SINGULAR OPERATORS WITH KERNELS IN Llog LαSn −1H.. AL-QASSEM Received 15 November 2005; Revised 20 May 2006; Accepted 28 May 2006 We establish the L p-boundedness for a class of singular
Trang 1SINGULAR OPERATORS WITH KERNELS IN L(log L)α(Sn −1)
H M AL-QASSEM
Received 15 November 2005; Revised 20 May 2006; Accepted 28 May 2006
We establish the L p-boundedness for a class of singular integral operators and a class
of related maximal operators when their singular kernels are given by functions Ω in
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 letx denotex/ | x |forx ∈ R n \{0}and letp denote 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 ∈ R n } be the surface of revolution generated by a suitable functionφ : [0, ∞) → R Let K Ω,h(y) be a Calder ´on-Zygmund-type kernel of the form
K Ω,h(y) = h
whereh : [0, ∞) → Cis a measurable function, andΩ is an integrable function over Sn −1, satisfying
LetL(logL) α(Sn −1) (forα > 0) denote the space of all those measurable functionsΩ on
Sn −1which satisfy
Ω L(logL) α(Sn −1 )=
Sn −1
Ω(y)logα
2 +Ω(y)dσ(y) < ∞ . (1.3)
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 96732, Pages 1 16
DOI 10.1155/JIA/2006/96732
Trang 2Forγ > 1, defineΔγ(R +) to be the set of all measurable functionsh onR +satisfying the condition
sup
R>0
R −1
R
0
h(t)γ dt
1/γ
and defineΔ∞(R +)= L ∞(R +) Also, forγ ≥1, defineᏴγ(R +) to be the set of all measur-able functionsh on R +satisfying the condition h L γ( R + ,dr/r) =(
R +| h(r) | γ dr/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 theL pmapping properties of singular integral operatorsT φ, Ω,hinRnalong the surface of revolutionΓφdefined for (x, x n+1)∈ R n × R =
Rn+1by
T φ, Ω,h f
x, x n+1
=p.v
Rn f
x − y, x n+1 − φ
| y |K Ω,h(y)d y. (1.5)
Also, we are interested in studying theL p-boundedness of the related maximal operator
(φ, γ)Ωgiven by
(φ, γ)Ωf
x, x n+1
h ∈Ᏼγ( R + )
T φ, Ω,h f
Wheneverφ(t) ≡0 andh ≡1, thenT φ, Ω,hessentially is the classical Calder ´on-Zygmund singular integral operatorTΩgiven by
TΩf (x) =p.v
Rn f (x − y) Ω(y )|y | − n d y. (1.7)
Ifφ(t) ≡0, we will denoteT φ, Ω,hbyT Ω,hand (φ, γ)Ωby (Ωγ)
The investigation of the L p-boundedness problem of the operatorTΩ began with Calder ´on and Zygmund in their well-known papers [7,8] The operatorT φ, 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 theL p-boundedness
ofTΩholds for 1< p < ∞ifΩ∈ L log L(S n −1) Moreover, the conditionΩ∈ L log L(S n −1) turns out to be the most desirable size condition for the L p-boundedness ofTΩ This was made clear by Calder ´on and Zygmund where it was shown thatTΩ may fail to be bounded on L p for any p if the conditionΩ∈ L log L(S n −1) is replaced by any weaker metric conditionΩ∈ L ϕ(Sn −1) with aϕ satisfying ϕ(t) = o(t log t) as t → ∞(e.g.,ϕ(t) = L(logL)1− ε(Sn −1), 0< ε < 1) (see [8])
Trang 3Kim et al [18] studied theL p-boundedness of T φ, Ω,h as described in the following theorem
Theorem 1.1 Let T φ, Ω,h be given as in ( 1.5 ) and h ≡ 1 Assume thatΩ∈ C ∞(Sn −1) and satisfies ( 1.2 ) Assume also that φ( ·) is in C2of [0, ∞), convex, and increasing Then T φ,Ω,1
is bounded on L p(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 onΩ from Ω∈ C ∞(Sn −1) to
Ω∈ L q(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∈ L q(Sn −1) is replaced by the weaker conditionΩ∈
L log L(S n −1) In fact, they proved the following
Theorem 1.2 Let φ be a C2, convex, and increasing function satisfying φ(0) = 0 IfΩ∈
L log L(S n −1) and h ∈Δγ(R +) for some γ > 1, then the operator T φ, Ω,h in ( 1.2 )–( 1.5 ) is bounded on L p(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 as γ
