PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE
Nouvelle série, tome 92(106) (2012), 165–176
DOI: 10.2298/PIM1206165H
SHARP FUNCTION INEQUALITIES AND BOUNDNESS
FOR TOEPLITZ TYPE OPERATOR RELATED TO
GENERAL FRACTIONAL SINGULAR
INTEGRAL OPERATOR
Chuangxia Huang and Lanzhe Liu
Communicated by Stevan Pilipović
Abstract. We establish some sharp maximal function inequalities for the
Toeplitz type operator, which is related to certain fractional singular integral operator with general kernel. These results are helpful to investigate the
boundedness of the operator on Lebesgue, Morrey and Triebel–Lizorkin spaces
respectively.
1. Introduction
In recent decades, commutators have attracted a rapidly growing attention of
the researchers in the field of harmonic analysis and have been widely studied by
many authors [7, 19, 20]. In [4, 17, 18], the authors prove that the commutators
generated by the singular integral operators and BMO functions are bounded on
𝐿𝑝 (𝑅𝑛 ) for 1 < 𝑝 < ∞. Chanillo proves a similar result when singular integral
operators are replaced by the fractional integral operators in [2]. The boundedness
for the commutators generated by the singular integral operators and Lipschitz
functions on Triebel–Lizorkin and 𝐿𝑝 (𝑅𝑛 ) (1 < 𝑝 < ∞) spaces are obtained in
[3, 8, 14]. Some singular integral operators with general kernel are introduced, and
the boundedness for the operators and their commutators generated by BMO and
Lipschitz functions are obtained [1, 11]. In [9, 10], some Toeplitz type operators
related to the singular integral operators and strongly singular integral operators
are introduced, and the boundedness for the operators generated by BMO and
Lipschitz functions are established. The main purpose of this paper is to study the
Toeplitz type operators generated by some fractional singular integral operators
with general kernel and the Lipschitz and BMO functions. We will prove the
2010 Mathematics Subject Classification: Primary 42B20; Secondary 42B25.
This work was jointly supported by the National Natural Science Foundation of China
(No. 11101053), the Key Project of Chinese Ministry of Education (No. 211118), the Excellent
Youth Foundation of Educational Committee of Hunan Provincial (No. 10B002).
165
166
HUANG AND LIU
sharp maximal inequalities for the Toeplitz type operator 𝑇𝛿𝑏 . These results are
helpful to investigate the the 𝐿𝑝 -norm inequality, the boundedness of the operator
on Lebesgue, Morrey and Triebel–Lizorkin spaces respectively.
2. Preliminaries
At first, we should introduce some notations in the following. Throughout this
paper, 𝑄 denotes a cube of 𝑅𝑛 with sides parallel to the axes. For any locally
integrable function 𝑓 , the sharp maximal function of 𝑓 is defined by
∫︁
1
𝑀 # (𝑓 )(𝑥) = sup
|𝑓 (𝑦) − 𝑓𝑄 |𝑑𝑦,
𝑄∋𝑥 |𝑄| 𝑄
∫︀
where, 𝑓𝑄 = |𝑄|−1 𝑄 𝑓 (𝑥) 𝑑𝑥. It is well known that [7, 19]
∫︁
1
𝑀 # (𝑓 )(𝑥) ≈ sup inf
|𝑓 (𝑦) − 𝑐|𝑑𝑦.
𝑄∋𝑥 𝑐∈𝐶 |𝑄| 𝑄
We say that 𝑓 belongs to BMO(𝑅𝑛 ) if 𝑀 # (𝑓 ) belongs to 𝐿∞ (𝑅𝑛 ) and define
‖𝑓 ‖BMO = ‖𝑀 # (𝑓 )‖𝐿∞ . It is known [19] that ‖𝑓 − 𝑓2𝑘 𝑄 ‖BMO 6 𝐶𝑘‖𝑓 ‖BMO . Let
∫︁
1
|𝑓 (𝑦)|𝑑𝑦.
𝑀 (𝑓 )(𝑥) = sup
𝑄∋𝑥 |𝑄| 𝑄
For 𝜂 > 0 we denote 𝑀𝜂 (𝑓 )(𝑥) = 𝑀 (|𝑓 |𝜂 )1/𝜂 (𝑥). For 0 < 𝜂 < 𝑛 and 1 6 𝑟 < ∞ set
(︂
)︂1/𝑟
∫︁
1
𝑟
𝑀𝜂,𝑟 (𝑓 )(𝑥) = sup
|𝑓 (𝑦)| 𝑑𝑦
.
1−𝑟𝜂/𝑛
𝑄∋𝑥 |𝑄|
𝑄
According to [7], the 𝐴𝑝 weight can be defined as follows, for 1 < 𝑝 < ∞,
}︃
{︃
)︂ (︂
)︂𝑝−1
(︂
∫︁
∫︁
1
1
𝑤(𝑥) 𝑑𝑥
𝑤(𝑥)−1/(𝑝−1) 𝑑𝑥
<∞
𝐴𝑝 = 𝑤 ∈ 𝐿1loc (𝑅𝑛 ) : sup
|𝑄| 𝑄
|𝑄| 𝑄
𝑄
and 𝐴1 = {𝑤 ∈ 𝐿𝑝loc (𝑅𝑛 ) : 𝑀 (𝑤)(𝑥) 6 𝐶𝑤(𝑥), a.e.}.
For 𝛽 > 0 and 𝑝 > 1 let 𝐹˙𝑝𝛽,∞ (𝑅𝑛 ) be the homogeneous Triebel–Lizorkin
space [14]. For 𝛽 > 0 the Lipschitz space Lip𝛽 (𝑅𝑛 ) is the space of functions 𝑓 such
that
|𝑓 (𝑥) − 𝑓 (𝑦)|
‖𝑓 ‖Lip𝛽 = sup
< ∞.
