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SINGULAR OPERATORS WITH KERNELS IN

L(log

L)

α

(S

n−1

)

H. M. AL-QASSEM

Received 15 November 2005; Revised 20 May 2006; Accepted 28 May 2006

We establish the Lp-boundedness for a class of singular integral operators and a class of related maximal operators when their singular kernels are given by functions Ωin

L(logL)α(Sn−1).

Copyright © 2006 H. M. Al-Qassem. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Throughout this paper, we letxdenotex/|x|forx∈Rn\{0}and letpdenote the

con-jugate index ofp; that is, 1/ p+ 1/ p=1. Also, we letRn,n2, denote then-dimensional Euclidean space and letSn−1denote the unit sphere inRnequipped with the normalized Lebesgue measuredσ=dσ(·).

Let Γφ= {(y,φ(|y|)) :y∈Rn} be the surface of revolution generated by a suitable functionφ: [0,)R. LetKΩ,h(y) be a Calder ´on-Zygmund-type kernel of the form

KΩ,h(y)=h

|y|Ω(y)|y|−n, (1.1)

whereh: [0,)Cis a measurable function, andΩis an integrable function overSn−1,

satisfying

Sn−1Ω(u)(u)=0. (1.2)

LetL(logL)α(Sn−1) (forα >0) denote the space of all those measurable functionsΩon

Sn−1which satisfy

ΩL(logL)α(Sn−1)=

Sn−1

Ω(y)logα2 +Ω(y)(y)<∞. (1.3)

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 96732, Pages1–16

(2)

Forγ >1, defineΔγ(R+) to be the set of all measurable functionshonR+satisfying the

condition

sup R>0

R−1

R

0

h(t)γ

dt 1

<∞ (1.4)

and defineΔ(R+)=L∞(R+). Also, forγ≥1, defineᏴγ(R+) to be the set of all

measur-able functionshon R+satisfying the conditionhLγ(R+,dr/r)=(R

+|h(r)|

γdr/r)11

and defineᏴ(R+)=L∞(R+,dt/t).

We remark at this point that Δ(R+)Δγ2(R+)Δγ1(R+) forγ1< γ2,Ᏼ(R+)= Δ(R+), andᏴγ(R+)Δγ(R+) for 1< γ <∞.

The purpose of this paper is to study theLpmapping properties of singular integral operators,Ω,hinRnalong the surface of revolutionΓφdefined for (x,xn+1)RR=

Rn+1by

,Ω,hfx,xn+1

=p.v.

Rn f

x−y,xn+1−φ

|y|KΩ,h(y)d y. (1.5)

Also, we are interested in studying theLp-boundedness of the related maximal operator (γ)

φ,Ωgiven by

(γ) φf

x,xn+1= sup h∈γ(R+)

Tφ,Ω,hf

x,xn+1. (1.6)

Wheneverφ(t)0 andh≡1, then,Ω,hessentially is the classical Calder ´on-Zygmund singular integral operatorTΩgiven by

TΩf(x)=p.v.

Rn f(x−y)Ω(y

)|y|−nd y. (1.7)

Ifφ(t)0, we will denote,Ω,hbyTΩ,hand (φγ,)Ωby (γ)

Ω .

The investigation of the Lp-boundedness problem of the operatorTΩ began with Calder ´on and Zygmund in their well-known papers [7,8]. The operator,Ω,h, whose singular kernel has the additional roughness in the radial direction due to the presence ofh, was first studied by Fefferman [17] and subsequently by other several well-known authors. For a sampling of past studies, see [2,4,9,13,15,20]. We will content ourselves here with recalling only the following pertinent results.

In their celebrated paper [8], Calder ´on and Zygmund showed that theLp-boundedness ofTΩholds for 1< p <∞ifΩ∈LlogL(Sn−1). Moreover, the conditionΩ∈LlogL(Sn−1)

turns out to be the most desirable size condition for the Lp-boundedness ofT

Ω. This

was made clear by Calder ´on and Zygmund where it was shown thatTΩ may fail to be

bounded on Lp for any p if the conditionΩLlogL(Sn−1) is replaced by any weaker

metric conditionΩ∈Lϕ(Sn−1) with aϕsatisfyingϕ(t)=o(tlogt) ast→ ∞(e.g.,ϕ(t)=

(3)

Kim et al. [18] studied theLp-boundedness of T

φ,Ω,h as described in the following theorem.

Theorem1.1. LetTφ,Ω,hbe given as in (1.5) andh≡1. Assume thatΩ∈C∞(Sn−1)and

satisfies (1.2). Assume also thatφ(·)is inC2of[0,), convex, and increasing. ThenT φ,Ω,1

is bounded onLp(Rn+1)for1< p <.

