Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 859680, 7 pages
http://dx.doi.org/10.1155/2013/859680
Research Article
Almost Everywhere Convergence of Riesz Means Related to
Schrödinger Operator with Constant Magnetic Fields
Liurui Deng1 and Bolin Ma2
1
2
College of Finance and Statistics, Hunan University, Changsha 410082, China
College of Science and Information Engineering, Jiaxing University, Jiaxing,
Zhejiang 314001, China
Correspondence should be addressed to Bolin Ma; blma@mail.zjxu.edu.cn
Received 25 September 2012; Accepted 28 January 2013
Academic Editor: Chuanzhi Bai
Copyright © 2013 L. Deng and B. Ma. 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.
We study almost everywhere convergence for Riesz means related to Schrödinger operator with constant magnetic fields. Through
researching the weighted norm estimates for the maximal operator with power-weight functions, we obtain the desired result, which
is similar to the work given by Anthony Carbery, Jose L. Rubio de Francia, and Luis Vega.
For 𝛽 > 0, set the Riesz means of order 𝛽 as
1. Introduction
The magnetic Schrödinger operator (MSO) with constant
magnetic fields 𝐻𝑏 in R𝑛 is of the form
𝐻𝑏 = −(∇ +
𝑖𝐵𝑧 2
),
2
𝑧 ∈ 𝑅𝑛 ,
(1)
where 𝐵 is a real antisymmetric matrix. If 𝐵 is degenerated
(this requires 𝑛 = 2𝑑 + 𝑙 to be odd, and 2𝑑 is the
rank of the matrix 𝐵), then its eigenvalues have the form
±𝑖𝑏𝑗 , 𝑗 = 1, 2, . . . , 𝑑. In properly chosen coordinates 𝑧 =
(𝑧1 , 𝑧2 , . . . , 𝑧𝑛 ) = (𝑥1 , 𝑦1 , 𝑥2 , 𝑦2 , . . . , 𝑥𝑑 , 𝑦𝑑 , 𝑧2𝑑+1 , . . . , 𝑧2𝑑+𝑙 ),
the operator becomes
2
𝑑
𝑏𝑗
𝑏𝑗
𝐻𝑏 = − ∑ ((𝜕𝑥𝑗 + 𝑖 𝑦𝑗 ) + (𝜕𝑦𝑗 + 𝑖 𝑥𝑗 )) − Δ 𝑙 ,
2
2
𝑗=1
(2)
2
where Δ 𝑙 = ∑𝑙𝑗=1 𝜕2 /𝜕𝑧2𝑑+𝑗
is the Laplacian in 𝑅𝑙 . The
spectrum of 𝐻𝑏 is in the semiaxis starting from the point
𝑐 = ∑ 𝑏𝑗 , and its spectral expansion is continuous (see [1]).
Let 𝑏𝑗 ’s be positive. Denote by 𝐸𝑡 the spectral function of
𝐻𝑏 . It is an integral operator with a kernel 𝑒𝑡 (𝑧, 𝑧󸀠 ), which
is skew-translation invariant; that is, 𝑒𝑡 (𝑧 + 𝜔, 𝑧󸀠 + 𝜔) =
𝑒𝑡 (𝑧, 𝑧󸀠 ) exp 𝑖⟨𝐵(𝑧 − 𝑧󸀠 ), 𝜔⟩.
∞
𝛽
𝑆𝜆 𝑓 (𝑥) = ∫ (1 −
0
∞
𝑡 𝛽
𝑡 𝛽
) 𝑑𝐸𝑡 𝑓 (𝑥) = ∫ (1 − ) 𝑑𝐸𝑡 𝑓 (𝑥) ,
𝜆 +
𝜆 +
𝑐
(3)
𝛽
and the kernel of 𝑆𝜆 as
∞
𝛽
𝑠𝜆 (𝑧, 𝑧󸀠 ) = 𝑠𝛽 (𝜆, 𝑧, 𝑧󸀠 ) = ∫ (1 −
0
𝑡 𝛽
) 𝑑𝑒 (𝑧, 𝑧󸀠 ) ,
𝜆 + 𝑡
(4)
with the same skew-translation invariance. We will denote by
𝑆∗𝛽 the corresponding maximal operator; that is,
󵄨󵄨 𝛽
󵄨󵄨
𝑆∗𝛽 𝑓 (𝑥) = sup 󵄨󵄨󵄨𝑆𝜆 𝑓 (𝑥)󵄨󵄨󵄨 .
