Volume 2010, Article ID 961502, 20 pagesdoi:10.1155/2010/961502 Research Article Commutators of Littlewood-Paley Operators on the Generalized Morrey Space 1 Department of Mathematics and
Trang 1Volume 2010, Article ID 961502, 20 pages
doi:10.1155/2010/961502
Research Article
Commutators of Littlewood-Paley Operators on the Generalized Morrey Space
1 Department of Mathematics and Mechanics, Applied Science School, University of Science and Technology Beijing, Beijing 100083, China
2 Laboratory of Mathematics and Complex Systems (BNU), School of Mathematical Sciences, Beijing Normal University, Ministry of Education, Beijing 100875, China
3 The College of Mathematics and System Science, Xinjiang University, Urumqi, Xinjiang 830046, China
Received 6 May 2010; Accepted 11 July 2010
Academic Editor: Shusen Ding
Copyrightq 2010 Yanping Chen et al 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
Let μΩ, μ
S , and μ ∗, λ denote the Marcinkiewicz integral, the parameterized area integral, and
the parameterized Littlewood-Paley g∗λ function, respectively In this paper, the authors give a
characterization of BMO space by the boundedness of the commutators of μΩ, μ
S , and μ ∗, λ on
the generalized Morrey space L p,ϕRn
1 Introduction
dσ Suppose thatΩ satisfies the following conditions
Ωμx
S n−1Ωx
dσ
x
Trang 2c Ω ∈ LipS n−1, that is,
Ωx
μΩ
f
x
0
|F Ω,t x|2dt
t3
1/2
where
F Ω,t x
|x−y|≤t
Ωx − y
x − yn−1 f
y
We refer to see1,2 for the properties of μΩ
μ
⎛
Γx
1
t
|y−z|<t
Ωy − z
y − zn − f zdz
2
dydt
t n 1
⎞
⎠
1/2
:|x − y| < t}, and
μ ∗, λ f x
⎛
Rn 1
t
t x − y
λn
1
t
|y−z|<t
Ωy − z
y − zn − f zdz
2
dydt
t n 1
⎞
⎠
1/2
3 8
b, μΩ
⎛
0
|x−y|≤t
Ωx − y
x − yn−1
b x − by
f
y
dy
2
dt
t3
⎞
⎠
1/2
Trang 3Let 0 < < n and λ > 1 The commutator b, μ
S of μ
S and the commutatorb, μ ∗, λ of μ ∗, λ
are defined, respectively, by
b, μ S
⎛
Γx
1
t
|y−z|≤t
Ωy − z
y − zn − bx − bzfzdz
2
dydt
t n 1
⎞
⎠
1/2
b, μ ∗, λ
⎛
Rn 1
t
t x − y
λn
×
1
t
|y−z|≤t
Ωy − z
y − zn − bx − bzfzdz
2
dydt
t n 1
⎞
⎠
1/2
.
1.10
Let b ∈ LlocRn It is said that b ∈ BMOR n if
B⊂Rn
M b, B |B|1
B
There are some results about the boundedness of the commutators formed by BMO
S , and μ ∗, λ see 7,9,10
Many important operators gave a characterization of BMO space In 1976, Coifman et
characterization of BMO space
The purpose of this paper is to give a characterization of BMO space by the
L p,ϕRn
Definition 1.1 Let 1 < p < ∞ Suppose that ϕ : 0, ∞ → 0, ∞ be such that ϕt is
by
L p,ϕRn f ∈ LlocRn :f
where
f
L p,ϕ sup
x∈Rn
r>0
1
ϕ |Bx, r|
1
|Bx, r|
B x,r
f
yp
dy
1/p
Trang 4We refer to see13,14 for the known results of the generalized Morrey space L p,ϕ
t λ/n−1/p 0 < λ < n, L p,ϕRn coincides with the Morrey space L p,λRn
The main result in this paper is as follows
Theorem 1.2 Assume that ϕt is nonincreasing and t 1/p ϕ t is nondecreasing Suppose that b, μΩ
is defined as1.8, Ω satisfies 1.1, 1.2, and
Ωx
log2/x− yγ , C1> 0, γ > 1, x, y∈ S n−1. 1.15
If b, μΩ is bounded on L p,ϕRn for some p 1 < p < ∞, then b ∈ BMOR n .