𝑛
|𝑥 − 𝑦|𝛽
𝑥,𝑦∈𝑅
𝑥̸=𝑦
Definition 2.1. Let 𝜙 be a positive, increasing function on 𝑅+ and there
exists a constant 𝐷 > 0 such that 𝜙(2𝑡) 6 𝐷𝜙(𝑡) for 𝑡 > 0. Let 𝑓 be a locally
integrable function on 𝑅𝑛 . Set, for 1 6 𝑝 < ∞,
(︂
)︂1/𝑝
∫︁
1
𝑝
|𝑓 (𝑦)| 𝑑𝑦
,
‖𝑓 ‖𝐿𝑝,𝜙 = sup
𝑥∈𝑅𝑛, 𝑑>0 𝜙(𝑑) 𝑄(𝑥,𝑑)
where 𝑄(𝑥, 𝑑) = {𝑦 ∈ 𝑅𝑛 : |𝑥 − 𝑦| < 𝑑}. As usual, the generalized Morrey space
can be defined by 𝐿𝑝,𝜙 (𝑅𝑛 ) = {𝑓 ∈ 𝐿1loc (𝑅𝑛 ) : ‖𝑓 ‖𝐿𝑝,𝜙 < ∞}.
BOUNDEDNESS FOR TOEPLITZ TYPE OPERATOR
167
If 𝜙(𝑑) = 𝑑𝛿 , 𝛿 > 0, then 𝐿𝑝,𝜙 (𝑅𝑛 ) = 𝐿𝑝,𝛿 (𝑅𝑛 ) are classical Morrey spaces
[15, 16]. If 𝜙(𝑑) = 1, then 𝐿𝑝,𝜙 (𝑅𝑛 ) = 𝐿𝑝 (𝑅𝑛 ) are Lebesgue spaces [13].
Since Morrey spaces can be regarded as extensions of Lebesgue spaces, it is
natural and important to study the boundedness of operators on the Morrey spaces
[5, 6, 12, 13]. Here we study some singular integral operators, defined as follows [1].
Definition 2.2. Fix 0 < 𝛿 < 𝑛. Let 𝑇𝛿 : 𝑆 → 𝑆 ′ be a linear operator
such that 𝑇𝛿 is bounded on 𝐿2 (𝑅𝑛 ) and has a kernel 𝐾, that is, there exists a
locally integrable ∫︀function 𝐾(𝑥, 𝑦) on 𝑅𝑛 × 𝑅𝑛 r {(𝑥, 𝑦) ∈ 𝑅𝑛 × 𝑅𝑛 : 𝑥 = 𝑦} such
that 𝑇𝛿 (𝑓 )(𝑥) = 𝑅𝑛 𝐾(𝑥, 𝑦)𝑓 (𝑦) 𝑑𝑦 for every bounded and compactly supported
function 𝑓 , where 𝐾 satisfies: there is a sequence of positive constant numbers
{𝐶𝑗 } such that for any 𝑗 > 1,
∫︁
(|𝐾(𝑥, 𝑦) − 𝐾(𝑥, 𝑧)| + |𝐾(𝑦, 𝑥) − 𝐾(𝑧, 𝑥)|) 𝑑𝑥 6 𝐶,
2|𝑦−𝑧|<|𝑥−𝑦|
(︂∫︁
(1)
(|𝐾(𝑥, 𝑦) − 𝐾(𝑥, 𝑧)| + |𝐾(𝑦, 𝑥) − 𝐾(𝑧, 𝑥)|)𝑞 𝑑𝑦
)︂1/𝑞
2𝑘 |𝑧−𝑦|6|𝑥−𝑦|<2𝑘+1 |𝑧−𝑦|
′
6 𝐶𝑘 (2𝑘 |𝑧 − 𝑦|)−𝑛/𝑞 +𝛿 ,
where 1 < 𝑞 ′ < 2 and 1/𝑞 + 1/𝑞 ′ = 1. We write 𝑇𝛿 ∈ GSIO(𝛿).
Moreover, if 𝑏 is a locally integrable function on 𝑅𝑛 , then the Toeplitz type op∑︀𝑚
erator related to 𝑇𝛿 can be defined by 𝑇𝛿𝑏 = 𝑘=1 𝑇𝛿𝑘,1 𝑀𝑏 𝑇 𝑘,2 , where 𝑇𝛿𝑘,1 are 𝑇𝛿 or
±𝐼 (the identity operator), 𝑇 𝑘,2 are the linear operators, 𝑘 = 1,..., 𝑚, 𝑀𝑏 (𝑓 ) = 𝑏𝑓 .
Remark 2.1. We should point out that the classical Calderón–Zygmund singular integral operator satisfies Definition 2.2 with 𝐶𝑗 = 2−𝑗𝛿 [7, 19].
Remark 2.2. It is obvious that the fractional integral operator with rough
kernel satisfies Definition 2.2 [3], that is, for 0 < 𝛿 < 𝑛, let 𝑇𝛿 be the fractional
integral operator with rough kernel defined by (see [3])
∫︁
Ω(𝑥 − 𝑦)
𝑓 (𝑦) 𝑑𝑦,
𝑇𝛿 𝑓 (𝑥) =
𝑛−𝛿
𝑅𝑛 |𝑥 − 𝑦|
∫︀
where Ω is homogeneous of degree zero on 𝑅𝑛 , 𝑆 𝑛−1 Ω(𝑥′ ) 𝑑𝜎(𝑥′ ) = 0 and Ω ∈
Lip𝜀 (𝑆 𝑛−1 ) for some 0 < 𝜀 6 1, and there exists a constant 𝑀 > 0 such that for
any 𝑥, 𝑦 ∈ 𝑆 𝑛−1 , |Ω(𝑥) − Ω(𝑦)| 6 𝑀 |𝑥 − 𝑦|𝜀 . When Ω ≡ 1, 𝑇𝛿 is the Riesz potential
(fractional integral operator) [2].
Remark 2.3. One can obtain that the commutator [𝑏, 𝑇𝛿 ](𝑓 ) = 𝑏𝑇𝛿 (𝑓 )−𝑇𝛿 (𝑏𝑓 )
is a particular operator of the Toeplitz type operator 𝑇𝑏 . The Toeplitz type operators 𝑇𝛿𝑏 are nontrivial generalizations of commutators.
3. Some Lemmas
We begin with some preliminary lemmas.
Lemma 3.1. [1] Let 𝑇𝛿 be the singular integral operator as Definition 1.2. Then
𝑇𝛿 is bounded from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) for 1 < 𝑝 < 𝑛/𝛿 and 1/𝑟 = 1/𝑝 − 𝛿/𝑛.