(5)
󵄨
󵄨
0<𝜆<∞
As an indispensable part in harmonic analysis, many
people investigate the convergence of Bochner-Riesz means
for Fourier transform in norm and almost everywhere, which
is defined as
𝛽
𝛽
∧
(𝑇𝑅 𝑓) (𝜉) = (1 −
𝛽
|𝜉|2 ̂
) 𝑓 (𝜉) .
𝑅 +
(6)
Since the convergence of 𝑇𝑅 in 𝐿𝑝 -norm is equivalent to
𝛽
the boundedness of 𝑇1 in 𝐿𝑃 (R𝑛 ), persons look for the 𝐿𝑝
2
Abstract and Applied Analysis
boundedness of it. For 2𝑛/(𝑛 + 1 + 2𝛽) < 𝑝 < 2𝑛/(𝑛 − 1 − 2𝛽)
and 0 ≤ 𝛽 ≤ (𝑛 − 1/2), Carleson, Cordoba, and Fefferman
turn out the boundedness in R2 (see [2–4]). When 𝑛 > 2, it
is only proven for 𝛽 ≥ (𝑛 − 1)/2(𝑛 + 1) (see [4, 5]). Returning
𝛽
the problem about almost everywhere convergence of 𝑇𝑅 ,
Carbery has finished it for 𝛽 > 0 and 2𝑛/(𝑛 + 1 + 2𝛽) <
𝑝 < 2𝑛/(𝑛 − 1 − 2𝛽) in two dimensions in 1983 (see [6]).
For higher dimensions, it is completed by Christ only for
𝛽 ≥ (𝑛 − 1)/2(𝑛 + 1) (see [7]). In the special case, 𝛽 = 0,
Fefferman studies the 𝐿2 -boundeness of 𝑆𝜆0 for 𝑛 ≥ 2 (see [8]).
Not until 1988 was it solved by Carbery et al. for 𝛽 > 0, 𝑛 ≥ 2,
and 2 ≤ 𝑝 < 2𝑛/(𝑛 − 1 − 2𝛽) (see [9]).
In [1], Rozenblum and Tashchiyan investigated the 𝐿𝑝 norm convergence for Riesz means for Schrödinger operator
with constant magnetic fields. They showed that under the
restriction theorem similar to one for Fourier transform in
[5], it is of the same results as Bochner-Riesz means in
R𝑛 . However, very few results are considered for almost
everywhere convergence of Riesz means for Schrödinger
operator with constant magnetic fields. In the paper, we are
interested in it. Through researching the boundedness of the
maximal operator in 𝐿(R𝑛 , |𝑥|𝛼 𝑑𝑥), we get the desired result.
where
󵄨
󵄨
𝜀𝛼 , 𝜀 󵄨log 𝜀󵄨 , 0 ≤ 𝛼 < 1,
{
{ 󵄨 󵄨󵄨 󵄨 󵄨󵄨
𝐵𝛼 (𝜀) = {𝜀 󵄨󵄨󵄨log 𝜀󵄨󵄨󵄨 ,
𝛼 = 1,
{
𝜀,
1 < 𝛼 < 𝑛.
{
For a small 𝛿 > 0, let 𝑚𝛿 (𝑡) be a 𝐶∞ function supported
in [1 − 𝛿, 1] and satisfy
If 𝑓 ∈ 𝐿 (R ) with 2 ≤ 𝑝 < 𝑝𝛽 , then
almost everywhere.
∀𝑘 ∈ N.
