# L

Authors: Yanping Chen; Kai Zhu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We establish theLp boundedness for some commutators of oscillatory singular integrals with the kernel condition which was introduced by Grafakos and Stefanov. Our theorems contain various conditions on the phase function.


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## Body

## 1. Introduction

The homogeneous singular integral operatorTΩ is defined by
(1)TΩf(x)=p.v.∫Rn‍Ω(x-y)|x-y|nf(y)dy,
where Ω∈L1(Sn-1) satisfies the following conditions.(a)
Ω is homogeneous function of degree zero on Rn∖{0}; that is,
(2)Ω(tx)=Ω(x)
for any t>0 and x∈Rn∖{0}.(b)
Ω has mean zero on Sn-1, the unit sphere in Rn; that is,
(3)∫Sn-1‍Ω(x′)dσ(x′)=0.The oscillatory singular integral we will consider here is defined by(4)Tϕf(x)=p.v.∫Rn‍eiϕ(y)Ω(y)|y|nf(x-y)dy.
If ϕ(x)≡0, the operator Tϕ becomes the singular integral operator TΩ.Whenϕ(x)=P(x) is a real polynomial, the Lp boundedness of Tϕ was first studied by Ricci and Stein [1] with Ω∈C1(Sn-1), and Hu and Pan [2] obtained the weighted H1 boundedness of Tϕ. When Ω∈Lr(Sn-1), r>1, Lu and Zhang proved the Lp boundedness [3] and this was extended to the case of Ω∈Lln⁡+L(Sn-1) by Ojanen [4] and the case of Ω∈H1(Sn-1) by Fan and Pan [5].Grafakos and Stefanov [6] introduced a class of kernel functions Fα(Sn-1) which contains all Ω(y)∈L1(Sn-1) satisfying (3) and
(5)sup⁡ξ∈Sn-1∫Sn-1‍|Ω(y)|(ln⁡|y·ξ|-1)1+αdσ(y)<∞,
where α>0 is a fixed constant. This kernel condition has been considered by many authors [7–13].The singular integral along surfaces which is defined by(6)Tϕ,Ωf(x,xn+1)=p.v.∫RnΩ(y)|y|nf(x-y,xn+1-ϕ(|y|))dy
was also studied by many authors [14–18]. Under the condition Ω∈Fα(Sn-1), Pan et al. [16] established the following Theorem.Theorem A (see [16]).
Letϕ(t)∈C1([0,∞)), ϕ(0)=ϕ′(0)=0, and ϕ′ is a convex increasing function for t>0, Ω∈Fα(Sn-1) for some α>0; then, Tϕ,Ω is bounded on Lp(Rn+1) for (2+2α)/(1+2α)<p<2+2α.Later, Cheng and Pan [14] improved the result for n=2 by removing the condition ϕ′(0)=0.Theorem B (see [14]).
Letϕ(t)∈C1([0,∞)), ϕ(0)=0, and ϕ′ is a convex increasing function for t>0, Ω∈Fα(Sn-1) for some α>0; then, Tϕ,Ω is bounded on Lp(R3) for (2+2α)/(1+2α)<p<2+2α.It has been proved that the boundedness ofTϕ on Lp(Rn) can be obtained from the Lp(Rn+1) boundedness of Tϕ,Ω (see [5]).For a functionb∈Lloc⁡(Rn), let A be a linear operator on some measurable function space; the commutator between A and b is defined by [b,A]f(x)∶=b(x)Af(x)-A(bf)(x).It has been proved by Hu [19] that Ω∈L(log⁡L)2(Sn-1) is a sufficient condition for the commutator to be bounded on Lp(Rn), which is defined by
(7)[b,TΩ]f(x)=p.v.∫Rn‍Ω(x-y)|x-y|n(b(x)-b(y))f(y)dy.
Recently, Chen and Ding [20] established the Lp boundedness of the commutator of singular integrals with the kernel condition Ω∈Fα(Sn-1).It is natural to ask whether the similar result holds for the commutators of oscillatory singular integrals, which is defined by(8)[b,Tϕ]f(x)=p.v.∫Rn‍eiϕ(y)Ω(y)|y|n(b(x)-b(x-y))dsdsdsdsd×f(x-y)dy.
