Open Access

L p Bounds for the parabolic singular integral operator

Journal of Inequalities and Applications20122012:121

https://doi.org/10.1186/1029-242X-2012-121

Received: 7 March 2012

Accepted: 30 May 2012

Published: 30 May 2012

Abstract

Let 1 < p < ∞ and n ≥ 2. The authors establish the L p (n+1) boundedness for a class of parabolic singular integral operators with rough kernels.

MR(2000) Subject Classification: 42B20; 42B25.

Keywords

parabolic singular integral operatorrough kernelsurfaces of revolution

1 Introduction

Let α1,..., α n be fixed real numbers, α i ≥ 1. For fixed x n , the function F x , ρ = i = 1 n x i 2 ρ 2 α i is a decreasing function in ρ > 0. We denote the unique solution of the equation F(x, ρ) = 1 by ρ(x). Fabes and Rivière [1] showed that ρ(x) is a metric on n , and ( n , ρ) is called the mixed homogeneity space related to α i i = 1 n .

For λ > 0, let A λ = λ α 1 0 0 λ α n . Suppose that Ω(x) is a real valued and measurable function defined on n . We say is Ω(x) is homogeneous of degree zero with respect to A λ , if for any λ > 0 and x n
Ω A λ x = Ω x .
(1.1)
Moreover, Ω(x) satisfies the following condition
S n - 1 Ω x J x d σ x = 0 ,
(1.2)

where J(x') is a function defined on the unit sphere Sn-1in n , which will be defined in Section 2.

In 1966, Fabes and Rivière [1] proved that if Ω C1(Sn-1) satisfying (1.1) and (1.2), then the parabolic singular integral operator TΩ is bounded on L p ( n ) for 1 < p < ∞, where TΩ is defined by
T Ω f x = p .v . n Ω y ρ y α f x - y d y and α = i = 1 n α i .

In 1976, Nagel et al. [2] improved the above result. They showed TΩ is still bounded on L p ( n ) for 1 < p < ∞ if replacing Ω C1(Sn-1) by a weaker condition Ω L log+L(Sn-1). Recently, Chen et al. [3] improve Theorem A, the result is

Theorem A. If Ω H1(Sn-1) satisfies (1.1) and (1.2); then the operator TΩis bounded on L p ( n ) for 1 < p < ∞.

For a suitable function ϕ on [0, 1), and Γ = {(y, ϕ(ρ(y)): y n }. Define the singular integral operator Tϕin n+1along Γ by
T ϕ , Ω f x , x n + 1 = p .v . n f x - y , x n + 1 - ϕ ρ y Ω y ρ y α d y ,

where (x, xn+1) n × = n+1.

On the other hand, we note that if α1 = ... = α n = 1, then ρ(x) = |x|, α = n and ( n , ρ) = ( n , |·|). In this case, Tϕis just the classical singular integral operator along surfaces of revolution, which was studied by the authors of [47].

The purpose of this article is to investigate the L p boundedness of the parabolic singular integral operator Tϕalong Γ when Ω F β (Sn-1). For a β > 0, F β (Sn-1) denotes the set of all Ω which are integrable over Sn-1and satisfies
sup ξ S n - 1 S n - 1 | Ω θ | ln 1 | θ ξ | 1 + β d θ < .
(1.3)
Condition (1.3) was introduced by Grafakos and Stefanov [8]. The examples in [8] show that there is the following relationship between F β (Sn-1) and H1(Sn-1):
β > 0 F β S n - 1 H 1 S n - 1 β > 0 F β S n - 1 .

We shall state our main results as follows:

Theorem 1 Let m . Suppose that ϕ is a polynomial of degree m and d α i ϕ t d t α i | t = 0 = 0 , where α i s are the all positive integers which is less than m in {α1,..., α n }. In addition, let Ω F β (Sn-1) for some β > 0 and satisfies (1.1) and (1.2), then Tϕis bounded on L p (n+1) for p 2 + 2 β 1 + 2 β , 2 + 2 β .

Corollary 1 Let m . Suppose that ϕ is a polynomial and d α i ϕ t d t α i | t = 0 = 0 , where α i s are the all positive integers which is less than m in {α1,..., α n }. In addition, let Ω β > 0 F β S n - 1 and satisfies (1.1) and (1.2), then T ϕ, Ω is bounded on L p (n+1) for 1 < p < ∞.

