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,n≥2, 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)dσ(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)dσ(y)<∞. (1.3)
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 96732, Pages1–16
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/γ
<∞ (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)1/γ≤1
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 operatorsTφ,Ω,hinRnalong the surface of revolutionΓφdefined for (x,xn+1)∈Rn×R=
Rn+1by
Tφ,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) φ,Ωf
x,xn+1= sup h∈Ᏼγ(R+)
Tφ,Ω,hf
x,xn+1. (1.6)
Wheneverφ(t)≡0 andh≡1, thenTφ,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 denoteTφ,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 operatorTφ,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 conditionΩ∈LlogL(Sn−1) is replaced by any weaker
metric conditionΩ∈Lϕ(Sn−1) with aϕsatisfyingϕ(t)=o(tlogt) ast→ ∞(e.g.,ϕ(t)=
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 h∈Δ
γ(R+) for some γ >1, then the operator Tφ,Ω,h in (1.2)–(1.5) is
bounded onLp(Rn+1)for|1/ p−1/2|<min{1/γ, 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/ p−1/2|< min{1/γ, 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 Tφ,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∈Ᏼγ(R+)for some1< γ≤ ∞and Ω∈L(logL)1/γ
(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/γ, 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.
Theorem1.5 [1]. Letn≥2and let (Ωγ)be given as in (1.6). Then
(a)ifΩ∈L(logL)1/γ
(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/γ(Sn−1)and satisfies (1.2). Then (γ)
φ,Ωis bounded onLp(Rn)forγ≤p <∞and1< γ≤2, 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−1⊂L(logL)Sn−1⊂H1Sn−1⊂L1Sn−1,
L(logL)βSn−1⊂L(logL)αSn−1 if 0< α < β,
L(logL)αSn−1⊂H1Sn−1 ∀α≥1,
(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∈Ᏼ∞(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 ofTφ,Ω,hifh∈Ᏼγ(R+) for all 1< γ≤ ∞.
(3) We notice that the singular integral operators TΩ,h are bounded onLp if Ω∈
L(logL)1/γ(Sn−1) andh∈Ᏼ
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∈Ᏼγ(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/γ
(Sn−1) for anyγ >1,Theorem 1.3represents an improvement overTheorem
1.2in the caseh∈Ᏼγ(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.
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−1→Rwithμ∈
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,μ∗f. (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
Iμ,k(ξ,η)= ak+1
μ
ak μ
Sn−1Ωμ(x)e
−i(tξ·x+ηφ(t))dσ(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,η∈R, 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
≤ aμ
1
Sn−1
e−i{ak
μtξ·x+ηφ(akμt)}−e−iηφ(akμt)Ω
μ(x)dσ(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
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−1Ωμ(x)e
−i(tak
μξ·x+ηφ(takμ))dσ(x)
2
=
Sn−1×Sn−1Ωμ(x)Ωμ(u)e
−iak
μtξ·(x−u)dσ(x)dσ(u),
(2.8)
which leads to
Iμ,k(ξ,η)2
=
Sn−1×Sn−1Ωμ(x)Ωμ(u)
aμ
1 e
−iak
μtξ·(x−u)dt
t
dσ(x)dσ(u). (2.9)
By employing integration by parts, we get
aμ
1 e
−iak
μtξ·(x−u)dt
t
≤Cmin(μ+ 1), (μ+ 1)aμkξ·(x−u)−1 (2.10)
and hence
aμ
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
aμ
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·(x−u)−1/2
dσ(x)dσ(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∈Ᏼγ(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−1Ωμ(x)e
−i(tξ·x+ηφ(t))dσ(x)h(t)dt
Then there exist positive constantsCindependent ofk,φ, andμsuch that Rk,μ(ξ,η)≤C(μ+ 1)1/γ
, (2.16)
Rk,μ(ξ,η)≤C(μ+ 1)1/γak 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/γ ak+1
μ
ak μ
Sn−1Ωμ(x)e
−i(tξ·x+ηφ(t))dσ(x)γ
dt t
1/γ
≤∞
0
h(t)γdt
t
1/γ ak+1
μ
ak μ
Sn−1Ωμ(x)e
−i(tξ·x+ηφ(t))dσ(x)γ
dt t
1/γ
≤ a
k+1
μ
ak μ
Sn−1Ωμ(x)e
−i(tξ·x+ηφ(t))dσ(x)γ
dt t
1/γ
.
(2.18)
Now, if 2≤γ<∞, by noticing that|Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)| ≤1, we get
Rk,μ(ξ,η)≤ ak+1
μ
ak μ
Sn−1Ωμ(x)e
−i(tξ·x+ηφ(t))dσ(x) 2
dt t
1/γ
(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,∞),
k∈Zσk∗gk 2
1/2
p0
≤CBλ
k∈Zgk 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
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∈Ᏼγ(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/γ
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/γ
Υ∗
μ
|f|γ1/γ≤CΥ∗
μ
|f|γ1/γ, (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∈Ᏼγ(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
k∈Zσφ,k,μ∗gk2
1/2
Lp(Rn+1)
≤Cp(μ+ 1)1/γ
k∈Zgk2
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,μ∗gk
x,xn+1γ
≤C ak+1
μ
ak μ
Sn−1
Ωμ(y)gk
x−yt,xn+1−φ(t)γ
dσ(y)dt
t .
Letd=p/γ. For{gk} ∈Ld(Rn+1,l2), by duality, there exists a nonnegative functionω∈
Ld
(Rn+1) such thatω
Ld≤1 and
k∈Zσφ,k,μ∗gkγ 1/γ γ p =
Rn+1
k∈Z
σφ,k,μ∗gk
x,xn+1γ
ωx,xn+1
dx dxn+1.