In this paper, we will give a positive answer to the above question by imposing some conditions on ϕ.We first prove the boundedness of the commutator of singular integral along surfaces, which is defined by(9)[b,Tϕ,Ω]f(x,xn+1)=p.v.∫Rn‍Ω(y)|y|n(b(x,xn+1)-b(x-y,xn+1-ϕ(|y|)))fsdfdsfdgfd×f(x-y,xn+1-ϕ(|y|))dy.Theorem 1.
LetΩ be a function in L1(Sn-1) satisfying (2) and (3), b∈BMO(Rn+1), radial function ϕ∈C1([0,∞)) with ϕ(0)=ϕ′(0)=0, and ϕ′ is a convex increasing function. If Ω∈Fα(Sn-1) for some α>1, then [b,Tϕ,Ω] is bounded on L2(Rn+1).Theorem 2.
LetΩ be a function in L1(S1) satisfying (2) and (3), b∈BMO(R3), radial function ϕ(|t|)=|t|. If Ω∈Fα(S1) for some α>1, then [b,Tϕ,Ω] is bounded on Lp(R3) for (α+1)/α<p<α+1.Remark 3.
However, forn≥3, we can not prove the Lp(Rn+1) boundedness of [b,Tϕ,Ω] by our method using Lemma 11, since the conditions imposed on ϕ in Theorem 1 conflict with Lemma 11. Only when n=2 by removing the condition ϕ′(0)=0 in Theorem 1 can we eliminate the conflict, and ϕ(|t|)=|t| is a feasible function. Also, by another method, it is hard to give the boundedness of the maximal operator defined by
(10)[b,Mϕ,Ω]f(x,xn+1)=sup⁡j∈Z|∫2j<|y|<2j+1Ω(y)|y|ndfdfdffdfdsdsdsf×((x-y,xn+1-ϕ(|y|))b(x,xn+1)-bfdfdfdfdfdfdfdfdff×(x-y,xn+1-ϕ(|y|)))dfdfdffdfdsdsdsf×f(x-y,xn+1-ϕ(|y|))dy∫2j<|y|<2j+1Ω(y)|y|n|.
Then we give the boundedness of the commutators of oscillatory singular integral[b,Tϕ].Letb(x)∈BMO(Rn), x~=(x,xn+1)∈Rn+1, B(x~)=b(x), and we have the following result.Theorem 4.
If
[
B
,
T
ϕ
,
Ω
]
is bounded on
L
p
(
R
n
+
1
)
with bound
C
∥
B
∥
B
M
O
(
R
n
+
1
)
, then
[
b
,
T
ϕ
]
is bounded on
L
p
(
R
n
)
with bound
C
∥
b
∥
B
M
O
(
R
n
)
.Combining Theorem4 with Theorems 1 and 2, respectively, we can get the following two theorems immediately.Theorem 5.
LetΩ be a function in L1(Sn-1) satisfying (2) and (3), b∈BMO(Rn), radial function ϕ∈C1([0,∞)) with ϕ(0)=ϕ′(0)=0, and ϕ′ is a convex increasing function. If Ω∈Fα(Sn-1) for some α>1, then [b,Tϕ] is bounded on L2(Rn).Theorem 6.
LetΩ be a function in L1(S1) satisfying (2) and (3), b∈BMO(R2), radial function ϕ(|t|)=|t|. If Ω∈Fα(S1) for some α>1, then [b,Tϕ] is bounded on Lp(R2) for (α+1)/α<p<α+1.In above theorems, the phase functions are radial. But when Ricci and Stein first studied the oscillatory singular integralTϕ, they take ϕ(x)=P(x), apparently nonradial. In Theorem 7, we will take ϕ(x)=P(x)=∑|α|/2=1m‍aαxα, and this condition was mentioned in [21].Theorem 7.
LetΩ be a function in L1(Sn-1) satisfying (2) and (3), b∈BMO(Rn). If Ω∈Fα(Sn-1) is an odd kernel for some α>1, ϕ(x)=∑|α|/2=1m‍aαxα is an even phase; then, [b,Tϕ] extends to a bounded operator from Lp(Rn) into itself for (α+1)/α<p<α+1.


## 2. Lemmas

We give some lemmas which will be used in the proof of Theorems1 and 2.Lemma 8.