Even though the authors of [18] imposed the conditionΩ∈C∞(Sn−1) inTheorem

1.1, the arguments employed in [18] can be modified to show that the conclusion in

Theorem 1.1remains valid when one weakens the condition onΩfromΩ∈C∞(Sn−1) to Ω∈Lq(Sn−1) for someq >1.

An improvement and extension over the above result was obtained by Al-Salman and Pan in [4], where the conditionΩ∈Lq(Sn−1) is replaced by the weaker conditionΩ

LlogL(Sn−1). In fact, they proved the following.

Theorem1.2. Letφbe aC2, convex, and increasing function satisfyingφ(0)=0. IfΩ

LlogL(Sn−1)and hΔ

γ(R+) for some γ >1, then the operator ,Ω,h in (1.2)–(1.5) is

bounded onLp(Rn+1)for|1/ p1/2|<min{1, 1/2}.

We remark that the range of p given inTheorem 1.2is the full range (1,) when-ever γ≥2. However, this range of p becomes a tiny open interval around 2 asγ ap-proaches 1. ForLp-boundedness results on singular integrals forpsatisfying|1/ p1/2|< min{1, 1/2}, we refer the readers to [2–4,14,15], among others.So, an unsolved prob-lem is whether theLp-boundedness ofT

φ,Ω,hholds forpoutside this range.

The main focus of this paper is to have a solution to the above problem. In fact, we have made some progress in resolving this problem by imposing a more restrictive condition onh. However, the price we paid in having a more restricted condition onhis compen-sated by the fact that we are able to prove our results under a much weaker condition on

Ω. More precisely we prove the following.

Theorem 1.3. Let ,Ω,h be given as in (1.2)–(1.5) and letφ be aC2, convex, and

in-creasing function satisfyingφ(0)=0. Suppose that h∈γ(R+)for some1< γ≤ ∞and Ω∈L(logL)1

(Sn−1). ThenT

φ,Ω,his bounded onLp(Rn)for1< p <∞.

At this point, it is worth mentioning that the proof ofTheorem 1.3cannot be obtained by a simple application of existing arguments on singular integrals. Even though we have a more restrictive condition onh, if we try to apply previously known arguments, then we can prove our result only for p satisfying|1/ p−1/2|<min{1, 1/2}. To be able to obtain theLp-boundedness for the full range 1< p <, a new maximal function that intervenes here in the proof ofTheorem 1.3is the maximal operator (φγ,Ω) defined in (1.6).

The study of the maximal operator (φγ,Ω) began by Chen and Lin in [10] and subsequently

by many other authors [1,12,19]. For example, Chen and Lin proved the following.

Theorem1.4 [10]. Assumen≥2,1≤γ≤2, andΩ∈C(Sn−1)satisfying (1.2). Then (γ)

Ω is bounded onLp(Rn)for(γn)< p <. Moreover, the range ofpis the best possible.

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Theorem1.5 [1]. Letn≥2and let (Ωγ)be given as in (1.6). Then

(a)ifΩ∈L(logL)1

(Sn−1)and satisfies (1.2), then (γ)

Ω is bounded onLp(Rn)forγ≤ p <∞;

(b)there exists anΩwhich lies inL(logL)1/2−ε(Sn−1)for allε >0and satisfies (1.2) such

that (2)Ω is not bounded onL2(Rn).

Our result regarding (φγ,Ω) is the following.

Theorem1.6. Let (φγ,Ω) be given as in (1.6) and letφbe aC2, convex and increasing function satisfyingφ(0)=0. SupposeΩ∈L(logL)1(Sn−1)and satisfies (1.2). Then (γ)

φis bounded onLp(Rn)forγp <and1< γ2, and it is bounded onL(Rn)forγ=1.

Remarks 1.7. (1) In order to clarify the relations between Theorems 1.1,1.2,1.4, and

1.5and theorems1.3and1.6, we remark that on the unit sphereSn−1, for anyq >1, the

following proper inclusions relations hold:

LqSn−1L(logL)Sn−1H1Sn−1L1Sn−1,

L(logL)βSn−1L(logL)αSn−1 if 0< α < β,

L(logL)αSn−1⊂H1Sn−1 ∀α1,

(1.8)

while

L(logL)αSn−1H1Sn−1L(logL)αSn−1 0< α <1. (1.9)

HereH1(Sn−1) is the Hardy space on the unit sphere in the sense of Coifman and Weiss

[11].

(2) For the case h∈(R+)=L∞(R+), the authors in [5] showed that there is a

function f ∈Lp such that the maximal operator acting on f (i.e., ()

Ω (f)) yields an

identically infinite function. It is still an open question whether the Lp-boundedness of (Ωγ) holds for 2< γ <∞. A point worth noting is thatTheorem 1.3implies theLp -boundedness of,Ω,hifh∈γ(R+) for all 1< γ≤ ∞.