2 Notations and lemmas

In this section, we give some notations and lemmas which will be used in the proof of Theorem 1. For any x n , set
x 1 = ρ α 1 cos φ 1 cos φ n - 2 cos φ n - 1 x 2 = ρ α 2 cos φ 1 cos φ n - 2 sin φ n - 1 x n - 1 = ρ α n - 1 cos φ 1 sin φ 2 x n = ρ α n sin φ 1 .

Then dx = ρα-1J (φ1,..., φn-1)dρdσ, where α = i = 1 n α i , is the element of area of Sn-1and ρα-1J(φ1,..., φn-1) is the Jacobian of the above transform. In [1], it was shown there exists a constant L ≥ 1 such that 1 ≤ J(φ1,..., φn-1) ≤ L and J(φ1,..., φn-1) C((0, 2π)n-2× (0, π)). So, it is easy to see that J is also a C function in the variable y' Sn-1. For simplicity, we denote still it by J(y').

In order to prove our theorems, we need the following lemmas:

Lemma 2.1. ([9]) Let d . Suppose that γ (t): + d satisfies γ t = M γ t t for a fixed matrix M, and assume γ(t) doesn't lie in an affine hyperplane. Then
1 2 e i γ t η d t C | η | 1 / d .
Lemma 2.2. ([9]) Suppose that λ j s and α j s are fixed real numbers, ϕ(t) is a polynomial and Γ t = λ 1 t α 1 , , λ n t α n , ϕ t is a function from +to n+1. For suitable f, the maximal function associated to the homogeneous curve Γ is defined by
M Γ f x = sup h 1 h 0 h | f x - Γ t | d t , h > 0 .
(2.1)
Then for 1 < p ≤ ∞, there is a constant C > 0, independent of λ j s , the coefficient of ϕ(t) and f, such that
| | M Γ f | | L p C | | f | | L p .
(2.2)
Lemma 2.3. Let L : n+1 n be a linear transformation. Suppose that {σ k }kis a sequence of uniformly bounded measures on d satisfying
| σ ^ k ξ | C  min | A 2 k L ξ | , ln | A 2 k L ξ | - 1 - β
(2.3)
for ξ n+1and k . For any 1 < p0<and A > 0
k | σ k * g k | 2 1 / 2 L p 0 A k | g k | 2 1 / 2 L p 0
(2.4)
holds for arbitrary functions {g k }kon n+1. Then for p 2 + 2 β 1 + 2 β , 2 + 2 β there exists a constant C p = C(p, n) which is independent of L such that
k σ k * f L p C p f L p
(2.5)
and
k | σ k * f | 2 1 / 2 L p C | | f | | L p
(2.6)

for every f L p (n+1).

Proof. The main idea of the proof is taken from [7, 8], we assume that = (ξ1,..., ξ n ) = ζ for ξ = (ξ1,..., ξ n , ξn+1) n+1. Choose a ψ C 0 such that 0·ψ ≤ 1, supp(ψ) (1/4, 4), and
j ψ 2 j t 2 1
(2.7)
For each j, we define Φ j in n by
Φ j ^ ζ = ψ 2 j ρ ζ f ^ ζ
for ξ = (ξ1,..., ξn+1) n+1. If we set
T f = k σ k * f ,
(2.8)
and let δ represent the Dirac delta on , then by (2.7), for any Schwartz function f,
T f = j T j f ,
where
T j f = k Φ j + k δ * σ k * Φ j + k δ * f .
By using (2.4) and Littlewood-Paley theory (as in [3]), one obtains that for any 1 < p0< ∞,
| | T j f | | L p 0 n + 1 C p 0 | | f | | L p 0 n + 1 .
(2.9)
On the other hand, by using Plancherel's theorem and (2.3), If j > 0, using the estimate | σ k ^ ξ | C | A 2 k ζ | we have
T j ( f ) L 2 ( n + 1 ) k 2 - j - k - 1 ρ ( ζ ) 2 - j - k + 1 | f ^ ( ξ ) | 2 | A 2 k ζ | 2 d ξ = k 2 - j - k - 1 ρ ( ζ ) 2 - j - k + 1 | f ^ ( ξ ) | 2 2 2 k α 1 ρ ( ζ ) 2 α 1 ( ζ 1 ) 2 + + 2 2 k α n ρ ( ζ ) 2 α n ( ζ n ) 2 d ξ 2 - 2 j  min { α j } k 2 - j - k - 1 ρ ( ζ ) 2 - j - k + 1 | f ^ ( ξ ) | 2 ( ζ 1 ) 2 + + ( ζ n ) 2 d ξ 2 - 2 j k 2 - j - k - 1 ρ ( ζ ) 2 - j - k + 1 | f ^ ( ξ ) | 2 d ξ = C 2 - 2 j | | f | | L 2 ( n + 1 ) .
(2.10)
Similar to the proof of (2.10), using Plancherel's theorem and (2.3), if j < 0 we get
| | T j f | | L 2 n + 1 C 1 + | j | - 1 + β | | f | | L 2 n + 1 .
(2.11)
In short
| | T j f | | L 2 n + 1 C 1 + | j | - 1 + β | | f | | L 2 n + 1 , f o r j .
(2.12)
By interpolating between (2.9) and (2.12), we obtain
| | T j f | | L p n + 1 C 1 + | j | - 1 + β | | f | | L p n + 1 .
(2.13)
for
p 2 + 2 β 1 + 2 β , 2 + 2 β