Letmδ(ξ~)∈C1(Rn+1)(0<δ<∞) be a family of multipliers such that supp⁡mδ⊂{ξ~:|ξ~|≤δ}, ∇ξmδ=(∂mδ/∂ξ1,...,∂mδ/∂ξn), and for some constants C,0<A≤1/2, and α>0(11)∥mδ∥∞≤Cmin⁡{Aδ,log⁡-(α+1)(2+δ)},∥∇ξmδ∥∞≤C.
Let Tδ be the multiplier operator defined by Tδf^(ξ~)=mδ(ξ~)f^(ξ~), ξ~=(ξ,ξn+1). For b∈BMO(Rn+1), denote by [b,Tδ] the commutator of Tδ. Then for any 0<ν<1, there exists a positive constant C=C(n,ν) such that
(12)∥[b,Tδ]f∥2≤C∥b∥BMO(Rn+1)(Aδ)νlog⁡(1A)∥f∥2,ifδ<10A;∥[b,Tδ]f∥2≤C∥b∥BMO(Rn+1)log⁡-(α+1)ν+1(2+δ)∥f∥2,ifδ>1A.Proof.
We assume that∥b∥BMO(Rn+1)=1. Let x~=(x,xn+1) and let Ψ(x~) be a radial function such that supp⁡Ψ⊂{x~:1/4≤|x~|≤4}, and
(13)∑l∈Z‍Ψ(2-lx~)=1
for |x~|>0. Set Ψ0(x~)=∑l=-∞0‍Ψ(2-lx~) and Ψl(x~)=Ψ(2-lx~) for positive integer l. Let Kδ(x~)=mδ∨(x~) the inverse Fourier transform of mδ. Split Kδ as
(14)Kδ(x~)=Kδ(x~)Ψ0(x~)+∑l=1∞‍Kδ(x~)Ψl(x~)=∑l=0∞‍Kδ,l(x~).
LetTδ,l be the convolution operator whose kernel is Kδ,l; that is, Tδ,lf=Kδ,l*f. Recall that supp⁡mδ⊂{ξ~:|ξ~|≤δ}. Trivial computation shows that ∥Kδ,l∥∞≤∥Kδ∥∞≤∥mδ∥1≤Cδn+1. This via the Young inequality says that
(15)∥Tδ,lf∥∞≤Cδn+1∥f∥1.
Note that ∫Rn+1‍Ψ^(η~)dη~=0. Thus
(16)∥Kδ,l^∥∞=∥∫Rn+1‍(mδ(ξ-2-lη,ξn+1-2-lηn+1)-mδdsdsdsddsd×(ξ,ξn+1-2-lηn+1))Ψ^(η~)dη~∫Rn+1‍(mδ(ξ-2-lη,ξn+1-2-lηn+1)-mδ∥∞≤C2-l∥∇ξmδ∥∞∫Rn+1|η||Ψ^(η~)|dη~≤C2-l∥∇ξmδ∥∞∫Rn+1|η~||Ψ^(η~)|dη~≤C2-l.
On the other hand, by the Yong inequality, we have
(17)∥Kδ,l^∥∞≤∥Kδ^∥∞∥Ψl^∥1≤Cmin⁡{Aδ,log⁡-(α+1)(2+δ)}.
Then, using the same argument of the proof of Lemma 2 in [22] we can prove Lemma 8.Let the measureσj on Rn+1 be defined by
(18)∫Rn+1‍f(y,yn+1)dσj=∫Rn‍f(y,ϕ(|y|))Ω(y′)|y|nχ{2j<|y|≤2j+1}dy
for all j∈Z. Define the maximal operator in Rn+1 by σ*f=sup⁡j∈Z|σj|*|f|.Lemma 9 (see [18]).
Supposeσ* is bounded on Lq(Rn+1) for all 1<q<∞. Then, for arbitrary functions gj, the following vector valued inequality:
(19)∥(∑j‍|σj*gj|2)1/2∥Lq(Rn+1)≤C∥(∑j‍|gj|2)1/2∥Lq(Rn+1)
holds with any 1<q<∞.The maximal function inR2 is defined by
(20)(Mϕf)(x1,x2)=sup⁡k∈Z12k∫2k2k+1‍|f(x1-t,x2-ϕ(t))|dt.