(3) We notice that the singular integral operators TΩ,h are bounded onLp if Ω

L(logL)1(Sn−1) andh

γ(R+) for someγ >1, while the classical Calder ´on-Zygmund

singular integral operatorTΩ=TΩ,1is bounded onLpifΩ∈L(logL)(Sn−1). The reason

for this new phenomenon on singular integrals is that the singular operatorsTΩ,h (with

h∈γ(R+) for some 1< γ <∞) have weaker singularities than the singular operators

TΩ,1due to the presence of the strong condition onh.

(4) We notice thatTheorem 1.6 represents an improvement and extension over the result inTheorem 1.4and it is an extension overTheorem 1.5. Also, sinceLlogL(Sn−1)

L(logL)1

(Sn−1) for anyγ >1,Theorem 1.3represents an improvement overTheorem

1.2in the caseh∈γ(R+) for some 1< γ <∞.

(5) The method employed in this paper is based in part on a combination of ideas and arguments from [2,13,15,16,19], among others.

(5)

2. Some basic lemmas

Let us begin this section with the following definition.

Definition 2.1. For an arbitrary functionφ(·) onR+,aμ>1, andΩμ:Sn−1Rwithμ∈

N∪ {0}, define the sequence of measures{σφ,k,μ:k∈Z}and the corresponding maximal operatorσφ,μonRn+1by

Rn+1 f dσφ,k,μ=

ak μ≤|u|<akμ+1

fu,φ|u|KΩμ,h(u)du,

σφ∗,μ(f)=sup k∈Z

σφ,k,μf. (2.1)

Now let us establish the following Fourier transform estimates that will be used in later sections. One of the key points in these Fourier transform estimates is that the radial na-ture of the hypersurfaceΓφ(x)=(x,φ(|x|)) allows one to obtain these estimates without any condition onφ.

Lemma2.2. Letμ∈N∪ {0},aμ=2(μ+1), and letφ(·)be an arbitrary function onR+. Let Ωμ(·)be a function onSn−1 satisfying the following conditions: (i)ΩμL2(Sn−1)≤a2μ, (ii)

ΩμL1(Sn−1)1, and (iii)Ωμsatisfies the cancellation conditions in (1.2) withΩreplaced

byΩμ. Let

,k(ξ,η)= ak+1

μ

ak μ

Sn−μ(x)e

−i(tξ·x+ηφ(t))(x) 2

dt t

1/2

. (2.2)

Then there exist positive constantsCandαsuch that

Iμ,k(ξ,η)C(μ+ 1)1/2, (2.3)

Iμ,k(ξ,η)C(μ+ 1)1/2ak μξ

±α/(μ+1)

, (2.4)

whereξ∈Rn,ηR, andt±α=inf{tα,t−α}. The constantsCandαare independent ofk,

μ,ξ,ηandφ(·).

Proof. First, by condition (ii) onΩμ, it is easy to see that (2.3) holds. Next, by the cancel-lation properties ofΩμand by a change of variable, we have

Iμ,k(ξ,η)2

1

Sn−1

e−i{ak

μtξ·x+ηφ(akμt)}−e−iηφ(akμt

μ(x)(x)

2

dt

t , (2.5)

which easily implies

Iμ,k(ξ,η)C(μ+ 1)1/2a

μakμξ. (2.6)

By combining both estimates in (2.3) and (2.6), we get

Iμ,k(ξ,η)C(μ+ 1)β/2aβ μakμξ

β

(6)

for any 0< β <1 and for some constantC >0. By the last estimate and by lettingβ=

1/8(μ+ 1), we get the estimate in (2.4) withα=1/8 and with a plus sign in the exponent. To get the second estimate, we notice that

Sn−μ(x)e

−i(tak

μξ·x+ηφ(takμ))(x)

2

=

Sn−1×Sn−μ(xμ(u)e

−iak

μtξ·(x−u)(x)(u),

(2.8)

which leads to

Iμ,k(ξ,η)2

=

Sn−1×Sn−μ(xμ(u)

1 e

−iak

μtξ·(x−u)dt

t

(x)(u). (2.9)

By employing integration by parts, we get

1 e

−iak

μtξ·(x−u)dt

t

≤Cmin(μ+ 1), (μ+ 1)aμkξ·(x−u)1 (2.10)

and hence

1 e

−iak

μtξ·(x−u)dt

t

≤C(μ+ 1)βakμξ− β

ξ·(x−u)−β(μ+ 1)(1−β) for any 0< β <1.

(2.11)

By the last estimate and by lettingβ=1/4, we obtain

1 e

−iak

μtξ·(x−u)dt

t

≤C(μ+ 1)ak μξ−

1/4

ξ·(x−u)1/4. (2.12)

By Schwarz’s inequality, condition (i) onΩμ, and (2.9)–(2.12), we get

Iμ,k(ξ,η)2

≤C(μ+ 1)a4μakμξ− 1/4

Sn−1×Sn−1

ξ·(xu)1/2

(x)(u)

1/2

.