and some β > 0. Thus, (2.5) follows from (2.13). One may then use a randomization argument to derive (2.6). Lemma 2.1 is proved.

3 Proof of Theorem 1

The main idea of the proof of Theorem 1 is taken from [10]and [11]. Let Ω satisfies (1.1), (1.2), and (1.3) for some β > 0. Let Φ(y) = (y, ϕ(ρ(y))), where ϕ t = j = 0 m a j t j , m . Let D k = {y n : 2 k < ρ(y) ≤ 2k+1} and define the family of measures σ k on n+1by
n + 1 f y , y n + 1 d σ k = D k f y , ϕ ρ y Ω y ρ y α d y ,
(3.1)

and σ* f(x) = supk(|σ k | * | f | (x).

It is easy to see that
| | σ k | | = D k | Ω y | ρ y α d y = S n - 1 2 k 2 k + 1 | Ω y | J y | d ρ ρ d σ y C .
(3.2)

In light of (3.2) and Lemma 2.3, it suffices to show that σ k satisfies (2.3) and (2.4).

For (ξ, ξn+1) n × , y' Sn-1, and λ . Let
I λ ξ , ξ n + 1 , y = 1 2 e i A λ ρ ξ y + ξ n + 1 ϕ λ ρ d ρ .
Set Λ = {α i : α i is the positive integers which is less than m in {α1,..., α n } and Λ ¯ = 1 , 2 , . . . , m \ Λ . Then d α i ϕ t d t α i | t = 0 = 0 , where α i Λ, and Λ ¯ is not a subset of {α1,..., α n }. Therefore, we get
A λ ρ ξ y + ξ n + 1 ϕ λ ρ = ρ α 1 λ α 1 ξ 1 y 1 + . . . + ρ α n λ α n ξ n y n + ξ n + 1 j Λ ¯ a j λ ρ j .
Without loss of generality, we may assume Λ consists of r distinct numbers and let Λ ¯ = i 1 , i 2 , . . . , i m - r If α j s are all distinct, by Lemma 2.1, we get immediately
| I λ ξ , ξ n + 1 , y | | λ α 1 ξ 1 y 1 | + + | λ α n ξ n y n | + m - - r | λ ξ n + 1 | - 1 / n + m - r | λ α 1 ξ 1 y 1 + + λ α n ξ n y n | - 1 / n + m - r = | A λ ξ y | - 1 / n + m - r .
(3.3)
If {αj} only consists of s distinct numbers, we suppose that
α 1 = α 2 = = α l 1 , α l 1 + 1 = = α l 1 + l 2 , α l 1 + + l s - 1 + 1 = = α n ,
where s is a positive integer with 1 ≤ sn, l1, l2,..., l s are positive integers such that l1 + l2 + ··· + l s = n and α 1 , α l 1 + l 2 , , α l 1 + + l s 1 , α n are distinct. Obviously,
γ t = t α 1 , t α l 1 + l 2 , , t α l 1 + + l s - 1 , t α n , t i 1 , t i 2 , , t i m - r
does not lie in an affine hyperplane in s+m-r. Then using Lemma 2.1 again, there exists C > 0 such that for any vector η = (η1,..., η n ) n ,
1 2 e 2 i η 1 + + η l 1 t α l 1 + η l 1 + 1 + + η l 1 + l 2 t α l 1 + l 2 + + η l 1 + + l s - 1 + 1 + + η n t α n + λ ξ n + 1 j Λ ¯ t j d t C | η 1 + + η l 1 | 2 + | η l 1 + 1 + + η l 1 + l 2 | 2 + + | η l 1 + + l s - 1 + 1 + + η n | 2 + m - r | λ ξ n + 1 | 2 - 1 / 2 s + m - r C | η 1 + + η l 1 | + | η l 1 + 1 + + η l 1 + l 2 | + + | η l 1 + + l s - 1 + 1 + + η n | - 1 / s + m - r C j = 1 n η j - 1 / s + m - r .
Let η j = λ α j ξ j y j , we have