Theorem 1.3 Let 0 < < n and 1 < p < ∞ Assume that ϕt is nonincreasing and t 1/p ϕ t is
nondecreasing Suppose that b, μ
S is defined as 1.9, Ω satisfies 1.1, 1.2, and 1.15 If b, μ
S
is a bounded operator on L p,ϕRn for some p 1 < p < ∞, then b ∈ BMOR n .
Theorem 1.4 Let 0 < < n, λ > 1, and 1 < p < ∞ Assume that ϕt is nonincreasing and
t 1/p ϕ t is nondecreasing Suppose that b, μ ∗,ϕ λ is defined as 1.10, Ω satisfies 1.1, 1.2, and
1.15 If b, μ ∗, λ is on L p,ϕRn for some p 1 < p < ∞, then b ∈ BMOR n .
Remark 1.5 It is easy to check that b, μ
Theorem 1.3forb, μ
S
Remark 1.6 It is easy to see that the condition1.15 is weaker than Lipβ S n−1 for 0 < β ≤ 1.
Marcinkiewicz integral and the parameterized Littlewood-Paley operators are neither the convolution operator nor the linear operators, hence, we need new ideas and nontrivial estimates in the proof
Let us begin with recalling some known conclusion
Lemma 2.1 If Ω satisfies conditions 1.1, 1.2, and 1.15, let β > 0, then for |x| > 2|y|, we have
Ωx − y
x − yβ − Ωx
|x| β
≤
C
|x| β
Trang 5We may assume thatb, μΩL p,ϕ →L p,ϕ 1 We want to prove that, for any x0∈ Rnand
N 1
|Bx0, r|
B x0,r
b
y
− a0dy ≤ A
holds, where a0 |Bx0, r|−1B x0,rb ydy Since b − a0, μΩ b, μΩ, we may assume that
a0 0 Let
f
y
sgn
b
y
− c0
χ B x0,r
y
where c0 1/|Bx0, r|B x
0,rsgnbydy Since 1/|Bx0, r|B x
0,rb ydy a0 0, we
f
Rn
f
y
f
y
b
y
> 0, y ∈ Bx0, r , 2.7 1
|Bx0, r|
Rn
f
y
b
y
A i 1 ≤ i < j Since Ω satisfies 1.2, then there exists an A1such that 0 < A1< 1 and
σ
x∈ S n−1: Ωx
condition1.15, it is easy to see that
Λ :
x∈ S n−1: Ωx
log2/A1γ
2.10
is a closed set We claim that
if x∈ Λ and y∈ S n−1, satisfying x− y ≤ A1, thenΩy
Trang 6In fact, since|Ωx − Ωy| ≤ C1/ log2/|x− y| γ ≤ C1/ log2/A1γ , note that Ωx ≥ 2C1/ log2/A1γ , we can get Ωy ≥ C1/ log2/A1γ Taking A2> 3/A1, let
b, μΩ
f x ≥ μΩ
bfx − |bx|μΩf x
⎧
⎪
⎪
0
|x−y|≤t
x − y
x − yn−1 b
y
f
y
dy
2
dt
t3
⎫
⎪
⎪
1/2
− |bx|
⎧
⎪
⎪
0
|x−y|≤t
x − y
x − yn−1 f
y
dy
2
dt
t3
⎫
⎪
⎪
1/2
: I1− I2.
