Jiecheng Chen,Shaoyong He and Xiangrong Zhu

Department of Mathematics,Zhejiang Normal University,Jinhua 321004,China

Oscillatory Strongly Singular Integral Associated to the Convex Surfaces of Revolution

Jiecheng Chen,Shaoyong He and Xiangrong Zhu?

Department of Mathematics,Zhejiang Normal University,Jinhua 321004,China

.Here we consider the following strongly singular integral

Oscillatory strongly rough singular integral,rough kernel,surfaces of revolution.

AMS SubjectClassif i cations:42B20,42B35

1 Introduction

The standard Hilbert transform along a curve is def i ned as

where Γ :(?1,1) → Rnis a continuous curve in Rn.The study of these operators was initiated by Fabes and Rivi`ere[7].In[18],Stein and Wainger proved that HΓis bounded on Lp(1＜ p＜∞)if Γ is well-curved in Rn.Here we say that Γ is well-curved,if Γ is smooth with Γ(0)=0 and a segment of the curve containing the origin lies in a subspace of Rnspanned by

When n=2,Γ(t)can be written as(t,γ(t)).If γ(t)is f l at at the origin,i.e.,

for k=1,2,···,then it is easy to see that Γ(t)is not well-curved in R2.The main contributions on the Hilbert transforms along the f l at curves were made by Wainger and his colleagues.Readers can see[1,12–15,19,21,22]among numerous references,in particular, the good survey papers[18]and[23].

Another interesting operator in harmonic analysis is the hyper Hilbert transform

We know that the operator Hαis bounded from the Sobolev spaceto the Lebesgue space Lpfor 1 ＜ p ＜ ∞,because of the mean zero of the kernel of Hα.One naturally expected that,without the assumption of the mean zero on the kernel,the worsened singularity of Hαnearthe origincan be counterbalanced by an oscillatory factor ei|t|?β(β＞0)as t approaches zero.This idea motivated the study of the oscillatory hyper Hilbert transforms(see[10])and the strongly singular integral operators in high dimensional spaces.More details of the strongly singular integral operators can be found in[8,9,16, 20].

Consider the following oscillatory hyper Hilbert transforms,

where Γ(0)=0,β ＞ α.

Zielinski[24]studied the L2?boundedness of HΓ,α,βalong the parabola(t,t2).In[2], for Γ(t)=(t,|t|q),q ≥ 2,Chandarana proved that HΓ,α,βis bounded on L2if and only if β ≥ 3α.When n=3,a similar result was proved in[3].

In[4]and[5],we generalized these results in Rnand removed all assumptions on the indexes.At the same time,Laghi and Lyall in[11]proved that if Γ(t)is well-curved,then HΓ,α,βis bounded on L2(Rn)if and only if β ≥ (n+1)α.

In[6]we studythe generalcase Γ(t)=(t,γ(t))in R2where γ is f l at on(0,1).We obtain some interesting results.In the same paper,we construct some examples to illustrate the complexity of this problem.

Here we considerthe following oscillatory strongly singular integral associated to the surfaces of revolution,

At f i rst,when α =0 and ? is mean zero on Sn?1,by virtue of the decay estimates of the Fourier transform of ? on Sn?1,T is bounded from L2to itself without any assumptions on γ.It is easy to see that this arguments does not work when α ＞ 0.

If Hγ,α,βis bounded from L2to itself when n=1,then T?,γ,α,βis bounded from L2to itself for any n ＞ 1 only if ? ∈ L1(Sn?1).But in[6]we have seen that,for general α,β ＞ 0 and a convex curve Γ(t)=(t,γ(t)),Hγ,α,βis not bounded from L2to itself in general.

Here we will see that for the general convex curve,the L2-boundedness of T?,γ,α,βis more simple,even if the kernel is rough.Our main results are stated as the following two theorems.

Theorem 1.1.Suppose that ? ∈ Lp(Sn?1),p ＞ 1,n ＞ 1,α ＞ 0 and γ is convex on(0,∞).There exists A(p,n)＞ 0 which depends only on p,n such that if β ＞ A(p,n)(1+ α),then we have

where C is independent of γ and f.