| I λ ξ , ξ n + 1 , y | | λ α 1 ξ 1 y 1 | + + | λ α n ξ n y n | - 1 / s + m - r | λ α 1 ξ 1 y 1 + + λ α n ξ n y n | - 1 / s + m - r = | A λ ξ y | - 1 / s + m - r .
(3.3a)
On the other hand, it is easy to see that
| I λ ξ , ξ n + 1 , y | 1 .
(3.4)
From (3.3), (3.3') and (3.4), we get
| I λ ξ , ξ n + 1 , y | C ln 1 / | η y | 1 + β ln | A λ ξ | 1 + β , for | A λ ξ | 2 ,
where η = A λ ξ | A λ ξ | . Thus, by (1.3), we get
S n - 1 | I λ ξ , ξ n + 1 , y Ω y | d σ y C ln | A λ ξ | - 1 + β .
Therefore,
| σ k ^ ξ , ξ n + 1 | = D k e i ξ y + ξ n + 1 ϕ ρ y Ω y ρ y α d y = S n - 1 2 k 2 k + 1 e i ξ A ρ y + ξ n + 1 ϕ ρ Ω y J y d ρ ρ d σ y = S n - 1 1 2 e i ξ A 2 k ρ y + ξ n + 1 ϕ 2 k ρ Ω y J y d ρ ρ d σ y C S n - 1 | I 2 k ξ , ξ n + 1 , y | | Ω y | d σ y C ln | A 2 k ξ | - 1 + β .
(3.5)
On the other hand, by (1.2), we can obtain
| σ k ^ ξ , ξ n + 1 | = D k e i ξ y + ξ n + 1 ϕ ρ y Ω y ρ y α d y = 2 k 2 k + 1 S n - 1 e i ξ A ρ y + ξ n + 1 ϕ ρ Ω y J y d σ y d ρ ρ = 2 k 2 k + 1 S n - 1 e i ξ A ρ y + ξ n + 1 ϕ ρ - e i ξ n + 1 ϕ ρ Ω y J y d σ y d ρ ρ C 2 k 2 k + 1 S n - 1 | e i ξ A ρ y + ξ n + 1 ϕ ρ - e i ξ n + 1 ϕ ρ | | Ω y | | J y | d σ y d ρ ρ C 2 k 2 k + 1 S n - 1 | ξ A ρ y | | Ω y | | J y | d σ y d ρ ρ C 2 k 2 k + 1 S n - 1 | A 2 k ξ y | | Ω y | | J y | d σ y d ρ ρ C | A 2 k ξ | 2 k 2 k + 1 S n - 1 | Ω y | | J y | A 2 k + 1 ξ A 2 k + 1 ξ y d σ y d ρ C | A 2 k ξ | .
(3.6)
Clearly, (3.5) and (3.6) imply (2.3) holds. Finally, we shall show that (2.4) holds.
σ k * f x = sup k | σ k | * | f | x = D k | f x - Φ y | | Ω y | ρ y α d y = S n - 1 2 k 2 k + 1 | f x - Φ A ρ y | | Ω y | d ρ ρ d σ y 1 2 k S n - 1 | Ω y | 2 k 2 k + 1 | f x - Φ A ρ y | d ρ d σ y C S n - 1 | Ω y | M Φ f x d σ y .

By Lemma 2.2, we obtain ||MΦ(f)|| p C||f|| p , where C > 0 is independent of k, the coefficient of ϕ(t) and f, since Ω is integrable on Sn-1, thus ||σ*(f)|| p C||f|| p . This shows (2.4) holds. This completes the proof of the Theorem 1.

Declarations

Acknowledgements

The research was supported by the NSF of China (Grant No. 10901017), NCET of China (Grant No. NCET-11-0574), and the Fundamental Research Funds for the Central Universities.

Authors’ Affiliations

(1)
Department of Applied Mathematics, School of Mathematics and Physics, University of Science and Technology Beijing
(2)
Department of Information and Computer Science, School of Mathematics and Physics, University of Science and Technology Beijing
(3)
Department of Mathematics and Mechanics, University of Science and Technology Beijing

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© Chen et al; licensee Springer. 2012

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