We know that the Lq(Rn+1) boundedness of σ* is deduced from the Lq(R2) boundedness of Mϕ by method of rotations, and if ϕ is as in Theorem 1 or Theorem 2, Mϕ is a bounded operator on Lq(R2) for all 1<q<∞ (see [23, 24]).Letφ∈S(Rn) be a radial function satisfying 0≤φ≤1 with its support in the unit ball and φ(ξ)=1 for |ξ|≤1/2. The function φ0(ξ)=φ(ξ/2)-φ(ξ)∈S(Rn) satisfies ∑j∈Z‍φ0(2-jξ)=1 for ξ≠0. For j∈Z, denote by Δj and Gj the convolution operators whose symbols are φ0(2-jξ) and φ(2-jξ), respectively.Lemma 10 (see [20]).
For the multiplierGk(k∈Z), b∈BMO(Rn), and any fixed 0<τ<1/2, we have
(21)|Gkb(x)-Gkb(y)|≤C2kττ|x-y|τ∥b∥BMO,
where C is independent of k and τ.Letξ~=(ξ,ξn+1)∈Rn+1 and let ψ(ξ~)∈C0∞(Rn+1) be a radial function such that 0≤ψ≤1, supp⁡ψ⊂{1/2≤|ξ~|≤2}, and ∑l∈Z‍ψ3(2-lξ~)=1, |ξ~|≠0. Define the multiplier operator Sl by Slf^(ξ~)=ψ(2-l|ξ~|)f^(ξ~).Lemma 11.
For anyj∈Z, define the operator Tj by Tjf=σj*f, and ϕ is monotonic and satisfies condition (1) or (2):(1)
|ϕ(|y|)|≤C|y|;(2)
|ϕ(|y|)|≥C|y|,|ϕ(a)ϕ(b)|≤C|ϕ(ab)|for∀a,b>0, and|ϕ(|y|)|≤C|y|k1,k1>1if|y|>1,|ϕ(|y|)|≤C|y|k2,0<k2<1if|y|≤1. Let b∈BMO(Rn+1), and denote by [b,Sl-jTjSl-j2] the commutator of Sl-jTjSl-j2. Suppose Ω∈L1(Sn-1) satisfying (2). Then for any fixed 0<τ<1/2, 1<p<∞,
(22)∥∑j∈Z‍[b,Sl-jTjSl-j2]f(x~)∥Lp≤C∥b∥BMOmax⁡{2τlτ,2τk1lτ,2τk2lτ,2}∥f∥Lp.Proof.
We prove it by using arguments which are essentially the same as those in the proof of Lemma 3.7 in [20]. Two things must be modified:(i)
instead of Lemma 3.6 in [20], we use Lemma 9;(ii)
In [20], M1=∥∑j∈ZSl-j‍[π(TjSl-j2f)(b)-Tj(π(Sl-j2f)(b))]∥Lp, and πf(g)=∑j∈Z‍(Δjf)(Gj-3g) is the paraproduct of Bony [25] between two functions f and g. In the estimate of M1, we will use the following formulas:
(23)|[Gi-3b,Tj](ΔiSl-j2f)(x,xn+1)|=|Gi-3b(x,xn+1)Tj(ΔiSl-j2f)(x,xn+1)rfdsf-Tj((Gi-3b)(ΔiSl-j2f))(x,xn+1)|=|∫2j<|y|≤2j+1‍Ω(y)|y|ngfgfgffdfdfd×((x-y,xn+1-ϕ(|y|))Gi-3b(x,xn+1)-Gi-3bhghgghhggfdfdff×(x-y,xn+1-ϕ(|y|)))fdfdfffdfdfdfdf·ΔiSl-j2f(x-y,xn+1-ϕ(|y|))dy∫2j<|y|≤2j+1‍Ω(y)|y|n|≤C∫2j<|y|≤2j+1‍|Ω(y)||y|nfgdfgdfgffdfdfdfff×|Gi-3b(x,xn+1)-Gi-3bfdfdffdfdddsdsdsd×(x-y,xn+1-ϕ(|y|))|fgdfgdfgffdfdfdfff·|ΔiSl-j2f(x-y,xn+1-ϕ(|y|))|dy,by Lemma 10,
(24)|Gi-3b(x,xn+1)-Gi-3b(x-y,xn+1-ϕ(|y|))|≤C2iττ|(y,ϕ(|y|))|τ∥b∥BMO=C2iττ|y|2+ϕ2(|y|)τ∥b∥BMO.
