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 let Sn−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 over Sn−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 DOI 10.1155/JIA/2006/96732
Forγ >1, defineΔγ(R+) to be the set of all measurable functionshonR+satisfying the condition
sup
R>0
R−1
R
0
h(t)γdt 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 fx−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,Ωfx,xn+1
= sup
h∈Ᏼγ(R+)
Tφ,Ω,hfx,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 (γ)φ,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 ofTΩ. 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)= L(logL)1−ε(Sn−1), 0< ε <1) (see [8]).
Kim et al. [18] studied theLp-boundedness of Tφ,Ω,h as described in the following theorem.
Theorem 1.1. 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. ThenTφ,Ω,1 is bounded onLp(Rn+1) for 1< 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.
Theorem 1.2. Letφbe aC2, convex, and increasing function satisfyingφ(0)=0. IfΩ∈ LlogL(Sn−1) and h∈Δγ(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 ofTφ,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 thath∈Ᏼγ(R+) for some 1< γ≤ ∞and Ω∈L(logL)1/γ(Sn−1). ThenTφ,Ω,his bounded onLp(Rn) for 1< 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 (γ)φ,Ωdefined in (1.6).
The study of the maximal operator (γ)φ,Ωbegan by Chen and Lin in [10] and subsequently by many other authors [1,12,19]. For example, Chen and Lin proved the following.
Theorem 1.4 [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.
Very recently, Al-Qassem improved the result inTheorem 1.4as described in the fol- lowing theorem.
Theorem 1.5 [1]. Letn≥2 and let (γ)Ω 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 >0 and satisfies (1.2) such that (2)Ω is not bounded onL2(Rn).
Our result regarding (γ)φ,Ωis the following.
Theorem 1.6. Let (γ)φ,Ωbe 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 <∞and 1< γ≤2, and it is bounded onL∞(Rn) forγ=1.
Remarks 1.7. (1) In order to clarify the relations between Theorems1.1,1.2,1.4, and 1.5and theorems1.3and1.6, we remark that on the unit sphere Sn−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 (γ)Ω 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∈Ᏼγ(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.
Throughout the rest of the paper, the letterCwill stand for a constant but not neces- sarily the same one in each occurrence.
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,μ∗ onRn+1by
Rn+1 f dσφ,k,μ=
akμ≤|u|<ak+1μ 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.
Lemma 2.2. Letμ∈N∪ {0},aμ=2(μ+1), and letφ(·) be an arbitrary function onR+. Let Ωμ(·) be a function on Sn−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)= a
k+1μ
akμ
Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)
2dt 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)1/2akμξ±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{akμtξ·x+ηφ(akμt)}−e−iηφ(akμt)Ωμ(x)dσ(x) 2
dt
t , (2.5) which easily implies
Iμ,k(ξ,η)≤C(μ+ 1)1/2aμakμξ. (2.6) By combining both estimates in (2.3) and (2.6), we get
Iμ,k(ξ,η)≤C(μ+ 1)β/2aβμakμξβ(μ+ 1)(1−β)/2 (2.7)
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(takμξ·x+ηφ(takμ))dσ(x)
2
=
Sn−1×Sn−1Ωμ(x)Ωμ(u)e−iakμtξ·(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−iakμtξ·(x−u)dt t
dσ(x)dσ(u). (2.9) By employing integration by parts, we get
aμ
1 e−iakμtξ·(x−u)dt t
≤Cmin(μ+ 1), (μ+ 1)akμξ·(x−u)−1 (2.10)
and hence aμ
1 e−iakμtξ·(x−u)dt t
≤C(μ+ 1)βakμξ−βξ·(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−iakμtξ·(x−u)dt t
≤C(μ+ 1)akμξ−1/4ξ·(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)−1/2dσ(x)dσ(u) 1/2
. (2.13) Since the last integral is finite, we get
Iμ,k(ξ,η)≤C(μ+ 1)1/2a2μakμξ−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.