ap-proaches 1 ForL p-boundedness results on singular integrals forp satisfying |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 the L p -boundedness of T φ, Ω,h holds for p outside 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 on h is
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 T φ, Ω,h be given as in ( 1.2 )–( 1.5 ) and let φ be a C2, convex, and in-creasing function satisfying φ(0) = 0 Suppose that h ∈Ᏼγ(R +) for some 1 < γ ≤ ∞ and
Ω∈ L(logL)1/γ
(Sn −1) Then T φ, Ω,h is bounded on L p(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 theL p-boundedness for the full range 1< p < ∞, a new maximal function that
intervenes here in the proof ofTheorem 1.3is the maximal operator (φ, γ)Ωdefined in (1.6) The study of the maximal operator (φ, γ)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] Assume n ≥ 2, 1 ≤ γ ≤ 2, andΩ∈ C(S n −1) satisfying ( 1.2 ) Then (Ωγ)
is bounded on L p(Rn ) for ( γn) < p < ∞ Moreover, the range of p is the best possible.
Very recently, Al-Qassem improved the result inTheorem 1.4as described in the fol-lowing theorem
Trang 4Theorem 1.5 [1] Let n ≥ 2 and let (Ωγ) be given as in ( 1.6 ) Then
(a) ifΩ∈ L(logL)1/γ
(Sn −1) and satisfies ( 1.2 ), then (Ωγ) is bounded on L p(Rn ) for γ ≤
p < ∞ ;
(b) there exists an Ω which lies in L(logL)1/2 − ε(Sn −1) for all ε > 0 and satisfies ( 1.2 ) such that (2)
Ω is not bounded on L2(Rn ).
Our result regarding (φ, γ)Ωis the following
Theorem 1.6 Let (φ, γ)Ωbe given as in ( 1.6 ) and let φ be a C2, convex and increasing function satisfying φ(0) = 0 SupposeΩ∈ L(logL)1/γ (Sn −1) and satisfies ( 1.2 ) Then (φ, γ)Ωis bounded
on L p(Rn ) for γ ≤ p < ∞ and 1 < γ ≤ 2, and it is bounded on L ∞(Rn ) for γ = 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:
L q
Sn −1
⊂ L(logL)
Sn −1
⊂ H1
Sn −1
⊂ L1
Sn −1
,
L(logL) β
Sn −1
⊂ L(logL) α
Sn −1
if 0< α < β, L(logL) α
Sn −1
⊂ H1
Sn −1
∀ α ≥1,
(1.8)
while
L(logL) α
Sn −1
H1
Sn −1
L(logL) α
Sn −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 ∈ L p 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 L p-boundedness
of (Ωγ) holds for 2< γ < ∞ A point worth noting is thatTheorem 1.3implies theL p -boundedness ofT φ, Ω,hifh ∈Ᏼγ(R +) for all 1< γ ≤ ∞.
(3) We notice that the singular integral operators T Ω,h are bounded onL p 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 onL pifΩ∈ L(logL)(S n −1) The reason for this new phenomenon on singular integrals is that the singular operatorsT Ω,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, sinceL log L(S n −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 letterC will stand for a constant but not
neces-sarily the same one in each occurrence
Trang 52 Some basic lemmas
Let us begin this section with the following definition
Definition 2.1 For an arbitrary function φ( ·) onR +,a μ > 1, andΩμ: Sn −1→ Rwithμ ∈
N ∪ {0}, define the sequence of measures{ σ φ,k,μ:k ∈Z}and the corresponding maximal operatorσ φ,μ ∗ onRn+1by
Rn+1 f dσ φ,k,μ =
a k ≤| u | <a k+1 μ
f
u, φ
| u |KΩμ,h(u)du,
σ φ,μ ∗ (f ) =sup
k ∈Z
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 Let μ ∈ N ∪ {0} , a μ =2(μ+1) , and let φ( ·) be an arbitrary function onR + Let
Ωμ(·) be a function on Sn −1 satisfying the following conditions: (i) Ω μ L2(Sn −1 )≤ a2
μ , (ii)
Ω μ L1(Sn −1 )≤ 1, and (iii)Ωμ satisfies the cancellation conditions in ( 1.2 ) with Ω replaced
byΩμ Let
I μ,k(ξ, η) =
a k+1 μ
a k
Sn −1Ωμ(x)e − i(tξ · x+ηφ(t)) dσ(x)
2dt t
1/2
Then there exist positive constants C and α such that
I μ,k(ξ, η) ≤ C(μ + 1)1/2, (2.3)
I μ,k(ξ, η) ≤ C(μ + 1)1/2a k
μ ξ± α/(μ+1)
where ξ ∈ R n , η ∈ R , and t ± α =inf{t α,t − α } The constants C and α are independent of k,
μ, ξ, η and φ( ·).