(2.13)

Since the last integral is finite, we get

Iμ,k(ξ,η)C(μ+ 1)1/2a2 μakμξ−

1/8

. (2.14)

As above, by combining (2.14) with (2.3), we obtain the second estimate in (2.4).Lemma

2.2is proved.

Lemma2.3. Letμ∈N∪ {0},aμ=2(μ+1),h∈γ(R+)for some1< γ <∞, and letφ(·)

be an arbitrary function on R+. LetΩμ(·)be a function onSn−1 satisfying the following

conditions: (i)ΩμL2(Sn−1)≤aμ2, and (ii)ΩμL1(Sn−1)1. Let

Rk,μ(ξ,η)=

ak+1

μ

ak μ

Sn−μ(x)e

−i(tξ·x+ηφ(t))(x)h(t)dt

(7)

Then there exist positive constantsCindependent ofk,φ, andμsuch that Rk,μ(ξ,η)C(μ+ 1)1

, (2.16)

Rk,μ(ξ,η)C(μ+ 1)1ak μξ

±α/γ(μ+1)

. (2.17)

Proof. By H¨older’s inequality, we have

Rk,μ(ξ,η)

a

k+1

μ

ak μ

h(t)γdt

t

1 ak+1

μ

ak μ

Sn−μ(x)e

−i(tξ·x+ηφ(t))(x)γ

dt t

1

≤∞

0

h(t)γdt

t

1 ak+1

μ

ak μ

Sn−μ(x)e

−i(tξ·x+ηφ(t))(x)γ

dt t

1

a

k+1

μ

ak μ

Sn−μ(x)e

−i(tξ·x+ηφ(t))(x)γ

dt t

1

.

(2.18)

Now, if 2≤γ<∞, by noticing that|Sn−μ(x)e−i(tξ·x+ηφ(t))(x)| ≤1, we get

Rk,μ(ξ,η) ak+1

μ

ak μ

Sn−μ(x)e

−i(tξ·x+ηφ(t))(x) 2

dt t

1

(2.19)

and hence byLemma 2.2we easily get (2.17). On the other hand, if 1< γ<2, (2.17) follows byLemma 2.2and H¨older’s inequality. This finishes the proof ofLemma 2.3.

We will need the following lemma which has its roots in [2,13,15]. A proof of this lemma can be obtained by the same proof (with only minor modifications) as that of [2, Lemma 3.2]. We omit the details.

Lemma2.4. Let{σk:k∈Z}be a sequence of Borel measures onRnand letL:Rn→Rdbe a

linear transformation. Suppose that for allk∈Z,ξ∈Rn, for some a[2,),λ >0,α >0,

C >0, and for someB >1,

(i)σk ≤CBλ;

(ii)|σk(ξ)| ≤CBλ(akB|L(ξ)|)±α/B; (iii)for somep0(2,),

kZσk∗gk 2

1/2

p0

≤CBλ

kZgk 2

1/2

p0

(2.20)

holds for arbitrary functions{gk}onRn. Then forp0< p < p0there exists a positive constant

Cpsuch that

k∈Z

σk∗f

p

(8)

holds for all f inLp(Rn). The constantC

p is independent ofBand the linear

transforma-tionL.

Our aim now is to establish the following result.

Lemma2.5. Leth∈γ(R+)for someγ >1and letΩμ be a function satisfying conditions

(i) and (ii) inLemma 2.2. Assumeφis inC2([0,)), convex, and increasing. Then forγ< p≤ ∞andf ∈Lp(Rn+1), there exists a positive constantC

pwhich is independent ofμsuch

that

σ

φ,μ(f)p≤Cp(μ+ 1)1

fp. (2.22)

Proof. Without loss of generality, we may assume thatΩμ≥0 andh≥0. By H¨older’s inequality, we have

σφ,μ(f) ak+1

μ

ak μ

h(t)γdt

t 1

Υ

μ

|f|γ1CΥ

μ

|f|γ1, (2.23)

whereRn+1 f dΥk,μ=

ak

μ≤|u|< a

k+1

μ f(u,φ(|u|))|u|−nΩμ(u)duandΥ∗μ(f)=supk∈Z||Υk,μ|∗

f|. Therefore, in order to prove (2.22), it suffices to prove that

Υ

μ(f)Lp(Rn+1)≤Cp(μ+ 1)fLp(Rn+1) for 1< p≤ ∞. (2.24)

However, the proof of (2.24) follows by the same argument employed in the proof of [4,

Lemma 4.7] and hence the proof ofLemma 2.5is complete.