If ϕ satisfies condition (1), we have
(25)|Gi-3b(x,xn+1)-Gi-3b(x-y,xn+1-ϕ(|y|))|≤C2iττ|y|τ∥b∥BMO.
Thus
(26)|[Gi-3b,Tj](ΔiSl-j2f)(x,xn+1)|≤C2iττ∥b∥BMO×∫2j<|y|≤2j+1‍|Ω(y)||y|n|y|τdfdfdffdfdfdfdfdff×|ΔiSl-j2f(x-y,xn+1-ϕ(|y|))|dy≤C2(i+j)ττ∥b∥BMO×∫2j<|y|≤2j+1‍|Ω(y)||y|ndffdfdfdfdfdf×|ΔiSl-j2f(x-y,xn+1-ϕ(|y|))|dy=C2(i+j)ττ∥b∥BMOT|Ω|,j(|ΔiSl-j2f|)(x,xn+1).
If ϕ satisfies condition (2), we have
(27)|Gi-3b(x,xn+1)-Gi-3b(x-y,xn+1-ϕ(|y|))|≤C2iττ|ϕτ(|y|)|∥b∥BMO≤C|ϕτ(2i)|τ|ϕτ(|y|)|∥b∥BMO.
Thus if |y|>1,
(28)|[Gi-3b,Tj](ΔiSl-j2f)(x,xn+1)|≤C|ϕτ(2(i+j))|τ∥b∥BMO×∫2j<|y|≤2j+1‍|Ω(y)||y|nfdfddfffdffdfdfdfd×|ΔiSl-j2f(x-y,xn+1-ϕ(|y|))|dy≤C2(i+j)k1ττ∥b∥BMOT|Ω|,j(|ΔiSl-j2f|)(x,xn+1),
and if |y|≤1,
(29)|[Gi-3b,Tj](ΔiSl-j2f)(x,xn+1)|≤C2(i+j)k2ττ∥b∥BMOT|Ω|,j(|ΔiSl-j2f|)(x,xn+1).


## 3. The Proof of Theorems1 and 2

Proof of Theorem1.
Letξ~=(ξ,ξn+1)∈Rn+1 and let ψ(ξ~)∈C0∞(Rn+1) be a radial function such that 0≤ψ≤1, supp⁡ψ⊂{1/2≤|ξ~|≤2}, and
(30)∑l∈Zψ3(2-lξ~)=1,|ξ~|≠0.
Define the multiplier operator Sl by
(31)Slf^(ξ~)=ψ(2-l|ξ~|)f^(ξ~).
Let the measure σj on Rn+1 be defined by
(32)∫Rn+1‍f(y,yn+1)dσj=∫Rn‍f(y,ϕ(|y|))Ω(y′)|y|nχ{2j<|y|≤2j+1}dy
for all j∈Z. Since
(33)σj*f=∫Rn‍f(x-y,xn+1-ϕ(|y|))Ω(y′)|y|nχ{2j<|y|≤2j+1}dy,Tϕ,Ωf=∫Rn‍f(x-y,xn+1-ϕ(|y|))Ω(y′)|y|ndy,
we get
(34)Tϕ,Ωf=∑j∈Z‍σj*f.
Define the operator Tjf(x~)=σj*f(x~), where x~=(x,xn+1)∈Rn+1 and the multiplier
(35)Tjlf^(ξ~)=TjSl-jf^(ξ~)=ψ(2j-l|ξ~|)σj^(ξ~)f^(ξ~).
From the above notation, it is easy to see that
(36)[b,Tϕ,Ω]f(x~)=∑l∈Z‍∑j∈Z‍[b,Sl-jTjSl-j2]f(x~)=∑l∈Z‍∑j∈Z‍[b,Sl-jTjlSl-j]f(x~):=∑l∈Z‍Vlf(x~),
where
(37)Vlf(x~)=∑j∈Z‍[b,Sl-jTjlSl-j]f(x~).
