1 ) ≤
Ω y − z − Ω x0 − z |y − z| |x0 − z|
cd c 2d + n+1 n |x0 − z| |x0 − z| |x0 − z|
cd . |z − x0 |n+1
Here note |z − y| ≥ 2|y − x0 |. Thus, it follows from (2.4.10) that
(2.4.10)
123
2.4 Commutators of singular integral operators
Ω(y − z) Ω(x0 − z) |b(z) − bQ ||f (z)|dz |a3 (y) − a3 (x0 )| ≤ − √ n |x0 − z|n Rn \4 nQ |y − z| Z
≤ Cd
×
Z
! 10
|f (z)|s dz |z − x0 |n+1
!1
0
√ Rn \4 nQ
Z
|b(z) − bQ |s dz |z − x0 |n+1
√ Rn \4 nQ
s
s
h i1 s ≤ CkbkBMO M (|f |s )(x) .
(2.4.11) Thus from (2.1.20) and (2.4.8), (2.4.9) and (2.4.11) it follows that (2.4.7) holds. To complete the proof of Theorem 2.4.1 (i), we still need to introduce a result of Fefferman and Stein. Its proof can be found in Fefferman and Stein [FeS2]. Lemma 2.4.2 (Fefferman-Stein) Let 1 ≤ p 0 < ∞. Then for any p0 ≤ p < ∞, there exists a constant C, independent of f , such that kM f kp ≤ CkM ] f kp
(2.4.12)
for any function f satisfying M f ∈ Lp0 (Rn ). Actually the proof of Theorem 2.4.1 (i) is a direct consequence of Lemma 2.4.1, Lemma 2.4.2, Theorem 1.1.1 and Theorem 2.1.2. Now we pay attention to the proof of Theorem 2.4.1 (ii). Without loss of generality, we may assume
= 1.
[b, TΩ ] p p L
0 →L 0
We wish to prove that there exists a constant A = A(p 0 , Ω) such that Z 1 sup |b(x) − bQ |dx ≤ A. (2.4.13) Q |Q| Q By (2.4.4) and (2.4.5), we merely need to prove (2.4.13) in the case Q = Q 1 , √ where Q1 is a cube whose center is at the origin and side length 1/ n parallel
124
Chapter 2. Singular Integral Operators
to the coordinates. Moreover we notice that [b − b Q1 , TΩ ] = [b, TΩ ]. Thus we may assume bQ1 = 0. Now let ψ(x) = (sgn(b) − c0 )χQ1 (x), where c0 =
1 |Q1 |
Z
sgn(b(x))dx.
Q1
Then ψ satisfies the following properties: kψk∞ ≤ 2,
(2.4.14)
supp(ψ) ⊂ Q1 , Z ψ(x)dx = 0,
(2.4.15) (2.4.16)
Q1
1 |Q1 |
Z
ψ(x)b(x) ≥ 0, Z 1 ψ(x)b(x)dx = |b(x)|dx := B. |Q1 | Q1
(2.4.17) (2.4.18)
Since Ω satisfies (2.2.1) and (2.2.2), there exists 0 < A 1 < 1, such that σ x0 ∈ Sn−1 : Ω(x0 ) ≥ 2A1 > 0, (2.4.19)
where σ is a measure on Sn−1 induced by the Lesbesgue measure. Let Λ = x0 ∈ Sn−1 : Ω(x0 ) ≥ 2A1 .
Then Λ is a closed set. It is easy to see that, for any x 0 ∈ Λ and y 0 ∈ Sn−1 , when |x0 − y 0 | < A1 , we have Ω(y 0 ) ≥ A1 . Set 2 0 n + 1 and x ∈ Λ . G = x ∈ R : |x| > A2 = A1
Then when x ∈ G and y ∈ Q1 , we have 5 |x − y| > |x| 6 and |x| > 2|y|. Thus it implies by (2.1.11) that x x − y |y| 2 |x| − |x − y| ≤ 2 |x| ≤ |x| < A1 .
125
2.4 Commutators of singular integral operators Since x0 ∈ Λ, we have Ω follows that
x−y |x−y|
≥ A1 . Thus from (2.4.17) and (2.4.18), it
Z |TΩ (bψ)(x)| =
Ω(x − y) b(y)ψ(y)dy n Q1 |x − y| ≥ A3 |x|−n B,
(2.4.20)
where A3 depends on A1 and n. On the other hand, when x ∈ G and y ∈ Q1 , it follows from the condition of Ω and (2.1.12) that Ω(x − y) Ω(x) 1 0 |x − y|n − |x|n ≤ A4 |x|n+1 ,
(2.4.21)
where A04 only depends on n and Ω. Thus by (2.4.14), (2.4.16) and (2.4.21) we conclude that Ω(x − y) Ω(x) |ψ(y)|dy − |b(x)TΩ (ψ)(x)| ≤ |b(x)| n |x|n Q1 |x − y| ≤ A4 |b(x)||x|−(n+1) , Z
(2.4.22)
where A4 depends only on n and Ω. Therefore when x ∈ G, from (2.4.20) and (2.4.22) it follows that |[b, TΩ ](ψ)(x)| ≥ |TΩ (bψ)(x)| − |b(x)TΩ (ψ)(x)|
(2.4.23)
≥ A3 |x|−n B − A4 |b(x)||x|−(n+1) . Let F =
x ∈ G : |b(x)| >
It follows from (2.4.23) that
BA3 2A4
|x| and |x| < B
p00 n
.
126
Chapter 2. Singular Integral Operators
kψkpp00 ≥
Z
≥
Z
=
|[b, TΩ ](ψ)(x)|p0 dx p0 Z 1 −n ( ) ≥ A3 B|x| dx p00 2 (G\F )∩ |x|
=
Rn
(
1
n A5 (|F |+An 2 ) <|x|
A3 B 2 A3 B 2
p0 p0
σ(Λ)
Z
B
p0 0 n
)
∩G
1 A3 B|x|−n 2
p0
dx
p00 n
1 n A5 (|F |+An 2)
t−np0 +n−1 dt
i σ(Λ) h p00 (1−p0 ) n(1−p0 ) n 1−p0 . B − A5 (|F | + A2 ) n(1 − p0 )
So there exists A6 such that 0
0 (|F | + An2 )1−p0 ≤ A1−p B p0 (1−p0 ) . 6
Therefore, if 1 0
p0 B > (2An2 A−1 6 ) ,
then
0
A 6 B p0 ≥ |F | ≥ A6 B − . (2.4.24) 2 Now take g(x) = sgn(b(x))χF (x). Denote the conjugate operator of T Ω by TΩ0 . For x ∈ Q1 we then have that Z Z Ω(y − x) [b, TΩ0 ]g(x) ≥ Ω(y − x) |b(y)|dy − |b(x)| g(y)dy . n n Rn |y − x| F |y − x| (2.4.25) Since y ∈ F , we have BA3 |y|. |b(y)| > 2A4 p00
An2
Moreover, since x ∈ Q1 , applying the same method as in the estimate of (2.4.20), we know that there exists A 7 such that Z Z Ω(y − x) −n BA3 |y| |b(y)|dy ≥ A7 |y|dy. n 2A4 F F |y − x|
127
2.4 Commutators of singular integral operators From (2.4.24) it follows that there exists A 8 such that Z p0 Ω(y − x) ≥ A8 B 1+ n0 . |b(y)|dy n F |y − x|
(2.4.26)
On the other hand, there exists A9 such that Z Z Ω(y − x) |Ω(y − x)| dy n |y − x|n g(y)dy ≤ 2n |y|n R F Z dy kΩkL∞ (Sn−1 ) ≤ 2−n p0 0 |y|n A2 <|y|
which together with (2.4.25) and (2.4.26) yields that p0 [b, TΩ0 ]g(x) ≥ A8 B 1+ n0 − A9 |b(x)| log B.
(2.4.27)
Since [b, TΩ0 ] is the conjugate operator of [b, TΩ ], we have k[b, TΩ0 ]k
0
0
Lp0 →Lp0
= 1.
So it follows from definition of F and (2.4.27) that A10 B ≥ kgkp00 ≥ k[b, TΩ0 ]gkp00 ≥
Z
≥
Z
Q1
Q1
[b, TΩ0 ]g(x) dx
A8 B
1+
p00 n
(2.4.28)
− A9 |b(x)| log B dx
p00
= A8 B 1+ n |Q1 | − A9 |Q1 |B log B. By (2.4.28), it is clear to have B ≤ A(n, Ω, p 0 ). Thus we complete the proof of Theorem 2.4.1 (ii). Theorem 2.4.1 shows that Lp boundedness of the commutator of singular integral operator TΩ (Ω is a Lipschitz function) can be used to characterize BMO function. Next we will show that L p boundedness of the commutator of a linear operator T with BMO function can be derived from the weighted Lp boundedness of T .
128
Chapter 2. Singular Integral Operators
Theorem 2.4.2 Suppose that 1 < p < ∞. The operator T is defined by Z K(x, y)f (y)dy, T f (x) = Rn
and the commutator [b, T ] of T and b is defined by Z K(x, y)[b(x) − b(y)]f (y)dy. [b, T ]f (x) = Rn
For ω ∈ As (1 < s < ∞), if T is bounded on Lp (ω), then [b, T ] is bounded on Lp (Rn ) for every b ∈ BMO(Rn ). To prove Theorem 2.4.2, let us first introduce some lemmas. Lemma 2.4.3 Let 1 < s < ∞, λ > 0 and b ∈ BMO(R n ). Then there exists η = η(λ, s) > 0, when kbkBMO < η, such that eλb(x) ∈ As (Rn ). By the John-Nirenberg inequality, the Reverse H¨older inequality of A p can be used to prove Lemma 2.4.3. For the details, see Garcia-Cuerva & Rubio de Francia [GaR]. Now we turn back to the proof of Theorem 2.4.2. Take λ = p and b ∈ BMO(Rn ). Without loss of generality, we may assume that kbk BMO < η. By Lemma 2.4.3 we have epb ∈ As (Rn ). On the other hand, for every θ ∈ [0, 2π], b cos θ ∈ BMO(Rn ) and kb cos θkBMO = kbkBMO < η. Thus epb cos θ ∈ As (Rn ). Now for z ∈ C,
g(z) = ez[b(x)−b(y)]
is analytic on C. Thus by the Cauchy integral formula we get Z g(z) 1 0 b(x) − b(y) = g (0) = dz 2πi |z|=1 |z|2 Z 2π 1 iθ = ee [b(x)−b(y)] e−iθ dθ. 2π 0 For θ ∈ [0, 2π], let
iθ
hθ (x) = f (x)e−b(x)e .
(2.4.29)
129
2.4 Commutators of singular integral operators Since f ∈ Lp , we have hθ ∈ Lp (epb cos θ )
and khθ kLp (epb cos θ ) = kf kLp .
Applying (2.4.29), it implies that Z 2π Z 1 eiθ [b(x)−b(y)] −iθ e e dθ f (y)dy K(x, y) [b, T ](f )(x) = 2π 0 Rn Z 2π 1 iθ = T (hθ )(x)eb(x)e e−iθ dθ. 2π 0 Applying Minkowski’s inequality and the L p (ω) boundedness of T , we have Z 2π 1 p kT (hθ )kLp (epb cos θ ) dθ ≤ Ckf kLp . k[b, T ]f kL ≤ 2π 0 This finishes the proof of Theorem 2.4.2. We can immediately obtain the Lp boundedness of commutator [b, TΩ ] by Theorem 2.4.2 and Theorem 2.3.9. Corollary 2.4.1 Suppose that Ω ∈ Lq (Sn−1 ) (1 < q ≤ ∞) satisfies (2.2.1) and (2.2.2). If b ∈ BMO(Rn ), then [b, TΩ ] is bounded on Lp for 1 < p < ∞. Remark 2.4.2 Corollary 2.4.1 improves the conclusion (i) of Theorem 2.4.1. Next we will give the weighted boundedness of commutator [b, T Ω ]. In fact, similar to the conclusion of Theorem 2.4.2, weighted L p boundedness of [b, TΩ ] can also be attribute to weighted L p boundedness of TΩ . Theorem 2.4.3 Let 1 < p < ∞. Suppose that T is defined as in Theorem 2.4.2 and satisfies kT f kLp (ω) ≤ Ckf kLp (ω) for ω ∈ As (1 < s < ∞). Then [b, T ] is bounded on Lp (ω). Proof. The idea in the proof is essentially the same as that of Theorem 2.4.2. Since ω ∈ As , by Proposition 1.4.2 (x) we know that there exists ε > 0 such that ω 1+ε ∈ As . So kT f kLp (ω1+ε ) ≤ Ckf kLp (ω1+ε ) . Let λ =
p(1+ε) . ε
(2.4.30)
By Lemma 2.4.3 we have that there is η > 0 such that epb(x)(1+ε)/ε ∈ As (Rn )
130
Chapter 2. Singular Integral Operators
whenever kbkBMO < η. Thus for every θ ∈ [0, 2π], epb(1+ε) cos θ/ε ∈ As (Rn ) still holds. By the weighted boundedness of T we get kT f kLp (epb(1+ε) cos θ/ε ) ≤ Ckf kLp (epb(1+ε) cos θ/ε ) .
(2.4.31)
Applying the Stein-Weiss interpolation theorem with change of measure (Lemma 2.2.5) to (2.4.30) and (2.4.31) we have kT f kLp (ωepb cos θ ) ≤ Ckf kLp (ωepb cos θ ) .
(2.4.32)
iθ
Now for θ ∈ [0, 2π], denote hθ = f e−be . Then by f ∈ Lp (ω) we know hθ ∈ Lp (ωepb cos θ )
and khθ kLp (ωepb cos θ ) = kf kLp (ω) .
(2.4.33)
It follows from (2.4.29) that 1 [b, T ](f )(x) = 2π
Z
2π
iθ
T (hθ )(x)eb(x)e e−iθ dθ.
0
Thus by Minkowski’s inequality and (2.4.32) as well as (2.4.33), we have k[b, T ]f kLp (ω)
1 ≤ 2π
Z
2π 0
kT (hθ )kLp (ωepb cos θ ) dθ ≤ Ckf kLp (ω) .
This finishes the proof of Theorem 2.4.3. Lp
Applying Theorem 2.4.3 and Theorem 2.3.9, it is easy to get the weighted boundedness of commutator [b, TΩ ].
Theorem 2.4.4 Suppose that Ω ∈ Lq (Sn−1 ) (q > 1) satisfies (2.2.1) and (2.2.2). If b ∈ BMO(Rn ) and p, q, ω satisfy one of the following conditions, then [b, TΩ ] is bounded on Lp (ω): (i) q 0 ≤ p < ∞, p 6= 1 and ω ∈ A p0 ; q
0
(ii) 1 < p ≤ q, p 6= ∞ and ω 1−p ∈ A p0 ; q0
0
(iii) 1 < p < ∞ and ω q ∈ Ap .
131
2.5 Notes and references
Proof. The result under the condition (i) is a direct consequence of Theorem 2.4.3 and Theorem 2.3.9. Using the conclusion under the condition (i) and dual method (see the proof of Theorem 2.2.3) we can obtain the conclusion under the condition (ii). The conclusion under the condition (iii) is a corollary of the conclusions under the condition (i) and (ii) (for the details, see the proof of Theorem 2.2.4). Remark 2.4.3 1. Using the method in the proof of Theorem 2.4.2 and Theorem 2.4.3 as well as induction, we can get the similar result about commutators of degree k of linear operator T and b. Here the commutator of degree k is defined by Z k K(x, y)[b(x) − b(y)]k f (y)dy. [b, T ] f (x) = [b, · · · , [b, T ] · · · ]f (x) = Rn
2. In the same way, we can obtain the weighted L p boundedness of commutator of degree k generated by a singular integral operator with rough kernel TΩ and a function b.
2.5
Notes and references
Lemma 2.1.1 is due to Cotlar [Cot]. Lemma 2.1.4 and Theorem 2.1.9 are due to Benedek, Calder´on and Panzone [BeCP]. Theorem 2.1.11 comes from Garcia-Cuerva and Rubio de Francia [GaR]. The idea of the proof in Lemma 2.2.1 comes from Duoandikoetxea and Rubio de Francia [DuR]. The idea on the decomposition in the proof of Theorem 2.2.2 comes from Christ and Rubio de Francia [ChrR]. For L q Dini condition in Theorem 2.2.3 and Lemma 2.2.2, we refer to Kurtz and Wheeden [KuW]. Lemma 2.3.1 is due to Colzani [Col]. Lemma 2.3.2 was first obtained by Stefanov [Ste]. Here the idea of the proof ia taken from Pan, Wu and Yang [PaWY]. For the proof of Lemma 2.3.3, see Grafakos and Stefanov [GrS]. Theorem 2.3.5 was first obtained by Connett [Con] and independently by Ricci and Weiss [RiW]. The proof of Theorem 2.3.5 given here comes from Grafakos and Stefanov [GrS]. Theorem 2.3.6 is due to Grafakos and Stefanov [GrS]. Generalizations of Theorem 2.3.5 and Theorem 2.3.6 to
132
Chapter 2. Singular Integral Operators
rough kernels supported by subvarieties were established by Fan and Pan [FaP2]. The idea of proving Theorem 2.3.8 is taken from Duoandikoetxea [Du1]. And Theorem 2.3.9 is obtained by Watson [Wat] and independently by Duoandikoetxea [Du1]. In 2003, Ding and Lin [DiLin] established weighted Lp boundedness for the rough operators T Ω , MΩ and TΩ∗ with different weights. In 1976, Coifman, Rochberg and Weiss [CoiRW] established the conclusion (i) in Theorem 2.4.1 for singular integral operators with standard Calder´on-Zygmund kernels and the conclusion (ii) in Theorem 2.4.1 for the Riesz transforms Rj , 1 ≤ j ≤ n. Theorem 2.4.1 is obtained by Janson [Ja] and independently by Uchiyama [Uc]. The proof of Theorem 2.4.1 given here comes from [Uc]. Theorem 2.4.2 is first established by Alvarez, Bagby, Kurtz and Perez [ABKP]. The proof of Theorem 2.4.2 here comes from Hu [Hu] and Ding, Lu [DiL4]. In addition, Corollary 2.4.1 has been improved by Lu and Wu [LuW3]in which the condition Ω ∈ L q (Sn−1 )(q > 1) in Corollary 2.4.1 wasSreplaced by a class of bigger block spaces B q0,0 (Sn−1 ), 1 < q < ∞. In fact, r>1 Lr (Sn−1 ) ⊂ Bq0,0 (Sn−1 ), 1 < q < ∞ (see Lu, Taibleson and Weiss [LuTW]).
Chapter 3
Fractional Integral Operators
In this chapter, we will investigate the boundedness of the Riesz potential and its general form (fractinal integral). If 0 < α < n, then |x| −n+α ∈ Lloc (Rn ). Thus under the sense of distribution, we have that \ 1 (ξ) = γ(α)(2π)−α |ξ|−α , |x|n−α α n − 2 n α where γ(α) = π 2 2 Γ /Γ . 2 2
The Riesz potential is an operator defined by Z 1 f (y) 1 1 dy = ∗ f (x). Iα (f )(x) = γ(α) Rn |x − y|n−α γ(α) | · |n−α
(3.0.1)
(3.0.2)
Let us now explain the relationship between the Riesz potential and the Laplacian operator of fractional degree. Let 4 be the Laplacian operator, that is, ∂2 ∂2 4= + · · · + . ∂x2n ∂x21 For any ϕ ∈ S (Rn ), by the property of Fourier transform, we have \ (−4ϕ)(ξ) = 4π 2 |ξ|2 ϕ(ξ) b = (2π|ξ|)2 ϕ(ξ). b 133
134
Chapter 3. Fractional Integral Operators
Thus, for ϕ ∈ S (Rn ) and 0 < α < n, it follows from (3.0.1) and (3.0.2) that −α \ (I ϕ(ξ) b = ((−4)−α/2 ϕb)(ξ). α ϕ)(ξ) = (2π|ξ|)
Note that under the sense of distribution, for f ∈ S (R n ), n n X X (I1 ∗ Rj ∂j f )(x), Rj ∂j f (x) = f (x) = I1 ∗
(3.0.3)
j=1
j=1
where Rj denote the Riesz transforms (see the definition in chapter 2). By the Lp -boundedness of the Riesz transform R j (1 < p < ∞, j = 1, 2, · · · , n) and the (Lp , Lq )-boundedness of the Riesz potential with 1-order, from ∂j f ∈ Lp (1 < p < n, j = 1, 2, · · · , n) and (3.0.3), it follows that f ∈ Lq (Rn ). Therefore the research to the Riesz potential is closely related to the theory of Sobolev space and the Laplacian operator of fracrional degree.
3.1
Riesz potential
In this chapter we will show the (Lp , Lq )-boundedness of the Riesz potential Iα , 0 < α < n. We first show that 1 α 1 = − q p n
(3.1.1)
is a necessary condition for ensuring kIα f kq ≤ Ckf kp .
(3.1.2)
Indeed, for δ > 0, we define the dilation transform τδ (f )(x) = f (δx). Then, for 0 < α < n, τδ−1 Iα τδ = δ −α Iα
(3.1.3)
and kτδ (f )kp = δ −n/p kf kp , kτδ−1 Iα (f )kq = δ n/q kIα (f )kq .
(3.1.4)
135
3.1 Riesz potential Thus, for δ > 0, it follows from (3.1.2), (3.1.3), and (3.1.4) that kIα f kq = δ α kτδ−1 Iα τδ (f )kq = δ α+n/q kIα τδ (f )kq ≤ Cδ α+n/q kτδ f kp = Cδ α+n/q−n/p kf kp .
(3.1.5)
Since (3.1.5) holds for all δ > 0, we conclude that (3.1.1) must hold. The following theorem shows that, when 1 < p < αn , (3.1.1) is also a sufficient condition for ensuring (3.1.2). We first formulate a lemma. Lemma 3.1.1 Let f ∈ Lp (Rn ), 1 ≤ p < αn . Then, for x ∈ Rn , we have i1− αp αp h n |Iα f (x)| ≤ Ckf kpn M f (x) ,
where M is the Hardy-Littlewood maximal function and C = C(α, p, n).
Proof. Fix x ∈ Rn , for some r > 0, Z Z |f (x − y)| f (x − y) dy + dy := J1 + J2 . |Iα f (x)| ≤ n−α |y|n−α |y|>r |y|≤r |y| For J1 , we write J1 =
∞ Z X
−j−1 r<|y|≤2−j r j=0 2 Z ∞ X
f (x − y) dy |y|n−α
1 |f (x − y)|dy (2−j−1 r)n−α 2−j−1 r<|y|≤2−j r j=0 Z ∞ X (2−j )α α ≤ Cr |f (x − y)|dy (2−j r)n |y|≤2−j r ≤
(3.1.6)
j=0
≤ Cr α M f (x).
As to J2 , if p = 1, then If 1 < p <
n α,
J2 ≤ r α−n kf k1 .
(3.1.7)
then H¨older’s inequality implies that !1/p0 Z J2 ≤
≤ Cr
0
|y|>r
α− n p
|y|(α−n)p dy
kf kp .
kf kp
(3.1.8)
136
Chapter 3. Fractional Integral Operators
Therefore, when 1 ≤ p < αn , from (3.1.6)- (3.1.8), for all x ∈ R n , it follows that n |Iα f (x)| ≤ C(r α M f (x) + r α− p kf kp ). kf k
p
p Now take r = ( M f (x) ) n , then
r α M f (x) = r
α− n p
αp
αp
kf kp = kf kpn M f (x)1− n .
Thus the lemma is proved. Theorem 3.1.1 (Hardy-Littlewood-Sobolev theorem) Suppose that 0 < α < n, 1 ≤ p < αn and 1q = 1p − αn . (i) If f ∈ Lp (Rn )(1 < p < αn ), then kIα f kq ≤ Ckf kp ; (ii) If f ∈ L1 (Rn ), then for all λ > 0, |{x ∈ Rn : |Iα f (x)| > λ}| ≤ n ( Cλ kf k1 ) n−α , where C = C(α, n, p). p Proof. (i) Note that (1 − αp n )q = p. By Lemma 3.1.1 and L -boundedness of the Hardy-Littlewood maximal operator M for 1 < p < ∞, we get αp
1− αp n
kIα f kq ≤ Ckf kpn kM f kp
≤ Ckf kp .
(ii) Using Lemma 3.1.1 again and the weak (1, 1)-boundedness of the operator M , we have that ! n n−α λ |{x ∈ Rn : |Iα f (x)| > λ}| = x ∈ Rn : M f (x) > α C||f || n 1
α n
n ! n−α
C||f ||1 λ n n−α C ≤ . ||f ||1 λ ≤ C1
||f ||1
Let us give some remarks on the Riesz potential I α .
Remark 3.1.1 The second result (ii) of Theorem 3.1.1 cannot be improved as n ≤ Ckf k1 . (3.1.9) kIα f k n−α
137
3.2 Weighted boundedness of Riesz potential
There is a counterexample for (3.1.9). Let f (y) = 1 when |y| ≤ 1 and f (y) = 0 when |y| > 1. Then Z 1 dy . Iα f (x) = γ(α) |y|≤1 |x − y|n−α For |x| > 1 and |y| ≤ 1, |x − y| ≤ |x| + |y| ≤ |x| + 1 < 2|x|, and hence |Iα f (x)| ≥
C |x|n−α
for
|x| > 1,
which implies Z
Rn
n
|Iα f (x)| n−α dx ≥ C
Z
|x|>1
dx = ∞. |x|n
Remark 3.1.2 When p = αn , the first result (i) of Theorem 3.1.1 is not true. Example: − α (1+ε) n −α 1 |x| log |x| , |x| ≤ 21 f (x) = 0, |x| > 12 n
where ε is sufficiently small. Obviously, f ∈ L α (Rn ). But if we take any ε satisfies αn (1 + ε) ≤ 1, then 1 Iα (f )(0) = γ(α)
Z
|x|≤ 21
|x|
−n
1 log |x|
− α (1+ε) n
and Iα (f ) is not essentially bounded near the origin. L∞ (Rn ).
3.2
dx = ∞, Hence I α f (x) ∈ /
Weighted boundedness of Riesz potential
To study the weighted boundedness of I α , we need to introduce the following fractional maximal operator M α . For 0 < α < n and f ∈ Lloc (Rn ), define Mα by Z 1 Mα (f )(x) = sup n−α |f (x − y)|dy. (3.2.1) r>0 r |y|≤r
138
Chapter 3. Fractional Integral Operators
An equivalent definition of Mα is 1 Mα (f )(x) = sup 1− α n Qx |Qx |
Z
Qx
|f (y)|dy,
(3.2.2)
where the supremum is taken over all cubes Q x in Rn with the center at x and with the sides parallel to the axes. The fractional maximal operator Mα will be dominated by Iα in some sense. That is, for 0 < α < n, f ∈ Lloc (Rn ) and x ∈ Rn , we have Mα (f )(x) ≤ γ(α)Iα (|f |)(x).
(3.2.3)
In fact, for fixed x ∈ Rn and r > 0, we have that Z |f (x − y)| 1 dy Iα (|f |)(x) = γ(α) n−α ZRn |y| |f (x − y)| 1 ≥ γ(α) dy n−α |y|≤rZ |y| 1 1 ≥ γ(α) |f (x − y)|dy. r n−α
(3.2.4)
|y|≤r
The desired consequence follows from taking supremum for r > 0 on both sides of (3.2.4).
Theorem 3.2.1 Suppose that 0 < α < n, 1 ≤ p ≤ n , then (i) If f ∈ Lp (Rn ) 1 < p ≤ α
n α
and
1 1 α = − . q p n
kMα f kq ≤ Ckf kp .
(ii) If f ∈ L1 (Rn ), then for any λ > 0, n
|{x ∈ R : Mα f (x) > λ}| ≤
C kf k1 λ
n n−α
.
The above constant C only depends on α, n, p. Proof. From (3.2.3) and Theorem 3.1.1, when 1 ≤ p < αn , Theorem 3.2.1 follows immediately. For p = αn , H¨older’s inequality implies that M α n is bounded from L α (Rn ) to L∞ (Rn ). Let us now consider the weighted boundedness of the fractional maximal operator Mα . First we formulate a definition on the class of A(p, q) and the
139
3.2 Weighted boundedness of Riesz potential relation between the class of A(p, q) and the class of A p .
Suppose that ω(x) is a nonnegative locally integrable function on R n . Define ω ∈ A(p, q)(1 < p, q < ∞), if there exists a constant C > 0, such that for any cube Q in Rn , sup Q
1 |Q|
Z
q
ω(x) dx
Q
1 q
Z
1 |Q|
ω(x)
−p0
dx
Q
10 p
≤ C < ∞;
(3.2.5)
and say ω ∈ A(1, q)(1 < q < ∞), if there exists a constant C > 0, such that for any cube Q, sup Q
1 |Q|
Z
ω(x)q dx
Q
1
1 ess sup Q ω(x)
q
!
≤ C < ∞.
(3.2.6)
Theorem 3.2.2 (Relation between A(p, q) and A p ) Suppose that 0 < α < n, 1 ≤ p < αn and 1q = p1 − αn . (i) If p > 1, then ω ∈ A(p, q) ⇐⇒ ω q ∈ Aq n−α ⇐⇒ ω q ∈ A1+ q0 ⇐⇒ p
n
0
ω −p ∈ A1+ p0 ; q
(ii) If p > 1, then ω ∈ A(p, q) =⇒ ω q ∈ Aq and ω p ∈ Ap ; (iii) If p = 1, then ω ∈ A(1, q) ⇐⇒ ω q ∈ A1 .
Theorem 3.2.3 Suppose that 0 < α < n, 1 ≤ p < αn , 1q = 1p − αn and ω ∈ A(p, q). Then for ∀λ > 0, there exists a constant C > 0 such that for ∀f ∈ Lp (Rn , ω p ), Z
!1 q
q
ω(x) dx {x∈Rn :Mα f (x)>λ}
C ≤ λ
Z
p
Rn
|f (x)ω(x)| dx
1
p
.
(3.2.7)
Proof. For λ > 0 and K > 0, set Eλ = {x ∈ Rn : Mα f (x) > λ},
Eλ,K = Eλ ∩ B(0, K), where B(0, K) = {x ∈ Rn : |x| < K}.
Thus, for ∀x ∈ Eλ,K , by the definition of Mα , there exists a Qx such that α
|Qx |−1+ n
Z
Qx
|f (y)|dy > λ.
(3.2.8)
140
Chapter 3. Fractional Integral Operators
Since Eλ,K ⊂
[
x∈Eλ,K
Qx , there exists {xj } ⊂ Eλ,K , such that Eλ,K ⊂
{Qxj } is bounded overlapping, i.e., ∃ C = C(n), we have
X j
[
Q xj .
j
χQxj (x) ≤
C(n), ∀x ∈ Rn . ( Here we apply the Besicovitch overlapping theorem, to see [GaR]). Thus, Z For
p q
!p q
q
ω(y) dy Eλ,K
XZ ≤
q
ω(y) dy . q
Qx j
j
p
(3.2.9)
< 1, the right side of (3.2.9) is dominated by X Z j
!p q
ω(y)q dy Qx j
.