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
I μ,k(ξ, η) 2
≤
a μ
1
Sn −1
e − i { a k tξ · x+ηφ(a k t) } − e − iηφ(a k t)Ω
μ(x)dσ(x)2
dt
t , (2.5)
which easily implies
I μ,k(ξ, η) ≤ C(μ + 1)1/2 a μa k ξ. (2.6)
By combining both estimates in (2.3) and (2.6), we get
I μ,k(ξ, η) ≤ C(μ + 1) β/2 a β μa k ξβ
(μ + 1)(1− β)/2 (2.7)
Trang 6for any 0< β < 1 and for some constant C > 0 By the last estimate and by letting β =
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(ta k ξ · x+ηφ(ta k))dσ(x)
2=
Sn −1×S n −1Ωμ(x)Ωμ(u)e − ia k tξ ·(x − u) dσ(x)dσ(u),
(2.8) which leads to
I μ,k(ξ, η) 2
=
Sn −1×Sn −1Ωμ(x)Ωμ(u)
a μ
1 e − ia k tξ ·(x − u) dt
t
dσ(x)dσ(u). (2.9)
By employing integration by parts, we get
a μ
1 e − ia k tξ ·(x − u) dt
t
≤ C min
(μ + 1), (μ + 1)a k
μ ξ ·(x − u)−1
(2.10)
and hence
a μ
1 e − ia k tξ ·(x − u) dt
t
≤ C (μ + 1) βa k ξ− 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
a μ
1 e − ia k tξ ·(x − u) dt
t
≤ C(μ + 1)a k ξ−1/4ξ ·(x − u)−1/4
By Schwarz’s inequality, condition (i) onΩμ, and (2.9)–(2.12), we get
I μ,k(ξ, η) 2
≤ C(μ + 1)a4μa k ξ−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
I μ,k(ξ, η) ≤ C(μ + 1)1/2 a2
μa k ξ−1/8
As above, by combining (2.14) with (2.3), we obtain the second estimate in (2.4).Lemma
Lemma 2.3 Let μ ∈ N ∪ {0} , a μ =2(μ+1) , h ∈Ᏼγ(R +) for some 1 < γ < ∞ , and let φ( ·)
be an arbitrary function on R + LetΩμ(·) be a function on Sn −1 satisfying the following conditions: (i) Ω μ L2(Sn −1 )≤ a2
μ , and (ii) Ω μ L1(Sn −1 )≤ 1 Let
R k,μ(ξ, η) =
a k+1 μ
a k
Sn −1Ωμ(x)e − i(tξ · x+ηφ(t)) dσ(x)
h(t)dt
Trang 7Then there exist positive constants C independent of k, φ, and μ such that
R k,μ(ξ, η) ≤ C(μ + 1)1/γ
R k,μ(ξ, η) ≤ C(μ + 1)1/γ a k ξ± α/γ (μ+1)
Proof By H¨older’s inequality, we have
R k,μ(ξ, η)
≤
a k+1
μ
a k
h(t)γ dt
t
1/γ a k+1 μ
a k
Sn −1Ωμ(x)e − i(tξ · x+ηφ(t)) dσ(x)
γ dt t 1/γ
≤∞
0
h(t)γ dt
t
1/γ a k+1
μ
a k
Sn −1Ωμ(x)e − i(tξ · x+ηφ(t)) dσ(x)
γ dt t
1/γ
≤
a k+1
μ
a k
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
R k,μ(ξ, η) ≤ a k+1
μ
a k
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 { σ k:k ∈Z} be a sequence of Borel measures onRn and let L :Rn → R d be a linear transformation Suppose that for all k ∈ Z, ξ ∈ R n , for some a ∈[2,∞), λ > 0, α > 0,
C > 0, and for some B > 1,
(i) σ k ≤ CB λ ;
(ii)| σ k(ξ) | ≤ CB λ(a kB | L(ξ) |) ± α/B ;
(iii) for some p0∈(2,∞),
k ∈Z
σ k ∗ g k 2
1/2
p0
≤ CB λ
k ∈Z
g k 2
1/2
p0
(2.20)
holds for arbitrary functions { g k } onRn Then for p 0< p < p0there exists a positive constant
C p such that
k ∈Z
σ k ∗ f
p
Trang 8holds for all f in L p(Rn ) The constant C p is independent of B and the linear transforma-tion L.