Lemma 2.3. Letμ∈N∪ {0},aμ=2(μ+1),h∈Ᏼγ(R+) for some 1< γ <∞, and letφ(·) be an arbitrary function on R+. LetΩμ(·) be a function on Sn−1 satisfying the following conditions: (i)ΩμL2(Sn−1)≤a2μ, and (ii)ΩμL1(Sn−1)≤1. Let
Rk,μ(ξ,η)= ak+1μ
akμ
Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)h(t)dt
t . (2.15)
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/γakμξ±α/γ(μ+1). (2.17) Proof. By H¨older’s inequality, we have
Rk,μ(ξ,η)
≤ a
k+1μ
akμ
h(t)γdt t
1/γ ak+1μ akμ
Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)
γdt t
1/γ
≤∞
0
h(t)γdt t
1/γ ak+1μ akμ
Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)
γdt t
1/γ
≤ a
k+1μ
akμ
Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)
γdt 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)≤ a
k+1μ
akμ
Sn−1Ωμ(x)e−i(tξ·x+ηφ(t))dσ(x)
2dt 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.
Lemma 2.4. 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∈Z
σk∗gk2 1/2
p0
≤CBλ
k∈Z
gk2 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
≤CpBλfp (2.21)
holds for all f inLp(Rn). The constantCp is independent ofBand the linear transforma- tionL.
Our aim now is to establish the following result.
Lemma 2.5. Leth∈Ᏼγ(R+) for someγ >1 and 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 constantCpwhich is independent ofμsuch that
σφ,μ∗ (f)p≤Cp(μ+ 1)1/γfp. (2.22)
Proof. Without loss of generality, we may assume thatΩμ≥0 andh≥0. By H¨older’s inequality, we have
σφ,μ∗ (f)≤ a
k+1μ
akμ
h(t)γdt t
1/γ
Υ∗μ
|f|γ1/γ
≤CΥ∗μ
|f|γ1/γ
, (2.23)
whereRn+1 f dΥk,μ=
akμ≤|u|< ak+1μ 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.
Lemma 2.6. Leth∈Ᏼγ(R+) for someγ≥2 and letΩμ be a function on Sn−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∈Z
σφ,k,μ∗gk21/2
Lp(Rn+1)
≤Cp(μ+ 1)1/γ
k∈Z
gk21/2
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∗gkx,xn+1γ≤C ak+1μ
akμ
Sn−1
Ωμ(y)gkx−yt,xn+1−φ(t)γdσ(y)dt t . (2.26)
Letd=p/γ. For{gk} ∈Ld(Rn+1,l2), by duality, there exists a nonnegative functionω∈ Ld(Rn+1) such thatωLd≤1 and
k∈Z
σφ,k,μ∗gkγ1/γ
γ
p
=
Rn+1
k∈Z
σφ,k,μ∗gk
x,xn+1γωx,xn+1
dx dxn+1. (2.27) Therefore, by (2.27) and a change of variable, we get
k∈Z
σφ,k,μ∗gkγ1/γ
γ
p
≤C
Rn+1
k∈Z
gk
x,xn+1γMμωx,xn+1
dx dxn+1, (2.28) where
Mμωx,xn+1=sup
k∈Z
akμ≤|y|<ak+1μ ωx+y,xn+1+φ|y|Ωμ(y)|y|−nd y. (2.29) By H¨older’s inequality, we obtain
k∈Z
σφ,k,μ∗gkγ1/γ
γ
p
≤C
k∈Z
|gk|γ 1/γ
γ
p
Mμωd. (2.30)
By [4, Lemma 4.7], we haveMμωLd ≤Cp(μ+ 1) which in turn implies
k∈Z
σφ,k,μ∗gkγ1/γ
p
≤C(μ+ 1)1/γ
k∈Z
gkγ1/γ
p
. (2.31)
Moreover, again byLemma 2.5, we have
sup
k∈Z
σφ,k,μ∗gk
p
≤ σφ,μ∗
sup
k∈Z
gk
p
≤Cp(μ+ 1)1/γ sup
k∈Z
gk
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 1≤γ≤2. Forμ∈N, let Jμbe the set of pointsx∈Sn−1which satisfy 2μ≤ |Ω(x)|<2μ+1.