(3.2.10)
Since all Qxj satisfy (3.2.8), combining (3.2.9) with (3.2.10) yields that Z
!p q
q
ω(y) dy Eλ,K
≤
X Z j
!p q
q
ω(y) dy Qx j
1 α
λ|Qxj |1− n
Z
Qx j
|f (y)|dy
!p
.
(3.2.11)
If p > 1, applying H¨older’s inequality and (3.2.5), we conclude that Z
!p q
ω(y)q dy Eλ,K
≤
X Z j
!p q
ω(y)q dy Qx j
Z
p
· λ−p |Qxj |1−p− q Z
|f (y)ω(y)|p dy ω(y) Qx j Qx j Z X ≤ Cλ−p |f (y)ω(y)|p dy Q xj Zj −p ≤ Cλ |f (x)ω(y)|p dy. ×
−p0
! p0 p
dy
Rn
(3.2.12) The last inequality of (3.2.12) is due to the property of {Q xj } having finite overlap. Note that the constant C in (3.2.12) is independent of K, so the monotone convergence theorem leads to (3.2.7).
141
3.2 Weighted boundedness of Riesz potential If p = 1, then Z Z |f (y)|dy =
|f (y)ω(y)|ω(y)−1 dy Z −1 ≤ ess sup ω(y) |f (y)ω(y)|dy.
Qx j
Qx j
Qx j
(3.2.13)
Qx j
By (3.2.11), (3.2.13) and (3.2.6), we have that Z
!1 q
ω(y)q dy
Eλ,K
X1 ≤ λ j
Z
!1 q
ω(y)q dy Qx j
1 1
|Qxj | q
! Z 1 |f (y)ω(y)|dy × ess sup Qxj ω(y) Qx j Z 1X ≤C |f (y)ω(y)|dy λ Qx j j Z C ≤ |f (y)w(y)|dy. λ Rn
Thus, let k → ∞, we obtain that (3.2.7) still holds for p = 1.
Theorem 3.2.3 shows that A(p, q) is a sufficient condition for ensuring that Mα is weighted weak (p, q)-bounded. The following theorem will illustrate that A(p, q) is also sufficient for M α being a weighted (p, q)-bounded operator with 1 < p < αn .
Theorem 3.2.4 Suppose that 0 < α < n, 1 < p < ω ∈ A(p, q), then Z
q
Rn
[Mα f (x)ω(x)] dx
1 q
≤C
Z
Rn
n α
and
1 q
p
1
,
|f (x)ω(x)| dx
p
=
1 p
− αn . If
(3.2.14)
where C is independent of f . Proof.
Since ω ∈ A(p, q), Theorem 3.2.2 (i) leads to ω q ∈ A1+ q0 . By
the elementary property of Ap -weight, there exists 1 < s < 1 + ω q ∈ As . Now we take p1 , q such that 1 < p1 < p,
1 α 1 q1 = − and s = 1 + 0 . q1 p1 n p1
q p0
p
such that
(3.2.15)
142
Chapter 3. Fractional Integral Operators q
From (3.2.15) and (i) in Theorem 3.2.2, it follows that ω q1 ∈ A(p1 , q1 ). Using Theorem 3.2.3, for any λ > 0, we have that Z
Eλ
h
ω(x)
q q1
i q1
dx
p1 q1
≤ Cλ
−p1
Z
Rn
|f (x)|
p1
ω(x)
q q1
p1
dx .
That is equivalent to Z
v(x)dx Eλ
p1 q1
≤ Cλ−p1
Z
p1
Rn
|f (x)|p1 v(x) q1 dx,
(3.2.16)
where v(x) = ω q (x). α
Now let f (x) = g(x)v(x) n and define a sublinear operator α
T (g)(x) = Mα (gv n )(x). Thus (3.2.16) is equivalent to Z
{x∈Rn :T (g)(x)>λ}
v(x)dx ≤ Cλ−q1
Z
Rn
|g(x)|p1 v(x)dx
On the other hand, we take p2 such that p < p2 < then q q2 1 + 0 > 1 + 0. p2 p
n α
q1
p1
and let
.
(3.2.17)
1 q2
=
1 p2
− αn ,
q
By Theorem 3.2.2 (i), it implies that ω q2 ∈ A(p2 , q2 ). Similar to (3.2.17), we have Z q2 Z p2 −q2 p2 . (3.2.18) v(x)dx ≤ Cλ |g(x)| v(x)dx {x∈Rn :T (g)(x)>λ}
Rn
Applying the Marcinkiewicz interpolation theorem to (3.2.17) and (3.2.18), we obtain that Z 1 Z 1 q p q p [T (g)(x)] v(x)dx ≤C . (3.2.19) |g(x)| v(x)dx Rn
Rn
Set g(x) = f (x)v(x) Z
−α n
and v(x) = ω(x)q , then we have that q
Rn
[Mα (f )(x)ω(x)] dx
1 q
≤C
Z
p
Rn
|f (x)ω(x)| dx
1
p
.
143
3.2 Weighted boundedness of Riesz potential This finishes the proof of Theorem 3.2.4.
From the weighted boundedness of Mα , we can deduce the weighted boundedness of the Riesz potential I α . Before stating the next lemma, we will give a definition on A∞ . A nonnegative function ω(x) on Rn satisfies A∞ condition, if for a given ε > 0, there exists a δ > 0, such that for any cube Q, and a measurable subset E of Q satisfying |E| ≤ δ|Q|, Z Z ω(x)dx. (3.2.20) ω(x)dx ≤ ε Q
E
We call all functions satisfying the A ∞ condition as the class of A∞ . The relation between A∞ and Ap (1 ≤ p < ∞) is [ A∞ = Ap . (3.2.21) p≥1
The next lemma will reveal another relation between I α and Mα . For its proof, we refer to [MuW]. Lemma 3.2.1 Suppose that 0 < α < n, 0 < q < ∞ and ω(x) ∈ A ∞ . Then there exists a constant C independent of f such that Z Z [Mα f (x)]q ω(x)dx |Iα f (x)|q ω(x)dx ≤ C Rn
Rn
and
sup λq λ>0
Z
{x:|Iα f (x)|>λ}
ω(x)dx ≤ C sup λq λ>0
Z
ω(x)dx.
{x:Mα f (x)>λ}
Applying Lemma 3.2.1, we can get the weighted boundedness of I α . Theorem 3.2.5 Suppose that 0 < α < n, 1 ≤ p < αn , 1q = ω(x) ∈ A(p, q). (i) If 1 < p < ∞, then Z 1 Z 1 q p p q ≤C |f (x)ω(x)| dx ; |Iα f (x)ω(x)| dx Rn
−
α n
and
(3.2.22)
Rn
(ii) If p = 1, then for any λ > 0, q Z Z C q ω(x) dx ≤ q |f (x)ω(x)|dx , λ {x:|Iα f (x)|>λ} Rn where C is independent of f, λ.
1 p
(3.2.23)
144
Chapter 3. Fractional Integral Operators
The proof of Theorem 3.2.5 is simple. In fact, by Theorem 3.2.2, ω(x) q ∈ A1+ q0 (p > 1) or ω(x)q ∈ A1 (p = 1), therefore by (3.2.21), we have ω(x) q ∈ p A∞ . Applying Lemma 3.2.1, Theorem 3.2.3 and Theorem 3.2.4, we obtain that both (3.2.22) and (3.2.23) hold.
3.3
Fractional integral operator with homogeneous kernels
In this chapter we will discuss the (L p , Lq )-boundedness and weighted boundedness of fractional integral operators which is more general than the Riesz potential Iα . Assume that Ω(x) is a homogeneous function with degree zero on R n , i.e. for ∀λ > 0, ∀x ∈ Rn , Ω(λx) = Ω(x), (3.3.1) n
as well as Ω ∈ L n−α (Sn−1 ), where Sn−1 denotes the unit sphere {x ∈ Rn : |x| = 1}, 0 < α < n. Then the fractional integral operator with homogeneous kernel is defined by Z Ω(x − y) f (y)dy. (3.3.2) TΩ,α f (x) = n−α Rn |x − y| It is obvious that when Ω ≡ 1, TΩ,α is the same as the Riesz potential Iα except for a constant. On the other hand, if α = 0 and Ω satisfies the vanishing moment condition on Sn−1 : Z Ω(x0 )dσ(x0 ) = 0, (3.3.3) Sn−1
then TΩ,α becomes a Calder´on-Zygmund singular integral operator (in the sense of principal value Cauchy integral ). The following result shows that the Hardy-Littlewood-Sobolev theorem still holds for TΩ,α . n
Theorem 3.3.1 Suppose that 0 < α < n, Ω ∈ L n−α (Sn−1 ) satisfies (3.3.1).
145
3.3 Fractional integral operator with homogeneous kernels (i) If f ∈ L1 (Rn ), then for ∀λ > 0, |{x ∈ Rn : |TΩ,α f (x)| > λ}| ≤
C kf k1 λ
n n−α
.
(ii) If f ∈ Lp (Rn )(1 < p < αn ), then kTΩ,α f kq ≤ Ckf kp , where and C = C(n, α, p).
1 q
= p1 − αn
Proof. We will complete the proof by three steps. Set K(x) = Ω(x)/|x|n−α and E(s) = {x ∈ Rn : |K(x)| > s}.
First we will show that for ∀s > 0,
n
|E(s)| ≤ As− n−α ,
(3.3.4)
where A depends only on α, n. In fact, by (3.3.1), we have that Z |Ω(x)| 1 dx |E(s)| ≤ s E(s) |x|n−α 1 = s
Z
= As where A = α1 kΩk
Sn−1 n − n−α
n n−α n L n−α (Sn−1 )
|Ω(x0 )| ,
|Ω(x0 )| s
1 n−α
r α−1 drdσ(x0 )
0
.
Next we prove, for 1 ≤ p < Take a fixed µ > 0, let and
Z
n 1 α, q
= 1p − αn , that TΩ,α is of weak type (p, q).
K1 (x) = sgn(K(x)) (|K(x)| − µ) χE(µ) (x) K2 (x) = K(x) − K1 (x).
Thus if p = 1, then kK2 k∞ ≤ µ; if 1 < p < αn , from (3.3.4) it follows that Z Z µ 0 0 |K2 (x)|p dx = p0 sp −1 |E(s)|ds n R 0 Z µ n 0 0 ≤pA sp −1− n−α ds 0
n p0 A p0 − n−α = 0 n ·µ p − n−α 0 n p n−α Aqµ n−α q . = n
146
Chapter 3. Fractional Integral Operators
Thus, when 1 ≤ p <
n α,
we obtain that
kK2 kp0 ≤
10
µ (n−α)q .
10
µ (n−α)q kf kp .
n−α Aq n
p
n
(3.3.5)
So H¨older’s inequality implies that kK2 ∗ f k∞ ≤
Now for ∀λ > 0, set µ such that
n−α Aq n
n−α Aq n
10 p
p
n
n
µ (n−α)q kf kp =
λ , 2
then
Thus
x ∈ Rn : |K2 ∗ f (x)| > λ = 0. 2
x ∈ Rn : |TΩ,α f (x)| > λ}| ≤ |{x ∈ Rn : |K1 ∗ f (x)| > λ 2 p 2 kK1 ∗ f kp . ≤ λ
It follows from (3.3.4) that Z Z (|K(x)| − µ)dx |K1 (x)|dx = E(µ) Rn Z ∞ ≤ |E(t + µ)|dt 0Z ∞ n ≤A t− n−α dt =
(3.3.6)
(3.3.7)
µ α αA − n−α . n−α µ
For ∀f ∈ L∞ (Rn ), ∀x ∈ Rn , by (3.3.7), we conclude that Z α αA − n−α |K1 (x)|dx ≤ |K1 ∗ f (x)| ≤ kf k∞ µ kf k∞ . n − α Rn For ∀f ∈ L1 (Rn ), we have that Z Z α αA − n−α µ kf k1 . |K1 (x − y)||f (y)|dydx ≤ kK1 ∗ f k1 ≤ n−α Rn Rn
(3.3.8)
(3.3.9)
3.3 Fractional integral operator with homogeneous kernels
147
Thus (3.3.8) and (3.3.9) show that T 1 : f 7−→ K1 ∗ f is (∞, ∞)-type and (1, 1)-type. The Riesz-Th¨orin theorem leads to that T 1 is also (p, p)-type (1 < p < ∞), and α αA − n−α . (3.3.10) kT1 k(p,p) ≤ µ n−α Combining (3.3.6) with (3.3.10) yields that p −α 2 αA n−α n |{x ∈ R : |TΩ,α f (x)| > λ}| ≤ µ kf kp λn − α (3.3.11) q 1 =C kf kp , λ where C is independent of λ and f . Finally we will finish the proof of Theorem 3.3.1. (3.3.11) tells us that consequence (i) holds. Now we prove that (ii) also holds. For ∀ 1 < p < αn , set p0 such that p < p0 < αn , and let q0 satisfy 1 1 α = − . q0 p0 n Thus by the second step we know that T Ω,α is of weak type (p0 , q0 ). By (i) and the Marcinkiewicz interpolation theorem, we see that T Ω,α is (p, q)-type, where 1q = 1p − αn . In fact, there exists 0 < θ < 1 so that p1 = 1−θ p0 + θ and then
1 q
=
1−θ q0
+
(n−α)θ . n
By the proof of Theorem 3.3.1 we can get a more general consequence. Assume that K(x) is a measurable function on R n . For a measurable function f on Rn , let T f (x) = (K ∗ f )(x).
For 1 < r < ∞, if there exists a constant C > 0 such that for ∀s > 0, |{x ∈ Rn : |K(x)| > s}| ≤ Cs−r ,
then for 1 ≤ p < r 0 and p1 + 1r = 1q + 1, (i) when p = 1, T is weak type (1, r); (ii) when 1 < p < r 0 , T is (p, q)-type.
The following operator related to T Ω,α is a fractional maximal operator with homogeneous kernels. Its definition is Z 1 MΩ,α f (x) = sup n−α |Ω(y)||f (x − y)|dy. (3.3.12) r>0 r |y|≤r
148
Chapter 3. Fractional Integral Operators
By the idea of the proof of (3.2.3), we can get a relation between M Ω,α and T|Ω|,α . n
Lemma 3.3.1 Assume that 0 < α < n, Ω ∈ L n−α (Sn−1 ) satisfies (3.3.1). Then MΩ,α ≤ C(n, α)T|Ω|,α (|f |)(x). Thus by Lemma 3.3.1 and Theorem 3.3.1, we can obtain the (L p , Lq )boundedness of fractional maximal operators with homogeneous kernels. n
Theorem 3.3.2 Assume that 0 < α < n, Ω ∈ L n−α (Sn−1 ) satisfies (3.3.1). (i) If f ∈ L1 (Rn ), then for ∀λ > 0, n
|{x ∈ R : MΩ,α f (x) > λ}| ≤
C kf k1 λ
n n−α
.
(ii) If f ∈ Lp (Rn )(1 < p ≤ αn ), then kMΩ,α f kq ≤ Ckf kp , where and C = C(n, α, p).
1 q
= 1p − αn
Proof. Omitted.
3.4
Weighted boundedness of TΩ,α
In this section, we will extend the weighted boundedness of the Riesz potential Iα to the case of fractional integral operators with homogeneous kernels TΩ,α . To the end, we will first show weighted boundedness of the fractional maximal operator MΩ,α with homogeneous kernels. In this section, we always assume that Ω is a homogeneous with degree zero (i.e. Ω satisfies (3.3.1)). Theorem 3.4.1 Suppose that 0 < α < n, 1 ≤ s 0 < p < 0
n α
and
1 q
=
1 p
− αn .
If Ω ∈ Ls (Sn−1 ) and ω(x)s ∈ A( sp0 , sq0 ), then there exists a constant C independent of f such that Z
q
Rn
(MΩ,α f (x)ω(x)) dx
1 q
≤C
Z
p
Rn
|f (x)ω(x)| dx
1
p
.
3.4 Weighted boundedness of TΩ,α
149
Proof. From H¨older’s inequality, it follows that Z 1 MΩ,α f (x) = sup n−α |Ω(y)||f (x − y)|dy r>0 r |y|≤r !1 ! 10 Z Z s s 1 s s0 ≤ sup n−α |Ω(y)| dy |f (x − y)| dy r>0 r |y|≤r |y|≤r ! 10 Z s 1 s0 ≤ CkΩkLs (Sn−1 ) sup |f (x − y)| dy r n−αs0 |y|≤r r>0 10 0 s . = CkΩkLs (Sn−1 ) Mαs (|f |s )(x) (3.4.1) p n 1 0 The hypothesis conditions imply that 0 < αs < n, 1 < s0 < αs0 and (q/s 0) = 1 (p/s0 )
Z
Rn
−
αs0 n .
Applying Theorem 3.2.4 together with (3.4.1) yields that
q
(MΩ,α f (x)ω(x)) dx
1 q
Z ≤C
Z ≤C =C
Z
s0
Rn
s0
(Mαs0 (|f | )(x)ω(x) (x)) dx 0
p s0
0
(|f (x)|s ω(x)s (x)) dx Rn
Rn
q s0
p
|f (x)ω(x)| dx
1
p
s0 p
10 s
s0 q
10 s
.
This finishes the proof of Theorem 3.4.1. The next lemma gives a pointwise relationship between fractional integral operators with homogeneous kernels and fractional maximal operators with homogeneous kernels. Lemma 3.4.1 Suppose that ε > 0 satisfies 0 < α − ε < α + ε < n, x ∈ R n . Then 1
1
|TΩ,α f (x)| ≤ C(n, α, ε)(MΩ,α+ε f (x)) 2 (MΩ,α−ε f (x)) 2 . Proof. Given ε > 0 and x ∈ Rn , take δ > 0 such that δ 2ε =
MΩ,α+ε f (x) . MΩ,α−ε f (x)
(3.4.2)
150
Chapter 3. Fractional Integral Operators
Set TΩ,α f (x) =
Z
Ω(x − y) f (y)dy+ |x − y|n−α
|x−y|<δ
Z
|x−y|≥δ
Ω(x − y) f (y)dy := I1 +I2 . |x − y|n−α
Thus |I1 | ≤ ≤
∞ Z X
2−j−1 δ≤|x−y|<2−j δ
j=0
∞ X
(2
−j−1
δ)
−(n−α)
j=0 ε
Z
|Ω(x − y)| |f (y)|dy |x − y|n−α
|x−y|<2−j δ
|Ω(x − y)||f (y)|dy
≤ Cδ MΩ,α−ε f (x). Similarly, ∞ Z X
|Ω(x − y)| |f (y)|dy n−α j−1 δ≤|x−y|<2j δ |x − y| j=1 2 Z ∞ X 1 ≤C (2j δ)−ε j n−α−ε |Ω(x − y)||f (y)|dy (2 δ) |x−y|<2j δ
|I2 | ≤
j=1 −ε
≤ Cδ
MΩ,α+ε f (x).
Therefore |TΩ,α f (x)| ≤ C(δ ε MΩ,α−ε f (x) + δ −ε MΩ,α+ε f (x)) 1
1
= 2C(MΩ,α+ε f (x)) 2 (MΩ,α−ε f (x)) 2 .
To show the weighted boundedness of T Ω,α , we also need the following property of A(p, q).
Lemma 3.4.2 Suppose that 0 < α < n, 1 < p < Then there exists ε > 0 such that (i) ε < α < α + ε < n; 1 n−ε (ii) 1p > α+ε n , q < n ; (iii) ω ∈ A(p, qε ), ω ∈ A(p, qeε ) hold, where
n 1 α, q
=
1 α+ε 1 1 α−ε 1 = − = − , . qε p n qeε p n
1 p
− αn , ω ∈ A(p, q).
3.4 Weighted boundedness of TΩ,α Proof. and
151
Since α > 0 and q > 1, we may choose ε 1 > 0 such that ε1 < α
Let
1 ε1 + < 1. q n 1 α − ε1 1 1 ε1 = − = + , q ε1 p n q n
then q > qε1 > 1 and 1+
p0 p0 <1+ . q q ε1
Applying Theorem 3.2.2 together with the property of A p -weight lead to 0
ω −p ∈ A1+ p0 ⊂ A1+ q
p0 qε 1
.
Applying Theorem 3.2.2 again, it is equivalent to ω ∈ A(p, qε1 ).
(3.4.3)
On the other hand, by the property of A p , there exits η > 0, such that η < and 0 ω −p ∈ A1+p0 ( 1 −η) .
1 q
q
Again choose ε2 > 0, such that ε2 < min{α, n − α}, p1 > Let 1 α + ε2 1 = − , q ε2 p n then 0 < qε1 < 1 and q1ε = 2 2 property of Ap , we have
1 q
−
ε2 n
>
1 q
α+ε2 n
and
ε2 n
< η.
− η. By Theorem 3.2.2 and the
0
ω(x)−p ∈ A1+p0 ( 1 −η) ⊂ A1+ q
p0 qε 2
.
The last is equivalent to ω ∈ A(p, qε2 ).
(3.4.4)
Set ε = min{ε1 , ε2 }, then it is easy to prove that ε satisfies all the properties 1 1 α−ε of ε1 and ε2 . If we let q1ε = 1p − α+ε n , qeε = p − n , then by (3.4.3) and (3.4.4), we see that ω ∈ A(p, qε ) and ω ∈ A(p, qeε ).
152
Chapter 3. Fractional Integral Operators
Lemma 3.4.3 Suppose that 0 < α < n, 1 ≤ s 0 < p < s0
A( sp0 , sq0 ).
n 1 α, q
1 p
=
ω(x) ∈ Then there exists ε > 0 such that (i) ε < α < α + ε < n; 1 n−ε (ii) 1p > α+ε n , q < n ; 0
−
α n
and
0
(iii) ω(x)s ∈ A( sp0 , qsε0 ), ω(x)s ∈ A( sp0 , qseε0 ) hold, where qε and qeε are the same as in Lemma 3.4.2.
Proof. This consequence follows directly from Lemma 3.4.2.
Next we will state and prove the weighted (L p , Lq )-boundedness of TΩ,α . Theorem 3.4.2 Suppose that 0 < α < n, 1 ≤ s 0 < p < αn and 1q = p1 − αn . 0 If Ω ∈ Ls (Sn−1 ) and ω(x)s ∈ A( sp0 , sq0 ), then there exists a constant C independent of f such that Z
q
Rn
|TΩ,α f (x)ω(x)| dx
1 q
≤C
Z
p
Rn
|f (x)ω(x)| dx
1
p
.
Proof. Let ε > 0 be determined in Lemma 3.4.3. Let l1 =
2qeε 2qε , l2 = , q q
then l11 + l12 = 1. For the above ε, applying Lemma 3.4.1 and H¨older’s inequality, we have that Z ≤C ≤C =C
q
Rn
Z
|TΩ,α f (x)ω(x)| dx
Rn
Z
Rn
Z
Rn
1 q
q 2
q 2
(MΩ,α+ε f (x)ω(x)) (MΩ,α−ε f (x)ω(x)) dx (MΩ,α+ε f (x)ω(x))
ql1 2
qε
dx
(MΩ,α+ε f (x)ω(x)) dx
1 ql1
1 2qε
Z
Rn
Z
Rn
1 q
(MΩ,α−ε f (x)ω(x))
qeε
Rn
qε
(MΩ,α+ε f (x)ω(x)) dx
dx
(MΩ,α−ε f (x)ω(x)) dx
By Lemma 3.4.3 and Theorem 3.4.1, we have Z
ql2 2
1 2qε
1 2 ≤ Ckf kp,ω p
1 ql2
1 2f qε
.
3.4 Weighted boundedness of TΩ,α and
Z
Rn
Therefore
153
qeε
(MΩ,α−ε f (x)ω(x)) dx
1 2f qε
1 2 ≤ Ckf kp,ω p.
kTΩ,α f kq,ω(x)q ≤ Ckf kp,ωp .
Next we will state the dual form of Theorem 3.4.2. Theorem 3.4.3 Suppose that 0 < α < n, 1 < p < Ls (Sn−1 )
−s0
If Ω ∈ and ω(x) independent of f such that Z
Rn
q0
p0
n 1 α, q
=
1 p
−
α n
and s > q.
∈ A( s0 , s0 ), then there exists a constant C > 0 q
|TΩ,α f (x)ω(x)| dx
1 q
≤C
Z
p
Rn
|f (x)ω(x)| dx
1
p
.
e satisfies the conditions as Ω does. It is e Proof. Let Ω(x) = Ω(−x), then Ω easy to check that TΩ,α is the dual operator of TΩ,α . Hence e kTΩ,α f kq,ω(x)q =
sup
kgk
0 ≤1 q 0 ,ω −q
kgk
0 ≤1 q 0 ,ω −q
=
sup
≤ kf kp,ωp
Z
R
Z
R
TΩ,α f (x)g(x)dx n
f (x)TΩ,α e g(x)dx n
sup
kgk
q 0 ,ω −q
0 ≤1
kTΩ,α e g(x)kp0 ,ω −p0 .
By the given condition, we have 1 1 α n q 0 p0 0 0 −1 s0 = − , ). , s < q < and (ω ) ∈ A( p0 q0 n α s0 s0 From Theorem 3.4.2, it follows that kTΩ,α gkp0 ,ω−p0 ≤ Ckgkq0 ,ω−q0 . Thus kTΩ,α f kq,ωq ≤ Ckf kp,ωp .
154
Chapter 3. Fractional Integral Operators
As an immediate result of Theorem 3.4.3 and Lemma 3.3.1, we can get the dual form of Theorem 3.4.1: Theorem 3.4.4 Suppose that 0 < α < n, 1 < p < Ls (Sn−1 )
q0
0 ω(x)−s
p0
n 1 α, q
=
1 p
−
α n
and s > q.
If Ω ∈ and ∈ A( s0 , s0 ), then there exists a constant C > 0 independent of f such that Z 1 Z 1 q p q p (MΩ,α f (x)ω(x)) dx ≤C . |f (x)ω(x)| dx Rn
Rn
Note that in Theorem 3.4.2 and Theorem 3.4.4, the class of weight functions depend on s. The following theorem shows that this restriction can be dropped off. Theorem 3.4.5 Suppose that 0 < α < n, 1 < p < αn , and 1q = 1p − αn . If for some s satisfying αn + 1s < p1 < s10 , Ω ∈ Ls (Sn−1 ), furthermore, there exists 0 1 < r < s/( αn )0 such that ω(x)r ∈ A(p, q), then there exists a constant C independent of f such that Z 1 Z 1 q p q p ≤C |f (x)ω(x)| dx . |TΩ,α f (x)ω(x)| dx Rn
Rn
To prove Theorem 3.4.5, we need the following interpolation theorem with change of measure. This is a simple corollary of the Stein-Weiss Intepolation Theorem with change of measure (see [StW]). Lemma 3.4.4 Suppose that 0 < α < n, 1 < p 0 < p1 < 1 α 1 q1 = p1 − n . If a linear operator T satisfies kT f kLq0 (ωq0 ) ≤ C0 kf kLp0 (ωp0 ) 0
0
and kT f kLq1 (ωq1 ) ≤ C1 kf kLp1 (ωp1 ) , 1
then where
1
kT f kLq (ωq ) ≤ Ckf kLp (ωp ) , 1 θ 1 1−θ 1 α + , = − , = p p0 p1 q p n
ω = ω01−θ ω1θ and C ≤ C01−θ C1θ (0 < θ < 1).
n 1 α , q0
=
1 p0
−
α n
and
3.4 Weighted boundedness of TΩ,α
155
Proof. Omitted. If we can show that, under the condition of Theorem 3.4.5, there exists a real number θ (0 < θ < 1), p0 , p1 , q0 , q1 , ω0 , ω1 such that the following conditions hold: n (3.4.5) 1 ≤ s 0 < p0 < p < p 1 < , α n < q0 < q < q1 < s, (3.4.6) n−α 1 1 α 1 1 α 1 1−θ θ = − , = − , = − , (3.4.7) q0 p0 n q1 p1 n p p0 p1 ω = ω01−θ ω1θ , (3.4.8) q1 p1 0 p0 q0 0 (3.4.9) ω0s ∈ A 0 , 0 , ω1−s ∈ A 0 , 0 , s s s s then applying Theorem 3.4.2, Theorem 3.4.4 and Lemma 3.4.4, we can obtain the consequence of Theorem 3.4.5. Now we will validate the conditions of (3.4.5)-(3.4.9), respectively. Let us recall a property on Ap weight: ω ∈ Ap if and only if there exist u, v ∈ A1 , such that ω(x) = u(x)v(x)1−p . A nonnegative locally integrable function u(x) is called Ap weight, if there exists a constant C > 0 such that for any square cube Q, Z 1 u(y)dy ≤ Cu(x), a.e. x ∈ Q. |Q| Q 0
0
Since ω r ∈ A(p, q), Theorem 3.2.2 (i) implies ω r q ∈ Aq n−α . Thus there n exist u, v ∈ A1 , such that 0
ω(x)r q = u(x)v(x)1−q That is,
1
1
ω(x) = u(x) r0 q v(x) r0 q
n−α n
.
− n−α nr 0
.
So write ω = (uτ v β )1−θ (uγ v δ )θ ,
(3.4.10)
where τ, β, γ and δ satisfies τ (1 − θ) + γθ =
1 1 n−α , β(1 − θ) + δθ = 0 − 0 . r0q rq rn
(3.4.11)
156
Chapter 3. Fractional Integral Operators
Now write ω0 (x) = u(x)τ v(x)β and ω1 (x) = u(x)γ v(x)δ . Notice that if 1 ≤ s0 < p0 < p < αn , q10 = p10 − αn , and set τ=
1 1 , β = − 0 p0 0 , q0 s ( s0 )
then we have
p0 q0 , ). s0 s0 In fact, since u, v ∈ A1 , for any cube Q, we have that 0
ω0s ∈ A(
1/( p00 )0 Z p s 1 s0 −( s00 )0 [ω0 (x) ] dx dx |Q| Q Q s0 1/( p00 )0 Z Z q0 p s p 1 1 −τ s0 ( s00 )0 −βs0 ( s00 )0 q0 τ q0 β u(x) v(x) dx u(x) v(x) dx = |Q| Q |Q| Q s0 β s0 −s0 τ Z Z Z q0 1 1 1 q0 τ ≤C · v(x)dx u(x) dx u(x)dx |Q| Q |Q| Q |Q| Q 1/( p00 )0 Z p s 1 −βs0 ( s00 )0 v(x) dx |Q| Q
1 |Q|
Z
s0 q0
q
0 0 [ω0s (x)] s0
≤C.
Obviously, C is independent of θ. Applying the same method as the above, n < q < q1 < s and q11 = p11 − αn , then taking we can prove that if n−α q0
q0
0
γ = − p10 , δ = 1/s0 ( s10 )0 , we have ω(x)−s ∈ A( s10 , 1
p01 s0 ).