(2.27)
Therefore, by (2.27) and a change of variable, we get
k∈Zσφ,k,μ∗gkγ 1/γ 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
k∈Zσφ,k,μ∗gkγ 1/γ γ p ≤C
k∈Z|gk|γ 1/γ γ p
Mμω
d. (2.30)
By [4, Lemma 4.7], we haveMμωLd ≤Cp(μ+ 1) which in turn implies
k∈Zσφ,k,μ∗gkγ 1/γ
p
≤C(μ+ 1)1/γ
k∈Zgkγ 1/γ p . (2.31)
Moreover, again byLemma 2.5, we have
supk∈Zσφ,k,μ∗gk
p
≤σφ∗,μ
sup k∈Z
gk
p
≤Cp(μ+ 1)1/γ
supk∈Zgk
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/γ
(Sn−1) and
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)
bμ(x)−
μ∈{0}∪(N−I)
Sn−1bμ(x)dσ(x)
,
Ωμ(x)=
λμ
−1
bμ(x)−
Sn−1bμ(x)dσ(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/γ(Sn−1), (3.3)
Sn−1Ωμ(u)dσ(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)
φ,Ωμf(x,xn+1) and hence the proof of Theorem 1.6is completed if we can show that
(γ) φ,Ωμf
Lp(Rn+1)≤Cp(μ+ 1) 1/γ
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) φ,Ωμf
x,xn+1= sup h∈Ᏼγ(R+)
∞
0 h(t)
Sn−1f
x−tu,xn+1−φ(t)Ωμ(u)dσ(u)dt
t
= ∞
0
Sn−1 f
x−tu,xn+1−φ(t)
Ωμ(u)dσ(u)
2dtt 1/2
=
k∈Z
ak+1
μ
ak μ
Sn−1f
x−tu,xn+1−φ(t)Ωμ(u)dσ(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
Then for any f ∈(Rn+1) andl∈Zwe 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) φ,Ωμf
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)dσ(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))dσ(x)2dt
t f(ξ,η)
2
dξ dη
≤C(μ+ 1)2−2α|l| k∈Z
R
Δk+l
f(ξ,η)2dξ dη
≤C(μ+ 1)2−2α|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
On the other hand, we need to compute theLp-norm ofT
l,μ,Ωμ(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+12dσ(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) andh∈L1(R
+,dr/r), then
∞
0 h(t)
Sn−1f
x−tu,xn+1−φ(t)
Ωμ(u)dσ(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) φ,Ωμf
x,xn+1
≤CfL∞(Rn+1) (3.16)
for almost every (x,xn+1)∈Rn+1. Hence,
(1) φ,Ωμf
L∞(Rn+1)≤CfL∞(Rn+1). (3.17)
Case 3(1< γ <2). We will use an idea employed in [19]. By duality,
(γ) φ,Ωμf
x,xn+1
=
Sn−1f
x−tu,xn+1−φ(t)
Ωμ(u)dσ(u)
Thus,
(γ) φ,Ωμf
Lp(Rn+1)= S(f)Lp(Lγ(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(t)Ωμ(u)dσ(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/γ
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
Tφ,Ω,hfx,xn+1
=lim
ε→0Tεf
x,xn+1
, (3.24)
whereTεis 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)dσ(u)dt
t
≤ ∞
0
Sn−1 f
x−tu,xn+1−φ(t)
Ω(u)dσ(u) γ
dt t
1/γ
.
(3.26)
Therefore,
Tεf
p≤ (γ)
forγ≤p <∞and 1< γ≤2, andCis independent ofε. By a standard duality argument,
Tε is bounded onLp(Rn+1) for 1< p≤γ and 1< γ≤2. Passing to the limit as ε→0, Fatou’s lemma givesTφ,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 ofTφ,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/γ
fp for 1< p <∞. (3.28)
To this end, we argue as above by consideringTεinstead ofTφ,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/γ
fp, (3.29)
whereCpis independent ofε. In particular,
Tεf
p≤Cp(1 +μ)1/γ
fp (3.30)
for 2≤p <∞. By duality,
Tεf
p≤Cp(1 +μ)1/γ
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∈Ᏼγ(R+)for some1< γ≤ ∞. Letφbe aC2, convex, and increasing
function satisfyingφ(0)=0. LetΩ∈L(logL)1/γ
(Sn−1)and satisfy (1.2). For eachλ∈R
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 <∞.
Theorem4.2. Leth∈Ᏼγ(R+)for some1< γ≤ ∞. Suppose thatΩ∈L(logL)1/γ(Sn−1)
andφis aC2, convex, and increasing function satisfyingφ(0)=0. Then the maximal
oper-atorᏹφ,Ω,hgiven by
ᏹφ,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 ofᏹ
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 ofᏹφ,Ω,hunder the conditionsΩ∈LlogL(Sn−1) andh∈ Δγ(R+) for someγ >1.
Proof ofTheorem 4.2. By (3.4) we have
ᏹφ,Ω,hfx,xn+1≤
μ∈I∪{0}
λμᏹφ,|Ωμ|,h (4.3)
and henceTheorem 4.2follows by noticing that
ᏹφ,|Ωμ|,h(f)≤Cσφ∗,μ
|f| (4.4)
and invokingLemma 2.5.
As usual, the pointwise existence ofTφ,Ω,hf for f inLp spaces can be established by considering the following maximal truncated singular integral
Tφ,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∈Ᏼγ(R+)for some1< γ≤ ∞
andΩ∈L(logL)1/γ
(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.
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