2.13
For I1, noting that if y ∈ Bx0, r , then |x − x0| > A2|y − x0| for x ∈ G Thus, we have
x − y− x − x0 ≤ 2y − x0
|x − x0| ≤
2
A2
< A1. 2.14
Using2.11, we get Ωx − y ≥ C1/ log2/A1γ Noting that |x − x0
from2.5, 2.7, 2.8, and H¨older’s inequality that
I1≥
⎧
⎪
⎪
|x−x0 |
⎛
B x0,r
x − yb
y
f
y
x − yn−1 χ {|x−y|≤t}
y
dy
⎞
⎟
2
dt
t3
⎫
⎪
⎪
1/2
≥
⎛
|x−x0 |
B x0,r
x − yb
y
f
y
x − yn−1 χ {|x−y|≤t} dy dt
t3
⎞
|x−x0 |
dt
t3
−1/2
log2/A1γ |x − x0|
B x0,r
x − y−n 1 b
y
f
y
|x−x0|≤t
|x−y|≤t
dt
t3dy
log2/A1γ |x − x0|−n
B x0,rb
y
f
y
dy
A3Nr n |x − x0|−n
2.15
Trang 7For x ∈ G, by Ω ∈ L∞S n−1, 2.4, 2.5, 2.6, the Minkowski inequality, andLemma 2.1, we obtain
I2 |bx|
⎧
⎨
⎩
0
Rn
f
y Ωx − y
x − yn−1χ {|x−y|≤t}− Ωx − x0
|x − x0|n−1χ {|x−x0|≤t} dy
2
dt
t3
⎫
⎬
⎭
1/2
≤ |bx|
⎧
⎪
⎪
⎛
0
|x−y|≤t<|x−x0 |
Ωx − y
x − yn−1 f
ydy 2dt
t3
⎞
⎠
1/2
⎛
0
|x−x0|≤t<|x−y|
|Ωx − x0|
|x − x0|n−1f
ydy 2dt
t3
⎞
⎠
1/2
⎛
0
⎛
|x−x0|≤t
|x−y|≤t
Ωx − y
x − yn−1 − Ωx − x0
|x − x0|n−1
f
ydy⎞⎠
2
dt
t3
⎞
⎟
1/2⎫
⎪
⎪
≤ |bx|
⎧
⎨
⎩
B x0,r
Ωx − y
x − yn−1 f
y
|x−y|≤t<|x−x0 |
dt
t3
1/2
dy
B x0,r
|Ωx − x0|
|x − x0|n−1f
y
|x−y|>t≥|x−x0 |
dt
t3
1/2
dy
B x0,r
f
y
Ωx − y
x − yn−1− Ωx − x0
|x − x0|n−1
⎛
|x−y|≤t
|x−x0|≤t
dt
t3
⎞
⎠
1/2
dy
⎫
⎪
⎪
≤ C|bx|
r 1/2
B x0,r
f
y
|x − x0|n 1/2 dy
B x0,r
f
y
|x − x0|n
log|x − x0|/rγ dy
≤ A4|bx|r n |x − x0|−n
log|x − x0|
r
−γ
.
2.16
Let
F
!
x ∈ G : |bx| > A3N
2A4
log|x − x0|
r
γ
, |x − x0| < N 1/n r
"
Trang 8result Since ϕt is nonincreasing, it follows that ϕ|Bx0, N 1/n r | ≤ ϕ|Bx0, r | ϕr n By
2.13, 2.15, and 2.16, we have
fp
L p,ϕ ≥b, μΩ
fp
L p,ϕ
ϕB
x0, N 1/n rpB
x0, N 1/n r
|x−x0|<N 1/n r
b, μΩ
f xp
dx
ϕ r np Nr n
G\F∩{|x−x0|<N 1/n r}
1
2A3Nr
n |x − x0|−n
p
dx
ϕ r np
Nr n
{A5|F| A2rn1/n
< |x−x0|<N 1/n r }∩G
1
n |x − x0|−np
dx
ϕ r np
Nr n
A3Nr n
2
pN 1/n r
A5|F| A2rn1/n t −pn n−1 dt
≥ ω n−1
ϕ r np Nr np−1A3/2p
n − np
N1−pr n 1−p − A 1−pn5 |F| A2rn1−p
.