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HUANG AND LIU
Lemma 3.2. [14] For 0 < 𝛽 < 1 and 1 < 𝑝 < ∞, we have
⃦
⃦
⃦
⃦
∫︁
∫︁
⃦
⃦
⃦
⃦
1
1
⃦ ≈ ⃦sup inf
⃦ .
sup
‖𝑓 ‖𝐹˙𝑝𝛽,∞ ≈ ⃦
|𝑓
(𝑥)−𝑓
|
𝑑𝑥
|𝑓
(𝑥)−𝑐|
𝑑𝑥
𝑄
⃦ 𝑄∋· |𝑄|1+𝛽/𝑛
⃦ 𝑝 ⃦ 𝑄∋· 𝑐 |𝑄|1+𝛽/𝑛
⃦ 𝑝
𝑄
𝑄
𝐿
𝐿
⋃︀
Lemma 3.3. [7] Let 0 < 𝑝 < ∞ and 𝑤 ∈ 16𝑟<∞ 𝐴𝑟 . Then, for any smooth
function 𝑓 for which the left-hand side is finite,
∫︁
∫︁
𝑀 (𝑓 )(𝑥)𝑝 𝑤(𝑥) 𝑑𝑥 6 𝐶
𝑀 # (𝑓 )(𝑥)𝑝 𝑤(𝑥) 𝑑𝑥.
𝑅𝑛
𝑅𝑛
Lemma 3.4. [2] Suppose that 0 < 𝜂 < 𝑛, 1 6 𝑠 < 𝑝 < 𝑛/𝜂 and 1/𝑟 = 1/𝑝−𝜂/𝑛.
Then ‖𝑀𝜂,𝑠 (𝑓 )‖𝐿𝑟 6 𝐶‖𝑓 ‖𝐿𝑝 .
Lemma 3.5. Let 1 < 𝑝 < ∞, 0 < 𝐷 < 2𝑛 . Then, for any smooth function 𝑓
for which the left-hand side is finite, ‖𝑀 (𝑓 )‖𝐿𝑝,𝜙 6 𝐶‖𝑀 # (𝑓 )‖𝐿𝑝,𝜙 .
Proof. For any cube 𝑄 = 𝑄(𝑥0 , 𝑑) in 𝑅𝑛 , we know 𝑀 (𝜒𝑄 ) ∈ 𝐴1 for any cube
𝑄 = 𝑄(𝑥, 𝑑) by [7]. Noticing that 𝑀 (𝜒𝑄 ) 6 1 and 𝑀 (𝜒𝑄 )(𝑥) 6 𝑑𝑛 /(|𝑥 − 𝑥0 | − 𝑑)𝑛
if 𝑥 ∈ 𝑄𝑐 , by Lemma 3.3, we have, for 𝑓 ∈ 𝐿𝑝,𝜙 (𝑅𝑛 ),
∫︁
∫︁
𝑀 (𝑓 )(𝑥)𝑝 𝑑𝑥 =
𝑀 (𝑓 )(𝑥)𝑝 𝜒𝑄 (𝑥) 𝑑𝑥
𝑄
𝑅𝑛
∫︁
∫︁
6
𝑀 (𝑓 )(𝑥)𝑝 𝑀 (𝜒𝑄 )(𝑥) 𝑑𝑥 6 𝐶
𝑀 # (𝑓 )(𝑥)|𝑝 𝑀 (𝜒𝑄 )(𝑥) 𝑑𝑥
𝑅𝑛
(︂ ∫︁
𝑝
#
=𝐶
𝑅𝑛
∞ ∫︁
∑︁
)︂
𝑀 (𝑓 )(𝑥) 𝑀 (𝜒𝑄 )(𝑥) 𝑑𝑥
#
𝑀 (𝑓 )(𝑥) 𝑀 (𝜒𝑄 )(𝑥) 𝑑𝑥 +
𝑄
𝑘=0
(︂ ∫︁
𝑝
2𝑘+1 𝑄r2𝑘 𝑄
∞ ∫︁
∑︁
)︂
|𝑄|
6𝐶
𝑀 (𝑓 )(𝑥) 𝑑𝑥 +
𝑀 (𝑓 )(𝑥) 𝑘+1 𝑑𝑥
|2
𝑄|
𝑘+1 𝑄r2𝑘 𝑄
𝑄
𝑘=0 2
)︂
(︂ ∫︁
∫︁
∞
∑︁
𝑀 # (𝑓 )(𝑥)𝑝 2−𝑘𝑛 𝑑𝑦
6𝐶
𝑀 # (𝑓 )(𝑥)𝑝 𝑑𝑥 +
𝑝
#
𝑄
6 𝐶‖𝑀
#
𝑘=0
(𝑓 )‖𝑝𝐿𝑝,𝜙
∞
∑︁
−𝑘𝑛
2
𝜙(2
#
𝑝
2𝑘+1 𝑄
𝑘+1
𝑑) 6 𝐶‖𝑀
#
(𝑓 )‖𝑝𝐿𝑝,𝜙
𝑘=0
∞
∑︁
(2−𝑛 𝐷)𝑘 𝜙(𝑑)
𝑘=0
6 𝐶‖𝑀 # (𝑓 )‖𝑝𝐿𝑝,𝜙 𝜙(𝑑),
thus
(︂
1
𝜙(𝑑)
)︂1/𝑝
(︂
)︂1/𝑝
∫︁
1
𝑀 (𝑓 )(𝑥)𝑝 𝑑𝑥
6𝐶
𝑀 # (𝑓 )(𝑥)𝑝 𝑑𝑥
𝜙(𝑑) 𝑄
𝑄
∫︁
and ‖𝑀 (𝑓 )‖𝐿𝑝,𝜙 6 𝐶‖𝑀 # (𝑓 )‖𝐿𝑝,𝜙 . This finishes the proof.
Lemma 3.6. Let 0 < 𝐷 < 2𝑛 , 1 6 𝑠 < 𝑝 < 𝑛/𝜂 and 1/𝑟 = 1/𝑝 − 𝜂/𝑛. Then
‖𝑀𝜂,𝑠 (𝑓 )‖𝐿𝑟,𝜙 6 𝐶‖𝑓 ‖𝐿𝑝,𝜙 .
The proof of Lemma 3.6 is similar to that of Lemma 3.5 by Lemma 3.4, we
omit the details.