(10)
Define
𝐾𝜆𝛿 𝑓 (𝑥) = ∫ 𝑚𝛿 (𝜆𝑡) 𝑑𝐸𝑡 𝑓 (𝑥) ,
󵄨
󵄨
𝐾∗𝛿 𝑓 (𝑥) = sup 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨 .
(11)
𝜆>0
∞
Let 𝜙(𝑡) ∈ 𝐶 (R) supported in [−1, 2] and 𝜙(𝑡) = 1 for 0 ≤
𝑡 ≤ 1. Set
𝜙 (𝑡) ,
𝑙 = 0,
ℎ𝑙 (𝑡) = {
−𝑙
1−𝑙
𝜙 (2 𝑡) − 𝜙 (2 𝑡) , 𝑙 ≥ 1.
(12)
Denoting by 𝜒𝐸 the character function of the set 𝐸, write
󵄨
󵄨
𝑗 = 0,
𝜒
(󵄨󵄨𝑥 − 𝑦󵄨󵄨) ,
𝐻𝑗 (𝑥, 𝑦) = { [0,1) 󵄨 󵄨󵄨 󵄨 󵄨󵄨
𝜒[2𝑗−1 ,2𝑗 ) (󵄨󵄨𝑥 − 𝑦󵄨󵄨) , 𝑗 ≥ 1.
Theorem 1. Set 𝛽 > 0 and 𝑛 ≥ 2. Write 𝑝𝛽 = 2𝑛/(𝑛 − 1 − 2𝛽).
𝑛
󵄨󵄨 𝑘 𝛿 󵄨󵄨
󵄨󵄨𝐷 𝑚 (𝑡)󵄨󵄨 ≤ 𝐶𝛿−𝑘
󵄨
󵄨
0 ≤ 𝑚𝛿 (𝑡) ≤ 1,
2. Main Results
𝑝
(9)
𝛽
lim𝜆 → ∞ 𝑆𝜆 𝑓(𝑥)
= 𝑓(𝑥)
Usually, we replace the almost everywhere convergence
of Riesz means with 𝐿𝑝 estimates for the maximal operator.
However, we only need to think about weighted 𝐿2 estimates
for the maximal operator 𝑆∗𝛽 , as follows. In fact, based on the
idea in [9], for 2 ≤ 𝑝 < 𝑝𝛽 , there exists a number 𝛼 with
0 ≤ 𝛼 < 1 + 2𝛽 such that 𝐿𝑝 ⊆ 𝐿2 + 𝐿2 (|𝑥|𝛼 ). We have gotten
𝐿2 boundedness of the maximal operator in another paper.
Theorem 2. Suppose that 𝛽 > 0 and 0 ≤ 𝛼 < 1 + 2𝛽 ≤ 𝑛.
Then,
󵄨2
󵄨
󵄨2
󵄨
∫ 󵄨󵄨󵄨󵄨𝑆∗𝛽 𝑓 (𝑥)󵄨󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥 ≤ 𝐶𝛼,𝛽 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥.
Let 𝑘𝛿 (𝑡) be a function with Fourier transform as
𝛿
𝑘̂𝜆 (𝑡) = 𝑚𝛿 (𝜆𝑡) .
𝛿,𝑙,𝑗
𝑘𝜆
𝑗
𝐸𝑡 𝑓 (𝑥)
(𝑡) = 𝑘𝜆𝛿 (𝑡) ℎ𝑙 (
𝛿2 2−𝑗 𝑡
),
𝜆
R𝑛
𝛿,𝑙,𝑗
as
𝛿,𝑙,𝑗
𝛿,𝑙,𝑗
𝑗
𝐾𝜆 𝑓 (𝑥) = ∫ 𝑘̂𝜆 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥) ,
where 𝑘̂𝜆
𝛿,𝑙,𝑗
𝛿
𝐾𝜆𝛿 𝑓 (𝑥) = ∫ 𝑘̂𝜆 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥) ,
(17)
we decompose
𝛿
𝐾𝜆𝛿 𝑓 (𝑥) = ∫ 𝑘̂𝜆 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)
∞ ∞
𝛿,𝑙,𝑗
(18)
𝑗=0 𝑙=0
∞ ∞
(8)
(16)
is the Fourier transform of 𝑘𝜆 . Apparently, since
= ∑ ∑𝐾𝜆 𝑓 (𝑥)
Lemma 3 (see [9]). If 0 < 𝜀 < 1/2, then
(15)
= ∫ 𝑒𝑡 (𝑥, 𝑦) 𝐻𝑗 (𝑥, 𝑦) 𝑓 (𝑦) 𝑑𝑦.