Then by the Minkowski inequality, we get
(38)∥[b,Tϕ,Ω]f∥L2(Rn+1)≤∥∑l=-∞[log⁡2]‍Vlf∥L2(Rn+1)+∥∑l=[log⁡2]+1∞‍Vlf∥L2(Rn+1).
For ∥∑l=-∞[log⁡2]Vlf∥L2(Rn+1), we recall
(39)σj^(ξ,ξn+1)=∫Sn-1‍Ω(θ)∫2j2j+1‍e-i(sθ·ξ+ϕ(|s|)ξn+1)dssdσ(θ).
By Lemma 2.3 of [16], we have
(40)|σj^(ξ,ξn+1)|≤C∥Ω∥L1|2jξ|.
Denote by ∇ξσj^ the before n components truncation of ∇σj^; that is,
(41)∇ξσj^=(∂σj^∂ξ1,...,∂σj^∂ξn).
Since
(42)σj^(ξ,ξn+1)=∫Rn‍Ω(y′)|y|nχ{2j<|y|≤2j+1}e-i(y·ξ+ϕ(|y|)ξn+1)dy,
we get
(43)|∇ξσj^|≤C2j∥Ω∥L1.
Set mj(ξ~)=σj^(ξ~), mjl(ξ~)=mj(ξ~)ψ(2j-l|ξ~|). Recall that Tjl by Tjlf^(ξ~)=mjl(ξ~)f^(ξ~). Straightforward computations lead to
(44)∥mjl(2-jξ~)∥L∞≤C∥Ω∥L12l.
Since
(45)supp⁡{mjl(2-jξ~)}⊂{|ξ~|≤2l+2},
we get
(46)∥∇ξmjl(2-jξ~)∥L∞≤C∥Ω∥L1.
Let T~jl be the operator defined by T~jlf^(ξ~)=mjl(2-jξ~)f^(ξ~). Denote by Tj,b,1lf=[b,Tjl]f and Tj,b,0lf=Tjlf. Similarly, denote by T~j,b,1lf=[b,T~jl]f and T~j,b,0lf=T~jlf. Thus via the Plancherel theorem and Lemma 8 it is stated that for any fixed 0<v<1, k∈{0,1},
(47)∥T~j,b,klf∥L2≤C∥b∥BMO(Rn+1)k∥Ω∥L12vl∥f∥L2,fdfddfdfdfdfl≤[log⁡2].
Dilation-invariance says that
(48)∥Tj,b,klf∥L2≤C∥b∥BMO(Rn+1)k∥Ω∥L12vl∥f∥L2,fdfdfdfdfdfl≤[log⁡2].
By the proof of Theorem 1 in [20], we can get
(49)∥Vlf∥L2≤C∥b∥BMO(Rn+1)2vl∥Ω∥L1∥f∥L2,fdsfdfdsdfdsfsl≤[log⁡2].
So, we have
(50)∥∑l=-∞[log⁡2]‍Vlf∥L2(Rn+1)≤C∑l=-∞[log⁡2]‍2vl∥b∥BMO(Rn+1)∥f∥L2(Rn+1)≤C∥b∥BMO(Rn+1)∥f∥L2(Rn+1).
For ∥∑l=1+[log⁡2]∞Vlf∥L2(Rn+1), by Lemma 2.3 of [16], if ϕ satisfies the hypotheses in Theorem 1, we have
(51)|σj^(ξ,ξn+1)|≤Clog⁡-α-1(|2jξ|+2),|∇ξσj^|≤C2j.
When ϕ(|t|)=|t|, if n=2, we also have the above estimates (see [14]). Set mj(ξ~)=σj^(ξ~), mjl(ξ~)=mj(ξ~)ψ(2j-l|ξ~|). Recall Tjl by Tjlf^(ξ~)=mjl(ξ~)f^(ξ~). Straightforward computations lead to
(52)∥mjl(2-jξ~)∥L∞≤Clog⁡-α-1(2+2l),∥∇ξmjl(2-jξ~)∥L∞≤C,supp⁡{mjl(2-jξ~)}⊂{|ξ~|≤2l+2}.