Lemma2.6. Leth∈γ(R+)for some γ≥2 and letΩμ be a function onSn−1satisfying

conditions (i) and (ii) in Lemma 2.2. Letφ be inC2([0,)), convex, and an increasing

function withφ(0)=0. Then, forγ< p <∞, there exists a positive constantCp which is

independent ofμsuch that

kZσφ,k,μ∗gk2

1/2

Lp(Rn+1)

≤Cp(μ+ 1)1

kZgk2

1/2

Lp(Rn+1)

(2.25)

holds for arbitrary measurable functions{gk}onRn+1.

Proof. We follow a similar argument as in [6]. Letγ< p <∞. By H¨older’s inequality and the condition onh, we get

σφ,k,μgk

x,xn+1γ

≤C ak+1

μ

ak μ

Sn−1

Ωμ(y)gk

x−yt,xn+1−φ(t)γ

(y)dt

t .

(9)

Letd=p/γ. For{gk} ∈Ld(Rn+1,l2), by duality, there exists a nonnegative functionω∈

Ld

(Rn+1) such thatω

Ld≤1 and

kZσφ,k,μ∗gkγ 1 γ p =

Rn+1

k∈Z

σφ,k,μgk

x,xn+1γ

ωx,xn+1

dx dxn+1.

(2.27)

Therefore, by (2.27) and a change of variable, we get

kZσφ,k,μ∗gkγ 1 γ p ≤C

Rn+1

k∈Z

gk

x,xn+1γ

Mμω

x,xn+1

dx dxn+1,

(2.28)

where

Mμω

x,xn+1

=sup

k∈Z

ak μ≤|y|<akμ+1

ωx+y,xn+1+φ

|y|Ωμ(y)|y|−nd y. (2.29)

By H¨older’s inequality, we obtain

kZσφ,k,μ∗gkγ 1 γ p ≤C

kZ|gk|γ 1 γ p

Mμω

d. (2.30)

By [4, Lemma 4.7], we haveMμωLd ≤Cp(μ+ 1) which in turn implies

kZσφ,k,μ∗gkγ 1

p

≤C(μ+ 1)1

kZgkγ 1 p . (2.31)

Moreover, again byLemma 2.5, we have

supkZσφ,k,μ∗gk

p

σφ,μ

sup k∈Z

gk

p

≤Cp(μ+ 1)1

supkZgk

p

. (2.32)

By using the operator interpolation theorem between (2.31) and (2.32) and sinceγ∈

[1, 2], we get (2.25) which concludes the proof of the lemma.

We are now ready to present the proofs of our main results.

3. Proofs of Theorems1.3and1.6

Since the proof ofTheorem 1.3will rely heavily onTheorem 1.6as well as on its proof, we start by provingTheorem 1.6.

Proof ofTheorem 1.6. Assume thatΩsatisfies (1.2) and belongs toL(logL)1

(Sn−1) and

(10)

Also, we letJ0 be the set of all those pointsx∈Sn−1 which satisfy|Ω(x)|<2. Forμ∈

N∪ {0}, setbμ=ΩχJμ,λ0=1, andλμ= bμ1forμ∈N. SetI= {μ∈N:λμ≥2

2μ}and

define the sequence of functions{Ωμ}μ∈I∪{0}by

Ω0(x)=

μ∈{0}∪(N−I)

(x)

μ∈{0}∪(N−I)

Sn−1(x)(x)

,

Ωμ(x)=

λμ

1

(x)

Sn−1(x)(x)

forμ∈I.

(3.1)

Forμ∈I∪ {0}, setaμ=2(μ+1). Then one can easily verify that the following hold for all

μ∈I∪ {0}and for some positive constantC:

Ωμ

2≤Ca2μ, Ωμ1≤C, (3.2)

μ∈I∪{0}

(μ+ 1)1/γλμ≤CΩL(logL)1(Sn−1), (3.3)

Sn−μ(u)(u)=0; Ω=

μ∈I∪{0}

λμΩμ. (3.4)

By (3.4), we have (φγ,Ω) f(x,xn+1)

μ∈I∪{0}λμ (γ)

φμf(x,xn+1) and hence the proof of Theorem 1.6is completed if we can show that

(γ) φμf

Lp(Rn+1)≤Cp(μ+ 1) 1

fLp(Rn+1) (3.5)

holds for γ≤p <∞ if 1< γ≤2 and for p= ∞if γ=1. To prove (3.5), we need to consider three cases. We first prove (3.5) for the caseγ=2.

Case 1(γ=2). By duality, we have

(2) φμf

x,xn+1= sup h∈γ(R+)

0 h(t)

Sn−1f

x−tu,xn+1−φ(tμ(u)(u)dt

t

=

0

Sn−1 f

x−tu,xn+1−φ(t)

Ωμ(u)(u)

2dtt 1/2

=

k∈Z

ak+1

μ

ak μ

Sn−1f

x−tu,xn+1−φ(tμ(u)(u)

2dtt

1/2

.