2.18
Thus,
|F| A2rn1−p
≤ A6N1−pr n 1−p
1 ϕr npfp
L p,ϕ
Now, we claim that
f
L p,ϕ≤ C
where C is independent of r In fact,
f
L p,ϕ sup
x∈Rn
t>0
1
ϕ |Bx, t|
1
|Bx, t|
B x,t
f
yp
dy
1/p
Case 1 t > r Since s 1/p ϕ s is nondecreasing in s, then
1
ϕ |Bx, t|
1
ϕ r n
1
Trang 9f
L p,ϕ≤ sup
x∈Rn
t>0
1
ϕ r n
1
r n/p
B x,t
f
yp
dy
1/p
sup
x∈Rn
t>0
1
ϕ r n
1
r n/p
B x,t∩Bx0,r
f
yp
dy
1/p
ϕ r n.
2.23
Case 2 t ≤ r Since ϕs is nonincreasing in s, then
1
ϕ |Bx, t|≤
1
Thus,
f
L p,ϕ≤ sup
x∈Rn
t>0
1
ϕ r n
1
|Bx, t|
B x,t
f
yp
dy
1/p
ϕ r n.
2.25
7 A n2, thenTheorem 1.2is proved If N > 2A−17 A n2, then
Let gy χ B x0,ry For x ∈ F, we have
b, μΩ
g x ≥ |bx|
⎧
⎪
⎪
0
|x−y|≤t
x − y
x − yn−1 g
y
dy
2
dt
t3
⎫
⎪
⎪
1/2
−
⎧
⎪
⎪
0
|x−y|≤t
x − y
x − yn−1 b
y
g
y
dy
2
dt
t3
⎫
⎪
⎪
1/2
: K1− K2.
2.28
Trang 10Noting that if y ∈ Bx0, r and x ∈ F, we get |x − y− x − x0| ≤ A1 Applying2.11, we
follows that
⎧
⎪
⎪
|x−x0 |
⎛
B x0,r
x − y
x − yn−1 χ {|x−y|≤t}
y
dy
⎞
⎟
2
dt
t3
⎫
⎪
⎪
1/2
≥ |bx|
⎛
|x−x0 |
B x0,r
x − y
x − yn−1 χ {|x−y|≤t} dy dt
t3
⎞
|x−x0 |
dt
t3
−1/2
log2/A1γ |x − x0|
B x0,r
x − y−n 1
|x−x0|≤t
|x−y|≤t
dt
t3dy
≥ A8|bx||x − x0|−n
B x0,rdy
A8r n |bx||x − x0|−n
2.29
we have
K2≤ C
B x0,r
b
y
x − yn dy
≤ A9|x − x0|−n
B x0,r
b
ydy
A9Nr n |x − x0|−n
2.30
Thus, by2.28, 2.29, and 2.30, we get, for x ∈ F,
b, μΩ
Trang 11ϕ Nr n ≤ ϕr n , and |bx| > NA3/2A4log|x − x0|/r γ when x ∈ F, we have
A10
ϕ r n ≥g
L p,ϕ≥b, μΩ
g
L p,ϕ
ϕ Nr n Nr n1/p
|x−x0|<N 1/n r
b, μΩ
g xp
dx
1/p
ϕ r n Nr n1/p
|x−x0|<N 1/n r
b, μΩ
g xdx
|x−x0|<N 1/n r
dx
−1/p
ϕ r n Nr n
F
b, μΩ
g xdx
ϕ r n Nr n
F
|bx||x − x0|−n dx− A9Nr n
ϕ r n Nr n
F
|x − x0|−n dx
ϕ r n
F
log|x − x0|
r
γ
|x − x0|−n dx
ϕ r n
F
|x − x0|−n dx
: L1− L2.
2.32
We first estimate L2 Since A2r < |x − x0| < N 1/n r for x ∈ F, we have
L2≤ A9ω n−1
ϕ r n
N 1/n r
A2r
ρ−1dρ≤ A12
Case 1 γ ≥ n Since the function log s/s is decreasing for s ≥ 3 and 3r < A2r < |x − x0| <
N 1/n r for x ∈ F, by 2.27, we get
L1 A11r −n
ϕ r n
F
log|x − x0|/r
|x − x0|/r
n log|x − x0|
r
γ −n
dx
≥ A7A11
2ϕr n
log A2γ −n
N
N 1/n
n
ϕ r n
log Nn
.