BOUNDEDNESS FOR TOEPLITZ TYPE OPERATOR
169
4. Theorems and Their Proofs
We shall prove the following theorems.
Theorem 4.1. Let the sequence {𝐶𝑗 } ∈ 𝑙1 , 0 < 𝛽 < 1, 𝑞 ′ 6 𝑠 < ∞ and
𝑏 ∈ Lip𝛽 (𝑅𝑛 ). Suppose 𝑇𝛿 is a bounded linear operator from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) for
any 𝑝, 𝑟 with 1 < 𝑝 < 𝑛/𝛿 and 1/𝑟 = 1/𝑝 − 𝛿/𝑛, and has a kernel 𝐾 satisfying (1).
If 𝑇𝛿1 (𝑔) = 0 for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞), then there exists a constant 𝐶 > 0
such that, for any 𝑓 ∈ 𝐶0∞ (𝑅𝑛 ) and 𝑥
˜ ∈ 𝑅𝑛 ,
𝑀 # (𝑇𝛿𝑏 (𝑓 ))(˜
𝑥) 6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝑘=1
𝐶0∞ (𝑅𝑛 )
Proof. It suffices to prove for 𝑓 ∈
and some constant 𝐶0 , the following inequality holds:
∫︁
𝑚
∑︁
⃒ 𝑏
⃒
1
⃒𝑇𝛿 (𝑓 )(𝑥) − 𝐶0 ⃒ 𝑑𝑥 6 𝐶‖𝑏‖Lip
𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝛽
|𝑄| 𝑄
𝑘=1
Without loss of generality, we may assume 𝑇𝛿𝑘,1 are 𝑇𝛿 (𝑘 = 1, . . . , 𝑚). Fix a cube
𝑄 = 𝑄(𝑥0 , 𝑑) and 𝑥
˜ ∈ 𝑄. Write, for 𝑓1 = 𝑓 𝜒2𝑄 and 𝑓2 = 𝑓 𝜒(2𝑄)𝑐 ,
𝑏−𝑏𝑄
𝑇𝛿𝑏 (𝑓 )(𝑥) = 𝑇𝛿
(𝑏−𝑏𝑄 )𝜒2𝑄
(𝑓 )(𝑥) = 𝑇𝛿
(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐
(𝑓 )(𝑥) + 𝑇𝛿
(𝑓 )(𝑥) = 𝑔(𝑥) + ℎ(𝑥).
Then
∫︁
∫︁
∫︁
⃒ 𝑏
⃒
1
1
⃒𝑇𝛿 (𝑓 )(𝑥) − ℎ(𝑥0 )⃒ 𝑑𝑥 6 1
|𝑔(𝑥)|𝑑𝑥 +
|ℎ(𝑥) − ℎ(𝑥0 )|𝑑𝑥 = 𝐼1 + 𝐼2 .
|𝑄| 𝑄
|𝑄| 𝑄
|𝑄| 𝑄
For 𝐼1 , choose 1 < 𝑟 < ∞ such that 1/𝑟 = 1/𝑠 − 𝛿/𝑛, by (𝐿𝑠 , 𝐿𝑟 )-boundedness of
𝑇𝛿 and Hölder’s inequality, we obtain
∫︁
1
|𝑇 𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 𝑘,2 (𝑓 )(𝑥)|𝑑𝑥
|𝑄| 𝑄 𝛿
(︂
)︂1/𝑟
∫︁
1
𝑘,1
𝑘,2
𝑟
6
|𝑇 𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 (𝑓 )(𝑥)| 𝑑𝑥
|𝑄| 𝑅𝑛 𝛿
(︂∫︁
)︂1/𝑠
6 𝐶|𝑄|−1/𝑟
|𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 𝑘,2 (𝑓 )(𝑥)|𝑠 𝑑𝑥
𝑅𝑛
−1/𝑟
(︂∫︁
(|𝑏(𝑥) − 𝑏𝑄 ‖𝑇
6 𝐶|𝑄|
𝑘,2
)︂1/𝑠
(𝑓 )(𝑥)|) 𝑑𝑥
𝑠
2𝑄
6 𝐶|𝑄|−1/𝑟 ‖𝑏‖Lip𝛽 |2𝑄|𝛽/𝑛 |2𝑄|1/𝑠−(𝛽+𝛿)/𝑛
(︂
)︂1/𝑠
∫︁
1
𝑘,2
𝑠
×
|𝑇
(𝑓
)(𝑥)|
𝑑𝑥
|2𝑄|1−𝑠(𝛽+𝛿)/𝑛 2𝑄
6 𝐶‖𝑏‖Lip𝛽 𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥),
170
HUANG AND LIU
thus
∫︁
𝑚
∑︁
1
𝐼1 6
|𝑇 𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 𝑘,2 (𝑓 )(𝑥)|𝑑𝑥
|𝑄| 𝑄 𝛿
𝑘=1
6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝑘=1
′
For 𝐼2 , recalling that 𝑠 > 𝑞 , we get, for 𝑥 ∈ 𝑄,
|𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥) − 𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥0 )|
∫︁
6
|𝑏(𝑦) − 𝑏2𝑄 ‖𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)‖𝑇 𝑘,2 (𝑓 )(𝑦)|𝑑𝑦
=
(2𝑄)𝑐
∞
∑︁ ∫︁
𝑗=1
|𝑏(𝑦) − 𝑏2𝑄 ‖𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)‖𝑇 𝑘,2 (𝑓 )(𝑦)|𝑑𝑦
2𝑗 𝑑6|𝑦−𝑥0 |<2𝑗+1 𝑑
6 𝐶‖𝑏‖Lip𝛽
∞
∑︁
|2𝑗+1 𝑄|𝛽/𝑛
(︂∫︁
|𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)|𝑞 𝑑𝑦
)︂1/𝑞
2𝑗 𝑑6|𝑦−𝑥0 |<2𝑗+1 𝑑
𝑗=1
(︂∫︁
×
|𝑇
𝑘,2
𝑞′
)︂1/𝑞′
(𝑓 )(𝑦)| 𝑑𝑦
2𝑗+1 𝑄
6 𝐶‖𝑏‖Lip𝛽
∞
∑︁
′
′
|2𝑗+1 𝑄|𝛽/𝑛 𝐶𝑗 (2𝑗 𝑑)−𝑛/𝑞 +𝛿 |2𝑗+1 𝑄|1/𝑞 −(𝛽+𝛿)/𝑛
𝑗=1
(︂
1
∫︁
𝑘,2
𝑠
)︂1/𝑠
|𝑇 (𝑓 )(𝑦)| 𝑑𝑦
|2𝑗+1 𝑄|1−𝑠(𝛽+𝛿)/𝑛 2𝑗+1 𝑄
∞
∑︁
6 𝐶‖𝑏‖Lip𝛽 𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥)
𝐶𝑗 6 𝐶‖𝑏‖Lip𝛽 𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥),
×
𝑗=1
thus
𝐼2 6
1
|𝑄|
∫︁ ∑︁
𝑚
|𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥) − 𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥0 )|𝑑𝑥
𝑄 𝑘=1
𝑚
∑︁
6 𝐶‖𝑏‖Lip𝛽
𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝑘=1
This completes the proof of Theorem 4.1.