(3.6)

Let{ψk,μ}∞−∞ be a smooth partition of unity in (0,) adapted to the intervalEk,μ= [a−k−1

μ ,a−μk+1]. To be precise, we require the following:ψk,μ∈C∞, 0≤ψk,μ≤1,kψk,μ(t)= 1, suppψk,μ⊆Ek,μ,|dsψk,μ(t)/dts| ≤Cs/ts, whereCs is independent of the lacunary se-quence{ak

μ:k∈Z}. Define the multiplier operatorsSk,μinRn+1by

Sk,μf

(ξ,η)=ψk,μ

(11)

Then for any f ᏿(Rn+1) andlZwe have f(x,x

n+1)=k∈Z(Sk+l,μf)(x,xn+1). Thus,

by (3.6) and applying Minkowski’s inequality, we get

(2) φμf

x,xn+1

k∈Z

ak+1

μ ak μ

l∈Z

Hk+l,t,μfx,xn+1

2 dt t

1/2

l∈Z

k∈Z

ak+1

μ

ak μ

Hk+l,t,μf

x,xn+12dt

t 1/2

=

l∈Z

Tl,μf

x,xn+1

,

(3.8)

where

Hl,t,μf

x,xn+1

=

Sn−1

Sl,μf

x−tu,xn+1−φ(t)

Ωμ(u)(u),

Tl,μfx,xn+1=

k∈Z

ak+1

μ

ak μ

Hk+l,t,μf

x,xn+12dt

t 1/2

.

(3.9)

Therefore, to prove (3.5) forγ=2, it suffices to prove

Tl,μ(f)

Lp(Rn+1)≤Cp(μ+ 1)1/22−θp|l|fLp(Rn+1) (3.10)

for some positive constantsCp,θp, and for all 2≤p <∞.

The proof of (3.10) follows by interpolation between a sharpL2estimate and a cruder

Lpestimate ofT

l,μ(f). First, theL2-boundedness ofTl,μ(f) is provided by a simple appli-cation of Plancherel’s theorem, Fubini’s theorem and usingLemma 2.2:

Tl,μ(f)2 2= R Rn

k∈Z

ak+1

μ

ak μ

Hk+l,t,μf

x,xn+1 2dt

t dx dxn+1

k∈Z

R

Δk+l

ak+1

μ

ak μ

Ωμ(x)e−i(tξ·x+ηφ(t))(x)2dt

t f(ξ,η)

2

dξ dη

≤C(μ+ 1)22α|l| k∈Z

R

Δk+l

f(ξ,η)2dξ dη

≤C(μ+ 1)22α|l|f2 2,

(3.11)

whereΔk= {ξ∈Rn:|ξ| ∈Ek,μ}. The last inequality holds since the setsΔk are finitely overlapping. Therefore, we have

Tl,μ,Ω

μ(f)2≤C(μ+ 1)

1/22−α|l|f

(12)

On the other hand, we need to compute theLp-norm ofT

l,μμ(f) forp >2. By duality,

there is a functionginL(p/2)

(Rn+1) withg

(p/2)1 such that

Tl,μ,Ω

μ(f)

2 p=

k∈Z

Rn+1

ak+1

μ

ak μ

Hk+l,t,Ω

μf

x,xn+1 2dt

t g

x,xn+1dx dxn+1

Ωμ1

k∈Z

Rn+1

ak+1

μ

ak μ

Sn−1

Ωμ(u)g

x+tu,xn+1+φ(t)

×Sk+l,μf

x,xn+12(u)dt

t dx dxn+1

≤C

k∈Z

Rn+1

Sk+l,μf

x,xn+12σφ,μ(g)

−x,−xn+1dx dxn+1

≤C

k∈Z

Sk+l,μf2

(p/2)

σ

φ,μ(g)(p/2),

(3.13)

whereg(x,xn+1)=g(−x,−xn+1). By usingLemma 2.5, the Littlewood-Paley theory, and

[21, Theorem 3] along with the remark that follows its statement in [21, page 96], we have

Tl,μ,Ω

μ(f)p≤Cp(μ+ 1)1/2fp for 2≤p <∞. (3.14)

By interpolation between (3.12) and (3.14), we get (3.10) which ends the proof of (3.5) in the caseγ=2.