Accordingly, define the operator 𝐾𝜆
𝛿,𝑙,𝑗
󵄨2
󵄨2 𝛼
󵄨󵄨
󵄨
∫
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥 ≤ 𝐶𝛼 𝐵𝛼 (𝜀) ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥| 𝑑𝑥,
||𝑥|−1|≤𝜀
(14)
Set
(7)
In order to prove the theorem, we introduce the following
lemmas, which are the essential tools. In the following lemmas, we reduce the maximal operator to the one generated
by a multiplier with compact support. Since it is easy to see
that the boundedness of the maximal operator generated by
the multipliers is independent of the index 𝛽, the dimension
𝑛 = 2𝑑 + 𝑙 will not play an important role in the following
estimates.
(13)
𝛿,𝑙,𝑗
= ∑ ∑ ∫ 𝑘̂𝜆
𝑗=0 𝑙=0
𝑗
(𝑡) 𝑑𝐸𝑡 𝑓 (𝑥) .
Abstract and Applied Analysis
3
Lemma 4. For 𝜆 > 0, one has
󵄩2
󵄩󵄩 𝛿,𝑙,𝑗
󵄩󵄩𝐾 𝑓 (𝑥)󵄩󵄩󵄩 ≤ 𝐶2−2𝑀(𝑗+𝑙) 𝛿2𝑀󵄩󵄩󵄩𝑓󵄩󵄩󵄩2 ,
󵄩󵄩2
󵄩󵄩 𝜆
󵄩 󵄩2
We have
(19)
where the constant 𝐶 is independent of 𝜆 and 𝛿.
Proof. With the method similar to the proof of Lemma 4 in
[9], we write ℎ(𝑡) = 𝜙(𝑡) − 𝜙(2𝑡) and expand 𝑚𝛿 into a Taylor
series around 𝜆𝑡. Then,
󵄨󵄨
= 󵄨󵄨󵄨󵄨∫ 𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓1 (𝑦) 𝑑𝑦
󵄨 R𝑛
󵄨󵄨
+ ∫ 𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓2 (𝑦) 𝑑𝑦󵄨󵄨󵄨󵄨
𝑛
󵄨
R
󵄨󵄨
󵄨󵄨
≤ 󵄨󵄨󵄨󵄨∫ 𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓1 (𝑦) 𝑑𝑦󵄨󵄨󵄨󵄨
󵄨 R𝑛
󵄨
󵄨󵄨
󵄨󵄨
+ 󵄨󵄨󵄨󵄨∫ 𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓2 (𝑦) 𝑑𝑦󵄨󵄨󵄨󵄨