Theorem 4.2. Let the sequence {2𝑗𝛽 𝐶𝑗 } ∈ 𝑙1 , 0 < 𝛽 < 1, 𝑞 ′ 6 𝑠 < ∞ and
𝑏 ∈ Lip𝛽 (𝑅𝑛 ). Suppose 𝑇𝛿 is a bounded linear operator from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) for
any 𝑝, 𝑟 with 1 < 𝑝 < 𝑛/𝛿 and 1/𝑟 = 1/𝑝 − 𝛿/𝑛, and has a kernel 𝐾 satisfying (1).
If 𝑇𝛿1 (𝑔) = 0 for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞), then there exists a constant 𝐶 > 0
such that, for any 𝑓 ∈ 𝐶0∞ (𝑅𝑛 ) and 𝑥
˜ ∈ 𝑅𝑛 ,
∫︁
𝑚
∑︁
⃒ 𝑏
⃒
1
⃒𝑇𝛿 (𝑓 )(𝑥) − 𝐶0 ⃒ 𝑑𝑥 6 𝐶‖𝑏‖Lip
sup
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝛽
1+𝛽/𝑛
𝑄∋˜
𝑥 |𝑄|
𝑄
𝑘=1
BOUNDEDNESS FOR TOEPLITZ TYPE OPERATOR
171
Proof. It suffices to prove for 𝑓 ∈ 𝐶0∞ (𝑅𝑛 ) and some constant 𝐶0 , the following inequality holds:
∫︁
𝑚
∑︁
⃒ 𝑏
⃒
1
⃒𝑇𝛿 (𝑓 )(𝑥) − 𝐶0 ⃒ 𝑑𝑥 6 𝐶‖𝑏‖Lip
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝛽
|𝑄|1+𝛽/𝑛 𝑄
𝑘=1
Without loss of generality, we may assume 𝑇𝛿𝑘,1 are 𝑇𝛿 (𝑘 = 1, . . . , 𝑚). Fix a cube
𝑄 = 𝑄(𝑥0 , 𝑑) and 𝑥
˜ ∈ 𝑄. For 𝑓1 = 𝑓 𝜒2𝑄 and 𝑓2 = 𝑓 𝜒(2𝑄)𝑐 , write
𝑏−𝑏𝑄
𝑇𝛿𝑏 (𝑓 )(𝑥) = 𝑇𝛿
(𝑏−𝑏𝑄 )𝜒2𝑄
(𝑓 )(𝑥) = 𝑇𝛿
(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐
(𝑓 )(𝑥) + 𝑇𝛿
(𝑓 )(𝑥) = 𝑔(𝑥) + ℎ(𝑥)
and
∫︁
1
|𝑄|1+𝛽/𝑛
∫︁
1
⃒ 𝑏
⃒
⃒𝑇𝛿 (𝑓 )(𝑥) − ℎ(𝑥0 )⃒ 𝑑𝑥 6
|𝑔(𝑥)|𝑑𝑥
|𝑄|1+𝛽/𝑛 𝑄
∫︁
1
+
|ℎ(𝑥) − ℎ(𝑥0 )|𝑑𝑥 = 𝐼3 + 𝐼4 .
|𝑄|1+𝛽/𝑛 𝑄
𝑄
By using the same argument as in the proof of Theorem 4.1, we get, for 1 < 𝑟 < ∞
with 1/𝑟 = 1/𝑠 − 𝛿/𝑛,
(︂∫︁
)︂1/𝑟
𝑚
∑︁
𝐶
𝑘,1
1−1/𝑟
𝑘,2
𝑟
𝐼3 6
|𝑄|
|𝑇
𝑀
𝑇
(𝑓
)(𝑥)|
𝑑𝑥
(𝑏−𝑏𝑄 )𝜒2𝑄
𝛿
|𝑄|1+𝛽/𝑛
2𝑄
𝑘=1
(︂∫︁
)︂1/𝑠
𝑚
∑︁
𝐶
−1/𝑟
𝑘,2
𝑠
6
|𝑄|
|𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 (𝑓 )(𝑥)| 𝑑𝑥
|𝑄|𝛽/𝑛
2𝑄
𝑘=1
(︂∫︁
)︂1/𝑠
𝑚
∑︁
𝐶
−1/𝑟
𝛽/𝑛
𝑘,2
𝑠
|𝑄|
‖𝑏‖
|2𝑄|
|𝑇
(𝑓
)(𝑥)|
𝑑𝑥
6
Lip𝛽
|𝑄|𝛽/𝑛
2𝑄
𝑘=1
(︂
)︂1/𝑠
∫︁
𝑚
∑︁
1
𝑘,2
𝑠
6 𝐶‖𝑏‖Lip𝛽
|𝑇 (𝑓 )(𝑥)| 𝑑𝑥
|2𝑄|1−𝑠𝛿/𝑛 2𝑄
𝑘=1
6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥),
𝑘=1
𝐼4 6
𝑚
∑︁
𝑘=1
6
𝑚
∑︁
𝑘=1
∫︁ ∑︁
∞ ∫︁
1
|𝑄|1+𝛽/𝑛
𝐶
|𝑄|1+𝛽/𝑛
𝑄 𝑗=1
|𝑏(𝑦) − 𝑏2𝑄 |
2𝑗 𝑑6|𝑦−𝑥0 |<2𝑗+1 𝑑
× |𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)‖𝑇 𝑘,2 (𝑓 )(𝑦)|𝑑𝑦𝑑𝑥
(︂∫︁
)︂1/𝑞′
∫︁ ∑︁
∞
𝑗+1
𝛽/𝑛
𝑘,2
𝑞′
‖𝑏‖Lip𝛽 |2 𝑄|
|𝑇 (𝑓 )(𝑦)| 𝑑𝑦
2𝑗+1 𝑄
𝑄 𝑗=1
(︂∫︁
𝑞
×
6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝑘=1
|𝑄|
)︂1/𝑞
|𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)| 𝑑𝑦
2𝑗 𝑑6|𝑦−𝑥0 |<2𝑗+1 𝑑
∞
∑︁
−𝛽/𝑛
𝑗+1
𝛽/𝑛
|2
𝑗=1
𝑄|
′
𝑑𝑥
′
𝐶𝑗 (2𝑗 𝑑)−𝑛/𝑞 +𝛿 |2𝑗+1 𝑄|1/𝑞 −𝛿/𝑛
172
HUANG AND LIU
)︂1/𝑠
∫︁
1
𝑘,2
𝑠
|𝑇
(𝑓
)(𝑦)|
𝑑𝑦
|2𝑗+1 𝑄|1−𝑠𝛿/𝑛 2𝑗+1 𝑄
𝑚
∞
∑︁
∑︁
6 𝐶‖𝑏‖Lip𝛽
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥)
2𝑗𝛽 𝐶𝑗
(︂
×
6 𝐶‖𝑏‖Lip𝛽
𝑘=1
𝑚
∑︁
𝑗=1
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝑘=1
This completes the proof of Theorem 4.2.