Now let us calculate θ by (3.4.11). Note p −1 1 1 0 β = − s0 ( 0 )0 − = s p0 s0 and δ= thus θ= Obviously θ < 1.
q0 s ( 10 )0 s 0
τ − β − n−α r0 n = δ−γ−β+τ
−1 1 s0
,
− αn − n−α r0 n . 1 α 2( s0 − n )
(3.4.12)
3.4 Weighted boundedness of TΩ,α
157
n Since 1 < r < s/( αn )0 , there exists ε > 0 such that 1r = n−α ( 1s + ε). Thus n 1 α n−α 1 α n−α 1 1− − − 0 = 0− − ( + ε) = ε. s0 n rn s n n n−α s
By (3.4.12), we have θ=
2( s10
ε > 0. − αn )
Thus we have explained that if θ was determined by (3.4.12), and (3.4.5)(3.4.7) hold, then both (3.4.8) and (3.4.9) hold. Hence it remains to choose p0 , p1 , q0 , q1 such that (3.4.5)-(3.4.7) hold. Since
1 p
>
α n
+
1 s
and θ > 0, we have 1 αθ θ 1 − − > . p(1 − θ) n(1 − θ) s(1 − θ) p
By (3.4.13) and the fact that
1 p
<
1 s0 ,
(3.4.13)
we choose p0 satisfies
1 1 1 1 αθ θ < min{ 0 , < − − }. p p0 s p(1 − θ) n(1 − θ) s(1 − θ) From this it follows that s0 < p0 < p and p1 > 1−θ p0 + Choose σ > 0 such that 1 − θ α 1 = + + σ θ. p p0 n
(3.4.14)
αθ n .
(3.4.15)
Now let p11 = αn + σ. Then by p1 < αn and p > p0 we see that s0 < p0 < p < p1 < αn . So (3.4.5) holds. By (3.4.15) we see that (3.4.7) also holds. Let 1 1 α 1 1 α q0 = p0 − n , q1 = p1 − n , then it is obvious that n < q0 < q < q 1 . n−α
Finally we will show q1 < s. By (3.4.14) we can get 1−θ α 1 1 − − > . pθ p0 θ n s This is equivalent to
1 α 1 1 1 − > , i.e. > . p1 n s q1 s
Thus (3.4.6) holds. This finishes the proof of Theorem 3.4.5.
158
Chapter 3. Fractional Integral Operators
3.5
Commutators of Riesz potential
In this section we will study (Lp , Lq )-boundedness of commutators of the Riesz potential Iα . We will also illustrate that boundedness of commutators of Iα can characterize BMO(Rn ) spaces. First we will give some definitions and related results. Suppose that b ∈ Lloc (Rn ), then the commutator generated by b and the Riesz potential Iα is defined by Z [b(x) − b(y)] f (y)dy. (3.5.1) [b, Iα ]f (x) = b(x)Iα f (x) − Iα (bf )(x) = |x − y|n−α n R Let us first formulate the following consequences. Theorem 3.5.1 Suppose that 0 < α < n, 1 < p < αn and 1q = p1 − αn . Then [b, Iα ] is bounded from Lp (Rn ) to Lq (Rn ) if and only if b ∈ BMO(Rn ). To prove the theorem, we will prove the following lemma first. Lemma 3.5.1 Suppose that 0 < α < n, 1 < p < αn and 1q = 1p − αn . If 1 < r, s < p, b ∈ BMO(Rn ), then there exists a constant C independent of b such that, for any f ∈ Lp (Rn ), h 1 1 i r ] s r M [b, Iα ]f (x) ≤ CkbkBMO M (|Iα f | )(x) + Mαs (|f | )(x) s , a.e. x ∈ Rn .
(3.5.2)
Proof. Fix a cube Q and set [b, Iα ]f (x) = (b(x)−bQ )Iα f (x)−Iα ((b−bQ )f χ2Q )(x)−Iα ((b−bQ )f χ(2Q)c )(x) := a1 (x) − a2 (x) − a3 (x). Applying H¨older’s inequality and (2.4.2), we have that 1 |Q|
Z
Q
|a1 (x)|dx ≤
1 |Q|
Z
r0
10
|b(x) − bQ | dx Q 1 ≤ CkbkBMO M (|Iα f |r ) r .
r
1 |Q|
Z
r
Q
|Iα f (x)| dx
1 r
(3.5.3)
159
3.5 Commutators of Riesz potential
Since 1 < s < p, we can choose γ > 1, β > 1 such that γβ = s. Thus it implies 1 < β < p < αn . Then there exists u > β such that u1 = β1 − αn . Applying the Hardy-Littlewood-Sobolev theorem (Theorem 3.1.1), we have that
1 |Q|
Z
Q
|a2 (x)|dx ≤
≤C ≤
Z
1 |Q| 1 |Q|
1 u
C 11 |Q| u
u
Q
|Iα ((b − bQ )f χ2Q )(x)| dx
Z
Z
β
2Q
2Q
= CkbkBMO
|b(x) − bQ | |f (x)| dx
|b(x) − bQ |
≤ CkbkBMO |Q|
β
1
α n
1 |Q|
1 |Q|
1− αs n
Z
Z
γ0β
dx
1 βγ 0
s
2Q
2Q
|f (x)| dx
1
β
Z
|f (x)| dx s
u
1
γβ
2Q
|f (x)| dx
1 γβ
s
1
s
1
≤ CkbkBMO (Mαs (|f |s )(x)) s . Denote x0 as the center of Q, then for x ∈ Q and y ∈ estimate: |x − y| ∼ |x0 − y|. Note that
(2Q) c ,
|Iα ((b − bQ )f χ(2Q)c )(x) − Iα ((b − bQ )f χ(2Q)c )(x0 )| Z 1 1 |b(y) − bQ ||f (y)|dy ≤ − n−α |x0 − y|n−α Rn \2Q |x − y| Z |x − x0 | ≤C |b(y) − bQ ||f (y)|dy n−α+1 Rn \2Q |x0 − y| ! 10 Z s |x − x0 | 0 s |b(y) − b | dy ≤C Q n+1 Rn \2Q |x0 − y| !1 Z s |x − x0 | s × |f (y)| dy n+1−αs Rn \2Q |x0 − y| 1
≤ CkbkBMO (Mαs (|f |s )(x)) s .
(3.5.4) we have the
(3.5.5)
160
Chapter 3. Fractional Integral Operators
From (3.5.3)-(3.5.5) it follows that Z 1 [b, Iα ](f )(x) − Iα ((b − bQ )f χ(2Q)c )(x0 ) dx |Q| Q Z Z 1 1 |a1 (x)|dx + |a2 (x)|dx ≤ |Q| Q |Q| Q Z 1 + |a3 (x) − Iα ((b − bQ )f χ(2Q)c )(x0 )|dx |Q| Q h 1 i 1 ≤ CkbkBMO M (|Iα f |r )(x) r + Mαs (|f |s )(x) s . Note that C is independent of Q, thus (3.5.2) holds.
Let us now return to the proof of Theorem 3.5.1. First suppose b ∈ BMO(Rn ), by Lemma 2.4.2 and Lemma 3.5.1, we have that k[b, Iα ]f kq ≤ kM ([b, Iα ]f )kq ≤ C(n, q)kM] ([b, Iα ]f )kq ≤ CkbkBMO k(M (|Iα
1 f |r )) r k
q
+ k(Mαs
1 (|f |s )) s k
q
(3.5.6) .
Note that 1 < r < p < q. Since M is of ( rq , qr )-type, by Theorem 3.1.1, we have
1
r 1r
(M (|Iα f | )) ≤ C k(Iα f )r k rq ≤ Ckf kp . q
r
As long as we choose s > 1 such that 0 < α < αs < n, then we have n 1 1 and q/s 1 < ps < αs = p/s − αs n . Applying Theorem 3.2.1 yields that
s 1s
(Mαs (|f | )) ≤ Ckf kp . q
Thus it follows from (3.5.6) that
k[b, Iα ]f kq ≤ CkbkBMO kf kp .
Next we will give the proof of the necessity. Choose z 0 ∈ Rn , δ > 0 such √ that in the neighborhood {z : |z − z0 | < δ n }, function |z|n−α can be represented as a Fourier series which absolutely converges. That is |z|
n−α
=
∞ X
m=0
am ei
161
3.5 Commutators of Riesz potential
Let z1 = zδ0 . For any cube Q = Q(x0 , r), let y0 = x0 − rz1 , Q0 = Q0 (y0 , r). Then for x ∈ Q, y ∈ Q0 , we have that x − x0 y − y0 √ x − y r − z1 ≤ r + r ≤ n. Now set s(x) = sgn[b(x) − bQ0 ], then Z |b(x) − bQ0 |dx Q Z = (b(x) − bQ0 )s(x)dx Q Z Z 0 −1 = |Q | (b(x) − b(y))s(x)dydx Q
Q0
b(x) − b(y) δ(x − y) n−α s(x)χQ (x)χQ0 (y)dydx =δ r n−α r Rn Rn |x − y| Z Z X b(x) − b(y) i
Z
Z
Set
δ
fm (y) = e−i r
δ
gm (x) = ei r
Q
|b(x) − bQ0 |dx ≤ Cr −α ≤ Cr −α ≤ Cr −α
X
am
m
X m
X m
Z
|am | |am |
Rn
Z
Z
Z
Rn
Q
Rn
b(x) − b(y) fm (y)dygm (x)dx |x − y|n−α
|[b, Iα ](fm )(x)||gm (x)|dx
|[b, Iα ](fm )(x)|dx.
Applying H¨older’s inequality and (L p , Lq )-boundedness of [b, Iα ], we conclude that Z 1 Z X q 1 −α q 0 |am ||Q| q |b(x) − bQ0 |dx ≤ Cr |[b, Iα ](fm )(x)| dx Q
Q
m
≤ Cr −α
≤ C|Q|.
X m
1 q0
1
|am ||Q| |Q0 | p
162
Chapter 3. Fractional Integral Operators
This is equivalent to b ∈ BMO(Rn ), and therefore the proof of Theorem 3.5.1 is finished. For the commutator [b, Mα ] of fractional maximal operator Mα , there are some results parallel to Theorem 3.5.1. Here [b, M α ] is defined by Z 1 |b(x) − b(y)||f (y)|dy. [b, Mα ](f )(x) = sup n−α r>0 r |y−x|≤r Theorem 3.5.2 Suppose that 0 < α < n, 1 < p < αn and 1q = 1p − αn . Then commutator [b, Mα ] is bounded from Lp (Rn ) to Lq (Rn ) if and only if b ∈ BMO(Rn ). Proof. Applying the same method in the proof of (3.2.3), we obtain that [b, Mα ](f )(x) ≤ γ(α)[b, Iα ](|f |)(x).
∀x ∈ Rn .
(3.5.7)
Thus when b ∈ BMO(Rn ), by (3.5.7) and Theorem 3.5.1 we deduce that [b, Mα ] is bounded from Lp (Rn ) to Lq (Rn ). On the other hand, suppose that [b, M α ] is bounded from Lp (Rn ) to q L (Rn ). Choose any cube Q in Rn , Z Z Z 1 1 |b(x) − bQ |dx ≤ |b(x) − b(y)|dydx |Q| Q |Q|2 Q Q Z Z 1 1 = |b(x) − b(y)|χQ (y)dydx α α |Q|1+ n Q |Q|1− n Q Z 1 ≤ [b, Mα ](χQ )(x)dx α |Q|1+ n Q Z 1 q 1 1 q 0 q ≤ ([b, M ](χ )(x)) dx α |Q| α Q |Q|1+ n Q 1 1 1 ≤C |Q| q0 |Q| p = C. 1+ α |Q| n Thus b ∈ BMO(Rn ).
3.6
Commutators of fractional integrals with rough kernels
In this section we will discuss the weighted (L p , Lq )-boundedness of commutators generated by fractional integral operators with rough kernels
163
3.6 Commutators of fractional integrals with rough kernels
TΩ,α and BMO functions. First we will give their definitions. Suppose b ∈ BMO(Rn ). Then the commutator generated by T Ω,α and b is defined by [b, TΩ,α ](f )(x) = b(x)TΩ,α f (x) − TΩ,α (bf )(x) Z [b(x) − b(y)] = Ω(x − y)f (y)dy. |x − y|n−α n R The commutator [b, MΩ,α ] generated by MΩ,α and b is defined by [b, MΩ,α ](f )(x) = sup r>0
1 r n−α
Z
|x−y|≤r
|b(x) − b(y)||Ω(x − y)||f (y)|dy.
We begin with indicating some relations between A(p, q)-weight and BMO(R n ).
Lemma 3.6.1 Suppose that 0 < α < n, 1 < p < αn and 1q = ω ∈ A(p, q), then there exists ε > 0, such that ω 1+ε ∈ A(p, q). Proof.
1 p
−
α n.
If
Applying Theorem 3.2.2, we have ω(x) q ∈ A q(n−α) . By the propn
erty of Ap , there exists ε > 0 such that ω q(1+ε) ∈ A q(n−α) . Then we can get n
the desired consequence by applying Theorem 3.2.2 again.
Lemma 3.6.2 Suppose that 0 < α < n, 1 < p < αn and 1q = b ∈ BMO(Rn ), then there exists λ > 0 such that eλb ∈ A(p, q).
1 p
−
α n.
If
Proof. By the John-Nirenberg inequality and the inverse H¨older’s inequality of Ap , there exists a λ0 > 0, such that eλ0 b ∈ A q(n−α) (see Garci´a-Cuerva and Rubio de Francia [GaR]). Now take λ = Theorem 3.2.2, this leads to e λb ∈ A(p, q).
λ0 q ,
n
then eλbq ∈ A q(n−α) . By n
Lemma 3.6.3 Suppose that 0 < α < n, 1 < p < αn and 1q = 1p − αn . For λ > 0, there exists a η > 0 such that if b ∈ BMO and kbk ∗ < η, then eλb ∈ A(p, q). Proof. It is not difficult to show that if we set C C[q(n − α)/n − 1] η = min , , λq λq
164
Chapter 3. Fractional Integral Operators
here C is the absolute constant in the John-Nirenberg inequality, then when kbkBMO < η we have (see [GaR]) eλqb ∈ A q(n−α) . n
By Theorem 3.2.2, this is equivalent to e λb(x) ∈ A(p, q).
Now let us turn our attention to the weighted (L p , Lq )-boundedness of [b, TΩ,α ].
Theorem 3.6.1 Suppose that 0 < α < n, 1 < p < αn and 1q = 1p − αn . Let Ω ∈ Ls (Sn−1 ), s > 1 and b ∈ BMO(Rn ). If p, q, s, ω satisfy one of the following three conditions, then k[b, T Ω,α ]f kq,ωq ≤ Ckf kp,ωp : 0 (i) s0 < p and ω(x)s ∈ A( sp0 , sq0 ); 0 0 0 (ii) q < s and ω −s ∈ A( qs0 , ps0 ); 0 (iii) αn + 1s < 1p < s10 , 1 < r < s/( αn )0 and ω(x)r ∈ A(p, q). Proof. First we will give the proof of (i). By (2.4.29), we obtain that Z b(x) − b(y) |[b, TΩ,α ]f (x)| = Ω(x − y)f (y)dy n−α Rn |x − y| Z Z 2π 1 Ω(x − y) eiθ [b(x)−b(y)] −iθ = f (y)dy e e dθ n−α |x − y| Rn 2π 0 Z 2π Z 1 |Ω(x − y)| ≤ |f (y)|e−b(y) cos θ dy · eb(x) cos θ dθ. n−α 2π 0 Rn |x − y| Let gθ (y) = f (x)e−b(y) cos θ , θ ∈ [0, 2π]. Since f ∈ Lp (ω p ), we have, for any θ ∈ [0, 2π], that gθ ∈ Lp (ω p epb cos θ ) Thus
kgθ kp,ωp epb cos θ = kf kp,ωp
and
1 |[b, TΩ,α ]f (x)| ≤ 2π
Z
2π 0
(3.6.1)
T|Ω| (|gθ |)(x) eb(x) cos θ dθ.
From the Minkowski’s inequality, it follows that Z 2π 1 kT|Ω|,α (|gθ |)kq,ωq eqb cos θ dθ. k[b, TΩ,α ]f kq,ωq ≤ 2π 0 By the given conditions, 1 < αs0 n
1 (p/s0 )
<
n αs0 , 0
< αs0 < n and
(3.6.2) 1 (q/s0 )
=
1 (p/s0 )
−
together with Lemma 3.6.1, we see that there exists ε > 0 such that
3.6 Commutators of fractional integrals with rough kernels
165
0
ω s (1+ε) ∈ A( sp0 , sq0 ). Then by Theorem 3.4.1, for any ϕ ∈ L p (ω p(1+ε) ), we have that kT|Ω|,α (ϕ)kq,ωq(1+ε) ≤ C1 kϕkp,ωp(1+ε) . (3.6.3) 0
On the other hand, if we choose λ = s (1+ε) , then by Lemma 3.6.3, there ε exists η > 0 such that, for b ∈ BMO(Rn ) and kbkBMO < η, eλb ∈ A( sp0 , sq0 ), s0 i.e. e(1+ε)b/ε ∈ A( sp0 , sq0 ). Notice that if b ∈ BMO(Rn ) and for all |t| ≤ 1 then tb ∈ BMO and ktbkBMO ≤ kbkBMO . Thus for all |t| ≤ 1, we have s0 p q (3.6.4) etb(1+ε)/ε ∈ A 0, 0 . s s Thus, without loss of generality, we can prove Theorem 3.6.1 (i) only in the case kbkBMO < η. Applying Theorem 3.4.1 again and (3.6.4), for any θ ∈ [0, 2π] and ϕ ∈ Lp (epb(1+ε) cos θ/ε ), we have that kT|Ω|,α (ϕ)kq,eqb(1+ε) cos θ/ε ≤ C2 kϕkp,epb(1+ε) cos θ /ε .
(3.6.5)
Here C2 is independent of θ. By (3.6.3) and (3.6.5), applying the Stein-Weiss Intepolation Theorem with change measure (Lemma 3.4.4), for any θ ∈ [0, 2π] and ϕ ∈ L p (ω p epb cos θ ), we obtain that kT|Ω|,α (ϕ)kq,ωq eqb cos θ ≤ Ckϕkp,ωp epb cos θ , 1
(3.6.6)
ε
where C ≤ C11+ε C21+ε and C is independent of θ, ϕ. (3.6.1), (3.6.2) and (3.6.6) yield that Z 2π 1 k[b, TΩ,α ](f )kq,ωq ≤ Ckgθ kp,ωpb cos θ dθ = Ckf kp,ωp . 2π 0 This finishes the proof of consequence (i). Using the consequence (i) and the method in the proof of Theorem 3.4.3, e we can easily get the consequence (ii). Note that if we set Ω(x) = −Ω(−x), then [b, TΩ,α e ] is the dual operator of [b, T Ω,α ].
By the consequence (i)(ii) as well as the interpolation theorem with change of measure (Lemma 3.4.4) and using the method in the proof of Theorem 3.4.5, we can get the proof of the consequence (iii), therefore we omit the details here.
166
Chapter 3. Fractional Integral Operators
Remark 3.6.1 Let m ∈ N. The commutator of degree m generated by T Ω,α and b is defined by Z (b(x) − b(y))m m [b, TΩ,α ] (f )(x) = [b, · · · , [b, TΩ,α ]] = Ω(x − y)f (y)dy. |x − y|n−α Rn Using the method in the proof of Theorem 3.6.1 and the mathematical induction, we can prove that, under the conditions of Theorem 3.6.1, the consequences of weighted boundedness hold for [b, T Ω,α ]m . Remark 3.6.2 Using the method in the proof of (3.2.3) we have that [b, MΩ,α ]f (x) ≤ γ(α)[b, T|Ω|,α ](|f |)(x),
∀x ∈ Rn .
(3.6.7)
By (3.6.7) we see that under the conditions of Theorem 3.6.1, the consequences of weighted boundedness still hold for [b, M Ω,α ]. If m ∈ N, and the commutator of degree m generated by M Ω,α and b is defined by Z 1 m |b(x) − b(y)|m |Ω(x − y)||f (y)|dy. [b, MΩ,α ] (f )(x) = sup n−α r>0 r |x−y|≤r Similarly, it is easy to prove that under the conditions of Theorem 3.6.1, the consequences of weighted boundedness hold for [b, M Ω,α ]m .
3.7
Notes and references
Lemma 3.1.1 was proved by Hedberg [He]. The formula (3.2.3) shows that the fractional maximal operator can be dominated by the Riesz potential in pointwise sense, which is a special case of Lemma 3.3.1( i.e. Ω ≡ 1). And Lemma 3.3.1 was proved by Ding [Di3]. The definition of A(p, q) weights was first introduced by Muckenhoupt and Wheeden [MuW2]. Theorem 3.2.3, Theorem 3.2.4 and Theorem 3.2.5 come from [MuW2]. The fractional integral operator with homogeneous kernel was first introduced by Muckenhoupt and Wheeden [MuW1].In this paper, they obtained the power weighted (L p , Lq )-boundedness of TΩ,α (1 < p < αn ). The theorems presented in this section are generalizations of their consequences. As a complement of Muckenhoupt and Wheeden’s consen quences mentioned above, Ding [Di3] gave the power weighted weak (1, n−α )boundedness of TΩ,α .
167
3.7 Notes and references n
In 1988, Adams (see [Ad]) proved that if 0 < α < n and f ∈ L α (B), B = {x ∈ Rn : |x| < 1}, then n Z I (f )(x) n−α 1 α dx ≤ C, exp n kf k n |B| B α
where C = C(α, n). This result can be viewed as a replacement for the boundedness of Iα when p = αn . In 1996, Ding and Lu [DiL1] obtained a similar result for fractional integral operators with rough kernels. All theorems in Section 4 of this chapter are taken from Ding and Lu [DiL3]. And Lemma 3.4.1 is also taken from [DiL3]. Its idea of the proof comes from Welland [We]. The necessity of Theorem 3.5.1 with n − α being an even and the sufficiency of the theorem were first proved by Chanillo [Cha] in 1982. The proof given here is simpler which comes from Ding [Di2].
There are counterexamples showing that when b ∈ BMO, the commutan ,∞ 1 n−α )-type. However Ding, Lu and Zhang [DiLZ], tor [b, Iα ] is not (L , L as well as Cruz-Uribe and Fiorenza [CrF] independently, proved that when b ∈ BMO(Rn ) and p = 1, the commutator [b, Iα ] satisfies an estimate of weak L log + L type. Theorem 3.6.1 was proved by Ding [Di1]. In 1999, Ding, Lu [DiL4] established weighted norm inequalities for commutators generated by fractional integral operators with rough kernels and BM O functions. A two-weight weak-type norm inequality for the commutator generated by the Riesz potential and BM O function was obtained by Liu and Lu [LiL]. As space is limited, we do not mention the boundedness of homogeneous fractional integrals on Hardy spaces in this chapter. For related results, we refer to Ding, Lu [DiL5] and Lu, Wu [LuWu]. In [LuWu], the authors established some equivalent characterizations for (H 1 , Ln/(n−α) ) type boundedness of commutators of fractional integrals.
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Chapter 4
Oscillatory Singular Integrals Oscillatory integrals have been an essential part of harmonic analysis. Many important operators in harmonic analysis are some versions of oscillatory integrals, such as the Fourier transform, Bochner-Riesz means, Radon transform and so on. However, the object we study in this chapter is a class of oscillatory singular integrals with polynomial phases which is closely related to the Radon transform.
4.1
Oscillatory singular integrals with homogeneous smooth kernels
Suppose that K is a homogeneous Calder´on-Zygmund kernel in R n . Precisely, K satisfies the following conditions: (i) K is a C 1 function away from the origin, (ii) K is homogeneous of degree −n, and (iii) the mean-value of K on the unit sphere vanishes. Let P (x, y) be a real-valued polynomial on R n ×Rn . Consider an operator T of the following form Z eiP (x, y) K(x − y)f (y)dy. (4.1.1) T f (x) = p· v· Rn
Let us first begin with Lp -boundedness of T in this section. 169
170
Chapter 4. Oscillatory Singular Integrals
Theorem 4.1.1 Suppose that K satisfies (i), (ii) and (iii). Then T defined by (4.1.1) is bounded on Lp (Rn ), with 1 < p < ∞, and bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). In order to prove Theorem 4.1.1, we need to establish some lemmas on some general inequalities for polynomials in R n . Lemma 4.1.1 Suppose that P (x) = d, and ε < 1/d. Then sup
y∈Rn
Z
|x|≤1
P
|α|≤d aα x
|P (x − y)|−ε dx ≤ Aε
α
is a polynomial of degree
X
|α|=d
ε
|aα | ,
where bound Aε depends only on ε (and the dimension n), but not on the coefficients aα . P Lemma 4.1.2 Suppose that P (x) = |α|≤d aα xα is a homogeneous polynomial of degree d in Rn and ε < 1/d. Then ε Z X |aα | , |P (x)|−ε dσ(x) ≤ Aε Sn−1
|α|=d
which bound Aε depends only on ε (and the dimension n), but not on the coefficients aα .
Lemma 4.1.3 (van der Corput) Suppose that a real-valued φ ∈ C k [a, b], k > 1, and |φ(k) (t)| ≥ 1 for all t ∈ (a, b). Then Z b iλφ(t) e dt ≤ Ck λ−1/k , a
where λ ∈ R, and Ck is independent of a, b and φ.
Let us consider an operator T of more general form Z T f (x) = p· v· K(x, y)f (y)dy,
(4.1.2)
Rn
where K is a distribution, and for x 6= y, that is a function satisfying |K(x, y)| ≤
A , |x − y|n
x 6= y.
(4.1.3)
4.1 Oscillatory singular integrals with homogeneous smooth kernels For every ε > 0, consider the truncated operator T ε defined by Z Tε f (x) = K(x, y)f (y)dy.
171
(4.1.4)
|x−y|<ε
Lemma 4.1.4 Fix p, 1 ≤ p < ∞. If T is bounded on L p (Rn ), then so is each Tε . Moreover if k · k denotes the operator norm in L p , we have the estimate kTε k ≤ C{kT k + A}
with A arising in (4.1.3) and C independent of K. Proof. For any fixed h ∈ Rn , we f1 (y), f (y), f (y) = 2 f3 (y),
split f into three parts as |y − h| < ε/2, ε/2 ≤ |y − h| < 5 ε/4, otherwise.
It is easy to see that if |x − h| < ε/4, then Tε f1 (x) = T f1 (x). So by our assumption on K, we have that p Z Z Z p dx K(x, y)f (y)dy |Tε f1 (x)| dx = 1 |x−h|<ε/4 Rn |x−h|<ε/4 Z |T f1 (x)|p dx ≤ Rn
≤kT kp kf1 kp Z p =kT k
|y−h|<ε/2
|f (x)|p dx.
And if |x − h| < 1/4, ε/2 ≤ |y − h| < 5 ε/4, then ε/4 < |x − y| < 3 ε/2, and then (4.1.3) gives trivially that Z |Tε f2 (x)|p dx ≤ Akf2 kp . |x−h|<ε/4
Finally, it is obvious that Tε f3 (x) = 0
from |x − h| < ε/4 and |y − h| < 5 ε/4. Combining those three estimates together yields that Z Z p p |Tε f (x)| dx ≤ C(A + kT k) |f (y)|p dy |x−h|<ε/4
|y−h|<5 ε/4
172
Chapter 4. Oscillatory Singular Integrals
holds uniformly for h ∈ Rn . This means kTε kp ≤ C(A + kT k)kf kp . Proof of Theorem 4.1.1 We wish to obtain kT f kp ≤ C(degP, n)kf kp ,
1 < p < ∞.
(4.1.5)
We shall carry out the argument by a double induction on the degree in x and y of the polynomial P . If P (x, y) depends only on x or only on y, it is obvious that these cases are trivial. Let k and l be two positive integers and let P (x, y) have degree k in x and l in y. We assume that (4.1.5) holds for all polynomials which are sum of monomials of degree less than k in x times monomials of any degree in y, together with monomials which are of degree k in x times monomials which are of degree less than l in y. We proceed to the proof of the inductive step. Rewrite X P (x, y) = aα,β xα y β − y α+β + P0 (x, y) |α|=k,|β|=l
and also write P (x, y) =
X
xα Qα (y) + R(x, y),
(4.1.6)
|α|=k
where P0 (x, y) satisfies the inductive hypothesis. By dilation-invariance, we may assume that X |aα,β | = 1. |α|=u,|β|=v
It is easy to see that when |x| < 1 and |y| < 2, we have iP (x,y) − eiP0 (x,y) ≤ C|x − y|. e Decompose T as Z Z iP (x,y) e K(x − y)f (y)dy + T f (x) = |x−y|<1
:=T0 f (x) + T∞ f (x).
|x−y|≥1
eiP (x,y) K(x − y)f (y)dy
We first consider T0 . We shall show that kT0 f kp ≤ C(degP, n)kf kp .
(4.1.7)
4.1 Oscillatory singular integrals with homogeneous smooth kernels Put h ∈ Rn , and write
X
P (x, y) =
|α|=k,|β|=l
173
aα,β (x − h)α (y − h)β + R(x, y, h),
where the polynomial R(x, y, h) satisfies the inductive hypothesis, and the coefficients of R(x, y, h) depend on h. It follows that |T0 f (x)| Z P i[ |α|=k,|β|=l aα,β (y−h)α+β +R(x,y,h)] e K(x − y)f (y)dy ≤ |x−y|<1 Z P α+β +R(x,y,h) i a (y−h) ] K(x − y)f (y)dy + eiP (x,y) − e [ |α|=k,|β|=l α,β |x−y|<1 := |T01 f (x)| + |T02 f (x)|.
From the inductive hypothesis and Lemma 4.1.4 for truncated integral operator, we conclude that the operator T 01 is bounded on Lp (Rn ), and the norm of T01 is independent of P and h. When |x − h| < 1/4, |x − y| < 1, we have P α+β iP (x,y) − ei[ |α|=k,|β|=l aα,β (y−h) +R(x,y,h)] ≤ C|x − y|. e Thus, applying the properties of the kernel K, we conclude that Z |f (y)| 2 |T0 f (x)| ≤ A dy n−1 |x−y|<1 |x − y| Z |f (x − y)χB(h,5/4) (x − y)| ≤A dy. |y|n−1 |y|<1
By Minkowski’s inequality, we obtain Z Z 2 p p |T0 f (x)| dx ≤ A |x−h|<1/4
|y−h|<5/4
|f (y)|p dy.
Therefore kT02 f kp ≤ CA(degP, n)kf kp . Hence kT0 f kp ≤ C(A + 1)(degP, n)kf kp , where C is independent of the coefficients of P (x, y).
174
Chapter 4. Oscillatory Singular Integrals Let us now turn our attention to T∞ . Denote K∞ =
∞ X
Ψj ,
j=0
where Ψ0 is supported in {x : 12 ≤ |x| ≤ 1} and is bounded, while Ψj (x) = 2−nj Ψ(2−j x), j ≥ 1, with Ψ an appropriate function of class C 1 supported in {x : 21 ≤ |x| ≤ 1}. We set Z eiP (x,y) Ψj (x − y)f (y)dy. (4.1.8) Tj f (x) = Rn
Since estimating T0 is trivial, we will consider Tj for j ≥ 1, and claim that the L2 -operator norms satisfy the estimate kTj k2 ≤ C2−jε ,
for some ε > 0.