Let T~jl be the operator defined by T~jlf^(ξ~)=mjl(2-jξ~)f^(ξ~). Denote by Tj,b,1lf=[b,Tjl]f and Tj,b,0lf=Tjlf. Similarly, denote by T~j,b,1lf=[b,T~jl]f and T~j,b,0lf=T~jlf. Thus via the Plancherel theorem and Lemma 8 it is stated that for any fixed 0<v<1, k∈{0,1},
(53)∥T~j,b,klf∥L2≤C∥b∥BMO(Rn+1)klog⁡(-α-1)v+1(2+2l)∥f∥L2,fdfdfdfdfdfdffdffffdfdfdl≥1+[log⁡2].
Dilation-invariance says that
(54)∥Tj,b,klf∥L2≤C∥b∥BMO(Rn+1)klog⁡(-α-1)v+1(2+2l)∥f∥L2,fdfdfdfdfdfdfddfdffdfdfl≥1+[log⁡2].
By the proof of Theorem 1 in [20], we can get
(55)∥Vlf∥L2≤C∥b∥BMO(Rn+1)log⁡(-α-1)v+1(2+2l)∥f∥L2,gfdgfgdfgdfdfgdfdfgfgfdl≥1+[log⁡2].
So take v→1, and we have
(56)∥∑l=1+[log⁡2]∞‍Vlf∥L2(Rn+1)≤C∥b∥BMO(Rn+1)∑l=1+[log⁡2]∞‍l(-α-1)v+1∥f∥L2(Rn+1)≤C∥b∥BMO(Rn+1)∥f∥L2(Rn+1).
Then, by (50) and (56) we obtain Theorem 1.Proof of Theorem2.
By (36), we have
(57)∥[b,Tϕ,Ω]f∥Lp(R3)≤∥∑l=-∞[log⁡2]‍Vlf∥Lp(R3)+∥∑l=[log⁡2]+1∞‍Vlf∥Lp(R3).
For ∥∑l=-∞[log⁡2]Vlf∥Lp(R3), recall Tjlf(x~)=TjSl-jf(x~); then, Vlf(x~)=∑j∈Z[b,Sl-jTjSl-j2]f(x~). ϕ(|t|)=|t|, and applying Lemma 11, we get for 1<p<∞(58)∥Vlf∥Lp≤C∥b∥BMO(Rn+1)∥f∥Lp,l≤[log⁡2].
Interpolating between (49) and (58) with n=2, as the proof of Theorem 1 in [20], we can get
(59)∥∑l=-∞[log⁡2]‍Vlf∥Lp(R3)≤C∥b∥BMO(R3)∥f∥Lp(R3).
For ∥∑l=1+[log⁡2]∞Vlf∥Lp(R3), ϕ(|t|)=|t|, and applying Lemma 11, we get for any fixed 0<τ<1/2, 1<p<∞,
(60)∥Vlf∥Lp≤C∥b∥BMO(Rn+1)2τlτ∥f∥Lp,l≥1+[log⁡2].
Take τ=1/l; then, we get
(61)∥Vlf∥Lp≤Cl∥b∥BMO(Rn+1)∥f∥Lp,l≥1+[log⁡2].
For ϕ(|t|)=|t|, (55) can be established only when n=2, so interpolating between (55) and (61) with n=2, as the proof of Theorem 1 in [20], we get
(62)∥∑l=1+[log⁡2]∞Vlf∥Lp(R3)≤C∥b∥BMO(R3)∥f∥Lp(R3).
Then, by (59) and (62) we obtain Theorem 2.


## 4. The proof of Theorems4 and 7

We begin with a lemma, which plays an important role in proving Theorem4.Lemma 12.
Let
b
(
x
)
∈
B
M
O
(
R
n
)
,
x
~
=
(
x
,
x
n
+
1
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∈
R
n
+
1
, and
B
(
x
~
)
=
b
(
x
)
; then,
B
(
x
~
)
∈
B
M
O
(
R
n
+
1
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and
∥
B
∥
B
M
O
(
R
n
+
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=
∥
b
∥
B
M
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(
R
n
)
.Proof.