1
′
𝑛
Theorem 4.3. Let the sequence {𝑗𝐶𝑗 } ∈ 𝑙 , 𝑞 6 𝑠 < ∞ and 𝑏 ∈ BMO(𝑅 ).
Suppose 𝑇𝛿 is a bounded linear operator from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) for any 𝑝, 𝑟 with
1 < 𝑝 < 𝑛/𝛿 and 1/𝑟 = 1/𝑝 − 𝛿/𝑛, and has a kernel 𝐾 satisfying (1). If 𝑇𝛿1 (𝑔) = 0
for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞), then there exists a constant 𝐶 > 0 such that, for
any 𝑓 ∈ 𝐶0∞ (𝑅𝑛 ) and 𝑥
˜ ∈ 𝑅𝑛 ,
𝑚
∑︁
𝑀 # (𝑇𝛿𝑏 (𝑓 ))(˜
𝑥) 6 𝐶‖𝑏‖BMO
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝑘=1
∞
𝐶0 (𝑅𝑛 )
and some constant 𝐶0 , the followProof. It suffices to prove for 𝑓 ∈
ing inequality holds:
∫︁
𝑚
∑︁
⃒ 𝑏
⃒
1
⃒𝑇𝛿 (𝑓 )(𝑥) − 𝐶0 ⃒ 𝑑𝑥 6 𝐶‖𝑏‖BMO
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
|𝑄| 𝑄
𝑘=1
Without loss of generality, we may
𝑄 = 𝑄(𝑥0 , 𝑑) and 𝑥
˜ ∈ 𝑄. For 𝑓1 =
Theorem 4.1, we have
𝑏−𝑏
assume 𝑇𝛿𝑘,1
𝑓 𝜒2𝑄 and 𝑓2
are 𝑇𝛿 (𝑘 = 1, . . . , 𝑚). Fix a cube
= 𝑓 𝜒(2𝑄)𝑐 , similar to the proof of
(𝑏−𝑏𝑄 )𝜒
(𝑏−𝑏 )𝜒
𝑐
(2𝑄)
𝑇𝛿𝑏 (𝑓 )(𝑥) = 𝑇𝛿 𝑄 (𝑓 )(𝑥) = 𝑇𝛿 𝑄 2𝑄 (𝑓 )(𝑥) + 𝑇𝛿
(𝑓 )(𝑥) = 𝑔(𝑥) + ℎ(𝑥),
∫︁
∫︁
∫︁
⃒
⃒
1
1
⃒𝑇 𝑏 (𝑓 )(𝑥)−ℎ(𝑥0 )⃒ 𝑑𝑥 6 1
|𝑔(𝑥)|𝑑𝑥 +
|ℎ(𝑥)−ℎ(𝑥0 )|𝑑𝑥 = 𝐼5 +𝐼6 .