Furthermore,if ? ∈ C∞(Sn?1),we have

Theorem 1.2.Assume that ? ∈ C∞(Sn?1),n ＞ 1,α ＞ 0 and γ is convex on(0,∞).If β ＞ 2α, then we have

Throughout this note the letters C and c always denote two positive constants which depend only on α, β,p and n.They maybe vary in different cases.In general,we take C big enough and c small enough.

2 Proof of the main theorems

Firstly,we need the following Van der Corput lemma.

Lemma 2.1(Van der Corput”s Lemma).If ψ,φ are two smooth functions on the interval(a,b) and|ψ(k)(t)|≥ λ ＞ 1 on t∈ (a,b)for some k ∈ N,then we have

or

The lemma and its proof can be found in[17].When k=1,the result can be checked directly by the integral by parts.

We also need the following decay estimates of the Fourier transform of ? on Sn?1.

Lemma 2.2(see[17]).If ? ∈ C∞(Sn?1)and dμ = ?dθ,then we have the following estimate

Lemma 2.3(see[17]).If ? ∈ Lp(Sn?1)for some p ＞ 1 and dμ = ?dθ,then there exists two positive constants C and ∈ (＜ 1)such that

Proof of Theorem 1.1.Now we begin to prove Theorem 1.1.Using the inverse of Fourier transform we have

which implies that

where

To prove the L2boundednessof T?,γ,α,β,by Plancherel equality,we only need to show that m(ξ,η)is a bounded function.

It is easy to check that when α ＞ 0,

Set ψ(r)=r?β?rθ·ξ?γ(r)η.For any(ξ,η)∈ Rn× R,we estimate the bounds of ˉm(ξ,η) in two case.

Case 1:η ≤ 0.As γ′′(r)≥ 0 on(0,∞),there holds

If β ＞ 2α,by Van der Corput’s Lemma,we can get that

which implies that

where C does not depend on ξ,η and γ.

Case 2:η ＞ 0.In this case we divide ˉm(ξ,η)into two parts.Set

Then one have

Set dμ = ?(θ)dθ.For the term m1(ξ,η),by Lemma 2.3 we can obtain that

So(2.6)yields that

On the other hand,for the term m2(ξ,η),as

we get that

So for the derivative of ψ(r)when r＜ t(ξ)we have

Now,by Lemma 2.1,(2.8)and the convexity of γ one can obtain that

So we have

(2.5),(2.7)and(2.10)yield that when η ＞ 0,there holds

At last let everything together.When β ＞(α +1)and γ is convex,from(2.3),(2.4)and (2.11)we can obtain that

which means that the operator T?,γ,α,βis bounded from L2to itself.So we complete the proof of Theorem 1.1.

Proof of Theorem 1.2.Now we turn to prove Theorem 1.2.In the proof of Theorem 1.1, it is easy to see that the estimates(2.3),(2.4)and(2.10)remain true only if β ＞ 2α.When? ∈ C∞(Sn?1),n ＞ 1,we check that the estimate(2.7)holds only if β ＞ 2α.By Lemma 2.2 and the similar computations as in(2.6),we get that

At last,by the same arguments as in the proof of Theorem 1.1 we can show that the operator T?,γ,α,βis bounded from L2to itself.So we prove Theorem 1.2.

Acknowledgments

The authors are supported by NSFC(Nos.11471288,11371136 and 11671363),NSFZJ (LY14A010015)and China Scholarship Council.

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Received 6 September 2015;Accepted(in revised version)23 September 2016

?Correspondingauthor.Email addresses:jcchen@zjnu.edu.cn(J.C.Chen),17855868773@163.com(S.Y.He), zxr@zjnu.cn(X.R.Zhu)

where ? ∈ Lp(Sn?1),p ＞ 1,n ＞ 1,α ＞ 0 and γ is convex on(0,∞).

We prove that there exists A(p,n) ＞ 0 such that if β ＞ A(p,n)(1+ α),then T?,γ,α,βis bounded from L2(Rn+1)to itself and the constant is independent of γ.Furthermore, when ? ∈ C∞(Sn?1),we will show that T?,γ,α,βis bounded from L2(Rn+1)to itself only if β ＞ 2α and the constant is independent of γ.