2.34
Trang 12Case 2 1 < γ < n Since the function log s γ /s n is decreasing for s ≥ 3 and 3r < A2r <
|x − x0| < N 1/n r for x ∈ F, by 2.27, we have
L1 A11r −n
ϕ r n
F
log|x − x0|/rγ
≥ A7A11
2ϕr nN
log N 1/nγ
N
ϕ r n
log Nγ
.
2.35
L1≥ A15
ϕ r n
log Nτ
So by2.32, 2.33, and 2.36, we get
A10≥ A15
log Nτ
SL p,ϕ →L p,ϕ 1 We want to prove that, for any x0 ∈ Rnand
N |Bx1
0, r|
B x0,r
b
y
− a0dy ≤ B
holds, where a0 |Bx0, r|−1
B x0,rb ydy Since b − a0, μ S b, μ
a0 0 Let fy be as 2.3, then 2.4–2.8 hold In this proof for j 1, , 13, B jis a positive
exists a B1such that 0 < B1< 1 and
σ
x∈ S n−1: Ωx
Trang 13
where σ is the measure on S n−1which is induced from the Lebesgue measure onRn By the condition1.15, it is easy to see that
Λ :
x∈ S n−1:Ωx
log2/B1γ
3.3
if x∈ Λ and y∈ S n−1, satisfyingx− y ≤ B1, thenΩy
Taking B2> 3/B1 1, let
b, μ
S
f x ⎛⎝∞
0
|x−y|<t
|y−z|<t
Ωy− z
y − zn − bx − bzfzdz
2
dydt
t n 1 2
⎞
⎠
1/2
≥
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0 |
×
|y−z|<t
Ωy − z
y − zn − bx − bzfzdz
2
dydt
t n 1 2
⎞
⎠
1/2
≥
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0|, y−x0 ∈Λ
×
|y−z|<t
Ωy − z
y − zn − b zfzdz
2
dydt
t n 1 2
⎞
⎠
1/2
− |bx|
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0 |
|y−z|<t
Ωy − z
y − zn − f zdz
2
dydt
t n 1 2
⎞
⎠
1/2
: I1− I2.
3.6
Trang 14For I1, noting that if |z − x0| < r, |x − x0| > B2|z − x0|, and |y − x0| > 2B2|z − x0|, then we get
y − z−y − x0
≤ 2|z − x0|
y − x0 ≤ B12 < B1. 3.7
Then by3.4, we get Ωy − z ≥ C1/ log2/B1γ Since 4|x−x0| > |y−x0| |z−x0| ≥ |y−z| ≥
|y−x0|−|z−x0| > 2|x−x0|−|x−x0|/2 3|x−x0|/2 and 4|x−x0| > |x−y| ≥ |y−x0|−|x−x0| > |x−x0|,
we get 4|x − x0| ≥ |y − z| ≥ 3|x − x0|/2 and 4|x − x0| > |x − y| > |x − x0| Thus, by 2.5, 2.7,
I1≥ C
4|x−x 0 |
|x−y|<t, y−x0 ∈Λ 2|x−x 0|<|y−x0|<3|x−x0 |
B x0,r
Ωy − z
y − zn − b zfzχ {|y−z|<t} dz dydt
t n 1 2
×
⎛
4|x−x 0 |
|x−y|<t, y−x0 ∈Λ 2|x−x 0|<|y−x0|<3|x−x0 |
dydt
t n 1 2
⎞
⎠
−1/2
≥ C|x − x0|2−n
B x0,rb zfz
y−x0 ∈Λ 2|x−x 0|<|y−x0|<3|x−x0 |
4|x−x 0|<t, |x−y|<t
|y−z|<t
dtdy
t n 1 2 dz
C|x − x0|2−n
B x0,rb zfz
y−x0 ∈Λ 2|x−x 0|<|y−x0|<3|x−x0 |
4|x−x 0|<t
dtdy
t n 1 2 dz
≥ C|x − x0|−n
B x0,rb zfzdz