(4.1.9)
To do this we consider Tj∗ Tj , where Tj∗ is the adjoint operator of Tj . The ˜ j (y, z), is given by kernel of this operator, L Z ei(P (x,z)−P (x,y)) Ψj (x − z)Ψj (x − y)dx. Rn
Obviously, we will obtain the same norm if we replace ˜ j (y, z) by Lj (y, z) = 2nj L ˜ j (2j y, 2j z). L The result is then Lj (y, z) =
Z
Rn
ei(P (x,z)−P (x,y)) Ψ0 (x − z)Ψ0 (x − y)dx.
(4.1.10)
A trivial estimate for Lj that follows from (4.1.10) shows |Lj (y, z)| ≤ CχB2 (y − z), where χB2 is the characteristic function of a ball of radius 2. Now we make the changes of variables x → x+y in (4.1.10), and then get the x−integration in polar coordinates with x = rx0 , r = |x|, |x0 | = 1, dx = r n−1 drdσ(x). We also write, after invoking (4.1.6), X P (2j (x), 2j z) = 2kj xα Qα (2j z) + Rj (x, z), |α|=k
175
4.1 Oscillatory singular integrals with homogeneous smooth kernels where Rj has x−degree strictly less than k. Similarly X P (2j (x), 2j y) = 2kj xα Qα (2j y) + Rj (x, y). |α|=k
After these substitutions we get that Z 1 Z i(E+F ) ˜ Lj (y, z) = e Ψ(x, y, z)dr dσ(x0 ), S n−1
(4.1.11)
0
˜ where Ψ(x, y, z) = Ψ0 (x − z + y)Ψ0 (x)r n−1 with X (x0 )α (Qα (2j z) − Qα (2j y)) E = (r2j )k |α|=k
and F = Rj (x, z) − Rj (x, y). Note that F has degree strictly less than k, in r, while E is of degree k. Thus if we use the van der Corput Lemma (Lemma 4.1.3) for the integral inside (4.1.10), then we obtain
Since
−δk X Qα (2j z) − Qα (2j y) |Lj (y, z)|C ≤ 2−jδ χB2 (y − z). |α|=k
X
|α|=k,|β|=l
|aα,β | = 1,
there is an |α0 | = k and |β0 | = l, such that |aα0 ,β0 | ≥ c > 0. However X Qα0 (2j z) = 2jl α0 βz β + S(z), |β|=l
where S(z) has z−degree strictly less than l, so Z Z ˜ j (y, z)|dz ≤ C2−jδ |L |Qα0 (2j z) − Qα0 (2j y)|−δk dz, Rn
|z−y|≤2
and hence by Lemma 4.1.2 Z ˜ j (y, z)|dz ≤ C2−jδ 2−jlδ/k = C2−j2ε sup |L y
176
Chapter 4. Oscillatory Singular Integrals
if we choose δ sufficiently small to that δ/k < 1/l. Similarly Z ˜ j (y, z)|dy ≤ C2−j2ε . sup |L z
This proves that kT ∗ T k ≤ C2−j2ε , and therefore (4.1.9) is valid. It is obvious, however, that the norms of T j on L1 or L∞ are uniformly bounded. Thus by the interpolation with (4.1.9) we obtain that the L p norms of Tj are P exponential decreasing as j → ∞, if 1 < p < ∞, allowing us to sum Tj and concluding the proof of the theorem. Theorem 4.1.1 shows that the operator T defined by (4.1.1) is of type p (L , Lp ), where 1 < p < ∞. For the endpoint case (p = 1), we will formulate the following theorem. Theorem 4.1.2 The operator T defined by (4.1.1) is of weak type (L 1 , L1 ), with bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). The proof of Theorem 4.1.2 is more complicated and much longer. it will be convenient to divided it into several lemmas. For the purpose, we split the kernel K as K = K0 + K∞ , where K0 (x) = K(x) if |x| ≤ 1 and K∞ (x) = K(x) if |x| > 1. We consider the corresponding splitting T = T 0 + T∞ : Z eiP (x, y) K0 (x − y)f (y)dy, T0 f (x) = p· v· Rn
T∞ f (x) = p· v·
Z
Rn
eiP (x, y) K∞ (x − y)f (y)dy.
Next we shall prove for any λ > 0 and f ∈ L 1 that sup λ |{x ∈ Rn : |T0 f (x)| > λ}| ≤ Ckf k1
(4.1.12)
sup λ |{x ∈ Rn : |T∞ f (x)| > λ}| ≤ Ckf k1 .
(4.1.13)
λ>0
and λ>0
First, we shall prove that |{x ∈ B1 (0) : |T0 f (x)| > λ}| ≤ Cλ
−1
Z
|y|<2
|f (y)|dy,
(4.1.14)
where Br (x) denotes the closed ball with the center at x and radius r > 0. To prove the inequality (4.1.14), we first introduce a lemma.
177
4.1 Oscillatory singular integrals with homogeneous smooth kernels Lemma 4.1.5 Let K be an operator of the form: Z K(x, y)f (y)dy, Kf (x) = p· v· Rn
where the kernel K satisfies |K(x, y)| ≤ C|x − y| −n . For f ∈ L1 and ε > 0, set Z K(x, y)f (y)dy. Kε f (x) = p· v· |x−y|<ε
If sup λ |{x ∈ Rn : |Kf (x)| > λ}| ≤ Ckf k1 , λ>0
then, there exists a constant Cn depending only on the dimension n such that sup λ |{x ∈ Rn : |Kε f (x)| > λ}| ≤ Cn kf k1 . λ>0
Proof. Let us first show that |{x ∈ B(h, ε/4) : |Kε f (x)| > λ}| ≤ Cn λ−1
Z
|y−h|<5ε/4
|f (y)|dy
(4.1.15)
uniformly for h ∈ Rn . Integrating both sides of the inequality (4.1.15) with respect to h, we get the conclusion of Lemma 4.1.5. For any fixed h ∈ R n , we split f into three parts as f1 (y), |y − h| < ε/2, f (y), ε/2 ≤ |y − h| < 5 ε/4, f (y) = 2 f3 (y), otherwise.
Note that if |x − h| < ε/4, then
Kε f1 (x) = Kf1 (x). So by the assumption on K, |{x ∈ B(h, ε/4) : |Kε f1 (x)| > λ}| = |{x ∈ B(h, ε/4) : |Kf1 (x)| > λ}| = |{x : |Kf1 (x)| > λ}| ≤ Cn λ−1 kf1 k1 Z −1 ≤ Cn λ
|y−h|<5ε/4
|f (y)|dy.
178
Chapter 4. Oscillatory Singular Integrals
And if |x − h| < ε/4, ε/2 ≤ |y − h| < 5 ε/4, then ε/4 < |x − y| < 3 ε/2, and therefore Z |f (y)|dy. |Kε f2 (x)| ≤ Cn |y−h|<5ε/4
Hence, by Chebyshev’s inequality, |{x ∈ B(h, ε/4) : |Kε f2 (x)| > λ}| ≤ Cn λ
−1
Z
|y−h|<5ε/4
|f (y)|dy.
Finally, it follows that Kε f3 (x) = 0 from |x − h| < ε/4 and |y − h| < 5 ε/4. Combining those three estimates together yields (4.1.15). We turn to the proof of (4.1.14). The method of the proof of (4.1.14) is similar to that in Theorem 4.1.1. Rewrite X P (x, y) = aα,β xα y β − y α+β + P0 (x, y) |α|=k,|β|=l
where P0 (x, y) satisfies the inductive hypothesis. By dilation-invariance, we may assume that X |aα,β | = 1. |α|=u,|β|=v
If |x| < 1 and |y| < 2, then iP (x,y) − eiP0 (x,y) ≤ C|x − y|. e Hence,
Z |T0 f (x)| ≤
e Rn
iP0 (x, y)
Z K0 (x − y)f (y)dy + C
:= |U f (x)| + |V f (x)|.
|x−y|<1
|x − y|−n+1 |f (y)|dy
If |x| ≤ 1, then we have that Z eiP0 (x, y) K(x − y)(f χB(0,2) )(y)dy U f (x) = |x−y|<1
and V f (x) = V (f χB(0,2) )(x).
4.1 Oscillatory singular integrals with homogeneous smooth kernels
179
By the inductive hypothesis on P0 (x, y) and Lemma 4.1.5, it implies that U is bounded from L1 to L1,∞ . On the other hand, by Chebyshev’s inequality we get that Z −1 |{x ∈ B1 (0) : |V f (x)| > λ}| ≤ Cn λ |f (y)|dy. |y|<2
Combining these estimates of U and V implies (4.1.14). Similarly by the translation-invariance, for any h ∈ R n , we can easily get that Z −1 |{x ∈ B1 (h) : |T0 f (x)| > λ}| ≤ Cλ |f (y)|dy. (4.1.16) |y−h|<2
Integrating both sides of the inequality in (4.1.16) with respect to h, we get (4.1.12). Next we shall prove the inequality (4.1.13). Take a nonnegative ϕ ∈ such that
C0∞ (Rn )
∞ X
supp(ϕ) ⊂ {1/2 ≤ |x| ≤ 2},
j=0
ϕ(2−j x) = 1 if |x| ≥ 1.
Put Kj (x, y) = ϕ(2−j (x−y))K∞ (x, y), where K∞ (x, y) = eiP (x,y) K∞ (x−y) and decompose ∞ X Kj (x, y). K∞ (x, y) = j=0
Define Vj f (x) = and
Z
Rn
Kj (x, y)f (y)dy
V f (x) =
∞ X
for
j≥0
Vj f (x).
j=1
Then T∞ = V0 + V. Evidently V0 is bounded on L1 , so it suffices to deal with V. By the Calder´on-Zygmund decomposition at a positive λ, we have a collection {Q} of non-overlapping closed dyadic cubes and functions g and b such that f = g + b; Z 1 λ≤ |f | ≤ cλ; |Q| Q
180
Chapter 4. Oscillatory Singular Integrals | ∪ Q| ≤ kgk∞
ckf k1 ; λ ≤ cλ;
kgk1 ≤ ckf k1 ; X b= bQ ; Q
supp(bQ ) ⊂ Q; Z bQ = 0
and
kbQ k1 ≤ cλ|Q|.
We set Bi = S
X
bQ
|Q|=2in
(i ≥ 1),
B0 =
X
bQ .
|Q|≤1
˜ where Q ˜ denotes the cube with the same center as Q and Put U = Q, with sidelength 4 times that of Q. When x ∈ R n \ U, we note that ∞ X X V b(x) = Vj Bi (x) j=1
=
i≥0
∞ XZ X j=1 i≥0
=
Rn
∞ Z XX i≥0 j=1
Rn
Kj (x, y)Bi (y)dy Kj (x, y)Bi (y)dy.
By the definitions of Kj (x, y) , Bi and x ∈ Rn \ U, if i ≥ j ≥ 1, then it easily implies that Z Rn
Therefore, we have that
V b(x) =
Kj (x, y)Bi (y)dy = 0.
X X Z
n i≥0 j≥i+1 R
=
XXZ k≥1 j≥k
=
XX k≥1 j≥k
Rn
Kj (x, y)Bi (y)dy
Kj (x, y)Bj−k (y)dy
Vj (Bj−k )(x).
4.1 Oscillatory singular integrals with homogeneous smooth kernels
181
To prove the inequality (4.1.13), we introduce Lemma 4.1.6.
Lemma 4.1.6 There exists an ε > 0 such that, for any positive integer k,
2
X
−εk
Vj (Bj−k )
≤ c2 λkf k1 .
j≥k
(4.1.17)
2
By (4.1.8), we have, for some ε > 0 and for all j ≥ 1, that kVj k2 ≤ C2−jε . So we get that V is bounded on L2 (Rn ). Assume Lemma 4.1.6 is valid. We now prove that the inequality (4.1.13) holds. {x ∈ Rn \ U : |V f (x)| > λ} λ λ n n + x ∈ R \ U : |V b(x)| > ≤ x ∈ R \ U : |V g(x)| > 2 2
2
X X −1 −2 Vj (Bj−k ) ≤ Cλ kf k1 + Cλ
k≥1 j≥k 2 2 X 1/2 ≤ Cλ−1 kf k1 + Cλ−2 λ1/2 2−εk/2 kf k1 k≥1
≤ Cλ
−1
kf k1 .
Thus we obtain that {x ∈ Rn \ U : |V f (x)| > λ} ≤ Cλ−1 kf k1 .
(4.1.18)
On the other hand, from the definition of U it follows that |U| ≤ Cλ−1 kf k1 .
(4.1.19)
Combining (4.1.18) with (4.1.19) yields the boundedness of V from L 1 to L1,∞ . This is just our desired inequality (4.1.13).
182
Chapter 4. Oscillatory Singular Integrals It remains to show that the inequality (4.1.17) holds. For k, m ≥ 1, put Hkm (x, y) Z = K k (z, x)Km (z, y)dz Rn Z e−iP (z,x)+iP (z,y) K ∞ (z − x)K∞ (z − y)ϕk (z − x)ϕm (z − y)dz. = Rn
(4.1.20)
Then Vk∗ Vm f (x) =
Z
Rn
Hkm (x, y)dy,
where Vk∗ denotes the adjoint of Vk . We need to discuss the property of kernel Hkm furthermore. To the end, we formulate the following Lemma 4.1.7. Lemma 4.1.7 Let k ≥ m ≥ 1. Then Hkm (x, y) = 0, if |x − y| > 2k+2 ; and (i) |Hkm (x, y)| ≤ c2−kn , (ii) |Hkm (x, y)| ≤ c2−kn 2−m |q(x) − q(y)|−1/M . Proof. Suppose that |x − y| > 2k+2 . Since |z − x| + |z − y| ≥ |x − y| > 2k+2 , either of inequalities |z − x| > 2k+1
and
|z − y| > 2k+1
must hold. That means ϕk (z − x)ϕm (z − y) = 0. Now we prove the estimate of (ii) only. Note that M ∂ P (z, x) − P (z, y) = M ! q(x) − q(y) . ∂z
Hence, it follows from Lemma 4.1.3 that Z b −iP (z,x)+iP (z,y) e dz ≤ c|q(x) − q(y)|−1/M a
for any a and b. Therefore by the integration by parts in variable z in the formula of (4.1.20) and by using the properties of the kernel K, we easily get the desired conclusion. Let us give two definitions.
4.1 Oscillatory singular integrals with homogeneous smooth kernels Definition 4.1.1 For a real-valued polynomial P (x) = gree N , define kP k = max |aα |.
P
|α|≤N
183
aα xα of de-
|α|=N
Definition 4.1.2 For a real-valued polynomial P and β > 0, let R(P, β) = {x ∈ Rn : |P (x)| ≤ β}. Let d(E, F ) denote the distance between E and F and d(x, F ) = d({x}, F ). Let us state the following Lemma which will be used to prove the Lemma 4.1.6.
Lemma 4.1.8 Let s, m be integers and s ≥ m. Suppose N ≥ 1. Then, for any polynomial P of degree N satisfying kP k = 1 and for any γ > 0, there exists a positive constant Cn,N,γ depending only on n, N, γ such that Nm m x ∈ B2k (a) : d x, R(P, 2 ) ≤ γ2 ≤ Cn,N,γ 2(n−1)s 2m
uniformly in a ∈ Rn .
The proof of Lemma 4.1.8 needs to use geometrical properties of polynomials. For the details, one refers to [Sa] and we omit it here. Let λ > 0 and let {Bj }j≥0 be family of measurable functions such that Z
Qj
|Bj | ≤ λ|Qj |
for all cubes Qj in Rn with sidelength l(Qj ) = 2j .
Lemma 4.1.9 Let the kernels Hji be as in Lemma 4.1.7. Then we can find a constant c such that j X
Z sup n
i=k x∈R
R
Bj (y)Hji (x, y)dy ≤ cλ2−k n
for all integers j and k with 0 < k ≤ j.
184
Chapter 4. Oscillatory Singular Integrals
Proof. For m ∈ Z, let Dm be the family of all closed dyadic cubes Q with sidelength l(Q) = 2m . Fix x ∈ Rn . Let F = {Q ∈ Di−k : Q ∩ B2j+2 (x) 6= ∅} (0 < k ≤ i ≤ j). P Then clearly Q∈F |Q| ≤ c2jn . Decompose F = F0 ∪ F1 , where n
F0 = Q ∈ F : Q ∩ R q(·) − q(x), 2
L(i−k)
and F1 = F \ F0 . Then by Lemma 4.1.8 we have X |Q| ≤ c2j(n−1) 2i−k .
6= ∅
o
Q∈F0
It implies by (i) in Lemma 4.1.7 and above estimates that X Z X Z −jn |Bi−k (y)Hji (x, y)|dy ≤ c2 |Bi−k (y)|dy Q∈F0
Q
≤ c2
−jn
Q Q∈F 0 X
λ
Q∈F0
|Q| ≤ cλ2i−j−k .
From (ii) in Lemma 4.1.7 and above estimates, it follows that X Z |Bi−k (y)Hji (x, y)|dy Q Q∈F1 X Z −jn −i −L(i−k)/M |Bi−k (y)|dy ≤ c2 2 2 Q Q∈F X1
≤ c2−jn 2−i 2−L(i−k)/M λ
Q∈F
=
Q
X Z
Q∈F0
Q
(4.1.22)
|Q| ≤ cλ2−i 2−L(i−k)/M .
From (4.1.21) and (4.1.22) it follows that Z |Bi−k (y)Hji (x, y)|dy Rn XZ = |Bi−k (y)Hji (x, y)|dy Q∈F
(4.1.21)
|Bi−k (y)Hji (x, y)|dy +
≤ cλ 2i−j−k + 2−i 2−L(i−k)/M .
X Z
Q∈F1
Q
|Bi−k (y)Hji (x, y)|dy
4.1 Oscillatory singular integrals with homogeneous smooth kernels
185
Thus we see that Z j j X X |Bj (y)Hji (x, y)| dy ≤ sup cλ 2i−j−k + 2−i 2−L(i−k)/M ≤ cλ2−k . n i=k x∈R
Rn
i=k
This completes the proof of Lemma 4.1.9. Lemma 4.1.10 Let {Bj }j≥0 be as in Lemma 4.1.9. Suppose ∞. Then, for any positive integer k, we have
2
X X
−k
kBj k1 . Vj (Bj−k )
≤ cλ2
j≥k j≥0
X j≥0
kBj k1 <
2
Proof. Let h·, ·i denote the inner product in L 2 . Using Lemma 4.1.9, we note that
2
X j XX
Vj (Bj−k ) ≤ 2 |hVj (Bj−k ), Vi (Bi−k )i|
j≥k j≥s i=k 2
=2
j XX
Bj−k , Vj∗ Vi (Bi−k ) j≥s i=k
≤2
j XX j≥k i=k
≤ cλ2−k
kBj−k k1 Vj∗ Vi (Bi−k ) L∞
X j≥0
kBj k1 .
This completes the proof of Lemma 4.1.10. Set Bj = Bj , then we have that X X X kBj k1 = kBj k1 ≤ cλ j≥0
j≥0
Thus we obtain that
j≥1
X
|Qj
|=2jn
|Qj | + cλ
2
X
−k
Vj (Bj−k )
≤ cλ2 kf k1 ,
j≥k
X
|Q|≤1
|Q| ≤ ckf k1 .
2
which is just the inequality (4.1.17). This completes the proof of Theorem 4.1.2.
186
Chapter 4. Oscillatory Singular Integrals
4.2
Oscillatory singular integrals with rough kernels
In this section, we will investigate a class of oscillatory singular integral Ω(x0 ) operators with rough kernels. Suppose that K(x) = satisfies condi|x|n tions (1) Ω Z is homogeneous of degree 0; Ω(x0 )dσ(x0 ) = 0; and
(2)
Sn−1
(3) Ω ∈ Lq (Sn−1 ), for some q with 1 < q ≤ ∞. Let us introduce a truncated operator Z K(x − y)f (y)dy. Sf (x) =
(4.2.1)
|x−y|<1
Theorem 4.2.1 Suppose that K(x) satisfies (1) and (3). The operator T defined by (4.1.1) is bounded on Lp (Rn ) if and only if the truncated operator S is bounded on Lp (Rn ), 1 < p < ∞, with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). Proof. X Assume that a polynomial P (x, y) has degree k in x and l in y and |aαβ | = 1. Let us first consider the following operator T ∞ defined |α|=k,|β|=l
by
T∞ f (x) =
Z
|x−y|≥1
eiP (x,y) K(x − y)f (y)dy.
Write T∞ f (x) =
∞ Z X j=1
e
iP (x,y)
2j−1 ≤|x−y|<2j
K(x − y)f (y)dy :=
∞ X
Tj f (x),
j=1
where Tj f (x) = =
Z
Z
2j−1 ≤|y|<2j
Ω(y 0 ) Sn−1
eiP (x,x−y) K(y)f (x − y)dy
Z
0
eiP (x,x−ry ) 2j−1 ≤r<2j
f (x − ry 0 ) drdσ(y 0 ). r
For a fixed y 0 ∈ Sn−1 , let Y be the hyperplane through the origin orthogonal to y 0 . We have, for x ∈ Rn , x = z + sy 0 with s ∈ R and z ∈ Y ,
4.2 Oscillatory singular integrals with rough kernels
187
that Z
f (x − ry 0 ) dr r 2j−1 ≤r<2j Z 0 0 0 f (z + (s − r)y ) = eiP (z+sy ,z+(s−r)y ) dr r 2j−1 ≤r<2j Z 0 0 0 f (z + ty ) eiP (z+sy ,z+ty ) = dt s−t 2j−1 ≤s−t<2j 0
eiP (x,x−ry )
= Nj [f (z + ·y 0 )](s).
It is easy to see that Nj is a linear operator defined on L2 (R). Denote Nj∗ to be its adjoint operator. Next we will consider the operator N j∗ Nj with the kernel Z 0 0 0 0 ei[P (z+ry ,z+vy )−P (z+ry ,z+uy )] dr Mj (u, v) = (r − u)(r − v) 2j−1 ≤r−v,r−u<2j Z j 0 0 0 j 0 0 0 ei[P (z+2 ry +vy ,z+vy )−P (z+2 ry +vy ,z+uy )] = dr. 1 r(2j r + v − u) ≤r<1,2j−1 ≤r+v−u<2j 2
It is easy to check that 1 χ j−1 (|v − u|). 2j [0,2 )
|Mj (u, v)| ≤ Now rewrite P (x, y) as P (x, y) =
X
xα Qα (y) + R(x, y).
|α|=k
Thus we have that Mj (u, v) =
Z
E = (2j r)k
X
where
1 ≤r<1,2j−1 ≤r+v−u<2j 2
|α|=k
ei(E+F ) ψ(r)dr,
h i y 0α Qα (z + vy 0 ) − Qα (z + uy 0 ) ,
F with r−degree less than k, and ψ(r) =
1 . r(2j r + v − u)
(4.2.2)
188
Chapter 4. Oscillatory Singular Integrals
It follows from Lemma 4.1.3 that − 1 Z k h i X i(E+F ) jk 0α 0 0 e dr ≤ C 2 y Qα (z + vy ) − Qα (z + uy ) . 1 ≤r<1 |α|=k
2
It follows from integration by parts that
− 1 k h i X 0 0 0α −j jk 2 |Mj (u, v)| ≤ C2 y Qα (z + vy ) − Qα (z + uy ) . |α|=k (4.2.3) Combining the inequality (4.2.2) with (4.2.3) yields the following estimate
|Mj (u, v)|
− δ k X h i 0α 0 0 −j jk y Qα (z + vy ) − Qα (z + uy ) 2 χ[0,2j−1 ) (|v − u|) ≤ C2 |α|=k
holds uniformly in δ ∈ (0, 1]. Thus Z |Mj (u, v)|dv
− δ k h i X jk −j −jδ 0α 0 0 2 ≤ C2 2 y Qα (z + vy ) − Qα (z + uy ) dv |u−v|<2j |α|=k − δ k Z i h X u 0 0 jk −jδ 0α 2 ≤ C2 y Qα (z + (v + j y ) − Qα (z + uy ) dv. 2 |v|<1 |α|=k Z
Now we take δ ∈ (0, 1] such that δ/k < 1/l, then from Lemma 4.1.1 it follows that − δ k X Z 0α+β −jδ aαβ y |Mj (u, v)|dv ≤ C2 . |α|=k,|β|=l Thus, we have that
kNj∗ Nj |kL∞ (R)→L∞ (R)
− δ X k −jδ 0α+β ≤ C2 aαβ y |α|=k,|β|=l
4.2 Oscillatory singular integrals with rough kernels and kNj∗ Nj |kL1 (R)→L1 (R)
189
− δ k X −jδ 0α+β ≤ C2 a y αβ . |α|=k,|β|=l
By the Riesz-Th¨orin interpolation theorem, we obtain that
kNj∗ Nj |kL2 (R)→L2 (R) Since |Nj g(s)| ≤
Z
− δ k X 0α+β −jδ aαβ y ≤ C2 . |α|=k,|β|=l
2j−1 ≤s−t<2j
|g(t)| dt ≤ 4HL(g)(s), s−t
we have that kNj |kLp0 (R)→Lp0 (R) ≤ C(p0 ),
with
1 < p0 < ∞.
By the Riesz-Th¨orin interpolation theorem again, we obtain that
kNj |kLp (R)→Lp (R)
− θδ 2k X −θjδ/2 0α+β ≤ C2 aαβ y , |α|=k,|β|=l
(4.2.4)
where 0 < θ < 1. From Minkowski’s inequality and (4.2.4), it follows that kTj f kp Z Z 1/p Z ≤ |Ω(y 0 )| |Nj [f (z + ·y 0 )](s)|p dsdz dσ(y 0 ) Sn−1
Y
R
− θδ 2k X 0α+β 0 −θjδ/2 aαβ y |Ω(y )| dσ(y 0 ) ≤ C2 kf kp Sn−1 |α|=k,|β|=l 1/q0 − θδq0 2k Z X aαβ y 0α+β ≤ C2−θjδ/2 kf kp kΩkLq (Sn−1 ) dσ(y 0 ) . n−1 S |α|=k,|β|=l Z
We take δ ∈ (0, 1] such that δ < min{k/l, 2k/(k + l)q 0 }. By Lemma 4.1.2, we obtain that kTj f kp ≤ C2−θjδ/2 kΩkLq (Sn−1 ) kf kp .
190
Chapter 4. Oscillatory Singular Integrals
Thus X
If
|α|=k,|β|=l
kT∞ f kp ≤ CkΩkLq (Sn−1 ) kf kp . |aαβ | 6= 1, we denote
A=
X
|α|=k,|β|=l
We can write P (x, y) as X
P (x, y) =
|α|=k,|β|=l
1/(k+l)
|aαβ |
aαβ (Ax)α (Ay)β + R0 Ak+l
.
Ax Ay , A A
:= Q(Ax, Ay).
Thus T∞ f (x) = =
Z
|x−y|≥1
Z
eiQ(Ax,Ay) K(x − y)f (y)dy
y |≥1 |x− A
eiQ(Ax,y) K(Ax − y)f
y A
dy.
So we have that Z !1 p y p eiQ(Ax,y) K(Ax − y)f dy dx kT∞ f kp = y A |≥1 Rn |x− A !1 Z p Z y p = A−n eiQ(x,y) K (x − y) f dy dx |x−y|≥A A Rn
(·)
≤ CA−n/p
f A p Z
≤ CkΩkLq (Sn−1 ) kf kp .
Rewrite T f (x) = p· v·
Z
|x−y|<1
eiP (x, y) K(x−y)f (y)dy +T∞f (x) := T0 f (x)+T∞ f (x).
Note that we have proved that T∞ is a (Lp , Lp ) type operator. If T is a (Lp , Lp ) type operator, then T0 is also of (Lp , Lp ) type.
191
4.2 Oscillatory singular integrals with rough kernels
Let us now turn to prove that S is a (Lp , Lp ) type operator. To do this, we first introduce a notion. For a polynomial P , we say that P has property P, if it satisfies P (x, y) = P (x − h, y − h) + R0 (x, h) + R1 (y, h), where R0 , R1 are real polynomials. We take h ∈ Rn . For |x − h| < 1, we have T0 f (x) = T0 [f (·)χB2 (h) (·)](x). Thus Z
p
|x−h|<1
|T0 f (x)| dx
!1/p
Z
≤C
p
|y−h|<2
|f (y)| dx
!1/p
,
where C is independent of h. We choose polynomial P with the property P. It follows that Z
K(x − y)f (y)χB2 (h) (y)dy Z = e−iR0 (x,h) eiP (x,y) K(x − y)
Sf (x) =
|x−y|<1
|x−y|<1
×e
−iP (x−h,y−h) −iR1 (y,h)
e
f (y)χB2 (h) (y)dy.
Note that the Taylor expression of e iP (x−h,y−h) is
e−iP (x−h,y−h)
m ∞ X (−i)m X = aαβ (x − h)α (y − h)β m! m=0
α,β
∞ X 1 X = am,µ,ν (x − h)µ (y − h)ν . m! µ,ν, m=0
(4.2.5)
192
Chapter 4. Oscillatory Singular Integrals
Thus, if we set a = (1, 1, · · · , 1), b = (2, 2, · · · , 2), then we conclude that !1/p Z |x−h|<1
|Sf (x)|p dx
∞ Z X 1 X |am,µ,ν | |(x − h)µ |p ≤ m! µ,ν, |x−h|<1 m=0 p 1/p × T0 e−iR1 (·,h) f (·)χB(h,2) (·)(· − h)ν (x) dx ∞ X 1 X |am,µ,ν |aµ ≤ m! µ,ν, m=0
|y−h|<2
|f (y)|p |(y − h)ν |p dy
!1/p
!1/p Z ∞ X 1 X ≤ |f (y)|p dy |am,µ,ν |aµ bν m! |y−h|<2 µ,ν m=0 m !1/p Z ∞ X 1 X α β p ≤C |aαβ |a b |f (y)| dy m! |y−h|<2 m=0 α,β !1/p Z X ≤ C exp |f (y)|p dy |aαβ |aα bβ . |y−h|<2
α,β
Thus
Z
kSf kp ≤ Ckf kp . On the other hand, we assume that the truncated operator S is bounded on Lp (Rn ). Let us now turn to prove that operator T is also bounded on Lp (Rn ). As proved above, T∞ is bounded on Lp (Rn ). In fact, it suffices to check that T0 is also bounded on Lp (Rn ). We shall carry out the argument by a double induction on the degree in x and y of the polynomial P . If P (x, y) depends only on x or only on y, it is clear that there cases are trivial. Let k and l be two positive integers and let P (x, y) have degree k in x and l in y. We assume that T0 is bounded on Lp (Rn ) for all polynomial which are sums of monomials of degree less than k in x times monomials of any degree in y, together with monomials which are of degree k in x times monomials which are of degree less than l in y. We proceed now to the proof of the inductive step. Rewrite X P (x, y) = aα,β xα y β − y α+β + P0 (x, y) |α|=k,|β|=l
193
4.2 Oscillatory singular integrals with rough kernels and also write P (x, y) =
X
xα Qα (y) + P0 (x, y),
|α|=k
where P0 (x, y) satisfies the inductive hypothesis. By dilation-invariance, we may assume that X |aα,β | = 1. |α|=u,|β|=v
Take h ∈
Rn ,
and write X
P (x, y) =
|α|=k,|β|=l
aα,β (x − h)α (y − h)β + R(x, y, h),
where the polynomial R(x, y, h) satisfies the inductive hypothesis, and the coefficients of R(x, y, h) depend on h. It follows that |T0 f (x)| Z P α+β ei[ |α|=k,|β|=l aα,β (y−h) +R(x,y,h)] K(x − y)f (y)dy ≤ |x−y|<1 Z P α+β eiP (x,y) − ei[ |α|=k,|β|=l aα,β (y−h) +R(x,y,h)] K(x − y)f (y)dy + |x−y|<1 := |T01 f (x)| + |T02 f (x)|.