|𝑄| 𝑄 𝛿
|𝑄| 𝑄
|𝑄| 𝑄
For 𝐼5 , choose 1 < 𝑡 < 𝑠, by Hölder’s inequality and the boundedness of 𝑇𝛿 with
1 < 𝑟 < ∞ and 1/𝑟 = 1/𝑡 − 𝛿/𝑛, we obtain
∫︁
1
|𝑇 𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 𝑘,2 (𝑓 )(𝑥)|𝑑𝑥
|𝑄| 𝑄 𝛿
(︂
)︂1/𝑟
∫︁
1
6
|𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 𝑘,2 (𝑓 )(𝑥)|𝑟 𝑑𝑥
|𝑄| 𝑅𝑛
(︂∫︁
)︂1/𝑡
−1/𝑟
𝑘,2
𝑡
6 𝐶|𝑄|
|𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 (𝑓 )(𝑥)| 𝑑𝑥
𝑅𝑛
−1/𝑟
(︂∫︁
|𝑇
6 𝐶|𝑄|
𝑘,2
𝑠
(︂
|𝑏(𝑥) − 𝑏𝑄 |
(𝑓 )(𝑥)| 𝑑𝑥
2𝑄
6 𝐶‖𝑏‖BMO
)︂1/𝑠(︂∫︁
2𝑄
1
|2𝑄|1−𝑠𝛿/𝑛
∫︁
2𝑄
)︂1/𝑠
|𝑇 𝑘,2 (𝑓 )(𝑥)|𝑠 𝑑𝑥
𝑠𝑡/(𝑠−𝑡)
)︂(𝑠−𝑡)/𝑠𝑡
𝑑𝑥
BOUNDEDNESS FOR TOEPLITZ TYPE OPERATOR
173
6 𝐶‖𝑏‖BMO 𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥),
thus
𝐼5 6
∫︁
𝑙
𝑚
∑︁
∑︁
1
|𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒2𝑄 𝑇 𝑘,2 (𝑓 )(𝑥)|𝑑𝑥 6 𝐶‖𝑏‖BMO
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
|𝑄| 𝑄
𝑘=1
𝑘=1
′
For 𝐼6 , recalling that 𝑠 > 𝑞 , taking 1 < 𝑝 < ∞ with 1/𝑝 + 1/𝑞 + 1/𝑠 = 1, we get,
for 𝑥 ∈ 𝑄,
|𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥) − 𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥0 )|
∞ ∫︁
∑︁
6
|𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)‖𝑏(𝑦) − 𝑏2𝑄 ‖𝑇 𝑘,2 (𝑓 )(𝑦)|𝑑𝑦
𝑗=1
6
2𝑗 𝑑6|𝑦−𝑥0 |<2𝑗+1 𝑑
∞ (︂∫︁
∑︁
𝑗=1
|𝐾(𝑥, 𝑦) − 𝐾(𝑥0 , 𝑦)|𝑞 𝑑𝑦
)︂1/𝑞
2𝑗 𝑑6|𝑦−𝑥0 |<2𝑗+1 𝑑
(︂∫︁
|𝑏(𝑦) − 𝑏𝑄 |𝑝 𝑑𝑦
×
)︂1/𝑝 (︂∫︁
2𝑗+1 𝑄
6 𝐶‖𝑏‖BMO
|𝑇 𝑘,2 (𝑓 )(𝑦)|𝑠 𝑑𝑦
)︂1/𝑠
2𝑗+1 𝑄
∞
∑︁
′
𝐶𝑗 (2𝑗 𝑑)−𝑛/𝑞 +𝛿 𝑗(2𝑗 𝑑)𝑛/𝑝 (2𝑗 𝑑)𝑛/𝑠−𝛿
𝑗=1
(︂
×
1
|2𝑗+1 𝑄|1−𝑠𝛿/𝑛
∫︁
|𝑇
𝑘,2
𝑠
)︂1/𝑠
(𝑓 )(𝑦)| 𝑑𝑦
2𝑗+1 𝑄
6 𝐶‖𝑏‖BMO 𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥)
∞
∑︁
𝑗𝐶𝑗 6 𝐶‖𝑏‖BMO 𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥),
𝑗=1
thus
1
𝐼6 6
|𝑄|
∫︁ ∑︁
𝑙
𝑄 𝑘=1
|𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥) − 𝑇𝛿𝑘,1 𝑀(𝑏−𝑏𝑄 )𝜒(2𝑄)𝑐 𝑇 𝑘,2 (𝑓 )(𝑥0 )|𝑑𝑥
6 𝐶‖𝑏‖BMO
𝑙
∑︁
𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))(˜
𝑥).
𝑘=1
This completes the proof of Theorem 4.3.
Theorem 4.4. Let the sequence {𝐶𝑗 } ∈ 𝑙1 , 0 < 𝛽 < min(1, 𝑛 − 𝛿), 𝑞 ′ < 𝑝 <
𝑛/(𝛽 + 𝛿), 1/𝑟 = 1/𝑝 − (𝛽 + 𝛿)/𝑛 and 𝑏 ∈ Lip𝛽 (𝑅𝑛 ). Suppose 𝑇𝛿 is a bounded linear
operator from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) and has a kernel 𝐾 satisfying (1). If 𝑇𝛿1 (𝑔) = 0
for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞) and 𝑇 𝑘,2 are the bounded operators on 𝐿𝑝 (𝑅𝑛 )
for 1 < 𝑝 < ∞, 𝑘 = 1, . . . , 𝑚, then 𝑇𝛿𝑏 is bounded from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ).
Proof. Choose 𝑞 ′ < 𝑠 < 𝑝 in Theorem 4.1, we have, by Lemma 3.3 and 3.4,
‖𝑇𝛿𝑏 (𝑓 )‖𝐿𝑟 6 ‖𝑀 (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟 6 𝐶‖𝑀 # (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟
6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝑘=1
‖𝑀𝛽+𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))‖𝐿𝑟 6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝑘=1
‖𝑇 𝑘,2 (𝑓 )‖𝐿𝑝
174
HUANG AND LIU
6 𝐶‖𝑏‖Lip𝛽 ‖𝑓 ‖𝐿𝑝 .
This completes the proof.
Theorem 4.5. Let the sequence {𝐶𝑗 } ∈ 𝑙1 , 0 < 𝛽 < min(1, 𝑛 − 𝛿), 𝑞 ′ < 𝑝 <
𝑛/(𝛽 + 𝛿), 1/𝑟 = 1/𝑝 − (𝛽 + 𝛿)/𝑛, 0 < 𝐷 < 2𝑛 and 𝑏 ∈ Lip𝛽 (𝑅𝑛 ). Suppose 𝑇𝛿 is a
bounded linear operator from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) and has a kernel 𝐾 satisfying (1).
If 𝑇𝛿1 (𝑔) = 0 for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞) and 𝑇 𝑘,2 are the bounded operators
on 𝐿𝑝,𝜙 (𝑅𝑛 ) for 1 < 𝑝 < ∞, 𝑘 = 1, . . . , 𝑚, then 𝑇𝛿𝑏 is bounded from 𝐿𝑝,𝜙 (𝑅𝑛 ) to
𝐿𝑟,𝜙 (𝑅𝑛 ).
Proof. Choose 𝑞 ′ < 𝑠 < 𝑝 in Theorem 4.1, we have, by Lemma 3.5 and 3.6,
‖𝑇𝛿𝑏 (𝑓 )‖𝐿𝑟,𝜙 6 ‖𝑀 (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟,𝜙 6 𝐶‖𝑀 # (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟,𝜙
6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
‖𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))‖𝐿𝑟,𝜙 6 𝐶‖𝑏‖Lip𝛽
𝑘=1
𝑚
∑︁
‖𝑇 𝑘,2 (𝑓 )‖𝐿𝑝,𝜙
𝑘=1
6 𝐶‖𝑏‖Lip𝛽 ‖𝑓 ‖𝐿𝑝,𝜙 .
This completes the proof.