We know(63)∥b∥BMO(Rn)=sup⁡Q⊂Rn1|Q|∫Q‍|b(x)-bQ|dx,
where bQ=(1/|Q|)∫Q‍b(x)dx and Q is the square in Rn whose edges are parallel to the axis. So
(64)∥B∥BMO(Rn+1)=sup⁡Q~⊂Rn+11|Q~|∫Q~‍|B(x~)-BQ~|dx~,
where Q~ is the square in Rn+1 whose edges are parallel to the axis. Consider
(65)BQ~=1|Q~|∫Q~‍B(x~)dx~=1a|Q|∫mm+a‍∫Q‍b(x)dxdxn+1=1a|Q|∫Q‍b(x)dx∫mm+a‍dxn+1=1|Q|∫Q‍b(x)dx=bQ,
where Q is the projection on Rn of Q~ and a is the side length of Q~. Then
(66)∥B∥BMO(Rn+1)=sup⁡Q~⊂Rn+11|Q~|∫Q~‍|B(x~)-BQ~|dx~=sup⁡Q~⊂Rn+11a|Q|∫mm+a‍∫Q‍|b(x)-bQ|dxdxn+1=sup⁡Q~⊂Rn+11a|Q|∫Q‍|b(x)-bQ|dx∫mm+a‍dxn+1=sup⁡Q~⊂Rn+11|Q|∫Q‍|b(x)-bQ|dx=sup⁡Q⊂Rn1|Q|∫Q‍|b(x)-bQ|dx=∥b∥BMO(Rn).Proof of Theorem4.
By Lemma12, B∈BMO(Rn+1). Using the method in [5], for f∈Lp(Rn) and N∈N, let FN(x,xn+1)=f(x)e-ixn+1χ[-N,N](xn+1). Then by mean value theorem of integrals and Lemma 12, we have
(67)2N∫Rn‍|b(x)∫Rn‍Ω(y)|y|nf(x-y)eiϕ(|y|)χ[-N,N]dsdsdsdsdsdsdsd×(xn+1-ϕ(|y|))dydsdsdsd-∫Rn‍Ω(y)|y|nb(x-y)f(x-y)eiϕ(|y|)dsdsddsddsdd×χ[-N,N](xn+1-ϕ(|y|))dy∫Rn‍Ω(y)|y|n|pdx≤∫R‍∫Rn‍|b(x)∫Rn‍Ω(y)|y|nf(x-y)e-i(xn+1-ϕ(|y|))sdsdsdsdsdsdsdsd×χ[-N,N](xn+1-ϕ(|y|))dydsdsdsdsdsd-∫Rn‍Ω(y)|y|nb(x-y)f(x-y)fdfdfdfdffdsdsd×e-i(xn+1-ϕ(|y|))χ[-N,N]fdfdfdfdffdsdsd×(xn+1-ϕ(|y|))dy∫Rn‍Ω(y)|y|n|pdxdxn+1=∥[B,Tϕ,Ω]FN∥Lp(Rn+1)p≤C∥B∥BMO(Rn+1)p∥FN∥Lp(Rn+1)p=C2N∥b∥BMO(Rn)p∥f∥Lp(Rn)p.
Dividing both sides by 2N and letting N→∞, we obtain
(68)∥[b,Tϕ]f∥Lp(Rn)≤C∥b∥BMO(Rn)∥f∥Lp(Rn).
Thus, we obtain Theorem 4.Proof of Theorem7.
Theorem7 can be proved by using arguments which are essentially the same as the proof of Theorem 1 in [20]. Only the following two things must be modified.(i)
Instead ofKj(x) and Tjf(x), we use
(69)Kj,ϕ(x)=eiϕ(x)Ω(x)|x|nχ{2j<|x|≤2j+1},Tj,ϕf(x)=Kj,ϕ*f(x)=∫2j<|y|≤2j+1‍Ω(x)|x|neiϕ(x)f(x-y)dy.(ii)
SinceΩ(θ) is odd and ϕ(θt) is even with respect to θ, we get Ω(θ)eiϕ(θt) is odd and ∫Sn-1‍Ω(θ)eiϕ(θt)dσ(θ)=0. So we use the estimates in [21]: Consider
(70)|K^j,ϕ(ξ)|≤C∥Ω∥L1|2jξ|,|K^j,ϕ(ξ)|≤Clog⁡-α-1|2jξ+2|



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*Source: 393147-2014-08-05.xml*