󵄨 R𝑛
󵄨
󵄨
󵄨
= ∫ 󵄨󵄨󵄨󵄨𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓1 (𝑦)󵄨󵄨󵄨󵄨 𝑑𝑦
R𝑛
𝛿,𝑙,𝑗
2−(𝑗+𝑙) 𝛿2 𝑟
𝑘̂𝜆 (𝑡) = ∫ 𝑚𝛿 (𝜆 (𝑡 −
)) ℎ̂ (𝑟) 𝑑𝑟
𝜆
= ∫ 𝑚𝛿 (𝜆𝑡 − 2−(𝑗+𝑙) 𝛿2 𝑟) ℎ̂ (𝑟) 𝑑𝑟
󵄨󵄨
󵄨 󵄨󵄨
󵄨󵄨 𝑗
󵄨󵄨𝐸R 𝑓 (𝑥0 )󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ 𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓 (𝑦) 𝑑𝑦󵄨󵄨󵄨
󵄨󵄨
󵄨 󵄨󵄨 R𝑛
󵄨
󵄨
󵄨
+ ∫ 󵄨󵄨󵄨󵄨𝑒R (𝑥0 , 𝑦) 𝐻𝑗 (𝑥0 , 𝑦) 𝑓2 (𝑦)󵄨󵄨󵄨󵄨 𝑑𝑦
𝑛
R
(20)
(25)
󵄨
󵄨
≤ ∫ 󵄨󵄨󵄨𝑒R (𝑥0 , 𝑦) 𝑓1 (𝑦)󵄨󵄨󵄨 𝑑𝑦
R𝑛
= ∫ 𝑅𝑀 (𝑡, 𝑟) ℎ̂ (𝑟) 𝑑𝑟,
󵄨
󵄨
+ ∫ 󵄨󵄨󵄨𝑒R (𝑥0 , 𝑦) 𝑓2 (𝑦)󵄨󵄨󵄨 𝑑𝑦
R𝑛
where the remainder 𝑅𝑀 satisfies
󵄨󵄨
󵄨
󵄨󵄨𝑅𝑀 (𝑡, 𝑟)󵄨󵄨󵄨
󵄨
󵄨󵄨
󵄨𝑀
≤ 󵄨󵄨󵄨󵄨𝐷𝑀𝑚𝛿 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨2−(𝑗+𝑙) 𝛿2 𝑟󵄨󵄨󵄨󵄨 ≤ 2−𝑀(𝑙+𝑗) 𝛿𝑀|𝑟|𝑀.
(21)
(22)
(23)
Denote by 𝑒R (𝑥, 𝑦) the kernel of 𝐸R . For almost all 𝑥0 ∈
R𝑛 , we let 𝑆 = {𝑦 ∈ R𝑛 : 𝑒R (𝑥0 , 𝑦)𝑓(𝑦) > 0}. Decompose
𝑓 (𝑥) = 𝑓 (𝑥) 𝜒𝑆 (𝑥) + 𝑓 (𝑥) 𝜒R𝑛 /𝑆 (𝑥)
= 𝑓1 (𝑥) + 𝑓2 (𝑥) .
󵄩󵄩 𝛿,𝑙,𝑗
󵄩2
󵄩󵄩𝐾 𝑓 (𝑥)󵄩󵄩󵄩
󵄩󵄩 𝜆
󵄩󵄩2
󵄩󵄩 𝛿,𝑙,𝑗
󵄩󵄩2
𝑗
= 󵄩󵄩󵄩󵄩∫ 𝑘̂𝜆 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)󵄩󵄩󵄩󵄩
󵄩
󵄩2
𝛿,𝑙,𝑗
𝛿,𝑙,𝑗
𝑗
𝑗
= ⟨𝐸R (𝑘̂𝜆 (𝐻𝑏 ) 𝑓) , 𝐸R (𝑘̂𝜆 (𝐻𝑏 ) 𝑓)⟩
∞
0
It is easy to show
󵄩󵄩
󵄩󵄩2
𝛿,𝑙,𝑗
𝑗
= 󵄩󵄩󵄩󵄩∫ 𝑑𝐸𝑡 (𝑘̂𝜆 (𝐻𝑏 ) 𝑓 (𝑥))󵄩󵄩󵄩󵄩
󵄩
󵄩2
Since 𝐸R is a resolution of the identity, we see
𝐸R 𝑓 (𝑥) = ∫ 𝑑𝐸𝑡 𝑓 (𝑥) = 𝑓 (𝑥) .
󵄨
󵄨 󵄨
󵄨
≤ 󵄨󵄨󵄨𝑓1 (𝑥0 )󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓2 (𝑥0 )󵄨󵄨󵄨
󵄨
󵄨
≤ 𝐶 󵄨󵄨󵄨𝑓 (𝑥0 )󵄨󵄨󵄨 .
