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Lp Bounds for the parabolic singular integral operator

Abstract

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

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

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 dσ 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 Lp ( 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 Lp ( 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 Lp ( 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 Lp 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 sare 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 Lp (n+1) forp 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 sare the all positive integers which is less than m in {α1,..., α n }. In addition, letΩ β > 0 F β S n - 1 and satisfies (1.1) and (1.2), then T ϕ, Ω is bounded on Lp (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 dtC|η | 1 / d .

Lemma 2.2. ([9]) Suppose that λ j sand α j sare fixed real numbers, ϕ(t) is a polynomial andΓ 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 |dt,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 Lp (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 nby

Φ j ^ ζ =ψ 2 j ρ ζ f ^ ζ

for ξ = (ξ1,..., ξn+1) n+1. If we set

Tf= k σ k *f,
(2.8)

and let δ represent the Dirac delta on , then by (2.7), for any Schwartz function f,

Tf= 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.

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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.

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Correspondence to Yanping Chen.

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Authors' contributions

YC carried out the parabolic singular integral operator studies and drafted the manuscript. WY participated in the study of Littlewood-Paley theory. FW conceived of the study, and participated in its design and coordination. All authors read and approved the final manuscript.

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Chen, Y., Wang, F. & Yu, W. Lp Bounds for the parabolic singular integral operator. J Inequal Appl 2012, 121 (2012). https://doi.org/10.1186/1029-242X-2012-121

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