B3Nr n |x − x0|−n
3.8
By2.5 and 2.6, we have
I2 |bx|
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0 |
×
Rn
Ωy − z
y − zn − χ {|y−z|<t}− Ω
y − x0
y − x0n − χ {|y−x0|<t} f zdz
2
dydt
t n 1 2
⎞
⎠
1/2
Trang 15≤ |bx|
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0 |
×
|y−z|<t
|y−x0|<t
Ωy − z
y − zn − − Ω
y − x0
y − x0n − f zdz
2
dydt
t n 1 2
⎞
⎟
1/2
|bx|
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0 |
|y−z|<t
|y−x0|≥t
Ωy − z
y − zn − f zdz
2
dydt
t n 1 2
⎞
⎟
1/2
|bx|
⎛
4|x−x 0 |
|x−y|<t
2|x−x 0|<|y−x0|<3|x−x0 |
|y−z|≥t
|y−x0|<t
Ωy − x0
y − x0n − f zdz
2
dydt
t n 1 2
⎞
⎟
1/2
: I1
2 I2
2 I3
2.
3.9
In I22, we have t ≤ |y − x0| < 3|x − x0| and t ≥ 4|x − x0| In I3
2, we get t ≤ |y − z| < 4|x − x0|
and t ≥ 4|x − x0| It is easy to see that I2
2 I3
2, by Ω ∈ L∞S n−1, the
I21 ≤ C|bx|
B x0,r
f zdz⎛⎝
2|x−x 0|<|y−x0|<3|x−x0 |
4|x−x 0|≤t, |y−z|<t
|y−x0|<t, |x−y|≤t
×y − x02n− 1
logy − x0/r2γ
dtdy
t n 1 2
1/2
≤ B4|bx|r n |x − x0|−n
log|x − x0|
r
−γ
.
3.10
I2≤ B4|bx|r n |x − x0|−n
log|x − x0|
r
−γ
Let
F
!
x ∈ G : |bx| > B3N
2B4
log|x − x0|
r
γ
, |x − x0| < N 1/n r
"
Trang 16Without loss of generality, we may assume that N > B2 > 1, otherwise, we get the desired
result Since ϕt is nonincreasing, we have ϕ|Bx0, N 1/n r | ≤ ϕ|Bx0, r | ϕr n Then by,
3.6, 3.8, and 3.11, we get
fp
L p,ϕ≥b, μ
S
fp
L p,ϕ
ϕB
x0, N 1/n rpB
x0, N 1/n r
|x−x0|<N 1/n r
b, μ S
f xp
dx
ϕ r np
Nr n
G\F∩{|x−x0|<N 1/n r}
1
2B3Nr
n |x − x0|−np
dx
ϕ r np
Nr n
{B5|F| B2rn1/n < |x−x0|<N 1/n r }∩G
1
2B3Nr
n |x − x0|−n
p
dx
ϕ r np
Nr n
B3Nr n
2
pN 1/n r
B5|F| B2rn1/n t −pn n−1 dt
ΛJ
x
dσ
x
ϕ r np σ ΛNr np−1B3/2p
n − np
N1−pr n 1−p − B51−pn|F| B2rn1−p
.
3.13
Thus,
|F| B2rn1−p
≤ B6N1−pr n 1−p
1 ϕ r npfp
L p,λ
7 B n
2, thenTheorem 1.3is proved If N > 2B−17 B n
2, then
|F| ≥ B7
... Trang 4We refer to see13,14 for the known results of the generalized Morrey space L p,ϕ
t... integral and the parameterized Littlewood-Paley operators are neither the convolution operator nor the linear operators, hence, we need new ideas and nontrivial estimates in the proof
Let...
Trang 13where σ is the measure on S n−1which is induced from the Lebesgue