But ℎ̂ is a Schwartz function and can be integrated against
𝑀
|𝑟| . Hence,
󵄨󵄨 𝛿,𝑙,𝑗 󵄨󵄨
󵄨󵄨𝑘̂ (𝑡)󵄨󵄨 ≤ 𝐶 2−𝑀(𝑗+𝑙) 𝛿𝑀.
󵄨󵄨 𝜆
󵄨󵄨
𝑀
󵄨
󵄨
󵄨󵄨
󵄨󵄨
= 󵄨󵄨󵄨󵄨∫ 𝑒R (𝑥0 , 𝑦) 𝑓1 (𝑦) 𝑑𝑦󵄨󵄨󵄨󵄨
󵄨 R𝑛
󵄨
󵄨󵄨
󵄨󵄨
+ 󵄨󵄨󵄨󵄨∫ 𝑒R (𝑥0 , 𝑦) 𝑓2 (𝑦) 𝑑𝑦󵄨󵄨󵄨󵄨
󵄨 R𝑛
󵄨
󵄩󵄩 𝛿,𝑙,𝑗
󵄩󵄩2
≤ 󵄩󵄩󵄩󵄩𝑘̂𝜆 (𝐻𝑏 ) 𝑓󵄩󵄩󵄩󵄩
󵄩
󵄩2
∞ 󵄨 𝛿,𝑙,𝑗
󵄨󵄨2
󵄨
≤ ∫ 󵄨󵄨󵄨󵄨𝑘̂𝜆 (𝑡)󵄨󵄨󵄨󵄨 𝑑 (𝐸𝑡 𝑓, 𝑓)
󵄨
0 󵄨
∞
≤ 𝐶𝑀2−2𝑀(𝑗+𝑙) 𝛿2𝑀 ∫ 𝑑 (𝐸𝑡 𝑓, 𝑓)
0
󵄩2
󵄩󵄩𝑓󵄩󵄩󵄩2 .
−2𝑀(𝑗+𝑙) 2𝑀 󵄩
󵄩
≤ 𝐶𝑀2
(24)
𝛿
(26)
4
Abstract and Applied Analysis
Lemma 5. If 0 < 𝛼 < 𝑛 and 𝜆 > 0, then
󵄨2
󵄨󵄨 𝛿
󵄨 𝑑𝑥
󵄨󵄨2 𝑑𝑥
󵄨󵄨
,
∫ 󵄨󵄨󵄨󵄨∫ 𝑘̂𝜆 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)󵄨󵄨󵄨󵄨
𝛼 ≤ 𝐶𝛼 𝛿 ∫ 󵄨󵄨𝑓 (𝑥)󵄨󵄨
|𝑥|𝛼
󵄨 |𝑥|
󵄨
Choosing 𝑀 > 1, we get
󵄨󵄨2
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓2 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨
󵄨
(27)
where 𝐶𝛼 is independent of 𝛿 and 𝜆.