By the inductive hypothesis, we conclude that the operator T 01 is bounded on Lp (Rn ), and the norm of T01 is independent of P (x, y) and h. When |x − h| < 1/4, |x − y| < 1, we have P iP (x,y) i[ |α|=k,|β|=l aα,β (y−h)α+β +R(x,y,h)] − e e ≤ C|x − y|. Thus, we conclude that
|T02 f (x)| ≤ C ≤
Z
Z
|x−y|<1
|y|<1
|Ω((x − y)0 )| |f (y)|dy |x − y|n−1
|Ω(y 0 )| |f (x − y)χB(h,5/4) (x − y)|dy. |y|n−1
By Minkowski’s inequality, we obtain Z Z p 2 p |T0 f (x)| dx ≤ CkΩkLq (Sn−1 ) |x−h|<1/4
|x−h|<5/4
|f (x)|p dx.
194
Chapter 4. Oscillatory Singular Integrals
Therefore kT02 f kp ≤ CA(deg P, n)kf kp . Hence
kT0 f kp ≤ CkΩkpLq (Sn−1 ) kf kp ,
where C is independent of the coefficients of P (x, y). Theorem 4.2.1 tells us that the boundedness of an oscillatory singular integral operator is attributed to the boundedness of the corresponding truncated integral operator. The following theorem is an important application of Theorem 4.2.1. Theorem 4.2.2 Suppose that P (x, y) is a real-valued polynomial on R n ×Rn and K satisfies the condition (1),(2) and (3). Then the operator T defined by (4.1) is bounded on Lp (Rn ), 1 < p < ∞, with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). Proof. When K satisfies the condition (1), (2) and (3), By Theorem 2.3.7, we know that the operator Z K f (x) = K(x − y)f (y)dy Rn
is a (Lp , Lp ) type operator with 1 < p < ∞. It suffices to prove that the truncated operator S is a (Lp , Lp ) type operator with 1 < p < ∞. For any fixed h ∈ Rn , we split f into three parts as |y − h| < 1/2, f1 (y), f2 (y), 1/2 ≤ |y − h| < 5/4, f (y) = f3 (y), otherwise. It is easy to see that if |x − h| < 1/4, then we have Sf1 (x) = K f1 (x). So we have that Z
|x−h|<1/4
p
|Sf1 (x)| dx =
Z
|x−h|<1/4
Z
Z
R
p K(x − y)f1 (y)dy dx n
|K f1 (x)|p dx Z p ≤ kK k |f (x)|p dx.
≤
Rn
|y−h|<1/2
195
4.2 Oscillatory singular integrals with rough kernels
And if |x − h| < 1/4, 1/2 ≤ |y − h| < 5/4, then 1/4 < |x − y| < 3/2. It follows from Minkowski’s inequality that Z ≤
p
|x−h|<1/4
Z
|Sf2 (x)| dx
1/4<|y|≤1
|Ω(y 0 )| |y|n
≤ CkΩkpLq (Sn−1 )
Z
!1/p
Z
p
|x−h|<1/4
|x−h|<5/4
|f2 (x − y)| dx
|f (x)|p dx
!1/p
!1/p
dy
dy.
Finally, it is clear that Sf3 (x) = 0. The above estimates imply that kSf kp ≤ C(kK k + 1)kf kp . It immediately follows from Theorem 4.2.1 that operator T defined by (4.1) is bounded on Lp (Rn ), 1 < p < ∞, with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). Now we weaken the condition on the kernel Ω. In place of L q (Sn−1 ) with 1 < q ≤ ∞, assume that Ω satisfies (4)
Ω ∈ L log L+ (Sn−1 ).
Obviously, the following theorem will weaken the condition on the kernel Ω in Theorem 4.2.1. Theorem 4.2.3 Suppose that K satisfies (1) and (4). The operator T defined by (4.1.1) is bounded on Lp (Rn ) if and only if the truncated operator S is bounded on Lp (Rn ), 1 < p < ∞, with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). Proof. The idea and method of proving Theorem 4.2.3 are much similar to that of Theorem 4.2.1. We only give the different part. For Ω ∈ L log L+ (Sn−1 ), we decompose the kernel Ω as follows: Ωm (x0 ) = Ω(x0 )χEm (x0 ),
196
Chapter 4. Oscillatory Singular Integrals
where E0 = {x0 ∈ Sn−1 : |Ω(x0 )| < 1} and Em = {x0 ∈ Sn−1 : 2m−1 ≤ |Ω(x0 )| < 2m },
m ∈ N.
We split T∞ as follows: T∞ f (x) =
∞ Z ∞ X X
eiP (x,y)
j−1 ≤|x−y|<2j j=1 m=0 2
∞
X Ωm (x − y) f (y)dy := Tj,m f (x). |x − y|n j=1
By a method similar to that of proving Theorem 4.2.1, we can prove that there exists a δ > 0 such that kTj,m f kp ≤ C2−jδ kΩm kL∞ (Sn−1 ) kf kp ,
(4.2.6)
where C is independent of j and m. On the other hand, we obtain that Z |Ωm (y)| χEm (y)|f (x − y)|dy. |Tj,m f (x)| ≤ |y|n j−1 j 2 ≤|y|<2 It is clear that
|Ωm (y)| χEm (y) ∈ L1 (Rn ). |y|n
So we have that
kTj,m f kp ≤ CkΩm kL1 (Sn−1 ) kf kp .
(4.2.7)
Now we choose a positive integer M > δ −1 . Thus kT∞ f kp = =
∞ ∞ X X
j=1 m=0
∞ X j=1
kTj,m f kp
kTj,0 f kp +
∞ mM X X
m=1 j=1
kTj,m f kp +
∞ X X
m=1 j>mM
kTj,m f kp .
It follows from (4.2.6) that ∞ X j=1
kTj,0 f kp ≤ C2−jθδ kf kp
and ∞ X X
m=1 j>mM
kTj,m f kp ≤ C
∞ X X
m=1 j>mM
2−jθδ 2mθ kTj,m f kp ≤ Ckf kp .
4.3 Oscillatory singular integrals with standard kernels
197
By (4.2.7), we obtain that ∞ mM X X
m=1 j=1
kTj,m f kp ≤ C
∞ mM X X
m=1 j=1
≤ CM kf kp
kΩm kL1 (Sn−1 ) kf kp
∞ X
m=1
m2m |Em |
≤ CM kΩkL1 log+ L(Sn−1 ) kf kp ≤ Ckf kp .
Therefore we have that kT∞ f kp ≤ Ckf kp . The remainder of the proof is the same as that of Theorem 4.2.1, and we omit it here. The following Theorem is a direct consequence of Theorem 4.2.3. Theorem 4.2.4 Suppose that P (x, y) is a real-valued polynomial on R n ×Rn and K satisfies the condition (1),(2) and (4). The operator T defined by (4.1) is bounded on Lp (Rn ), 1 < p < ∞, with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y).
4.3
Oscillatory singular integrals with standard kernels
Let K(x, y) be a Calder´on-Zygmund standard kernel (also called distribution kernel). That means, K(x, y) satisfies C , |x − y|n
(a)
|K(x, y)| ≤
(b)
|∇x K(x, y)| + |∇y K(x, y)| ≤
x 6= y, C , x 6= y. |x − y|n+1
Clearly the kernel K does not have any homogeneous property. The oscillatory singular integral operator in this section is defined by Z eiP (x,y) K(x, y)f (y)dy, (4.3.1) T f (x) = p· v· Rn
where P (x, y) is a real-valued polynomial on R n ×Rn .
198
Chapter 4. Oscillatory Singular Integrals
Theorem 4.3.1 Suppose that K satisfies (a) and (b). If the singular integral operator K defined by Kf (x) = p· v·
Z
K(x, y)f (y)dy Rn
is bounded on L2 (Rn ) , then T defined by (4.3.1) is bounded on L p (Rn ), 1 < p < ∞, with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). We only state the above theorem without proof and will give the proof for its weighted version. In weighted case, we have the following theorem. Theorem 4.3.2 Under the same conditions on K as that of Theorem 4.3.1. If the operator K is bounded on L2 (Rn ), then for any ω ∈ Ap , 1 < p < ∞, we have kT f kp,ω ≤ Ckf kp,ω , (4.3.2) where Ap is the Muckenhoupt class, and C depends only on the total degree of P (x, y), but not on the coefficients of P (x, y). Proof. Using same double induction on the degree in x and y of the polynomial P as that of Theorem 4.1.1. We consider two cases. P Case 1. Assume |α|=k,|β|=l |aαβ | = 1. Decompose T f (x) =
Z
eiP (x,y) K(x, y)f (y)dy
|x−y|<1 ∞ Z X
eiP (x,y) K(x, y)f (y)dy
+
j=1
2j−1 ≤|x−y|<2j
:= T0 f (x) +
∞ X
Tj f (x).
j=1
For h ∈ Rn , P (x, y) can be written as P (x, y) =
X
|α|=k,|β|=l
aαβ (x − h)α (y − h)β + R(x, y, h),
where R(x, y, h) is a polynomial which satisfies the inductive hypothesis,
4.3 Oscillatory singular integrals with standard kernels
199
whose coefficients depend on h. Write Z n o X exp i R(x, y, h) + aαβ (y − h)α+β K(x, y)f (y)dy T0 f (x) = |x−y|<1 Z P α+β + eiP (x,y) − ei(R(x,y,h)+ aαβ (y−h) ) K(x, y)f (y)dy |x−y|<1
:= T01 f (x) + T02 f (x).
Now put f into three parts: f (y) =f (y)χ{|y−h|<1/2} (y) + f (y)χ{1/2≤|y−h|<5/4} (y) + f (y)χ{|y−h|≥5/4} (y) :=f1 (y) + f2 (y) + f3 (y).
When |x − h| < 1/4, it is easy to see that Z n o T01 f1 (x) = exp i(R(x, y, h) + Σaαβ (y − h)α+β ) K(x, y)f1 (y)dy. Rn
From the inductive hypothesis, it follows that Z Z p |T01 f1 (x)| ω(x)dx ≤ C |x−h|<1/4
|y−h|<1/2
|f (y)|p ω(y)dy,
(4.3.3)
where C does not depend on h and the coefficients of P (x, y). Note that when |x − h| < 1/4, 1/2 < |y − h| < 5/4, we have |x − y| > 1/4. Thus Z |f2 (y)| dy ≤ CM (f2 )(x). |T01 f2 (x)| ≤ C n 1/4<|x−y|<1 |x − y| Therefore, we have that Z Z |T01 f2 (x)|p ω(x)dx ≤ C |x−h|<1/4
|y−h|<5/4
|f (y)|p ω(y)dy,
(4.3.4)
where C does not depend on h and the coefficients of P (x, y). Also notice that when |x − h| < 1/4, |y − h| ≥ 5/4, we have |x − y| > 1. Hence T01 f3 (x) = 0. (4.3.5) By (4.3.3), (4.3.4) and (4.3.5), we obtain that Z Z p |T01 f (x)| ω(x)dx ≤ C |x−h|<1/4
|y−h|<5/4
|f (y)|p ω(y)dy,
(4.3.6)
200
Chapter 4. Oscillatory Singular Integrals
where C does not depend on h and the coefficients of P (x, y). Obviously, when |x − h| < 1/4, |x − y| < 1, we have P α+β iP (x,y) − ei(R(x,y,h)+ aαβ (y−h) ) ≤ C|x − y|. e
Therefore, when|x − h| < 1/4, we have that Z |f (y)| |T02 f (x)| ≤ C dy ≤ CM (f χB(h,5/4) )(x). n−1 |x−y|<1 |x − y| It follows that Z
|x−h|<1/4
|T02 f (x)|p ω(x)dx ≤ C
Z
|y−h|<5/4
|f (y)|p ω(y)dy,
(4.3.7)
where C does not depend on h and the coefficients of P (x, y). From (4.3.6) and (4.3.7), it follows that Z Z p |f (y)|p ω(y)dy |T0 f (x)| ω(x)dx ≤ C |x−h|<1/4
|y−h|<5/4
holds for every h ∈ Rn . This is equivalent to having that kT0 f kp,ω ≤ Ckf kp,ω ,
(4.3.8)
where C does not depend on the coefficients of P (x, y). For j ≥ 1, we have Z |f (y)| dy ≤ CM (f )(x), |Tj f (x)| ≤ |x − y|n j−1 j 2 ≤|x−y|<2 where C does not depend on j . By the reverse H¨older inequality, there exists an ε > 0 such that ω 1+ε ∈ Ap . Thus, we have kTj f kp,ω1+ε ≤ Ckf kp,ω1+ε .
(4.3.9)
On the other hand, applying a method similar to that of proving (4.1.9), we can get kTj k2 ≤ C2−jε .
And it is easy to see that the norm of T j on L1 or L∞ are uniformly bounded. Thus, we can deduce that kTj f kp ≤ C2−jδ kf kp ,
(4.3.10)
201
4.3 Oscillatory singular integrals with standard kernels
where C only depends on the total degree of P (x, y) and δ > 0. By (4.3.9), (4.3.10) and the Stein-Weiss interpolation theorem with change of measures, we obtain that kTj f kp,ω ≤ C2−jθδ kf kp,ω , (4.3.11) where 0 < θ < 1, θ is independent of j, and C only depends on the degree of P (x, y). It follows from (4.3.8) and (4.3.11) that kT f kp,ω ≤ Ckf kp,ω , where C depends only on the total degree of P (x, y), but not on the coefficients of P (x, y).
Case 2. Assume
X
|α|=k,|β|=l
then P (x, y) =
X
|aαβ | 6= 1. Denote b =
|α|=k,|β|=l
b−(k+l) aαβ (bx)α (by)β + R
|α|=k,|β|=l
Thus
X
bx by , b b
1 k+l
|aαβ |
,
:= Q(bx, by).
Z
eiQ(bx,by) K(x, y)f (y)dy Z y −n bx y iQ(bx,by) = p.v. f b dy , e K b b b Rn · := b−n Tb f (bx), b
T f (x) = p.v.
Rn
where
Tb f (x) = p.v. and
Z
Rn
eiQ(x,y) Kb (x, y)f (y)dy
x y , . b b It is not difficult to check that Kb satisfies (a) and (b), and then the operator Z f → p.v. Kb (x, y)f (y)dy Kb (x, y) = K
Rn
is of type (L2 , L2 ). The conclusion in the case 1 yields that kTb gkp,ω ≤ bn Ckgkp,ω ,
202
Chapter 4. Oscillatory Singular Integrals
where C onlydepends on the total degree of P (x, y). Noting that ω b· ∈ Ap , and C ω b· = C(ω), we have that Z Z p · p −np (bx) ω(x)dx |T f (x)| ω(x)dx = b Tb f b n n R ZR p x · −np (x) ω b−n dx =b Tb f b b n Z R p x −n x ≤C b dx f ω b b n ZR =C |f (x)|p ω(x)dx, Rn
where C depends only on the total degree of P (x, y), but not on the coefficients of P (x, y).
4.4
Multilinear oscillatory singular integrals with rough kernels
In this section, we will study a class of multilinear oscillatory singular integrals that is a generalization of commutators generated by oscillatory singular integrals and a BMO function. Precisely, The multilinear oscillatory singular integral operators TA is defined by Z Ω(x − y) Rm+1 (A; x, y)f (y)dy, (4.4.1) TA f (x) = p· v· eiP (x,y) |x − y|n+m Rn where P (x, y) is a real-valued polynomial on R n ×Rn , Rm+1 (A; x, y) denotes the m + 1-th order Taylor series remainder of A at x expanded about y, more precisely, X 1 Rm+1 (A; x, y) = A(x) − D α A(y)(x − y)α . α! |α|≤m
It is obvious to see that when m = 0, R1 (A; x, y) = A(x) − A(y). Thus we obtain that TA f = AT (f ) − T (Af ) = [A, T ]f. In this case, TA is just a commutator; The truncated operator corresponding to T A is defined by Z Ω(x − y) SA f (x) = p· v· R (A; x, y)f (y)dy, n+m m+1 |x−y|<1 |x − y|
(4.4.2)
4.4 Multilinear oscillatory singular integrals with rough kernels
203
In this section, we will investigate a class of multilinear oscillatory singular integral operators with rough kernels and their criterion on L p -Boundedness. Suppose that Ω satisfies conditions (1) Ω Z is homogeneous of degree 0; (2) Ω(x0 )dσ(x0 ) = 0; and Sn−1
(3) Ω ∈ Lq (Sn−1 ), for some q with 1 < q ≤ ∞.
Theorem 4.4.1 Suppose that Ω satisfies (1), (3) and a function A has derivatives of order m + 1-th in BMO(R n ). The operator TA defined by (4.4.1) is bounded on Lp (Rn ) if and only if the truncated operator S A defined by (4.4.2) is bounded on Lp (Rn ) (1 < p < ∞) with the bound depending only on the total degree of P (x, y), but not on the coefficients of P (x, y). To prove Theorem 4.4.1, let us first give some lemmas. Lemma 4.4.1 Let b(x) be a function on R n with derivatives of m-th order in Ls (Rn ) for some s, n < s ≤ ∞. Then |Rm (b; x, y)| ≤ Cm,n |x − y|m
1/s X 1 Z α s |D b(z)| dz , |Ixy | Ixy
|α|=m
where Ixy is the cube centered at x, with its sides paralled to the axes , whose √ diameter is 2 n|x − y|. Lemma 4.4.2 Suppose that Ω satisfies (1), (3) and a function A has derivatives of order m-th in BMO(Rn ). Denote MA f (x) = sup r>0
1 r n+m
Z
|x−y|
|Ω(x − y)Rm+1 (A; x, y)f (y)| dy.
(4.4.3)
Then MA is bounded on Lp (Rn ) with 1 < p < ∞. Proof. Define M A , a variant of MA , by M A f (x) = sup r>0
1 r n+m
Z
r/2≤|x−y|
|Ω(x − y)Rm+1 (A; x, y)f (y)| dy. (4.4.4)
204
Chapter 4. Oscillatory Singular Integrals
It follows that MA f (x) = sup r>0
≤ sup r>0
=
∞ X
k=0
=
∞ X
k=0
Z
1
r n+m |x−y|
sup r>0
r n+m
r/2k+1 ≤|x−y|
2−k(n+m) r n+m 2k
1
2k(n+m)
= CM A f (x).
|Ω(x − y)Rm+1 (A; x, y)f (y)| dy
sup r>0
Z
|Ω(x − y)Rm+1 (A; x, y)f (y)| dy
r/2k+1 ≤|x−y|
r 2k
1 n+m
Z
|Ω(x − y)Rm+1 (A; x, y)f (y)| dy
r/2k+1 ≤|x−y|
|Ω(x − y)Rm+1 (A; x, y)f (y)| dy
Hence if M A is bounded on Lp (Rn ) with 1 < p < ∞, then we immediately obtain that MA is also bounded on Lp (Rn ) with 1 < p < ∞. Therefore it suffices to prove Lemma 4.4.2 for M A . For fixed x ∈ Rn , √ r > 0, let Q(x, r) be the cube centered at x and have its sidelength 2 nr. Set X 1 m (D α A)y α , AQ (y) = A(y) − α! Q(x,r) |α|=m
where mQ(x,r) (D α A) denotes the mean value of D α A on Q(x, r). Clearly we have Rm+1 (AQ ; x, y) = Rm+1 (A; x, y). In the following, we will consider two cases: (i) When m = 0, in this case A ∈ BMO and Z 1 M A f (x) = sup n |Ω(x − y)(A(x) − A(y))f (y)| dy r>0 R r/2≤|x−y|
205
4.4 Multilinear oscillatory singular integrals with rough kernels It is easy to check that kP1 (f )kp ≤ CkΩkqLq (Sn−1 ) kf kp and kP2 (f )kp ≤ CkAkBMO kf kp .
(ii) When m ≥ 1, We have that M A f (x) 1
≤ sup
r n+m
r>0
+ sup r>0
Z
|Ω(x − y)Rm (AQ ; x, y)f (y)|dy Z X 1 α Q α dy Ω(x − y) D A (y)(x − y) f (y) α! r/2<|x−y|
r/2<|x−y|
1 r n+m
1
2
:= M A f (x) + M A f (x).
From Lemma 4.4.1, it follows that Z X 1 1 M A f (x) ≤ C sup n kD α AkBMO |f (y)|dy r>0 r r/2<|x−y|
where M (f )(x) denotes the Hardy-Littlewood maximal function of f . Then from the Lp − boundedness of Hardy-Littlewood maximal function, it follows that X 1 kM A (f )kp ≤ C kD α AkBMO kf kp . |α|=m−1
2
For 1 < p < ∞ , we choose a t such that 1 < t < p. For M A , we have that Z X 1 2 |D α A(y) − mQ(x,r) (D α A)kf (y)|dy M A f (x) ≤ C sup n R r>0 r/2<|x−y|
≤C
X
r>0
|α|=m
× sup r r>0
≤C
sup
X
−n
|α|=m−1
Z
1 Rn
Z
0
r/2<|x−y|
t
|x−y|
|f (y)| dy
|D α A(y) − mQ(x,r) (D α A)|t dy
!1/t
kD α AkBMO Mt (f )(x),
!1/t0
206
Chapter 4. Oscillatory Singular Integrals
where
h i1 t Mt (f )(x) = M (|f |t )(x) .
Thus we conclude that X kMA2 (f )kp ≤ C kD α AkBMO kMt (f )kp |α|=m
X
=C
|α|=m
≤C
X
|α|=m
α
kD AkBMO
"Z
Rn
t
[M (|f | )(x)]
p/t
dx
t/p #1/t
kD α AkBMO kf kp .
Therefore, for 1 < p < ∞, we have that X kM A f kp ≤ C kD α AkBMO kf kp . |α|=m
Proof of Theorem 4.4.1. Observe that if the operator T defined by Z Ω(x − y) T f (x) = p· v· R (A; x, y)f (y)dy n+m m+1 Rn |x − y|
is bounded on Lp (Rn ), then the truncated operator SA defined by (4.4.2) is also bounded on Lp (Rn ). By dilation invariance, we only treat the case of X kD α AkBMO = 1. |α|=m
Proof of sufficiency. Let k and l be positive integers. Let P (x, y) be a non-trivial real-valued polynomial with degree k in x and l in y. Write X P (x, y) = aα,β xα y β |α|≤k,|β|≤l
and decompose TA into Z TA f (x) =
eiP (x,y)
|x−y|<1
+ :=
Z
|x−y|≥1 0 TA f (x) +
Ω(x − y) Rm+1 (A; x, y)f (y)dy |x − y|n+m
eiP (x,y)
Ω(x − y) Rm+1 (A; x, y)f (y)dy |x − y|n+m
TA∞ f (x).
207
4.4 Multilinear oscillatory singular integrals with rough kernels
Let us now turn our attention to the operator T A∞ . For this purpose, we write ∞ X TA∞ f = TAd f, d=1
where
TAd f (x)
=
Z
eiP (x,y) 2d−1 ≤|x−y|<2d
Ω(x − y) Rm+1 (A; x, y)f (y)dy. |x − y|n+m
Let us first consider the operator TAd , d ≥ 1. Since Z Ω(x − y) d |TA f (x)| ≤ Rm+1 (A; x, y)f (y) dy n+m 2d−1 ≤|x−y|<2d |x − y| Z 2m |Ω(x, y)Rm+1 (A; x, y)f (y)| dy = 2m MA f (x). ≤ sup m+n r r>0 |x−y|
(4.4.5)
where C is independent of d, 1 < p < ∞. On the other hand, if we can prove that kTAd f k2 ≤ C2−θ1 d kf k2 ,
(4.4.6)
then by interpolating between (4.4.5) and (4.4.6), then we will obtain that kTAd f kp ≤ C2−θ2 d kf kp ,
1 < p < ∞,
(4.4.7)
where θ1 ,θ2 and C are independent of d and f . The remainder is to prove (4.4.6). To do this, we define Z Ω(x − y) d−1 d−1 d T A f (x) = Rm+1 (A; x, y)f (y)dy. eiP (2 x,2 y) |x − y|n+m 1≤|x−y|<2 We first prove that
d (4.4.8)
T A f ≤ C2−θd kf k2 . 2 S Decompose Rn into Rn = Ii , where each Ii is a cube with side of length 1, and the cubes have disjoint interiors. Set f i = f χi . From the definition of d d T A f and fi , it easily follows that the support of T A fi is contained in a fixed
208
Chapter 4. Oscillatory Singular Integrals d
multiple of Ii . Since the supports of the various terms T A fi are bounded overlapping, we have the ”almost orthogonality” property
X
d
d
T A f ≤ C
T A fi . 2
i
2
Therefore, it suffices to show that there exists a θ > 0 independent of d and fi (for all i) such that
d (4.4.9)
T A fi ≤ C2−θd kfi k2 . 2
For fixed i, denote I i = 100nIi . Let φi (x) ∈ C0∞ (Rn ) such that 0 ≤ φi ≤ 1 √ √ and φi is identically 1 on 10 nIi and vanishes outside of 50 nIi , kD γ φi k ≤ √ Cγ for all multi-index γ. Let x0 be a point on the boundary of 80 nIi . Denote X 1 m (D α A)(·)α ; y, x0 φi (y). Aφi (y) = Rm A(·) − α! I i |α|=m
A simple computation shows that Rm+1 (A; x, y) = Rm+1 (Aφi ; x, y). For multi-index α, define Z d−1 d−1 Ω(x − y) d,α (x − y)α f (y)dy. eiP (2 x,2 y) T A f (x) = n+m |x − y| 1≤|x−y|<2 It follows that d
d
T A fi (x) = T Aφi fi (x) Z Ω(x − y) d−1 d−1 Rm+1 (Aφi ; x, y)fi (y)dy eiP (2 x,2 y) = |x − y|n+m 1≤|x−y|<2 X 1 d,α d,0 = Aφi (x)T A fi (x) − T A D α Aφi fi (x) α! |α|<m X 1 d,α − T A D α Aφi fi (x) α! |α|=m
:= I + II + III. Now we will give the estimates of I, II and III respectively. For multi-index β, |β| < m, we have D β Aφi (y) =
X
β=µ+ν
X 1 m (D α A)(·)α ; y, x0 D ν φi (y). Cµ,ν Rm−|µ| D µ A(·) − α! I i
|α|=m
209
4.4 Multilinear oscillatory singular integrals with rough kernels d,α
By Lemma 4.4.1, we have |D β Aφi (y)| ≤ C. For the operator T A with multi-index α, it is not difficult to obtain the following estimate
d,α (4.4.10)
T A f ≤ C2−(θ+m−|α|)d kf kp , 1 < p < ∞. p
Thus it follows from (4.4.10) that
kIk2 ≤ C2−θd kfi k2 and kIIk2 ≤ C2−θd kfi k2 . It remains to estimate the third term III. For 0 < γ < n, we have Z d,α |Ω(x − y)f (y)|dy T A f (x) = C 1≤|x−y|<2
Z
0
|f (y)|q dy n−γ 1≤|x−y|<2 |x − y| h i1/q0 0 ≤ Cγ kΩkLq (Sn−1 ) Iγ (|f |q )(x) ,
≤ Cγ kΩkLq (Sn−1 )
where Iγ denotes the usual fractional integral of order γ. Choose p 1 ≥ q 0 , for σ > 0, we may take a γ such that 0 < γ < nq 0 /p1 and 1/(p1 + σ) = 1/p1 − γ/nq 0 . By the Hardy-Littlewood-Sobolev theorem, we get
d,α (4.4.11) ≤ Ckf kp1 .
T A f p1 +σ
By (4.4.10), for |α| = m, there exists δ 1 > 0 such that
d,α
T A f ≤ C2−δ1 d kf kp . p
(4.4.12)
If p ≥ q 0 , then we may set p = p1 . Interpolating between (4.4.11) and (4.4.12) yields that
d,α ≤ C2−δ2 d kf kp , (4.4.13)
T A f p+ρ
where δ2 > 0 and 0 < ρ ≤ σ; If 1 < p < q 0 , then interpolating between (4.4.11) and (4.4.12) yields that
d,α (4.4.14) ≤ C2−δ2 d kf kp2 ,
T A f p2 +ρp
210
Chapter 4. Oscillatory Singular Integrals
where p < p2 < p1 and 0 < ρp < σ. It follows from (4.4.13) and (4.4.14) that
d,α f ≤ C2−δ2 d kf kp , (4.4.15) T
A p+ρp
where 1 < p < ∞ and 0 < ρp < σ. On the other hand, if |β| = m, then D β Aφi (y) X =
Cµ,ν Rm−|µ|
β=µ+ν,|µ|<m
X 1 × D µ A(·) − m (D α A)(·)α ; y, x0 D ν φi (y) α! I i |α|=m X α + D A(y) − mI i (D α A) φi (y). |α|=m
Thus, it follows that X α β φi D A(y) − mI i (D α A) . D A (y) ≤ C 1 + |α|=m
and this shows for any t > 1 that
β φi A
D
≤ C. t
We choose η > 0 and 1 < t < ∞ such that
1 1 1 + = . 2 t 2−η It follows from (4.4.15) that kIIIk2 ≤ C2
−δ2 d
|α|=m
≤ C2−δ2 d ≤ C2
X
α φi
D A fi
−δ2 d
X
|α|=m
2−η
kD α Aφi kt kfi k2
kfi k2 .
All estimates above imply that (4.4.8) is valid. By the process of proving d,α (4.4.8), it is easy to see that inequality (4.4.8) also holds if T A f is replaced d,α by TA f . Thus, we obtain (4.4.7).
4.4 Multilinear oscillatory singular integrals with rough kernels
211
Let us now turn to prove that TA0 is also bounded on Lp (Rn ), i.e.,
0
TA f ≤ Ckf kp . (4.4.16) p
We will carry out an argument by a double induction on the degree in x and y of the polynomial P (x, y). If P (x, y) depends only on x or only on y, it is obvious that there cases are trivial. Let k and l be two positive integers and let P (x, y) have degree k in x and l in y. We assume that (4.4.16) holds for all polynomial which are sums of monomials of degree less than k in x times monomials of any degree in y, together with monomials which are of degree k in x times monomials which are of degree less than l in y. Rewrite X P (x, y) = aα,β xα y β − y α+β + P0 (x, y). |α|=k,|β|=l
Obviously P0 (x, y) satisfies the inductive hypothesis. It follows that Z Ω(x − y) eiP0 (x,y) TA0 f (x) = Rm+1 (A; x, y)f (y)dy |x − y|n+m |x−y|<1 Z Ω(x − y) + eiP (x,y) − eiP0 (x,y) Rm+1 (A; x, y)f (y)dy |x − y|n+m |x−y|<1 := TA01 f (x) + TA02 f (x).