Case 2(γ=1). Iff ∈L∞(Rn+1) andhL1(R

+,dr/r), then

0 h(t)

Sn−1f

x−tu,xn+1−φ(t)

Ωμ(u)(u)dt

t

≤CfL∞hL1(R+,dr/r) (3.15)

for every (x,xn+1). Taking the supremum on both sides of the above inequality over all

radial functionshwithhL1(R+,dr/r)1 yields

(1) φμf

x,xn+1

≤CfL∞(Rn+1) (3.16)

for almost every (x,xn+1)Rn+1. Hence,

(1) φμf

L∞(Rn+1)≤CfL∞(Rn+1). (3.17)

Case 3(1< γ <2). We will use an idea employed in [19]. By duality,

(γ) φμf

x,xn+1

=

Sn−1f

x−tu,xn+1−φ(t)

Ωμ(u)(u)

(13)

Thus,

(γ) φμf

Lp(Rn+1)= S(f)Lp((R+,dt/t),Rn+1), (3.19)

whereS:Lp(Rn+1)Lp(Lγ(R

+,dt/t),Rn+1) defined by

S(f)x,xn+1,t=

Sn−1f

x−tu,xn+1−φ(tμ(u)(u). (3.20)

By (3.5) (forγ=2) and (3.17), we interpret that

S(f)

Lp(L2(R+,dt/t),Rn+1)≤C(μ+ 1)1/2fLp(Rn+1) (3.21)

for 2< p <∞and

S(f)

L∞(L(R+,dt/t),Rn+1)≤CfL(Rn+1). (3.22)

Applying the real interpolation theorem for Lebesgue mixed normed spaces to the above results (see [6]), we conclude that

S(f)

Lp(Lγ(R+,dt/t),Rn+1)≤C(μ+ 1)1

fLp(Rn+1) (3.23)

forγ≤p <∞which in turn implies (3.5) for 1< γ <2. The proof ofTheorem 1.6 is

complete.

Proof ofTheorem 1.3. We divide the proof ofTheorem 1.3into three separate cases.

Case 1(1< γ≤2). As we pointed in the introduction, the proof of this case is based on

Theorem 1.6. To this end, we argue as follows. Notice that

,Ω,hfx,xn+1

=lim

ε→0Tεf

x,xn+1

, (3.24)

whereis the truncated singular integral operator given by

Tεfx,xn+1

=

|y|>εf

x−y,xn+1−φ

|y|KΩ,h(y)d y. (3.25)

Without loss of generality, we may assume thathLγ(R+,dr/r)=1. Notice that, by H¨older’s

inequality, we have

Tεf

x,xn+1

ε

h(t)

Sn−1f

x−tu,xn+1−φ(t)

Ω(u)(u)dt

t

0

Sn−1 f

x−tu,xn+1−φ(t)

Ω(u)(u) γ

dt t

1

.

(3.26)

Therefore,

Tεf

p≤ (γ)

(14)

forγ≤p <∞and 1< γ≤2, andCis independent ofε. By a standard duality argument,

is bounded onLp(Rn+1) for 1< p≤γ and 1< γ≤2. Passing to the limit as ε→0, Fatou’s lemma gives,Ω,hfp≤Cfpforγ≤p <∞and for 1< p≤γ. Ifγ=2, then we are done; otherwise an application of the real interpolation theorem gives theLp -boundedness of,Ω,hfor the remaining range ofp:γ < p < γ.

Case 2(2< γ <∞). As already seen in the proof ofTheorem 1.6, by using the decompo-sitionΩ=μ∈I∪{0}λμΩμ, it suffices to show that

Tφ,Ω

μ,hfp≤Cp(1 +μ)1

fp for 1< p <∞. (3.28)

To this end, we argue as above by consideringinstead of,Ω,h. WriteTεf=k∈Zσφ,k,μ∗

f, one can then apply Lemmas2.3,2.4, and2.6to obtain

Tεf

p≤Cp(1 +μ)1

fp, (3.29)

whereCpis independent ofε. In particular,

Tεf

p≤Cp(1 +μ)1

fp (3.30)

for 2≤p <∞. By duality,

Tεf

p≤Cp(1 +μ)1

fp (3.31)

for 1< p≤2. Thus, by Fatou’s lemma, we have (3.28).

Case 3(γ= ∞). Finally, it follows directly fromTheorem 1.2. This completes the proof

ofTheorem 1.3.

4. Some additional results

We will end the paper by presenting some additional results.

Theorem4.1. Leth∈γ(R+)for some1< γ≤ ∞. Letφbe aC2, convex, and increasing

function satisfyingφ(0)=0. LetΩ∈L(logL)1

(Sn−1)and satisfy (1.2). For eachλR

define the oscillatory singular integral operatorSλby

Sλf(x)=p.v.

Rne

iλφ(|x−y|)Ω(x−y)

|x−y|nh

|x−y|f(y)d y. (4.1)

Then the operators{Sλ}λ∈Rare uniformly bounded onLp(Rn)for1< p <∞.

(15)

Theorem4.2. Leth∈γ(R+)for some1< γ≤ ∞. Suppose thatΩ∈L(logL)1(Sn−1)

andφis aC2, convex, and increasing function satisfyingφ(0)=0. Then the maximal

oper-atorφ,Ω,hgiven by

φ,Ω,hf

x,xn+1

=sup

R>0

R−n

|y|<R

f

x−y,xn+1−φ

|y|h|y|Ω(y)d y

(4.2)

is bounded onLp(Rn+1)forγ< p≤ ∞.