Our aim now is to establish the following result
Lemma 2.5 Let h ∈Ᏼγ(R +) for some γ > 1 and letΩμ be a function satisfying conditions (i) and (ii) in Lemma 2.2 Assume φ is in C2([0,∞)), convex, and increasing Then forγ <
p ≤ ∞ and f ∈ L p(Rn+1 ), there exists a positive constant C p which is independent of μ such that
σ ∗
φ,μ(f )
p ≤ C p(μ + 1)1/γ
Proof Without loss of generality, we may assume thatΩμ ≥0 andh ≥0 By H¨older’s inequality, we have
σ φ,μ ∗ (f ) ≤
a k+1 μ
a k
h(t)γ dt
t
1/γ
Υ∗
μ
| f | γ 1/γ
≤ C
Υ∗
μ
| f | γ 1/γ
where
Rn+1 f dΥk,μ =a k ≤| u | < a k+1
μ f (u, φ( | u |))| u | − nΩμ(u )du andΥ∗
μ(f ) =sup k ∈Z ||Υ k,μ |∗
f | Therefore, in order to prove (2.22), it suffices to prove that
Υ∗
μ(f )
L p( Rn+1)≤ C p(μ + 1) f L p( 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 2.6 Let h ∈Ᏼγ(R +) for some γ ≥ 2 and letΩμ be a function on S n −1satisfying conditions (i) and (ii) in Lemma 2.2 Let φ be in C2([0,∞)), convex, and an increasing function with φ(0) = 0 Then, for γ < p < ∞ , there exists a positive constant C p which is independent of μ such that
k ∈Z
σ φ,k,μ ∗ g k 2
1/2
L p( Rn+1)
≤ C p(μ + 1)1/γ
k ∈Z
g k 2
1/2
L p( Rn+1)
(2.25)
holds for arbitrary measurable functions { g k } onRn+1
Proof We follow a similar argument as in [6] Letγ < p < ∞ By H¨older’s inequality and
the condition onh, we get
σ φ,k,μ ∗ g k
x, x n+1γ
≤ C
a k+1 μ
a k
Sn −1
Ωμ(y)g k
x − yt, x n+1 − φ(t)γ
dσ(y) dt
t .