By the inductive hypothesis, we conclude that the operator T A01 is bounded on Lp (Rn ). A direct computation shows that iP (x,y) − eiP0 (x,y) ≤ C|x − y|. e ˜ = f (y)χ{|y|≤2} . When |x| ≤ 1, we have that Denote f(y) Z 02 |Ω(x − y)| TA f (x) ≤ C |Rm+1 (A; x, y)||f˜(y)|dy n+m−1 |x − y| |x−y|<1 ˜ ≤ CMA f(x). By Lemma 4.4.2, we obtain Z Z 02 p |TA f (x)| dx ≤ C |x|≤1
Therefore, for any h ∈ Z
|x|≤2
Rn ,
|x−h|≤1
|f (x)|p dx.
it follows that
|TA02 f (x)|p dx
≤C
Z
|x−h|≤2
|f (x)|p dx,
212
Chapter 4. Oscillatory Singular Integrals
where C is independent of h. Integrating the last inequality with respect to h yields our desired result. Proof of necessity. To do this, we choose Q(x, y) such that it has the property P and decomposes Z
TA f (x) =
eiQ(x,y)
|x−y|<1
+ :=
Z
Ω(x − y) Rm+1 (A; x, y)f (y)dy |x − y|n+m
eiQ(x,y)
|x−y|≥1 0 TA f (x) +
Ω(x − y) Rm+1 (A; x, y)f (y)dy |x − y|n+m
TA∞ f (x).
The previous proof implies that TA∞ is bounded on Lp (Rn ), so is TA0 . It is easy to check that Z
|x−h|≤1
|TA0 f (x)|p dx
!1/p
≤C
Z
p
|x−h|≤2
|f (x)| dx
!1/p
,
where C is independent of h, 1 < p < ∞. Since Q(x, y) has the property P, when |x − h| < 1, we have that SA f (x) Z =
Ω(x − y) R (A; x, y)f (y)χB2 (h) (y)dy n+m m+1 |x−y|<1 |x − y| Z Ω(x − y) −iQ(x−h,y−h) −iR1 (y,h) −iR0 (x,h) =e eiQ(x,y) e e f (y)χB2 (h) (y)dy. |x − y|n+m |x−y|<1
Observe that the Taylor expansion of e iQ(x−h,y−h) is
e−iQ(x−h,y−h)
m ∞ m X X (−i) = aαβ (x − h)α (y − h)β m! m=0 α,β
=
∞ X
m=0
1 X am,µ,ν (x − h)µ (y − h)ν , m! µ,ν,
where u and v are multi-indice. Thus, if we set a = (1, 1, · · · , 1) and b =
4.5 Multilinear oscillatory singular integrals with standard kernels
213
(2, 2, · · · , 2), then we conclude that !1/p Z |x−h|<1
≤
|Sf (x)|p dx
∞ Z X 1 X |am,µ,ν | |(x − h)µ |p m! µ,ν, |x−h|<1 m=0 p 1/p × TA0 e−iR1 (·,h) f (·)χB(h,2) (·)(· − h)ν (x) dx
∞ X 1 X |am,µ,ν |aµ ≤ m! µ,ν, m=0
p
|y−h|<2
ν p
|f (y)| |(y − h) | dy
!1/p
!1/p Z ∞ X 1 X µ ν p ≤ |am,µ,ν |a b |f (y)| dy m! µ,ν |y−h|<2 m=0 m !1/p Z ∞ X 1 X |aαβ |aα bβ ≤C |f (y)|p dy m! |y−h|<2 m=0 α,β !1/p Z X ≤ C exp |f (y)|p dy |aαβ |aα bβ . |y−h|<2
α,β
Thus
Z
kSA f kp ≤ Ckf kp .
4.5
Multilinear oscillatory singular integrals with standard kernels
Suppose that K(x, y) is a Calder´on-Zygmund standard kernel (also called distribution kernel). That means, K(x, y) satisfies C , |x − y|n
x 6= y,
(a)
|K(x, y)| ≤
(b)
|∇x K(x, y)| + |∇y K(x, y)| ≤
C , x 6= y. |x − y|n+1
Obviously the kernel K does not have homogeneity. Let us study in this section an oscillatory singular integral operator defined by Z K(x, y) TA f (x) = p· v· eiP (x,y) Rm+1 (A; x, y)f (y)dy, (4.5.1) |x − y|m Rn
214
Chapter 4. Oscillatory Singular Integrals
where P (x, y) is a real-valued polynomial on R n ×Rn . Theorem 4.5.1 Suppose that K(x, y) satisfies (a) and (b). If A has derivatives of order m in BMO (Rn ), then for 1 < p < ∞, then the following two facts are equivalent: (i) If P (x, y) is a non-trivial real valued polynomial, then T A is bounded P α p p from L to L with the bound C(degP, n) |α|=m kD AkBMO , where deg P denotes the total degree of the polynomial P (x, y). (ii) The truncated operator SA f (x) = p· v·
Z
|x−y|<1
K(x, y) Rm+1 (A; x, y)f (y)dy |x − y|m
is bounded from Lp to Lp with the bound C
P
α Ak kD . BMO |α|=m
Proof of Theorem 4.5.1. By dilation invariance, we only deal with the case X kD α AkBMO = 1. |α|=m
(ii)⇒(i). Let k and l be positive integers. Let P (x, y) be a non-trivial real-valued polynomial with degree k in x and l in y. Write P (x, y) = P α β |α|≤k,|β|≤l aα,β x y and decompose TA into TA f (x) =
Z
eiP (x,y)
|x−y|<1
+
Z
K(x, y) Rm+1 (A; x, y)f (y)dy |x − y|m
eiP (x,y)
|x−y|≥1
K(x, y) Rm+1 (A; x, y)f (y)dy |x − y|m
:= TA0 f (x) + TA∞ f (x). We can easily obtain
kTA0 f kp ≤ Ckf kp . Let us now turn our attention to the operator T A∞ . For this purpose we write ∞ X ∞ TA f (x) = TAd f (x) , d=1
215
4.5 Multilinear oscillatory singular integrals with standard kernels where TAd f (x)
=
Z
eiP (x,y)
2d−1 ≤|x−y|<2d
K(x, y) Rm+1 (A; x, y)f (y)dy. |x − y|m
We first consider the operator TAd , d ≥ 1. Since Z K(x, y) d dy |TA f (x)| ≤ R (A; x, y)f (y) m+1 m 2d−1 ≤|x−y|<2d |x − y| Z m 2 |K(x, y)Rm+1 (A; x, y)f (y)| dy := MA f (x), ≤ sup m r>0 r |x−y|
(4.5.2)
where C is independent of d, 1 < p < ∞. On the other hand, if we can prove that kTAd f k2 ≤ C2−θ1 d kf k2 , (4.5.3) then by interpolating between (4.5.2) and (4.5.3), we will obtain that kTAd f kp ≤ C2−θ2 d kf kp ,
1 < p < ∞,
(4.5.4)
where θ1 ,θ2 and C are independent of d and f . The remainder is to prove (4.5.3). To do this, we define Z d−1 d−1 K(x, y) d T A f (x) = Rm+1 (A; x, y)f (y)dy. eiP (2 x,2 y) |x − y|m 1≤|x−y|<2 We shall prove that
d
kT A f k2 ≤ C2−θd kf k2 . (4.5.5) S n n Decompose R into R = Ii , where Ii is a cube with side of length 1, and the cubes have disjoint interiors. Set f i = f χi . From the definition of d d T A f and fi , it easily follows that the support of T A fi is contained in a fixed d multiple of Ii . Since the supports of the various terms T A fi have bounded overlaps, we have the ”almost orthogonality” property X d d kT A f k2 ≤ C kT A fi k2 . i
Therefore, it suffices to show that there exists a θ > 0 independent of d and fi (for all i) such that d
kT A fi k2 ≤ C2−θd kfi k2 .
(4.5.6)
216
Chapter 4. Oscillatory Singular Integrals
For fixed i, denote I i = 100nIi . Let φi (x) ∈ C0∞ (Rn ) such that 0 ≤ φi ≤ 1 √ √ and φi is identically 1 on 10 nIi and vanishes outside of 50 nIi , kD γ φi k∞ ≤ √ Cγ for all multi-index γ. Let x0 be a point on the boundary of 80 nIi . Denote X 1 m (D α A)(·)α ; y, x0 φi (y). Aφi (y) = Rm A(·) − α! I i |α|=m
A simple computation shows that Rm+1 (A; x, y) = Rm+1 (Aφi ; x, y). For multi-index α, define Z d−1 d−1 K(x, y) d,α eiP (2 x,2 y) (x − y)α f (y)dy. T A f (x) = |x − y|m 1≤|x−y|<2 It follows that d
d
T A fi (x) = T Aφi fi (x) Z K(x, y) d−1 d−1 R (Aφi ; x, y)fi (y)dy eiP (2 x,2 y) = m m+1 |x − y| 1≤|x−y|<2 X 1 d,α d,0 = Aφi (x)T A fi (x) − T A D α Aφi fi (x) α! |α|<m X 1 d,α α φi − T D A fi (x) α! A |α|=m
:= I + II + III. In the following, we will give the estimates of I, II and III respectively. For multi-index β, |β| < m, we have D β Aφi (y) =
X
β=µ+ν
X 1 m (D α A)(·)α ; y, x0 D ν φi (y). Cµ,ν Rm−|µ| D µ A(·) − α! I i
|α|=m
d,α
By Lemma 4.4.1, we have |D β Aφi (y)| ≤ C. For the operator T A with multi-index α, it is not difficult to obtain the following estimate
d,α (4.5.7)
T A f ≤ C2−(θ+m−|α|)d kf kp , 1 < p < ∞. p
Thus it follows from (4.5.7) that
kIk2 ≤ C2−θd kfi k2
217
4.5 Multilinear oscillatory singular integrals with standard kernels and kIIk2 ≤ C2−θd kfi k2 .
It remains to estimate the third term III . By the property (i) on K(x, y), for 0 < γ < n, we have Z d,α |f (y)|dy T A f (x) = C 1≤|x−y|<2
≤C
Z
1≤|x−y|<2
≤ CIγ (|f |)(x),
|f (y)| dy |x − y|n−γ
where Iγ denotes the usual fractional integral of order γ. For σ > 0, we may take a γ such that 0 < γ < n/p and 1/(p + σ) = 1/p − γ/n. By the Hardy-Littlewood-Sobolev theorem, we get
d,α ≤ Ckf kp . (4.5.8)
T A f p+σ
By (4.5.7), for |α| = m, there exists δ 1 > 0 such that
d,α
T A f ≤ C2−δ1 d kf kp .
(4.5.9)
p
Interpolating between inequalities (4.5.8) and (4.5.9) yields that
d,α ≤ C2−δ2 d kf kp ,
T A f p+ρ
(4.5.10)
where δ2 > 0 and 0 < ρ ≤ σ. On the other hand, if |β| = m, then D β Aφi (y) X =
Cµ,ν Rm−|µ|
β=µ+ν,|µ|<m
X 1 m (D α A)(·)α ; y, x0 D ν φi (y) × D µ A(·) − α! I i |α|=m X + D α A(y) − mI i (D α A) φi (y).
|α|=m
Thus, it follows that X α β φi D A(y) − mI i (D α A) . D A (y) ≤ C 1 + |α|=m
218
Chapter 4. Oscillatory Singular Integrals
And this shows for any t > 1 that kD β Aφi kt ≤ C We choose η > 0 and 1 < t < ∞ such that
It follows from (4.5.10) that
1 1 1 + = . 2 t 2−η
kIIIk2 ≤ C2−δ2 d ≤ C2−δ2 d ≤ C2
−δ2 d
X
α φi
D A fi
|α|=m
X
|α|=m
2−η
kD α Aφi kt kfi k2
kfi k2 .
All estimates above imply that (4.5.5) is true. By the process of proving d,α (4.5.5), it is easy to see that inequality (4.5.5) also holds if T A f is replaced by TAd,α f . Thus, we obtain (4.5.3). (ii)⇒(i). To do this, we choose Q(x, y) such that it has the property P and decomposes Z K(x, y) TA f (x) = eiQ(x,y) Rm+1 (A; x, y)f (y)dy |x − y|m |x−y|<1 Z K(x, y) eiQ(x,y) Rm+1 (A; x, y)f (y)dy + |x − y|m |x−y|≥1 := TA0 f (x) + TA∞ f (x).
The previous proof has proved that T A∞ is bounded on Lp (Rn ). Thus TA0 is also bounded on Lp (Rn ). It is easy to check that !1/p !1/p Z Z |x−h|≤1
|TA0 f (x)|p dx
≤C
|x−h|≤2
|f (x)|p dx
,
where C is independent of h, 1 < p < ∞. Since Q(x, y) has the property P, when |x − h| < 1, it follows that SA f (x) Z =
K(x, y) R (A; x, y)f (y)χB2 (h) (y)dy m m+1 |x−y|<1 |x − y| Z −iR0 (x,h) =e eiQ(x,y) K(x − y)e−iQ(x−h,y−h) e−iR1 (y,h) f (y)χB2 (h) (y)dy. |x−y|<1
4.5 Multilinear oscillatory singular integrals with standard kernels
219
Observe that the Taylor expansion of e iQ(x−h,y−h) is e−iQ(x−h,y−h)
m ∞ m X X (−i) = aαβ (x − h)α (y − h)β m! m=0
=
α,β
∞ X
1 X am,µ,ν (x − h)µ (y − h)ν , m! µ,ν, m=0
where u and v are multi-indice. Thus, if we set a = (1, 1, · · · , 1) and b = (2, 2, · · · , 2), then we conclude that Z
p
|x−h|<1
≤
|Sf (x)| dx
!1/p
∞ X 1 X |am,µ,ν | m! µ,ν,
m=0
×
Z
|x−h|<1
|(x −
h)µ |p TA0
∞ X 1 X ≤ |am,µ,ν |aµ m! µ,ν, m=0
Z
e
−iR1 (·,h)
f (·)χB(h,2) (·)(· − h) p
|y−h|<2
ν p
|f (y)| |(y − h) | dy
!1/p
ν
!1/p Z ∞ X 1 X ≤ |f (y)|p dy |am,µ,ν |aµ bν m! µ,ν |y−h|<2 m=0 m !1/p Z ∞ X 1 X p α β |f (y)| dy |aαβ |a b ≤C m! |y−h|<2 m=0 α,β !1/p Z X . |f (y)|p dy |aαβ |aα bβ ≤ C exp α,β
|y−h|<2
Thus kSA f kp ≤ Ckf kp , and then we finish the proof of Theorem 4.5.1. The following theorem is an application of Theorem 4.5.1.
p (x) dx
!1/p
220
Chapter 4. Oscillatory Singular Integrals
Theorem 4.5.2 Suppose that K(x, y) satisfies (a) and (b). Let A have derivatives of order m in BMO(Rn ). If the following two operators Z K(x, y)f (y)dy K f (x) = p· v· Rn
and T f (x) = p· v·
Z
Rn
K(x, y) Rm+1 (A; x, y)f (y)dy |x − y|m
(4.5.11)
are all bounded on L2 (Rn ), then TA defined is bounded on Lp (Rn ), P by (4.5.1) α 1 < p < ∞, with the bound C(degP, n) |α|=m kD AkBMO . Before proving Theorem 4.5.2, let us first introduce a class of auxiliary multilinear oscillatory singular integral operators defined by Z K(x, y) Qm+1 (A; x, y)f (y)dy (4.5.12) eiP (x,y) TA f (x) = p· v· |x − y|m Rn
and T f (x) = p· v· where
Z
K(x, y) Qm+1 (A; x, y)f (y)dy, |x − y|m
Rn
Qm+1 (A; x, y) = Rm (A; x, y) −
(4.5.13)
X 1 D α A(x)(x − y)α . α!
|α|=m
Obviously, Rm+1 (A; x, y) − Qm+1 (A; x, y) =
X 1 (x − y)α D α A(x) − D α A(y) , α!
|α|=m
where D α A ∈ BMO(Rn ). Thus we have TA f (x) − TA f (x) X 1 Z K(x, y)(x − y)α α α D A(x) − D A(y) f (y)dy eiP (x,y) = α! Rn |x − y|m |α|=m i X 1h = D α A, Kα f (x), α! |α|=m
where Kα f (x) =
Z
Rn
eiP (x,y)
K(x, y)(x − y)α f (y)dy |x − y|m
4.5 Multilinear oscillatory singular integrals with standard kernels
221
and [D α A, Kα ] is a commutator. Set Kα (x, y) =
K(x, y)(x − y)α . |x − y|m
It is easy to check that Kα (x, y) satisfies the condition (a) and (b). Define Z 1 K(x, y)Qm+1 (A; x, y)f (y) dy. MA f (x) = sup m r>0 r |x−y|
kMA f kp ≤ C
X
|α|=m
kD α A||BMO kf kp .
To prove Theorem 4.5.2, we need some lemmas. Lemma 4.5.1 Suppose that K(x, y) and A be the same as in Theorem 4.5.2. Moreover, b(x, y) ∈ L∞ (Rn × Rn ) , 1 < p < ∞. If the operator Z K(x, y) Tb f (x) = p· v· R (A; x, y)b(x, y)f (y)dy m m+1 Rn |x − y| P is bounded on Lp (Rn ) with the bound E |α|=m kD α AkBMO , then the truncated operator Z K(x, y) Sb f (x) = p· v· R (A; x, y)b(x, y)f (y)dy m m+1 |x−y|<1 |x − y| P α Ak is also bounded on Lp (Rn ) with the bound C(E+kbk∞ ) kD . BMO |α|=m If the operator Z K(x, y) Q (A; x, y)b(x, y)f (y)dy Tb f (x) = p· v· m m+1 Rn |x − y| P is bounded on Lp (Rn ) with the bound E |α|=m kD α AkBMO , then the truncated operator Z K(x, y) Qm+1 (A; x, y)b(x, y)f (y)dy Sb f (x) = p· v· |x − y|m |x−y|<1 P α Ak is also bounded on Lp (Rn ) with the bound C(E+kbk∞ ) kD BMO ; |α|=m
222
Chapter 4. Oscillatory Singular Integrals
The proof of Lemma 4.5.1 is similar to that of Theorem 4.2.2.
Lemma 4.5.2 Suppose that K(x, y) and A be the same as in Theorem 4.5.2. If the truncated operator S f (x) = p· v·
Z
|x−y|<1
K(x, y) Qm+1 (A; x, y)f (y)dy |x − y|m
is bounded on Lp (Rn ) 1 < p < ∞ with the bound
C(degP, n)
X
|α|=m
(4.5.14)
kD α AkBMO ,
then for any non-degenerate polynomial P (x, y), the operator T A defined by (4.5.12) is bounded on Lp (Rn ) with the bound
C(degP, n)
X
|α|=m
kD α AkBMO .
The proof of Lemma 4.5.2 is similar to that of Theorem 4.5.1.
Lemma 4.5.3 T Suppose that K(x, y) and A be the same as in Theorem 4.5.2. Let f ∈ L2 (Rn ) L∞ (Rn ). For ε > 0, set Iε =
Z
|y|≥ε
K(x0 , y) Rm+1 (A; x0 , y)f (y)dy, |x0 − y|m
p n where x0 ∈ {x : |x| = 3ε 5 }. If T defined by (4.5.11) is bounded on L (R ), then Z −n ε |T f (x) − Iε |dx |x|≤ε/2
Z
Z
K(y, x) R (A; x, y) m m+1 |x|≤ε/2 |y|≥ε |x − y| K(x0 , y) − R (A; x , y) |f (y)|dydx. m+1 0 |x0 − y|m
≤ CM2 f (0) + ε−n
4.5 Multilinear oscillatory singular integrals with standard kernels
223
Proof. Define f1 = f χ{|x|<ε} and f2 = f χ{|x|≥ε}. Then f = f1 + f2 . When |x| < ε/2, we have |T f (x) − Iε |
≤ |T f1 (x)| + |T f2 (x) − Iε |
≤ |T f1 (x)| Z K(x, y) K(x0 , y) |f (y)|dy. + R (A; x, y) − R (A; y, x ) m+1 0 |y − x|m m+1 |x0 − y|m |y|≥ε
So, it suffices to prove Z
|x|≤ε/2
|T f1 (x)|dx ≤ C2 εn M2 f (0). ..
Since T is bounded on L2 (Rn ), using Holder’s inequality implies the desired result. To prove Theorem 4.5.2, we also need two propositions. Proposition 4.5.1 Suppose that K(x, y) and A be the same as in Theorem 4.5.2. P If the operator T defined by (4.5.13) is bounded on L 2 (Rn ) with the bound |α|=m kD α AkBMO , then for any non-degenerate P (x, y), the operator TA defined (4.5.12) is bounded on Lp (Rn ), 1 < p ≤ 2, with the bound byP C(degP, n) 1 + |α|=m kD α AkBMO .
Proof. If we can prove that T is bounded from H 1 (Rn ) to L1 (Rn ), then by the well-known interpolation theorem we obtain that T is bounded on Lp (Rn )(1 < p ≤ 2). Lemma 4.5.1 directly implies that the truncated operator S defined by (4.5.14) is also bounded on L p (Rn )(1 < p ≤ 2). Lemma 4.5.2 shows us that the Lp − boundedness of TA immediately follows from Lp − boundedness of S . Let us now prove that T is bounded from H 1 (Rn ) to L1 (Rn ). By a standard argument, it suffices to show that there exists a constant C such that Z X |T a(x)|dx ≤ C kD α AkBMO Rn
|α|=m
H 1 (Rn )-atom
for each a, where C is independent of a. For this purpose, suppose that supp(a) ⊂ B for some ball B. Then Z Z Z |T a(x)|dx = |T a(x)|dx + |T a(x)|dx := I + II. Rn
2B
Rn \2B
224
Chapter 4. Oscillatory Singular Integrals
The estimate for I follows from the boundedness of T on L 2 (Rn ). Z I= |T a(x)|dx 2B
≤
Z
|T a(x)|2 dx
2B 1/2
≤ C|B|
1/2 Z
dx
2B
1/2
· kak2 ≤ C. ˜ = 1.2B, B ∗ = 5√nB. Let y0 be on the boundary To estimate II we set B ˜ mB ∗ (b) denotes the mean value of b on B ∗ . Set of B, X 1 ∗ AB (x) = A(x) − mB ∗ (D α A)xα . α! |α|=m ∗
Obviously, Qm+1 (A, x, y) = Qm+1 (AB , x, y), when x ∈ Rn \ 2B, Z K(x, y) ∗ Q (AB ; x, y)a(y)dy T a(x) = m m+1 |x − y| ZB ∗ K(x, y) Rm (AB ; x, y)a(y)dy = m B |x − y| X 1 Z K(x, y)(x − y)α ∗ − D α AB (x)a(y)dy α! B |x − y|m |α|=m
:= T 1 a(x) + T 2 a(x).
For T 1 a(x), when x ∈ Rn \ 2B, Z 1 K(x, y) K(x, y ) ∗ ∗ 0 B B T a(x) = Rm (A ; x, y) − Rm (A ; x, y0 ) a(y)dy m m |x − y0 | B |x − y| Z B∗ B∗ −n−m ≤ C|x − y0 | Rm (A ; x, y) − Rm (A ; x, y0 ) |a(y)|dy B Z K(x, y) K(x, y0 ) B∗ +C − R (A ; x, y )a(y) dy. m 0 |x − y|m |x − y0 |m B ∗
By the definition of Rm (AB ; x, y), we have X 1 ∗ ∗ ∗ Rm (AB ; x, y) − Rm (AB ; x, y0 ) = R (D β AB ; y0 , y)(x − y0 )β . β! m−|β| |β|<m
For |β| < m, applying Lemma 4.4.1 leads to X ∗ kD α AkBMO . Rm−|β| (D β AB ; y0 , y) ≤ C|y − y0 |m−|β| |α|=m
225
4.5 Multilinear oscillatory singular integrals with standard kernels Thus B∗ B∗ R (A ; x, y) − R (A ; x, y ) m m 0 X X ≤C |y − y0 |m−|β| |x − y0 ||β| kD α AkBMO . |β|<m |α|=m
Therefore Z ∗ ∗ |x − y0 | Rm (AB ; x, y) − Rm (AB ; x, y0 ) |a(y)|dy B Z X |y − y0 ||a(y)| dy, ≤C kD α AkBMO n+1 B |x − y| −n−m
(4.5.15)
|α|=m
where we use the following relation |x − y| ≈ |x − y 0 | ≥ |y − y0 |. When x ∈ Rn \ 2B, we have that
Z K(x, y) K(x, y0 ) B∗ R (A ; x, y )a(y) − dy m 0 m |x − y0 |m B |x − y| Z |y − y0 | ∗ |Rm (AB ; x, y0 )a(y)|dy ≤C m+n+1 |x − y| B X Z |y − y0 | =C |x − y|n+1 |α|=m B Z 1 α s 1/s α × |D A(z) − mB ∗ (D A | dz) |a(y)|dy |Ixy0 | Ixy0 X Z |y − y0 | ≤C |x − y|n+1 |α|=m B Z 1 α α s 1/s × |D A(z) − mC1 Ixy0 (D A | dz) |a(y)|dy |C1 Ixy0 | C1 Ixy0 X Z |y − y0 | α α y ∗ (D A) − mB (D A) |a(y)|dy. +C n+1 mC1 Ix 0 |x − y| B |α|=m
Note that B ∗ ∈ C1 Ixy0 . By a well-known inequality, if f ∈ BMO, then |mB1 (f ) − mB2 (f )| ≤ C log
|B2 | kf kBMO , B1 ⊂ B2 , 2|B1 | ≤ |B2 |, |B1 |
where Bi is cubes or balls (i = 1, 2), C1 is a constant. In fact, we may let
226
Chapter 4. Oscillatory Singular Integrals
√ C1 = 5 n and get Z K(x, y) K(x, y0 ) B∗ − R (A ; x, y )a(y) dy m 0 |x − y|m |x − y0 |m B Z X |x − y0 | |y − y0 ka(y)| ≤C kD α AkBMO log dy, n+1 rB B |x − y|
(4.5.16)
|α|=m
where rB denotes the radius of B. Combining (4.5.15) with (4.5.16) yields that Z X |y − y0 ka(y)| |x − y0 | 1 α dy. |T a(x)| ≤ C kD AkBMO log n+1 rB B |x − y| |α|=m
For the estimate of T 2 a(x), we have that |T 2 a(x)| Z K(x, y)(x − y)α K(x, y)(x − y )α X 1 0 α B∗ − ≤ D A (x) |a(y)|dy m m α! |x − y| |x − y| B |α|=m Z X 1 K(x, y)(x − y0 )α K(x, y0 )(x − y0 )α α B∗ |a(y)|dy. + − |D A (x)| α! |x − y|m |x − y0 |m B |α|=m
Since K(x, y) and Kα (x, y) satisfy (b), we conclude that Z X 1 |y − y0 | ∗ |T 2 a(x)| ≤ C |a(y)|dy. |D α AB (x)| α! |x − y|n+1 B |α|=m
Thus, from the previous estimates it follows that Z |T 1 a(x)|dx Rn \2B
≤C
X
|α|=m
kD α AkBMO
Z
B
|a(y)|dy
∞ Z X k=1
Ak
|y − y0 | |x − y0 | log dx, n+1 |x − y| rB
where Ak = 2k+1 B \ 2k B (k ≥ 1). Notice that y ∈ B, when x ∈ Ak . Then we have that |y − y0 | CrB Ck ≤ k = k . n+1 n+1 |x − y| (2 rB ) (2 rB )n 2k Thus
Z
Ak
|y − y0 | |x − y0 | Ck log dx ≤ k (2k rB )n ≤ Ck2−k . n+1 |x − y| rB (2 rB )n 2k
4.5 Multilinear oscillatory singular integrals with standard kernels Therefore Z
1
Rn \2B
|T a(x)|dx ≤ C ≤C
and
Z
Rn \2B
X
|α|=m
X
|α|=m
α
kD AkBMO kD α AkBMO
∞ X
k2
k=1
−k
Z
B
227
|a(y)|dy
|T 2 a(x)|dx
Z X 1 Z |y − y0 | ∗ ≤C |D α AB (x)ka(y)|dydx n+1 α! Rn \2B B |x − y| |α|=m
Z ∞ Z X 1 X |D α A(x) − mQ∗ (D α A)| ≤C |a(y)|dy dx α! (2k rB )n 2k B Ak ≤C
|α|=m
k=1
|α|=m
k=1
It follows that Z II ≤
∞ X X 1 X k2−k kD α AkBMO ≤ C kD α AkBMO . α!
Rn \2B
|α|=m
|T 2 a(x)|dx +
Z
Rn \2B
|T 2 a(x)|dx ≤ C
X
|α|=m
kD α AkBMO .
This finishes the proof of Proposition 4.5.1. Proposition 4.5.2 Suppose that K(x, y) and A be the same as in Theorem 4.5.2. P If the operator T defined by (4.5.11) is bounded on L 2 (Rn ) with the bound |α|=m kD α AkBMO , then for any non-degenerate P (x, y), the operator TA defined (4.5.1) is bounded on Lp (Rn ), 2 ≤ p < ∞, with the bound byP C(degP, n) 1 + |α|=m kD α AkBMO .