Proof. Suppose that 𝑓 is supported in {|𝑥| ≤ 𝐶2𝑗 }. Write 𝑓 =
𝜒{0≤|𝑥|≤1/4} (𝑥)𝑓(𝑥)+𝜒{𝐶<|𝑥|≤𝐶2𝑗 } (𝑥)𝑓(𝑥)+𝜒{1/4<|𝑥|≤𝐶} (𝑥)𝑓(𝑥) =
𝑓1 + 𝑓2 + 𝑓3 . If 𝐶 ≤ 1/4, then 𝑓3 = 0. Since
On the other hand, 𝐸𝑡 is self-adjoint. So,
⟨𝐸𝑡 𝑓, 𝑔⟩
󵄨2 𝑑𝑥 1/2
󵄨
)
(∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨
|𝑥|𝛼
= ∫ ∫ 𝑒𝑡 (𝑥, 𝑦) 𝑓 (𝑦) 𝑑𝑦𝑔 (𝑥) 𝑑𝑥
󵄨2 𝑑𝑥 1/2
󵄨
= (∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 (𝑓1 + 𝑓2 + 𝑓3 ) (𝑥)󵄨󵄨󵄨󵄨
)
|𝑥|𝛼
󵄨2 𝑑𝑥 1/2
󵄨
󵄨󵄨 𝛿 󵄨󵄨2 𝑑𝑥 1/2
󵄨󵄨𝐾𝜆 𝑓2 󵄨󵄨
≤ (∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓1 󵄨󵄨󵄨󵄨
)
+
(∫
󵄨 |𝑥|𝛼 )
󵄨
|𝑥|𝛼
= ∫ 𝑓 (𝑦) ∫ 𝑒𝑡 (𝑥, 𝑦) 𝑔 (𝑥) 𝑑𝑥𝑑𝑦
(28)
= ⟨𝑓, 𝐸𝑡 𝑔⟩
Hence,
we only need to prove that
󵄨󵄨 𝛿
󵄨󵄨2 𝑑𝑥
∫ 󵄨󵄨󵄨󵄨∫ 𝑘̂𝜆 (𝑡) 𝑑𝐸𝑡 𝑓𝑖 (𝑥)󵄨󵄨󵄨󵄨
󵄨
󵄨 |𝑥|𝛼
󵄨2 𝑑𝑥
󵄨
≤ 𝐶𝛼 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
,
|𝑥|𝛼
(29)
(𝑖 = 1, 2, 3) .
𝛿,𝑙,𝑗
󵄨2
󵄨󵄨
𝛿 ∫
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
𝐶≤|𝑥|≤𝐶2𝑗
and it implies that 𝐾𝜆
(30)
󵄨2 𝛼
󵄨󵄨
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥| 𝑑𝑥
𝐶2𝑘 <|𝑥|≤𝐶2𝑘+1
≤ 𝐶 ∑ 2−𝑘𝛼 ∫
(38)
󵄨󵄨2
󵄨󵄨 𝛿,𝑗,𝑙
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓2 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨
󵄨
󵄨2
󵄨
𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥|𝛼 𝑑𝑥.
(39)
󵄨2 𝑑𝑥
󵄨
≤ 𝐶𝛼 2−𝑙 2𝑗(𝛼−𝑀) 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
|𝑥|𝛼
(31)
Taking 𝑀 > 𝛼 + 1, we can establish the inequality
󵄨󵄨2 𝑑𝑥
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓2 (𝑥)󵄨󵄨󵄨
󵄨 |𝑥|𝛼
󵄨
(40)
󵄨2 𝑑𝑥
󵄨
≤ 𝐶𝛼 2 2 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
|𝑥|𝛼
−𝑙 −𝑗
Thus, we have
≤ 𝐶2 2
𝑗
󵄨2
󵄨
≤ 𝐶𝛼 2−𝑙 2−𝑗𝑀𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨2
󵄨
≤ 𝐶 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥|𝛼 𝑑𝑥.
−𝑙 −𝑗𝑀 𝑀
𝑗
⟨𝐸𝑡 𝑓, 𝑔⟩ = ⟨𝑓, 𝐸𝑡 𝑔⟩.
󵄨󵄨2 𝑑𝑥
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓2 (𝑥)󵄨󵄨󵄨
󵄨 |𝑥|𝛼
󵄨
󵄨2
󵄨󵄨
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥.
󵄨2
󵄨󵄨
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
𝐶2𝑘 <|𝑥|≤𝐶2𝑘+1
𝑘=0
is also self-adjoint; that is,
Therefore, by duality,
∑∫
𝑗−1
(37)
𝑗
= 𝑒𝑡 (𝑦, 𝑥) 𝐻𝑗 (𝑦, 𝑥) = 𝑒𝑡 (𝑦, 𝑥) ,
It is easy to see that
𝑘=0
(36)
𝑗
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀
𝑗−1
𝐻𝑗 (𝑥, 𝑦) = 𝐻𝑗 (𝑦, 𝑥) ,
𝑒𝑡 (𝑥, 𝑦) = 𝑒𝑡 (𝑥, 𝑦) 𝐻𝑗 (𝑥, 𝑦)
−𝑙 −𝑗𝑀 𝑀
𝑘
𝑘+1
𝑘=0 𝐶2 ≤|𝑥|≤𝐶2
(35)
we get
󵄩 󵄩2
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀󵄩󵄩󵄩𝑓2 󵄩󵄩󵄩2
×∑∫
𝑒𝑡 (𝑥, 𝑦) = 𝑒𝑡 (𝑦, 𝑥) .