(2.26)
Trang 9Letd = p/γ For{ g k } ∈ L d(Rn+1, 2), by duality, there exists a nonnegative functionω ∈
L d
(Rn+1) such that ω L d ≤1 and
k ∈Z
σ φ,k,μ ∗ g kγ
1/γ
γ
p
=
Rn+1
k ∈Z
σ φ,k,μ ∗ g k
x, x n+1γ
ω
x, x n+1
dx dx n+1
(2.27) Therefore, by (2.27) and a change of variable, we get
k ∈Z
σ φ,k,μ ∗ g kγ 1/γ
γ
p
≤ C
Rn+1
k ∈Z
g k
x, x n+1γ
M μ ω
x, x n+1
dx dx n+1,
(2.28) where
M μ ω
x, x n+1
=sup
k ∈Z
a k ≤| y | <a k+1 μ
ω
x + y, x n+1+φ
| y |Ωμ(y )| y | − n d y. (2.29)
By H¨older’s inequality, we obtain
k ∈Z
σ φ,k,μ ∗ g kγ 1/γ
γ
p
≤ C
k ∈Z
| g k | γ
1/γ
γ
p
M μ ω
d (2.30)
By [4, Lemma 4.7], we have M μ ω L d ≤ C p(μ + 1) which in turn implies
k ∈Z
σ φ,k,μ ∗ g kγ 1/γ
p
≤ C(μ + 1)1/γ
k ∈Z
g kγ 1/γ
p
Moreover, again byLemma 2.5, we have
supk ∈Zσ φ,k,μ ∗ g k
p
≤
σ φ,μ ∗
sup
k ∈Z
g k
p
≤ C p(μ + 1)1/γ
sup
k ∈Z
g k
p (2.32)
By using the operator interpolation theorem between (2.31) and (2.32) and sinceγ ∈
We are now ready to present the proofs of our main results
Since the proof ofTheorem 1.3will rely heavily onTheorem 1.6as well as on its proof,
we start by provingTheorem 1.6
Proof of Theorem 1.6 Assume thatΩ satisfies (1.2) and belongs toL(logL)1/γ
(Sn −1) and
1≤ γ ≤2 Forμ ∈ N, let J μbe the set of pointsx ∈Sn −1which satisfy 2μ ≤ |Ω(x)| < 2 μ+1
Trang 10Also, we let J0 be the set of all those pointsx ∈Sn −1 which satisfy|Ω(x)| < 2 For μ ∈
N ∪ {0}, setb μ = ΩχJμ,λ0=1, andλ μ = b μ 1forμ ∈ N Set I = { μ ∈ N:λ μ ≥2−2μ }and define the sequence of functions{Ω μ } μ ∈ I ∪{0}by
Ω0(x) =
μ ∈{0}∪(N− I)
b μ(x) −
μ ∈{0}∪(N− I)
Sn −1b μ(x)dσ(x)
,
Ωμ(x) =λ μ−1
b μ(x) −
Sn −1b μ(x)dσ(x)
forμ ∈ I.
(3.1)
Forμ ∈ I ∪ {0}, set a μ =2(μ+1) Then one can easily verify that the following hold for all
μ ∈ I ∪ {0}and for some positive constantC:
Ωμ
μ ∈ I ∪{0}
(μ + 1)1/γ λ μ ≤ C Ω L(logL)1/γ (Sn −1 ), (3.3)
Sn −1Ωμ(u)dσ(u) =0; Ω=
μ ∈ I ∪{0}
By (3.4), we have (φ, γ)Ωf (x, x n+1)≤μ ∈ I ∪{0} λ μ (φ, γ)Ωμ f (x, x n+1) and hence the proof of
Theorem 1.6is completed if we can show that
(γ)
φ,Ωμ f
L p( Rn+1)≤ C p(μ + 1)1/γ f L p( 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,Ωμ f
x, x n+1
h ∈Ᏼγ( R + )
∞
0 h(t)
Sn −1f
x − tu, x n+1 − φ(t)
Ωμ(u)dσ(u) dt
t
=
∞
0
Sn −1 f
x − tu, x n+1 − φ(t)
Ωμ(u)dσ(u)
2dt t 1/2
=
k ∈Z
a k+1 μ
a k
Sn −1f
x − tu, x n+1 − φ(t)
Ωμ(u)dσ(u)
2dt t
1/2
(3.6)
Let{ ψ k,μ } ∞
−∞ be a smooth partition of unity in (0,∞) adapted to the intervalE k,μ =
[a − k −1
μ ,a − k+1
μ ] To be precise, we require the following:ψ k,μ ∈ C ∞, 0≤ ψ k,μ ≤1,
k ψ k,μ(t) =
1, suppψ k,μ ⊆ E k,μ,| d s ψ k,μ(t)/dt s | ≤ C s /t s, whereC s is independent of the lacunary se-quence{ a k:k ∈Z} Define the multiplier operators S k,μinRn+1by
Sk,μ f(ξ, η) = ψ k,μ| ξ | f (ξ, η) for (ξ, η) ∈ R n × R (3.7)
...Since the proof ofTheorem 1.3will rely heavily onTheorem 1.6as well as on its proof,
we start by provingTheorem 1.6
Proof of Theorem 1.6 Assume thatΩ satisfies (1.2) and belongs... 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... h(t)dt
Trang 7Then there exist positive constants C independent of k, φ, and μ such that
R