Proof. By a method similar to that proving Proposition 4.5.1, it suffices to prove that T is bounded from L ∞ (Rn ) to BMO(Rn ). Assume T f ∈ L2 (Rn ) L∞ (Rn ) and ε > 0. By Lemma 4.5.3, we get that Z −n ε |T f (y) − Iε |dy |y|≤ε/2
Z
Z
K(x, y) R (A; x, y) m m+1 |y|≤ε/2 |x|≥ε |x − y| K(x, y0 ) − R (A; x, y ) m+1 0 |f (x)|dxdy. |x − y0 |m
≤ CM2 f (0) + ε−n
228
Chapter 4. Oscillatory Singular Integrals
√ Obviously, M2 (0) ≤ kf k∞ . Let B ε = {x : |x| ≤ 5 nε} and mB ε (b) denote the mean value of b on B ε . Set X 1 ε mB ε (D α A)xα . AB (x) = A(x) − α! |α|=m
Using the method in the proof of Proposition 4.5.1 , we can easily obtain that Z X ε−n |T f (y) − Iε |dy ≤ 1 + kD α AkBMO kf k∞ . |y|≤ε/2
|α|=m
By the translation invariance, we obtain that for any ball B, there exists IB , such that Z X 1 |T f (x) − IB |dx ≤ C kD α AkBMO kf k∞ . sup B |B| B |α|=m
Suppose f ∈ L∞ (Rn ). Set Bj = jB, j ∈ N, where B = {|x| ≤ 1}. If we can prove that there exist γj independent of f such that T f (x) = lim T (f χBj )(x) − γj (4.5.17) j→∞
exists, where the convergence is the meaning of local L 1 norm or almost everywhere, then by the definition of space BMO(R n ), we will prove that T is bounded from L∞ (Rn ) to BMO(Rn ). Actually, for some R > 0, if |x| ≤ R, then we take x0 with |x0 | = 6R/5. Set Z K(x0 , y) R (A; x0 , y)f (y)dy, γj = m m+1 2R≤|y−x0 |≤j |x0 − y| where the integral above exists obviously. Denote gj (x) = T (f χBj )(x) − γj . When j, l > 2R + 1 and j > l, we have that gj (x) = gl (x) + Slj (x), where Slj (x) Z =
l<|y−x0 |≤j
K(x, y) K(x0 , y) Rm+1 (A; x, y) − Rm+1 (A; x0 , y) f (y)dy. |x − y|m |x0 − y|m
4.5 Multilinear oscillatory singular integrals with standard kernels
229
By a well-known inequality on BM O before (4.5.16) and by an argument similar to that of estimating T 1 a(x) above, we obtain that Z X |y − x0 | |x − x0 | α log |Slj (x)| ≤ C kD AkBMO kf k∞ dy. n+1 |x − x0 | l<|y−x0 |≤j |x0 − y| |α|=m
For |x| ≤ R and |x0 | = 6R/5, when l, j → ∞, a simple computation yields that Z |y − x0 | |x − x0 | log dy n+1 |x − x0 | l<|y−x0 |≤j |x0 − y| converges to 0 in agreement. This means that the limit in (4.5.17) exists. Thus kT f kBMO ≤ lim kgj kBMO = lim kT (f χBj )kBMO ≤ Ckf k∞ . j→∞
j→∞
This finishes the proof of Proposition 4.5.2. Proof of Theorem 4.5.2. We first prove that operator Z Kα f (x) = Kα (x, y)f (y)dy Rn
is bounded on L2 (Rn ). For f ∈ L2 (Rn ), set fα =
f (·)(x − ·)α , |x − ·|m
then fα ∈ L2 (Rn ), where |α| = m. Thus
Z
kKα f k2 = K (·, y)f (y)dy
n α
R
Z
2
= K(·, y)fα (y)dy
Rn
2
= kK fα k2 ≤ Ckfα k2 = Ckf k2 .
Next we will prove that operator T defined by (4.5.13) is also bounded on L2 (Rn ). Since Kα (x, y) satisfies the condition (a) and (b), and K α is bounded on L2 (Rn ), we therefore deduce that Kα is bounded on Lpω (Rn ), where ω ∈ Ap is a weight function and 1 < p < ∞. Thus for D α A ∈
230
Chapter 4. Oscillatory Singular Integrals
BMO(Rn ), the corresponding commutator [D α A, Kα ] is bounded on Lp (Rn ). So X 1 [D α A, Kα ] KD m A = α! |α|=m
is also bounded on Lp (Rn ). Since T = T + KDm A and T is bounded on L2 (Rn ), it follows that T is bounded on L2 (Rn ). Thus it follows from Proposition 4.5.1 that TA is bounded on Lp (Rn ) with 1 < p ≤ 2. From Lemma 4.5.1 with D α A replacing A and R1 (D α A; x, y) = D α A(x)− α D A(y), it follows that the corresponding truncated operator X 1 Z m Kα (x, y) D α A(x) − D α A(y) f (y)dy SD A f (x) = α! |x−y|<1 |α|=m
is also bounded on Lp (Rn ). Hence by Theorem 4.5.1, we have that X 1 [D α A, Kα ] α!
|α|=m
is also bounded on Lp (Rn ). By TA = T A −
X 1 [D α A, Kα ], α!
|α|=m
we obtain that TA is bounded on Lp (Rn ) with 1 < p ≤ 2. This together with Proposition 4.5.2 yields our desired result.
4.6
Notes and references
Theorem 4.1.1 is due to Ricci and Stein [RiS]. Early form of Theorem 4.1.1 with P (x, y) being a bilinear form was first obtained by Phong and Stein [PhS1],[PhS2]. And Theorem 4.1.2 is due to Chanillo and Christ [ChaC]. The weighted version of Theorem 4.1.2 was established by Sato [Sa]. The proof of Theorem 4.1.2 is taken from [Sa]. For real Hardy spaces H p (Rn ) with 0 < p < 1, Pan [Pa] gave a counter example to show that there exists a non-trivial polynomial P (x, y) such that the corresponding oscillatory singular integral T is unbounded on H p (Rn ). For the case p = 1, we do not know whether T maps H 1 (Rn ) into L1 (Rn ). As a substitution result for p = 1, Lu and Yang [LuYa] showed that T maps
4.6 Notes and references
231
Herz type Hardy spaces into Herz spaces. Moreover, as a development of the result in [LuYa], Li [Li] gave a systematic work in the case 0 < p < 1. Theorem 4.2.1 and Theorem 4.2.2 are due to Lu and Zhang [LuZ1]. It should be pointed out that we do not need the condition (2) on the kernel in Theorem 4.2.1, but need it in Theorem 4.2.2. Theorem 4.2.3 and Theorem 4.2.4 are due to Jiang and Lu [JiL]. As we mentioned in [JiL], the condition (4) in Theorem 4.2.4 is sharp in the following sense: (4) can not be replaced by any Ω ∈ L(log + L)ε (Sn−1 ) with 0 < ε < 1. In addition, the weighted version of Theorem 4.2.4 was established by Ojanen [Oj]. Moreover, Lu and Wu [LuW1] proved that the conclusion of Theorem 4.2.4 still holds if the condition (4) in Theorem 4.2.4 is replaced by Ω ∈ B q0,0 (Sn−1 ), where Bq0,0 (Sn−1 ) is certain block space (see [LuTW]). In addition, Lu and Wu [LuW2] established a result similar to that of Lu and Wu [LuW1] for commutators of oscillatory singular integrals with rough kernels in certain block spaces. It should be also pointed out that whether the conclusion of Theorem 4.2.4 holds if the condition (4) in Theorem 4.2.4 is replaced by Ω ∈ H 1 (Sn−1 ) is still a problem, where H 1 (Sn−1 ) is the Hardy space on the unit sphere. However, Fan and Pan [FaP1] proved that the above conclusion holds when P (x, y) = P (x − y). Theorem 4.3.1 is due to Ricci and Stein [RiS]. As the weighted version of Theorem 4.3.1, Theorem 4.3.2 and its proof given here are due to Lu and Zhang [LuZ2]. Theorem 4.4.1 is due to Chen, Hu and Lu [CheHL]. Lemma 4.4.1 is due to Cohen and Gosselin [CohG1]. A part idea in the proof of Lemma 4.4.2 comes from [CohG2]. The weighted version of Theorem 4.4.1 was obtained by Ding, Lu and Yang [DiLYa]. There are two results that improved Theorem 4.4.1. One of them was obtained by Chen and Lu [CheL], where the condition (3) on the kernel in Theorem 4.4.1 was replaced by a weaker condition Ω ∈ L(log + L)2 (Sn−1 ). Another result was obtained by Lu and Wu [LuW2], where the condition (3) in the theorem was replaced by a weaker condition Ω ∈ Bq0,1 (Sn−1 ). Here Bq0,1 (Sn−1 ) is a block space on the unit sphere (see [LuTW]). The weighted boundedness for high order commutators of oscillatory singular integrals was established by Ding and Lu [DiL2]. Theorem 4.5.1 is due to Lu [Lu2]. As a direct application of Theorem 4.5.1, Theorem 4.5.2 was established by Lu and Yan [LuY]. For results on Lp boundedness of the maximal operator corresponding
232
Chapter 4. Oscillatory Singular Integrals
to oscillator singular integrals and multilinear oscillatory singular integrals, we did not mention them in this book, and refer to Lu and Zhang [LuZ1] and Yan, Meng and Lan [YML] respectively . For other related results on oscillatory singular integrals and multilinear oscillatory singular integrals, we refer to a survey paper by Lu [Lu3].
Chapter 5
Littlewood-Paley Operator We know that if f ∈ L2 (0, 2π), then by the Parseval equality we have 1 2π
Z
2π 0
|f (x)|2 dx =
where for k ∈ Z and ˆ = 1 f(k) 2π
Z
2π
2 X ˆ , f(k)
(5.0.1)
k∈Z
f (x)e−ikx dx
0
is called the Fourier coefficient of f . Similarly, by the Plancherel equality we have kf k2 = kfˆk2 (5.0.2) for f ∈ L2 (Rn ), where ˆ = 1 f(ξ) 2π
Z
f (x)e−2πix·ξ dx
Rn
is called Fourier transform of f . However, for 1 < p < ∞, p 6= 2, (5.0.1) (or (5.0.2)) does not hold for f ∈ Lp (0, 2π) (or f ∈ Lp (Rn )). The Littlewood-Paley theory, originated in the 1930s and developed in the late 1950s, is a very effective replacement. It has played a very important role in harmonic analysis, complex analysis and PDE. 233
234
5.1
Chapter 5. Littlewood-Paley Operator
Littlewood-Paley g function
For f ∈ Lp (Rn ), 1 < p < ∞, we set u(x, t) to be the Poisson integral of f , i.e., Z u(x, t) = Pt (y)f (x − y)dy, Rn
where
Pt (y) = cn
t (t2 + |y|2 )
n+1 ,
cn =
2
Γ( n+1 2 ) . π (n+1)/2
Then the Littlewood-Paley g function of f is defined by g(f )(x) = where
Z
∞ 0
2
|∇u(x, t)| tdt
1/2
,
2 n X ∂u ∂u 2 |∇u(x, t)| = + ∂xj . ∂t j=1
Stein proved that g function can characterize L p (Rn ) (1 < p < ∞). Theorem 5.1.1 Suppose that f ∈ Lp (Rn ), 1 < p < ∞. Then g(f ) ∈ Lp (Rn ), and there exist constants A, B such that Akf kp ≤ kg(f )kp ≤ Bkf kp
(5.1.1)
for every f ∈ Lp (Rn ). For the proof of this theorem, Stein [St3]. We shall discuss the properties of more general Littlewood-Paley g function. Definition 5.1.1 Suppose that ϕ(x) ∈ L 1 (Rn ) satisfies Z ϕ(x)dx = 0.
(5.1.2)
Rn
Then the generalized Littlewood-Paley g function g ϕ is defined by (it is still called as the Littlewood-Paley g function for convenience) gϕ (f ) =
Z
0
where ϕt (x) =
1 x ϕ . tn t
∞
|ϕt ∗ f (x)|
2 dt
t
1/2
,
235
5.1 Littlewood-Paley g function
In Section 2.1, we have obtained the following result by the boundedness of vector-valued singular integral operator (see Theorem 2.1.11), i.e., Theorem 5.1.2 Suppose that ϕ(x) ∈ L1 (Rn ) satisfies (5.1.2) and the following properties: |ϕ(x)| ≤ Z
Rn
C , (1 + |x|)n+α
x ∈ Rn
|ϕ(x + h) − ϕ(x)|dx ≤ C|h|α ,
h ∈ Rn
(5.1.3) (5.1.4)
where C and α > 0 are both independent of x and h. Then g ϕ is of type (p, p) (1 < p < ∞) and of weak type (1, 1). Next we will show that gϕ can also characterize Lp (Rn ) (1 < p < ∞). In other words, we can also obtain (5.1.1) for g ϕ . Theorem 5.1.3 Suppose that ϕ(x) ∈ L1 (Rn ) satisfies (5.1.2)-(5.1.4). Then there exist constants A and B such that Akf kp ≤ kgϕ (f )kp ≤ Bkf kp
(5.1.5)
for all f ∈ Lp (Rn ), 1 < p < ∞. Proof. Clearly we merely need to prove the first inequality in (5.1.5). Let us first begin with an identity. For all f, h ∈ L 2 (Rn ), we have Z Z dxdt ¯ ¯ ϕt ∗ f (x) · ϕt ∗ h(x) =C f (x) · h(x)dx. (5.1.6) t Rn+1 Rn Apply (2.1.43) and set C= where ξ ∈
Rn
Z
∞ 0
2 |ϕ(tξ)| b
dt , t
and C is independent of ξ. Then the left side of (5.1.6) is Z ∞Z dt ¯ lim ϕt ∗ f (x) · ϕt ∗ h(x)dx ε→0 ε t n Z ∞R Z dt 2b b ¯ f (ξ) · h(ξ)dξ = lim |ϕ(tξ)| b ε→0 ε t n R Z Z ∞ dt b ¯ b · h(ξ) · lim = f(ξ) |ϕ(tξ b 0 )| dξ ε→0 t Rn ε|ξ| Z b ¯ b · h(ξ)dξ, =C f(ξ) Rn
236
Chapter 5. Littlewood-Paley Operator
from which (5.1.6) follows. T Now we turn to the proof of (5.1.5). For f ∈ L 2 Lp (Rn ), we have Z ¯ , kf kp = sup f (x)h(x)dx Rn T 0 where the supremum is taken over all h ∈ L 2 Lp (Rn ) with khkp0 ≤ 1. Then (5.1.6) and the second inequality of (5.1.5) yield Z Z dxdt ¯ ¯ n f (x)h(x)dx = C n+1 ϕt ∗ f (x) · ϕt ∗ h(x) t R ZR ¯ gϕ (f )(x) · gϕ (h)(x)dx ≤C Rn
¯ p0 ≤ Ckgϕ (f )kp · kgϕ (h)k
≤ CBkgϕ (f )kp . T p p (Rn ). Since L2 Assume that f ∈ L L is dense in Lp , we can take {fk } ⊂ T L2 Lp such that fk converges to f in the sense of Lp norm. Hence we also have gϕ (fk ) → gϕ (f ) (k → ∞) in Lp and kf kp = lim kfk kp ≤ C lim kgϕ (fk )kp = CBkgϕ (f )kp . k→∞
k→∞
This finishes the proof of Theorem 5.1.3. In the following, we give the discrete form of Theorem 5.1.3. Theorem 5.1.4 Suppose that ψ ∈ S (Rn ) and ψ(0) = 0. For j ∈ Z, define an operator Sj by (Sj f )∧ (ξ) = ψ(2−j ξ)fb(ξ). Then
1/2
X
|Sj f |2 ≤ Cp kf kp
j
(5.1.7)
p
for 1 < p < ∞. Moreover, if X ψ(2−j ξ) 2 = C
(5.1.8)
j
for all ξ 6= 0, then
1/2
X
|Sj f |2 . kf kp ≤ Cp0
j p
(5.1.9)
237
5.1 Littlewood-Paley g function
Proof. By Theorem 2.1.11 and Remark 2.1.6 we know that (5.1.7) holds. On the other hand, by (5.1.8) we have Z X ¯ Sj f (x) · Sj h(x)dx Rn j XZ ¯ ∧ (ξ)dξ (Sj f )∧ (ξ) · (Sj h) = n R j XZ b ·b ¯h(ξ)dξ (5.1.10) |ψ(2−j ξ)|2 f(ξ) = n R jZ b b · h(ξ)dξ ¯ f(ξ) =C n ZR ¯ f (x)h(x)dx =C Rn
T T 0 for every f, h ∈ L2 (Rn ). When f ∈ L2 Lp (Rn ) and h ∈ L2 Lp (Rn ), by (5.1.10) we have Z Z X ¯ ¯ = C f (x)h(x)dx Sj f (x) · Sj h(x)dx Rn Rn j 1/2 1/2 Z X X 2 ¯ ≤ dx |Sj f (x)|2 |Sj h(x)| Rn
j
j
1/2 1/2
X
X
2 2 ¯ |Sj h| |Sj f | ≤
0
j
j p p
1/2
X
|Sj f |2 khkp0 , ≤ Cp0
j p
where (5.1.7) is used. Thus (5.1.9) holds for f ∈ L p cases, (5.1.9) follows immediately by taking limit.
T
L2 (Rn ). For general
Remark 5.1.1 Let ϕ(x) be a nonnegative radial Schwartz function such that ϕ(x) = 1 when |x| ≤ 12 and ϕ(x) = 0 when |x| ≥ 1. Define π − ϕ(x), ψ 2 (x) = ϕ 2 P −j 2 then j |ψ(2 )| = 1 for x 6= 0. Therefore, we can choose such a ψ to satisfy the condition of Theorem 5.1.4.
238
Chapter 5. Littlewood-Paley Operator
Next we give an application of the former result, which is very effective in dealing with operators with rough kernels. Theorem 5.1.5 Suppose that {σj }j∈Z is a sequence of finite Borel measures and satisfies the following conditions: (i) there exists C > 0 such that kσj k ≤ C for every j; (ii) there exists C, α > 0 such that α −α |σbj (ξ)| ≤ C min 2j ξ , 2j ξ
for every j; (iii) there exists 1 < p0 < ∞ such that the maximal operator σ ∗ (f ) = sup ||σj | ∗ f | j
is bounded on Lp0 . Then operators T (f ) =
X j
and
S(f ) =
X j
σj ∗ f 1/2
|σj ∗ f |2
1 1 1 . are both bounded on Lp for every p satisfying − < p 2 2p0
Proof. Take a nonnegative radial function ψ ∈ C 0∞ (Rn ) such that n 1 supp(ψ) ⊂ x ∈ R : ≤ |x| ≤ 2 2 and
X j
For f ∈
Then
S (Rn ),
ψ 2 (2−j x) = 1 (x 6= 0).
define Sj by (Sj f )∧ (ξ) = ψ 2−j ξ fb(ξ),
T f (x) =
X j
σj ∗
X k
2 Sj+k f
!
j ∈ Z.
(x) :=
X k
Tk f (x),
239
5.1 Littlewood-Paley g function where Tk f (x) =
X j
Sj+k (σj ∗ Sj+k f )(x).
By the Plancherel theorem and the condition (ii) we know (note the condition of the support of ψ)
kTk f k22
2
X
∧ = (Sj+k (σj ∗ Sj+k f )) (·)
j
2 2 Z X 2 −j−k b dξ f (ξ) σ b (ξ) · ψ 2 ξ = j Rn j Z X 2−2α|k| |fb(ξ)|2 dξ ≤C ≤
(5.1.11)
|ξ|∼2j+k j −2α|k| C2 kf k22 .
q 0 1 1 1 0 − = , then p0 = . Now we will show that, it 2 q0 2p0 2 implies {σj ∗ hj } ∈ Lq0 (l2 ) and Let q0 satisfy
1/2
X
|σj ∗ hj |2
j
q0
1/2
X
|hj |2 ≤C
j
(5.1.12)
q0
for every {hj } ∈ Lq0 (l2 ). In fact, by H¨older’s inequality it is easy to deduce that |σj ∗ hj (x)|2 ≤ kσj k · |σj | ∗ |hj |2 (x) . (5.1.13) Since
1/2
X
2 |σj ∗ hj |
j
q0
Z X 2 |σj ∗ hj (x)| u(x)dx , = sup u Rn j
240
Chapter 5. Littlewood-Paley Operator
where the supremum is taken over all functions u ∈ L p0 (Rn ) with kukp0 ≤ 1, then by (5.1.13) and H¨older’s inequality we have Z X XZ 2 |σ ∗ h (x)| u(x)dx ≤ C |σj | ∗ |hj |2 (x)|u(x)|dx j j n Rn R j j Z X |hj (x)|2 σ ∗ (u)(x)dx ≤C j
Rn
1/2
2
X
≤ C |hj |2 kσ ∗ (u)kp0 .
j
q0
Thus (5.1.12) follows from the condition (ii). 0 For any g ∈ Lp0 (Rn ), from (5.1.12), Theorem 5.1.4 and H¨older’s inequality it follows that Z Z X (σj ∗ Sj+k f )(x)(Sj+k g)(x)dx n Tk f (x)g(x)dx = R j Rn 1/2 1/2 Z X X ≤ |σj ∗ Sj+k f |2 |Sj+k g(x)|2 dx Rn
j
j
1/2
X
2 |σj ∗ Sj+k f | ≤C
kgkp00
j p0
1/2
X
≤ C |Sj+k f |2 kgkp00
j
p0
≤ Ckf kp0 kgkp00 .
Thus we have kTk f kp0 ≤ Ckf kp0 .
(5.1.14)
Applying the Riesz-Th¨orin interpolation theorem to (5.1.11) and (5.1.14), we have that kTk f kp ≤ C2−αθ|k|kf kp , 1 1 1 . Therefore we obtain that T where 0 < θ < 1 and p satisfies − < p 2 2p0 is of type (p, p).
241
5.2 Weighted Littlewood-Paley theory
Finally, we will prove that S is also of type X (p, p). Choose a sequence ε = {εj } such that εj = ±1. Set Tε f (x) = εj σj ∗ f (x), then the above j
discussion implies
kTε f kp ≤ Cp kf kp 1 1 1 for − < and Cp independent of ε. p 2 2p0 On the other hand, if we set F =
∞ X j=0
(5.1.15)
aj rj ∈ L2 ([0, 1]) ,
then F ∈ Lp ([0, 1]) (1 < p < ∞), where {rj (t)} is the Rademacher function system (see [St3]). Besides, there exist A p , Bp > 0 such that
Thus
Ap kF kp ≤ kF k2 =
∞ X j=0
1
2
|aj |
2
≤ Bp kF kp .
p p 2 p rj (t)σj ∗ f (x) dt. |σj ∗ f (x)| ≤ Bp S(f )(x) = 0 j j (5.1.16) Integrating both sides of (5.1.16) with respect to x and applying the Fubini theorem as well as (5.1.15), we can derive that S is also of type (p, p). This completes the proof of Theorem 5.1.5.
X
p/2
Z
1 X
The conclusions of the generalized Littlewood-Paley function g ϕ and its discrete form given in this section (see Theorem 5.1.2, Theorem 5.1.3 and Theorem 5.1.4) can be used to discuss the boundedness of Fourier multiplier.
5.2
Weighted Littlewood-Paley theory
In this section, we will discuss weighted Littlewood-Paley theory. That is, we will give the weighted form of Theorem 5.1.3 and Theorem 5.1.4. In order to derive the above conclusions, we will use the weighted normed inequality of vector-valued singular integral operator. More precisely, we will give the weighted form of Lp boundedness (Theorem 2.1.9) of the related vectorvalued singular integral operator in Section 2.1, then finish its proof by the
242
Chapter 5. Littlewood-Paley Operator
dual property. For the definitions and notations of vector-valued function and vector-valued singular integral operator, see Section 2.1. We say that a vector-valued kernel K satisfies (D r ) (1 ≤ r ≤ ∞) condition, if there exists ∞ X a sequence {ck }∞ such that ck = Dr (K) < ∞ and k=1 k=1
Z
Sk (|y|)
kK(x − y) − K(x)krL (A,B) dx
!1/r
≤ ck |Sk (|y|)|−1/r
0
(Dr )
for every k ≥ 1 and y ∈ Rn , where n o Sk (|y|) = x ∈ Rn : 2k |y| < |x| ≤ 2k+1 |y| .
Remark 5.2.1 (i) If r = 1, then the condition (D1 ) is just the H¨ ormander condition (2.1.38). (ii) If |x| > 2|y| and kK(x − y) − K(x)kL (A,B) ≤ C|y|α |x|−n−α (α > 0), then K satisfies the condition (D∞ ). (iii) H¨ older’s inequality implies that if 1 ≤ r ≤ s ≤ ∞, then (D s ) ⊂ (Dr ). Next we will give the weighted Lp boundedness of vector-valued singular integral operator. Theorem 5.2.1 Suppose that A and B are both reflexive Banach spaces. If there exists 1 < p0 < ∞ such that the vector-valued singular integral operator T is bounded from Lp0 (A) to Lp0 (B), then T can be extended to a bounded operator from Lpω (A) to Lpω (B) when K(x) satisfies the condition (Dr )(1 < r < ∞) and one of the following statements: (i) r 0 ≤ p < ∞ and ω ∈ Ap/r0 ; (ii) 1 < p ≤ r and ω ∈ Ap0 /r0 ; 0 (iii) 1 < p < ∞ and ω r ∈ Ap . Here, for 1 ≤ p < ∞ and nonnegative weight function ω, define L pω (A) by ) ( Z 1/p <∞ . Lpω (A) = f : kf kLpω (A) = kf (x)kpA ω(x)dx Rn
First we give the following lemma.
5.2 Weighted Littlewood-Paley theory
243
Lemma 5.2.1 Suppose that K satisfies the condition (D r ) (1 < r < ∞) and T is bounded from Lp0 (A) to Lp0 (B). Then there exists a constant C > 0 such that 1/r0 0 M ] (kT f kB )(x) ≤ C M kf krA (x) (5.2.1) n ∞ for every f ∈ L∞ c (A) and x ∈ R . Here, Lc (A) represents the set of all Avalued bounded functions with compact supports, M is the Hardy-Littlewood maximal operator and M ] is the Sharp maximal operator.
e = 2U. Set Proof. Choose x0 ∈ Rn . Let U be a ball centered at x0 and U Z b0 = K(x0 − y)f (y)dy ∈ B. e )c y∈(U
Put f1 = f χUe , f2 = f − f1 , then by Theorem 2.1.9 we know that T is also 0 0 bounded from Lr (A) to Lr (B). Thus Z 1 kT f (x) − bU kB dx |U | U Z Z 1 1 ≤ kT f1 (x)kB dx + kT f2 (y) − bU kB dy |U | U |U | U 1/r0 0 ≤ C M kf krA (x0 ) Z + sup kK(x − y) − K(x0 − y)kL (A,B) kf (y)kA dy. x∈U
e )c y∈(U
Since K satisfies (Dr ) condition, we have Z sup kK(x − y) − K(x0 − y)kL (A,B) kf (y)kA dy e )c y∈(U ∞ X
x∈U
≤
k=1
1/r0 0 . ck M kf krA (x0 )
By the discussion above we have proved (5.2.1). Hence this completes the proof of Lemma 5.2.1. Now we turn to the proof of Theorem 5.2.1. First we show this theorem holds when r 0 < p < ∞ and ω ∈ Ap/r0 . Using the method in the proof of Theorem 2.2.3, we can get M (kT f kB ) ∈ Lp (ω)
244
Chapter 5. Littlewood-Paley Operator
for f ∈ L∞ c (A). Thus, applying Lemma 2.1.3, Lemma 5.2.1 and Theorem 1.4.3 we have Z Z h ip p kT f (x)kB ω(x)dx ≤ C M ] (kT f kB )(x) ω(x)dx Rn Z Rn h ip/r0 0 (5.2.2) ≤C M kf krA (x) ω(x)dx Z Rn kf kpA ω(x)dx ≤C Rn
0 for every f ∈ L∞ c (A). If p = r , then ω ∈ A1 . Hence there exists p1 > p such that p1 ω p ∈ A1 ⊂ Ap1 /r0 .
Consequently, there exists ε > 0 such that ω p1 (1+ε)/p ∈ Ap1 /r0 . Now take ε and pe1 such that θ= 1+ε 1 θ 1−θ = + . p pe1 p1
It is clear that 1 < pe1 < p < p1 . Then (5.2.2) yields kT f kLpu1 (B) ≤ Ckf kLpu1 (A) ,
(5.2.3)
where u = ω p1 (1+ε)/p . On the other hand, by Theorem 2.1.9 we know that kT f kLpf1 (B) ≤ Ckf kLp1 (A) .
Applying the Stein-Weiss interpolation theorem with change of measure (Lemma 2.2.5) to the above formula and (5.2.3) we see that T is bounded from Lpω (A) to Lpω (B). Therefore we obtain the conclusion under the condition (i). Now we will show that it also holds under the condition (ii) by applying the dual method. Denote by Te the conjugate operator of T . Then e satisfies K(x) e e satisfies all conditions on K. K = K(−x). It is clear that K 0 Since both A and B are reflexive, it follows that (L p (A))∗ = Lp (A∗ ) and 0 (Lp (B))∗ = Lp (B ∗ ). Using the idea in the proof of Theorem 2.2.3, we can finish the proof of Theorem 5.2.1. Finally, the conclusion of Theorem 5.2.1 under the condition (iii) follows from the idea in the proof of Theorem 2.2.4. Thus we have finished the proof of Theorem 5.2.1. Remark 5.2.2 The restriction ”both A and B are reflexive” can be removed (see [RuRT]) .
245
5.2 Weighted Littlewood-Paley theory
In Section 5.1 we observe that, if ϕ ∈ L 1 satisfies (5.1.2)-(5.1.4), then the generalized Littlewood-Paley function g ϕ can characterize Lp (Rn ) (1 < p < ∞). It will be shown later that if we strengthen the condition of ϕ, then gϕ also characterizes weighted Lp spaces (1 < p < ∞). Theorem 5.2.2 Suppose that ϕ ∈ L1 (Rn ) satisfies (5.1.2) and the following conditions C |ϕ(x)| ≤ , (5.2.4) (1 + |x|)n+1 C |∇ϕ(x)| ≤ . (5.2.5) (1 + |x|)n+2
For 1 < p < ∞, if ω ∈ Ap , then there exist constants C1 and C2 , independent of f , such that C1 kf kp,ω ≤ kgϕ (f )kp,ω ≤ C2 kf kp,ω
(5.2.6)
for every f ∈ Lp (ω). Proof.
Let us first prove the second of (5.2.6) by applying inequality dt 2 . For 1 < q < ∞, it follows Theorem 5.2.1. Take A = C, B = L R+ , t q from Theorem 5.1.2 that gϕ is bounded from L (A) to Lq (B). Since ω ∈ Ap , there exists ε > 0 such that ω 1+ε ∈ Ap . If we set r 0 = 1 + ε, then 1 < r < ∞. Note that Z ∞ 1/2 2 dt |∇ϕt (x)| k∇ϕt (x)kL (A,B) = t 0 2 !1/2 Z |x| 1 dt x ≤ tn+1 ∇ϕ( t ) t 0 (5.2.7) 2 !1/2 Z ∞ dt 1 x + tn+1 ∇ϕ( t ) t |x| C . ≤ |x|n+1 By (5.2.9), we have that kϕt (x − y) − ϕt (x)kL (A,B) ≤ C
|y| |x|n+1
when |x| > 2|y|. Thus by Remark 5.2.1 (ii) & (iii) we know that ϕ t satisfies the condition (Dr ). Then the second inequality of (5.2.6) follows from Theorem 5.2.1.
246
Chapter 5. Littlewood-Paley Operator Now for every f ∈ L2
T
Lp (ω), we have Z ¯ , f (x)h(x)dx kf kp,ω = sup n
(5.2.8)
R
where the supremum is taken over all functions h ∈ L 2 fying khkp0 ,ω−p0 /p ≤ 1. The property of Ap implies that
T
0
0
Lp (ω −p /p ) satis-
0
ω ∈ Ap ⇐⇒ ω −p /p ∈ Ap0 . Applying (5.1.6) and the above conclusion, we have that Z Z Z ∞ dtdx ¯ ¯ ϕt ∗ f (x) · ϕt ∗ h(x) n f (x)h(x)dx = n C t R 0 R Z ¯ ≤C gϕ (f )(x)gϕ (h)(x)dx Rn
¯ 0 −p0 /p ≤ Ckgϕ (f )kp,ω kgϕ (h)k p ,ω ≤ Ckgϕ (f )kp,ω .