Theorem 4.6. Let the sequence {2𝑗𝛽 𝐶𝑗 } ∈ 𝑙1 , 0 < 𝛽 < 1, 𝑞 ′ < 𝑝 < 𝑛/𝛿,
1/𝑟 = 1/𝑝 − 𝛿/𝑛 and 𝑏 ∈ Lip𝛽 (𝑅𝑛 ). Suppose 𝑇𝛿 is a bounded linear operator
from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ) and has a kernel 𝐾 satisfying (1). If 𝑇𝛿1 (𝑔) = 0 for any
𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞) and 𝑇 𝑘,2 are the bounded operators on 𝐿𝑝 (𝑅𝑛 ) for
1 < 𝑝 < ∞, 𝑘 = 1, . . . , 𝑚, then 𝑇𝛿𝑏 is bounded from 𝐿𝑝 (𝑅𝑛 ) to 𝐹˙𝑟𝛽,∞ (𝑅𝑛 ).
Proof. Choose 𝑞 ′ < 𝑠 < 𝑝 in Theorem 4.2, we have, by Lemma 3.2 and 3.3,
⃦
⃦
∫︁
⃒ 𝑏
⃒ ⃦
⃦
1
𝑏
⃒
⃒
⃦
𝑇𝛿 (𝑓 )(𝑥) − 𝐶0 𝑑𝑥⃦
‖𝑇𝛿 (𝑓 )‖𝐹˙𝑟𝛽,∞ 6 𝐶 ⃦sup
⃦ 𝑟
1+𝛽/𝑛
𝑄∋· |𝑄|
𝑄
6 𝐶‖𝑏‖Lip𝛽
𝑚
∑︁
𝐿
‖𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))‖𝐿𝑟 6 𝐶‖𝑏‖Lip𝛽
𝑘=1
6 𝐶‖𝑏‖Lip𝛽 ‖𝑓 ‖
𝐿𝑝
𝑚
∑︁
‖𝑇 𝑘,2 (𝑓 )‖𝐿𝑝
𝑘=1
.
This completes the proof.
1
′
Theorem 4.7. Let the sequence {𝑗𝐶𝑗 } ∈ 𝑙 , 𝑞 < 𝑝 < 𝑛/𝛿, 1/𝑟 = 1/𝑝−𝛿/𝑛 and
𝑏 ∈ BMO(𝑅𝑛 ). Suppose 𝑇𝛿 is a bounded linear operator from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 )
and has a kernel 𝐾 satisfying (1). If 𝑇𝛿1 (𝑔) = 0 for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 ) (1 < 𝑢 < ∞)
and 𝑇 𝑘,2 are the bounded operators on 𝐿𝑝 (𝑅𝑛 ) for 1 < 𝑝 < ∞, 𝑘 = 1, . . . , 𝑚, then
𝑇𝛿𝑏 is bounded from 𝐿𝑝 (𝑅𝑛 ) to 𝐿𝑟 (𝑅𝑛 ).
Proof. Choose 𝑞 ′ < 𝑠 < 𝑝 in Theorem 4.3, we have, by Lemma 3.3 and 3.4,
‖𝑇𝛿𝑏 (𝑓 )‖𝐿𝑟 6 ‖𝑀 (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑝 6 𝐶‖𝑀 # (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟
6 𝐶‖𝑏‖BMO
𝑚
∑︁
‖𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))‖𝐿𝑟 6 𝐶‖𝑏‖BMO
𝑘=1
6 𝐶‖𝑏‖BMO ‖𝑓 ‖𝐿𝑝 .
𝑚
∑︁
𝑘=1
‖𝑇 𝑘,2 (𝑓 )‖𝐿𝑝
BOUNDEDNESS FOR TOEPLITZ TYPE OPERATOR
175
This completes the proof.
Theorem 4.8. Let the sequence {𝑗𝐶𝑗 } ∈ 𝑙1 , 𝑞 ′ < 𝑝 < 𝑛/𝛿, 1/𝑟 = 1/𝑝−𝛿/𝑛, 0 <
𝐷 < 2𝑛 and 𝑏 ∈ BMO(𝑅𝑛 ). Suppose 𝑇𝛿 is a bounded linear operator from 𝐿𝑝 (𝑅𝑛 )
to 𝐿𝑟 (𝑅𝑛 ) and has a kernel 𝐾 satisfying (1). If 𝑇𝛿1 (𝑔) = 0 for any 𝑔 ∈ 𝐿𝑢 (𝑅𝑛 )
(1 < 𝑢 < ∞) and 𝑇 𝑘,2 are the bounded operators on 𝐿𝑝,𝜙 (𝑅𝑛 ) for 1 < 𝑝 < ∞,
𝑘 = 1, . . . , 𝑚, then 𝑇𝛿𝑏 is bounded from 𝐿𝑝,𝜙 (𝑅𝑛 ) to 𝐿𝑟,𝜙 (𝑅𝑛 ).
Proof. Choose 𝑞 ′ < 𝑠 < 𝑝 in Theorem 4.3, we have, by Lemma 3.5 and 3.6,
‖𝑇𝛿𝑏 (𝑓 )‖𝐿𝑟,𝜙 6 ‖𝑀 (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟,𝜙 6 𝐶‖𝑀 # (𝑇𝛿𝑏 (𝑓 ))‖𝐿𝑟,𝜙
6 𝐶‖𝑏‖BMO
𝑚
∑︁
‖𝑀𝛿,𝑠 (𝑇 𝑘,2 (𝑓 ))‖𝐿𝑟,𝜙 6 𝐶‖𝑏‖BMO
𝑘=1
𝑚
∑︁
‖𝑇 𝑘,2 (𝑓 )‖𝐿𝑝,𝜙
𝑘=1
6 𝐶‖𝑏‖BMO ‖𝑓 ‖𝐿𝑝,𝜙 .
This completes the proof.
Corollary. Let 𝑇𝛿 ∈ GSIO(𝛿), that is 𝑇𝛿 is the singular integral operator as
Definition 1.2. Then Theorems 4.1–4.8 hold for 𝑇𝛿𝑏 .
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Department of Mathematics
Changsha University of Science and Technology
Changsha 410114
P. R. of China
cxiahuang@126.com
lanzheliu@163.com
(Received 23 07 2012)
(Revised 15 09 2012)