Earlier theLp(1< p≤ ∞) boundedness of

φ,Ω,1was proved in [18] under the stronger

condition thatΩ∈Lq(Sn−1) for someq >1 and later Al-Salman and Pan in [4] proved

Lp(γ< p≤ ∞) boundedness ofφ,Ω,hunder the conditionsΩLlogL(Sn−1) andh Δγ(R+) for someγ >1.

Proof ofTheorem 4.2. By (3.4) we have

φ,Ω,hfx,xn+1

μ∈I∪{0}

λμφ,|Ωμ|,h (4.3)

and henceTheorem 4.2follows by noticing that

φ,|Ωμ|,h(f)≤Cσφ∗,μ

|f| (4.4)

and invokingLemma 2.5.

As usual, the pointwise existence of,Ω,hf for f inLp spaces can be established by considering the following maximal truncated singular integral

,Ω,hf

x,xn+1

=sup

ε>0

Tεf

x,xn+1, (4.5)

whereTεf(x,xn+1) is defined as in (3.25).

Theorem4.3. LetTφ,Ω,hbe given as above. Suppose thath∈γ(R+)for some1< γ≤ ∞

andΩ∈L(logL)1

(Sn−1). ThenT

φ,Ω,his bounded onLp(Rn+1)forγ< p <∞.

A proof ofTheorem 4.3can be constructed by using the above estimates and the argu-ments in [2] (see also [13,15]). We omit the details.

Acknowledgment

The author is indebted to Professor Yibiao Pan for a helpful discussion.

References

[1] H. M. Al-Qassem,Lpbounds for a class of rough maximal operators, preprint.

[2] H. M. Al-Qassem and Y. Pan,Lpestimates for singular integrals with kernels belonging to certain

block spaces, Revista Matem´atica Iberoamericana18(2002), no. 3, 701–730.

(16)

[4] A. Al-Salman and Y. Pan,Singular integrals with rough kernels inLlogL(Sn−1), Journal of the

London Mathematical Society. Second Series66(2002), no. 1, 153–174.

[5] J. M. Ash, P. F. Ash, C. L. Fefferman, and R. L. Jones,Singular integral operators with complex homogeneity, Studia Mathematica65(1979), no. 1, 31–50.

[6] A. Benedek and R. Panzone,The spaceLp, with mixed norm, Duke Mathematical Journal28

(1961), no. 3, 301–324.

[7] A. P. Calder ´on and A. Zygmund,On the existence of certain singular integrals, Acta Mathematica 88(1952), no. 1, 85–139.

[8] ,On singular integrals, American Journal of Mathematics78(1956), 289–309. [9] L.-K. Chen,On a singular integral, Studia Mathematica85(1986), no. 1, 61–72, 1987.

[10] L.-K. Chen and H. Lin,A maximal operator related to a class of singular integrals, Illinois Journal of Mathematics34(1990), no. 1, 120–126.

[11] R. R. Coifman and G. Weiss,Extensions of Hardy spaces and their use in analysis, Bulletin of the American Mathematical Society83(1977), no. 4, 569–645.

[12] Y. Ding and Q. He,Weighted boundedness of a rough maximal operator, Acta Mathematica Sci-entia. English Edition20(2000), no. 3, 417–422.

[13] J. Duoandikoetxea and J. L. Rubio de Francia,Maximal and singular integral operators via Fourier transform estimates, Inventiones Mathematicae84(1986), no. 3, 541–561.

[14] D. Fan and Y. Pan,A singular integral operator with rough kernel, Proceedings of the American Mathematical Society125(1997), no. 12, 3695–3703.

[15] ,Singular integral operators with rough kernels supported by subvarieties, American Jour-nal of Mathematics119(1997), no. 4, 799–839.

[16] D. Fan, Y. Pan, and D. Yang,A weighted norm inequality for rough singular integrals, The Tohoku Mathematical Journal51(1999), no. 2, 141–161.

[17] R. Fefferman,A note on singular integrals, Proceedings of the American Mathematical Society74 (1979), no. 2, 266–270.

[18] W.-J. Kim, S. Wainger, J. Wright, and S. Ziesler,Singular integrals and maximal functions associ-ated to surfaces of revolution, The Bulletin of the London Mathematical Society28(1996), no. 3, 291–296.

[19] H. V. Le,Maximal operators and singular integral operators along submanifolds, Journal of Math-ematical Analysis and Applications296(2004), no. 1, 44–64.

[20] J. Namazi,A singular integral, Proceedings of the American Mathematical Society96(1986), no. 3, 421–424.

[21] E. M. Stein,Singular Integrals and Differentiability Properties of Functions, Princeton Mathemat-ical Series, no. 30, Princeton University Press, New Jersey, 1970.

References

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