By (5.2.8) and the density, we get the first inequality of (5.2.6), and then finish the proof of Theorem 5.2.2. The discrete form of Theorem 5.2.2 is the following conclusion, whose proof can be completed by applying Theorem 5.2.1 and the idea in the proof of Theorem 5.1.4. We omit it here. Theorem 5.2.3 Suppose that ψ ∈ C0∞ (Rn ) is nonnegative and satisfies n 1 supp(ψ) ⊂ x ∈ R : < |x| < 2 , 2 as well as
X j∈Z
ψ 2 2−j x = 1 (x 6= 0).
Set (Sj f )∧ (ξ) = ψ(2−j ξ)fb(ξ). Then there exist constants C1 and C2 such that
1
2
X
2 (5.2.9) C1 kf kp,ω ≤ |δj f (·)|
≤ C2 kf kp,ω
j∈Z
p,ω
for 1 < p < ∞ and ω ∈ Ap .
5.3 Littlewood-Paley g function with rough kernel
247
Remark 5.2.3 (i) If ϕ ∈ S (R) satisfies (5.1.2), then Theorem 5.2.2 holds clearly. (ii) The method in the proof of Corollary 2.1.5, together with Theorem 5.2.1 and Theorem 5.2.2, yields that the weighted vector-valued inequalities corresponding to (2.1.44) and (2.1.45) still hold for 1 < p < ∞, ω ∈ A p and f~ = (· · · , f−1 , f0 , f1 , · · · ) ∈ Lp (`2 )(Rn ). Here we omit the details.
5.3
Littlewood-Paley g function with rough kernel
In the first section of this chapter we have seen that, when ϕ satisfies the vanishing condition (5.1.2), the size condition and the smooth condition (5.1.4), the Littlewood-Paley g function g ϕ can characterize the space Lp (Rn ) (1 < p < ∞). Especially, gϕ is of type (p, p) (1 < p < ∞) and of weak type (1, 1). Note that gϕ can be regarded as a vector-valued singular integral operator. Thus the discussion about singular integral operators with rough kernels motivates us to put forward the following question: Can we weaken the size condition and the smooth condition on ϕ but still keep the (p, p) boundedness and the weak (1, 1) boundedness of g ϕ ? In this section we will consider these problems. Early in the 1958, while Stein presented and researched Littlewood-Paley g function in higher dimension, he defined the so called Marcinkiewicz integral. Let Ω(x) be a homogeneous function of degree zero on R n (n ≥ 2), i.e., Ω(λx) = Ω(x) (5.3.1) for every λ > 0 and x ∈ Rn , and it satisfies the vanishing condition Z Ω(x0 )dσ(x0 ) = 0. (5.3.2) Sn−1
If Ω(x0 ) ∈ L1 (Sn−1 ), then the Marcinkiewicz integral µ Ω is defined by Z µΩ (f )(x) =
0
∞ Z
|x−y|≤t
It is easy to verify that, if we take
2 21 dt Ω(x − y) f (y)dy 3 . n−1 t |x − y|
ϕ(x) = Ω(x)|x|−n+1 χB (x)
248
Chapter 5. Littlewood-Paley Operator
with B = {x ∈ Rn : |x| ≤ 1} and χB the characteristic function of B, then µΩ (f ) = gϕ (f ). Therefore µΩ is actually a generalized Littlewood-Paley g function. Stein first proved that, if Ω ∈ Lipα (Sn−1 ) (0 < α ≤ 1), then µΩ is of type (p, p) (1 < p ≤ 2) and of weak type (1, 1). By Theorem 2.3.2 and Theorem 2.3.5, we know that when Ω ∈ H 1 (Sn−1 ) (its definition see Section 2.3) satisfies (5.3.1) and (5.3.2), the singular integral operator T Ω related to Ω is of type (p, p) (1 < p < ∞). Thus, a natural question is, whether µ Ω is also of type (p, p) when Ω satisfies the above conditions. The following theorem gives a positive answer. Theorem 5.3.1 Suppose that Ω satisfies (5.3.1) and (5.3.2). If Ω ∈ H 1 (Sn−1 ), then µΩ is of type (p, p) (1 < p < ∞). Before proving the theorem, we state two lemmas first, their proof can be found in Fan & Pan [FaP2]. For the definition of H 1 (Sn−1 ) atom and H 1 (Rn ) atom can be found in Section 2.3. Now we present some notations. Let a(·) be a regular ∞-atom in H 1 (Sn−1 ) whose support satisfies supp(a) ⊂ Sn−1 When n ≥ 3, set Ea (s, ξ 0 ) = 1 − s2 and when n = 2, set ea (s, ξ 0 ) = 1 − s2
− 1 2
\
n−3 2
B(ξ 0 , ρ),
χ(−1,1) (s)
(0 < ϕ ≤ 2, ξ 0 ∈ Sn−1 ). Z
Sn−2
1 a s, 1 − s2 2 ye dσ(e y ),
h 1 1 i χ(−1,1) (s) a s, 1 − s2 2 + a s, − 1 − s2 2 .
Lemma 5.3.1 There exists a constant c > 0, independent of a, such that cEa (s, ξ 0 ) is an ∞-atom on R. That is, cEα satisfies kcEα k∞ ≤ 1/r(ξ 0 ), supp(Eα ) ⊂ ξ10 − 2r(ξ 0 ), ξ10 + 2r(ξ 0 ) and
Z
Ea (s, ξ 0 )ds = 0,
R
where r(ξ 0 ) = |ξ|−1 |Aρ ξ| and Aρ (ξ) = (ρ2 ξ1 , ρξ2 , · · · , ρξn ).
249
5.3 Littlewood-Paley g function with rough kernel
Lemma 5.3.2 For 1 < q < 2, there exists a constant c > 0, independent of a, such that cea (s, ξ 0 ) is a q-atom on R, the center of whose support is ξ 10 and the radius r(ξ 0 ) = |ξ|−1 (ρ4 ξ12 + ρ2 ξ22 )1/2 . Lemma 5.3.3 Suppose that a is a regular ∞-atom. Set x −t |x|−n+1 χ{|x|≤2t } (x) σ2t (x) = 2 a |x| and σ ∗ f (x) = sup ||σ2t | ∗ f (x)| . t∈R
Then there exists a constant C > 0, independent of t, a and f , such that kσ2t k1 ≤ 1 and kσ ∗ f kp ≤ Ckf kp for 1 < p ≤ ∞. Proof. Clearly we have kσ2t k1 ≤ 1. On the other hand, it is easy to see σ ∗ f (x) ≤ CMa f (x), where Ma is the maximal operator defined by (2.3.1) and C depends only on n. Since a ∈ L1 (Sn−1 ), kσ ∗ f kp ≤ Ckf kp follows from Theorem 2.3.3, where C is independent of t, a and f . Now we turn to the proof of Theorem 5.3.1. Since Ω ∈ H 1 (Sn−1 ) satisfies (5.3.2), by Lemma 2.3.1 we can write X Ω(x0 ) = λj aj (x0 ), j
where every aj is a regular ∞-atom and X |λj | ≤ ckΩkH 1 (Sn−1 ) . j
Thus we merely need to show that there exists a constant c > 0, for each regular ∞-atom a, such that
Z Z 2 1/2
∞ a(x − y)
dt kµa f kp = f (y)dy 3
≤ Ckf kp (5.3.3)
0 |x−y|≤t |x − y|n−1 t
p
250
Chapter 5. Littlewood-Paley Operator
for 1 < p ≤ ∞. Without loss of generality, we may assume \ supp(a) ⊂ Sn−1 B(`, ρ),
where ` = (1, 0, · · · , 0). Consider the case n ≥ 3 first. Chooose a nonnegative and radial C0∞ function ψ such that n 1 0 ≤ ψ ≤ 1, supp(ψ) ⊂ x ∈ R : ≤ |x| ≤ 2 2
and
Z
We define Φ and ∆ by
∞ 0
ψ(t) dt = 1. t
b Φ(ξ) = ψ(Aρ ξ)
and
b ∆(ξ) = ψ(ξ),
respectively. (Here Aρ ξ = (ρ2 ξ1 , ρξ2 , · · · , ρξn ), see Lemma 5.3.1). Write Φt (x) =
1 x Φ , tn t
1 x ∆ . tn t
∆t (x) =
Then it is easy to see that
ct (ξ) = ψ(tAρ ξ), Φ
and
Φt (x) =
1
ρn+1
ct (ξ) = ψ(tξ) ∆ A1 x
t−n ∆
ρ
t
!
.
Clearly for any f ∈ S (Rn ), we have that Z ∞ Z ∞ Φt ∗ f (x) f (x) ∼ dt ∼ Φ2t ∗ f (x)dt. t 0 −∞
(5.3.4)
Set gΦ (f )(x) =
Z
∞ −∞
2
|Φ2t ∗ f (x)| dt
1/2
∼
Z
0
∞
|Φt ∗ f (x)|
2 dt
t
1/2
.
Then there exists a constant C > 0, independent of ρ and f , such that kgΦ (f )kp ≤ Ckf kp .
(5.3.5)
251
5.3 Littlewood-Paley g function with rough kernel In fact, we see that Z y f Aρ A 1 x − y dy = ∆t ∗ h A 1 x , Φt ∗ f (x) ∼ t−n ∆ ρ ρ t Rn where h(x) = f (Aρ x). Note that Z b ∆(x)dx = ∆(0) = ψ(0) = 0. Rn
Applying Theorem 5.1.3 implies that
2 dt p/2 dx kgΦ (f )kp ≤ c ∆t ∗ h A 1 x ρ t Rn 0
Z
2 dt 1/2 ∞ n+1
= cρ p ∆t ∗ h A 1 x
ρ
0
t Z
Z
∞
!1/p
p
≤ cρ
n+1 p
khkp = ckf kp .
Notice that
µa f (x) = ≤c
Z
0
∞ Z
Z
∞
−∞
|x−y|≤t
2 1/2 dt a(x − y) f (y)dy 3 t |x − y|n 2
|σ2t ∗ f (x)| dt
1/2
.
By (5.3.4) and Minkowski’s inequality, we conclude that 1/2 Z ∞ Z ∞ 2 ds |Φ2s+t ∗ σ2t ∗ f (x)| dt µa f (x) ≤ c −∞ −∞ Z ∞ := c Hs f (x)ds. −∞
By Minkowski’s inequality again, we have Z ∞ kµa f kp ≤ c kHs f kp ds.
(5.3.6)
−∞
Under the condition of the theorem, if we constants c > 0 (independent of s, f and a), ( c2−θs kf kp kHs f kp ≤ c2ηs kf kp
can illustrate that there exist θ, η > 0, such that when s > 0, when s < 0,
(5.3.7)
252
Chapter 5. Littlewood-Paley Operator
then (5.3.3) follows from (5.3.6) and (5.3.7). Therefore, the proof of Theorem 5.3.1 can be reduced to the proof of (5.3.7), which is a direct consequence of two estimates below and the Riesz-Th¨orin interpolation theorem of sublinear operators (see Calder´on-Zygmund [CaZ2]): kHs f kp ≤ ckf kp
kHs f k2 ≤
(
c · 2−2s kf k2
(5.3.8) when s > 0,
c · 2s/4 kf k2
(5.3.9)
when s < 0.
In the above two formulas, c > 0 is independent on s, a and f. Let us first prove (5.3.8). For 1 < p < 2, since a is an ∞-atom, it follows that kσ2t k1 ≤ 1 from Lemma 5.3.3. Set Gs+t (x) = Φ2s+t ∗ f (x). Then we have
Z
∞
−∞
Z
σ2t ∗ Gs+t (·)dt ≤
1
∞
−∞
|Gt (·)|dt
.
On the other hand, applying Lemma 5.3.3 yields that
sup |σ2t ∗ Gs+t | ≤ σ ∗ sup |Gt | ≤ c sup |Gt |
t∈R
q
t∈R
q
(5.3.10)
1
t∈R
(5.3.11) q
for 1 < q < ∞. If we set
T Gs+t (x) = σ2t ∗ Gs+t (x),
then (5.3.10) and (5.3.11) show that T is bounded on L 1 L1 (R), Rn and Lq (L∞ (R), Rn ), respectively. For 1 < p < 2, take 1 2 = − 1. q p Then by applying the interpolation theorem mentioned above, we have that T is bounded on Lp (L2 (R), Rn ). That is,
Z
Z 1/2 1/2
∞ ∞
|Gt (·)|2 dt |σ2t ∗ Gs+t (·)|2 dt
,
≤C
−∞
−∞
p
p
5.3 Littlewood-Paley g function with rough kernel
253
where C is independent of s, a and f . This inequality together p 0 with (5.3.5) , then there implies that (5.3.8) holds for 1 < p < 2. If p > 2, set q = 2 q exists ω ∈ L with kωkq ≤ 1, such that Z Z ∞ 2 2 |Φ2s+t ∗ σ2t ∗ f (x)| ω(x)dtdx . kHs f kp = Rn
−∞
By H¨older’s inequality, Lemma 5.3.3 and (5.3.5), we have that kHs f k2p ≤ kgΦ (f )k2p kσ ∗ (|ω|)kq ≤ Ckf k2p .
This formula implies that (5.3.8) holds for p > 2, and C is independent of s, a and f . Now we estimate (5.3.9). By the Plancherel theorem, we have that Z ∞Z 2 |fb(ξ)|2 ψ(2s+t Aρ ξ)2 2−2t |Jt (ξ)|2 dξdt, kHs f k2 ≤ Rn
−∞
where
Jt (ξ) =
Z
|y|≤2t
a(y 0 ) −2πiξ·y e dy. |y|n−1
For ξ 6= 0, choose a rotation O such that O(ξ) = |ξ| · ` = |ξ|(1, 0, · · · , 0). Set y 0 = (u, y20 , · · · , yn0 ), then Jt (ξ) = O −1
Z
2t
Z
a(O −1 y 0 )e−2πir|ξ|u dσ(y 0 )dr,
Sn−1
0
−1 0 where denotes the reverse rotation T of 0O. Clearly, a(O y ) is a regular n−1 B(ξ , ρ). Thus ∞-atom whose support lies in S Z 2t Z Ea (u, ξ 0 )e−2πir|ξ|u dudr. Jt (ξ) = 0
R
By Lemma 5.3.1, we have that Z 2t |Jt (ξ)| ≤ c2 |ξ| |Ea (u, ξ 0 )||u − ξ10 |du ≤ c22t |Aρ ξ|. R
On the other hand, we obtain that Z 2t |ξ| Z dr 0 −2πiru |Jt (ξ)| ≤ C du Ea (u, ξ )e |ξ| R 2 Z 2t |ξ| dr c ≤C Ea (r) . |ξ| 2
(5.3.12)
254
Chapter 5. Littlewood-Paley Operator
Applying H¨older’s inequality, Hausdorff-Young’s inequality and Lemma 5.3.1, we know that 3/4 ≤ c23t/4 |Aρ ξ|−1/4 . |Jt (ξ)| ≤ C|ξ|−1 2t |ξ| kEn k 34 L (R)
This inequality together with (5.3.12) implies n o |Jt (ξ)| ≤ C min 23t/4 |Aρ ξ|−1/4 , 22t |Aρ ξ| .
(5.3.13)
It is clear that the constant C in (5.3.13) is independent of s, a and f. By (5.3.13), when s > 0, we have Z ∞Z b 2 2t kHs f k22 ≤ C f (ξ) 2 |Aρ ξ|2 dξdt ≤ C2
2−s−t−1 ≤|Aρ ξ|≤2−s−t+1 kf k22 .
−∞ −2s
When s < 0, by applying (5.3.13) we obtain that kHs f k22 ≤ c2s/2 kf k22 . Thus we get (5.3.9), and finish the proof of Theorem 5.3.1 when n ≥ 3. For n = 2, we just need to replace Lemma 5.3.1 by Lemma 5.3.2. Remark 5.3.1 (i) Since L(log + L)(Sn−1 ) ( H 1 (Sn−1 ), the conclusion of Theorem 5.3.1 still holds for Ω ∈ L(log + L)(Sn−1 ). (ii) When Ω ∈ H 1 (Sn−1 ) satisfies (5.3.1) and (5.3.2), whether µ Ω is of weak type (1, 1) is still a question. The following theorem will give another result on L p boundedness of Marcinkiewicz integral with rough kernel. Theorem 5.3.2 Suppose that Ω satisfies (5.3.1) and (5.3.2). If Ω ∈ L(log + L)1/2 (Sn−1 ), then µΩ is of type (p, p) (1 < p < ∞). First we state a lemma, which essentially is the continuous form of Theorem 5.1.5. Its proof will be put off after the proof of Theorem 5.3.2. Suppose τ = {τt : t ∈ R} is a family of measures on Rn . Define operators ∆τ and τ ∗ by Z 1/2 2 ∆τ (f )(x) = , (5.3.14) |τt ∗ f (x)| dt R
τ ∗ (f )(x) = sup (|τt | ∗ |f |) (x). t∈R
(5.3.15)
255
5.3 Littlewood-Paley g function with rough kernel
Lemma 5.3.4 Suppose that a ≥ 2, A > 0, γ > 0, q > 1 and C q > 0. Let a family of measures {τt : t ∈ R} on Rn satisfy (i) kτt k ≤ A, t ∈ R; n −1 oγ/ log a (ii) |b τt (ξ)| ≤ A min at |ξ|, at |ξ| , for every ξ ∈ Rn and t ∈ R; (iii) kτ ∗ (f )kq ≤ Cq Akf kq , for every f ∈ Lq (Rn ). Then for p satisfying 1 1 − < 1, p 2 2q there exists a constant Cp , independent of f and a, such that k∆τ (f )kp ≤ Cp kf kp for f ∈ Lp (Rn ). The proof of Theorem 5.3.2. For k ∈ N, Set n o Ek = y 0 ∈ Sn−1 : 2k−1 ≤ |Ω(y 0 )| < 2k
and
0
0
0
Ωk (y ) = Ω(y )χEk (y ) − Then
Z
Sn−1
for all k ∈ N. Let
^
Z
Ω(x0 )dσ(x0 ). Ek
Ωk (x0 )dσ(x0 ) = 0
(5.3.16)
n o = x ∈ N : |Ek | > 2−4k ,
where |Ek | represents the measure of Ek on Sn−1 induced by Lebesgue measure. Set X Ω0 = Ω − Ωk , k∈
V
then it is easy to verify that Ω0 ∈ L2 (Sn−1 ) and Z Ω0 (x0 )dσ(x0 ) = 0. Sn−1
For k ∈
V
, define a family of measures τ (k) = {τk,t : t ∈ R} on Rn by Z Z Ωk (y) −kt f (y)dy. f dτk,t = 2 n−1 Rn |y|≤2kt |y|
256
Chapter 5. Littlewood-Paley Operator
Now set k
ak = 2 , A k = 2
Z
Ek
|Ω(y 0 )|dσ(y 0 ), γ =
log 2 . 6
Then for t ∈ R, ξ ∈ Rn and p > 1, we have the following results: (i) kτk,t k ≤ Ak , γ/ log ak t , (ii) |τc k,t (ξ)| ≤ Ak ak |ξ| γ/ log ak t (iii) | τc , k,t (ξ)| ≤ CAk ak |ξ|
(k) ∗ (iv) τ
p p ≤ C p Ak .
(5.3.17)
L →L
In fact, (5.3.17) (i) is obvious. It follows from (5.3.16) that t |τc k,t (ξ)| ≤ Ak ak |ξ| .
This formula together with |τc k,t (ξ)| ≤ kτk,t k ≤ Ak implies (5.3.17) (ii) (Note that γ/ log ak < 1). Since Z ∗ Ωk y 0 My0 f (x)dσ(y 0 ), τ (k) (f )(x) ≤ Sn−1
where My0 is the Hardy-Littlewood maximal operator along y 0 , we have that (5.3.17) (iv) holds. Next we will show (5.3.17) (iii), and the method is analogous to that in the proof of (2.2.4). By H¨older’s inequality, we obtain that 2 Z 2kt Z 2 −kt 0 −2πirhy 0 ,ξi 0 |τc dσ(y ) dr k,t (ξ)| ≤ 2 n−1 Ωk (y )e S ZZ 0 Ωk (y 0 )Ωk (x0 ) ≤C n−1 n−1 S ×S Z 2kt+j 0 X dr 0 0 × 2j e−2πirhy −x ,ξi dσ(y 0 )dσ(x0 ). r 2kt+j−1 j=−∞
Set
I=
Z
2kt+j 2kt+j−1
0
0
e−2πirhy −x ,ξi
dr , r
then clearly we have |I| ≤ log 2. On the other hand, we see that −1
y 0 − x0 , ξ 0 −1 , |I| ≤ C 2kt+j−1 |ξ|
257
5.3 Littlewood-Paley g function with rough kernel where C is independent of k, i, j, x0 , y 0 and ξ. Thus −α
y 0 − x0 , ξ 0 −α |I| ≤ C 2kt+j−1 |ξ|
for 0 < α < 1. Note that ZZ
Sn−1 ×Sn−1
| hy 0
(5.3.18)
1 dσ(y 0 )dσ(x0 ) ≤ C1 . − x0 , ξ 0 i |α
Thus from (5.3.18) it follows that (5.3.17) (iii) holds. By Minkowski’s inequality, we have X µΩ (f ) ≤ µΩ0 (f ) + (k log 2)1/2 4τ (k) (f ). k∈
V
Since Theorem 5.3.1 implies that µΩ0 is of type (p, p) (1 < p < ∞), by (5.3.17) and Lemma 5.3.4 we obtain that X√ kAk kf kp kµΩ (f )kp ≤ Cp 1 + k∈
V
≤ Cp 1 + kΩkL(log+ L)1/2 kf kp .
Thus we finish the proof of Theorem 5.3.2. Next we give the proof of Lemma 5.3.4. The idea is analogous to that of (5.3.6). Here we merely give the main steps. First take a nonnegative radial function ψ in S (R n ), which satisfies 5a 4 n ≤ |x| ≤ 0 ≤ ψ ≤ 1, supp(ψ) ⊂ x ∈ R : 5a 4 and
Z
∞
ψ(t)
0
dt = 2 log a. t
b ct (ξ) = ψ(|tξ|2 ) and Set Ψ(ξ) = ψ(|ξ|2 ). Then for t > 0, Ψ Z ∞ Ψat ∗ f (x)dt f (x) = −∞
for f ∈ S (Rn ). By Minkowski’s inequality, we have ∆τ (f )(x) ≤
Z Z R
R
2
|Ψas+t ∗ τt ∗ f (x)| dt
1/2
ds.
258
Chapter 5. Littlewood-Paley Operator
If we set Hs (f )(x) =
Z
2
R
|Ψas+t ∗ τt ∗ f (x)| dt
1/2
,
then it suffices to prove that there exist constants C > 0, γ and θ > 0, independent of s and f , such that kHs f kp ≤
(
CAe−sγ kf kp sθ
CAe kf kp
if s > 0 if s < 0.
In order to prove this, we merely need to show kHs f kp ≤ CAkf kp and kHs f k2 ≤
(
CAe−sγ kf k2 sθ
CAe kf k2
(5.3.19)
if s > 0 if s < 0.
(5.3.20)
Applying the condition (i) and (iii) of Lemma 5.3.4, as well as the idea in the proof of (5.3.8), we can obtain (5.3.19). The proof of (5.3.20) can be completed by using the Plancherel theorem and the condition (ii). Remark 5.3.2 H 1 (Sn−1 ) and L(log + L)1/2 (Sn−1 ) both contain L log + L(Sn−1 ) as a subspace. However, there are examples showing that, neither H 1 (Sn−1 ) nor L(log + L)1/2 (Sn−1 ) contains the other. Finally, we state a result on the weighted L p boundedness of Marcinkiewicz integral with rough kernel. Theorem 5.3.3 Suppose that Ω ∈ Lq (Sn−1 ) (q > 1) satisfies (5.3.1) and (5.3.2). If p, q, ω satisfy one of the following conditions (a) q 0 < p < ∞ and ω ∈ Ap/q0 ; 0 (b) 1 < p < q and ω 1−p ∈ Ap0 /q0 ; 0 (c) 1 < p < ∞ and ω q ∈ Ap , then there exists a constant C, independent of f , such that kµ Ω f kp,ω ≤ Ckf kp,ω .
5.4 Notes and references
5.4
259
Notes and references
The Littlewood-Paley g function in one dimension was first introduced by Littlewood and Paley [LiP] in studying the dyadic decomposition of Fourier series. And this theory in high dimension was established by Stein [St1]. Theorem 5.1.1 is due to Stein. The conclusion of Theorem 5.2.1 under the condition (i) or (ii) comes from Rubio de Francia, Ruiz and Torrea [RuRT]. The assumption ”both A and B are reflexive spaces” in the theorem can be removed. The weighted Littlewood-Paley theory was first studied by Kurtz [Ku]. The proof of Theorem 5.2.3 can be found in Watson and Wheeden [WatW]. Theorem 5.3.1 and its proof are taken from Ding, Fan and Pan [DiFP2]. In 2003, Xu, Chen and Ying [XuCY] also gave the proof of Theorem 5.3.1 using another method. For the weak type (1,1) bounds of Marcinkiewicz integrals with rough kernels µΩ , Fan and Sato [FaS] proved that if Ω ∈ L(log + L)(Sn−1 ) satisfies (5.3.1) and (5.3.2), then µ Ω is of weak type (1,1). The conclusion of Theorem 5.3.2 for p = 2 was first proved by Walsh [Wal], and the proof for 1 < p < ∞ was given by Al-Salman, Al-Qassem, Cheng and Pan [AACP]. And Walsh [Wal] showed that if Ω ∈ L(log + L)1/2−ε (Sn−1 ) for 0 < ε < 1/2 and satisfies (5.3.2), then there exists an f ∈ L 2 (Rn ) such that µΩ f ∈ / L2 (Rn ). The weighted boundedness of Marcinkiewicz integral was first studied by Torchinsky and Wang [ToW]. They proved that if Ω ∈ Lipα (Sn−1 )(0 < α ≤ 1) and ω ∈ Ap (1 < p < ∞), then µΩ is bounded on Lp (ω). Theorem 5.3.3 was obtained by Ding, Fan and Pan [DiFP1], and independently by Duoandikoetxea and Seijo [DuS]. In this chapter, we do not mention the boundedness of the commutator generated by Marcinkiewicz integral µΩ and a BM O function b. Here we mere mention two results. Torchinsky and Wang [ToW] proved Lp -boundedness of [µΩ , b] with 1 < p < ∞ if Ω ∈ Lipα (Sn−1 )(0 < α ≤ 1). And Ding, Lu and Yabuta [DiLY] proved Lp -boundedness of the commutator if Ω ∈ L q (Sn−1 )(1 < q ≤ ∞). As space is limited, we do not mention the boundedness of Marcinkiewicz integral with homogeneous kernels on Hardy spaces and Companato spaces. For related results, we refer to Ding, Lu and Xue [DiLX1], [DiLX2].
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Index (Dr ) (1 ≤ r ≤ ∞) condition, 242 A(p, q), v, 138, 139, 141, 150, 163, 166 A1 weights, 31, 33, 34 Ap weights, v, 21, 22, 24, 26, 28, 30, 31, 33, 35, 36, 265 1 n H (R ), 120, 223, 230 H 1 (Sn−1 ), 96, 97, 231, 248, 258 L log+ L(Sn−1 ), 258 L∞ -Dini condition, 47, 79, 83 Lq -Dini condition, 79, 80, 84, 86, 90, 92, 131 Lipα (Sn−1 ), 259 BMO(Rn ), 121, 128
131, 244 dyadic cube, 9, 13, 14, 57, 58, 61, 179, 184 Fefferman-Stein inequality, 17, 19 Fourier transform, 106, 233, 264 fractional integral operator, v, 133, 144, 148, 162, 166, 167, 263, 264 generalized Littlewood-Paley function, 241 H¨older’s inequality, 22–24, 29 H¨ormander condition, 79, 80, 83, 92 Hardy-Littlewood maximal function, 1, 2, 57, 86, 135 Hardy-Littlewood Maximal Operator, 1 Hardy-Littlewood maximal operator, v, 2, 3, 6, 7, 10, 15, 28, 30, 31, 33, 49, 50, 53, 67, 94, 95, 103, 106, 121, 136, 256 Hardy-Littlewood-Sobolev theorem, 136, 144, 159, 209 Hilbert transform, 37–39, 54, 94, 119, 263 homogeneous kernel, v, 48, 78, 79, 92, 94, 106, 144, 147–149, 166, 259, 264
Banach space, 68, 120, 261 bounded overlapping, 140, 208 Calder´on-Zygmund decomposition, 9–11, 13, 14, 30, 35, 179 Calder´on-Zygmund singular integral operator, 37, 40, 41, 54, 67, 106, 119 Cauchy integral, 119, 128, 265 commutator, v, 119, 120, 127–131, 158, 162, 163, 166, 167, 202, 221, 230, 231, 259, 261–266 commutator of fractional integrals with, 162 Cotlar inequality, 49, 67 dual method, 83, 113, 116, 118, 271
272 Jones decomposition, 33 Kolmogorov inequality, 19 Laplacian operator, 38, 39, 133, 134 Lebesgue differentiation theorem, 7, 9, 29, 32 Lipschitz function, 119, 127 Littlewood-Paley g function, 234, 247, 248 Littlewood-Paley Operator, 233 Littlewood-Paley operator, v, 73, 74, 77 Marcinkiewicz integral, vi, 247, 254, 258, 259, 263, 264, 268, 269 Marcinkiewicz interpolation theorem, 15, 17, 19, 31, 36, 71, 83, 142, 147 Minkowski’s inequality, 129, 130, 164, 173, 189, 193, 195, 251, 257 multilinear oscillatory singular integral, 202, 203, 213 Oscillatory singular integral, 169 oscillatory singular integral, v, vi, 169, 186, 194, 197, 213, 231, 263–267 Poisson kernel, 37, 38 polynomial phases, v, 169 reflexive, 71, 242, 244, 259
INDEX Reverse H¨older inequality, 24, 27, 36, 88, 128 Riesz potential, 133, 134, 136, 137, 143, 144, 148, 158, 167 Riesz transform, 37, 39, 40, 54, 68, 96, 99, 132, 134 rotation method, 93–95 rough kernels, v, 132, 162, 167, 186, 202, 203, 231, 238, 247, 259, 262–267 Schwartz function, 237 sharp maximal function, 55, 86 smooth kernel, v, 169, 265 standard kernel, 197, 213 Stein-Weiss interpolation theorem with change of measure, 87, 111, 244 sublinear operator, 5, 15, 19, 87, 252, 261 van der Corput, 170, 175 vanishing condition, 75, 79, 82, 83, 92, 104, 109, 120, 247 vector-valued singular integral operator, 235, 241, 242, 247 Vitali type covering lemma, 3 weak Lp spaces, 5 weighted boundedness, 106, 129, 130, 137, 138, 143, 144, 148, 150, 166, 231, 259, 263