With
For the case of 𝑖 = 2, with Lemma 4, it follows that
󵄨󵄨2
󵄨󵄨 𝛿,𝑗,𝑙
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓2 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨
󵄨
𝑗−1
(34)
= ∫ 𝑓 (𝑦) ∫ 𝑒𝑡 (𝑦, 𝑥) 𝑔 (𝑥) 𝑑𝑥 𝑑𝑦.
󵄨2 𝑑𝑥 1/2
󵄨
+ (∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓3 󵄨󵄨󵄨󵄨
) ,
|𝑥|𝛼
≤ 𝐶2 2
(33)
󵄨2
󵄨
≤ 𝐶𝛼 2−𝑙 2−𝑗𝑀𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥|𝛼 𝑑𝑥.
𝛿,𝑙,𝑗
(32)
Nextly, we consider 𝑖 = 1. By the definition of 𝐾𝜆
𝛿,𝑙,𝑗
𝑓1 , we see that the kernel of 𝐾𝜆 is supported in
󵄨
󵄨
{(𝑥, 𝑦) : 2𝑗−1 ≤ 󵄨󵄨󵄨𝑥 − 𝑦󵄨󵄨󵄨 ≤ 2𝑗 }
and
(41)
Abstract and Applied Analysis
5
and 𝑓(𝑦) is supported in
Through we choose 𝑀 > 𝛼 + 1, it is not difficult to get
󵄨 󵄨 1
0 ≤ 󵄨󵄨󵄨𝑦󵄨󵄨󵄨 ≤ .
4
(42)
󵄨󵄨2 𝑑𝑥
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓3 (𝑥)󵄨󵄨󵄨
󵄨 |𝑥|𝛼
󵄨
𝛿,𝑙,𝑗
So, the support of 𝐾𝜆 𝑓(𝑥) is contained in
{𝑥 ∈ R𝑛 : 2𝑗 +
󵄨2 𝑑𝑥
󵄨
≤ 𝐶2 2 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
|𝑥|𝛼
1
1
1 1
≥ |𝑥| ≥ 2𝑗−1 − ≥ 20−1 − = } . (43)
4
4
4 4
With Lemma 4, we have
Combining (28), (40), (44) with (47), when 𝑓 is supported in
{|𝑥| ≤ 𝐶2𝑗 }, we come to the conclusion
󵄨󵄨2
󵄨󵄨 𝛿,𝑙,𝑗
𝑑𝑥
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓1 (𝑥)󵄨󵄨󵄨
󵄨 |𝑥|𝛼
󵄨
󵄨󵄨2 𝑑𝑥
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨
󵄨 |𝑥|𝛼
󵄨
󵄨󵄨2
󵄨󵄨 𝛿,𝑙,𝑗
≤ ∫ 󵄨󵄨󵄨𝐾𝜆 𝑓1 (𝑥)󵄨󵄨󵄨 4𝛼 𝑑𝑥
󵄨
󵄨
󵄨󵄨 𝛿,𝑙,𝑗
󵄨󵄨2
≤ 𝐶 ∫ 󵄨󵄨󵄨𝐾𝜆 𝑓1 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨
󵄨
󵄩 󵄩2
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀󵄩󵄩󵄩𝑓1 󵄩󵄩󵄩2
󵄨2 𝑑𝑥
󵄨
≤ 𝐶2−𝑙 2−𝑗 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
|𝑥|𝛼
Now, we hope to establish (48) for all 𝑓 ∈ 𝐿2 (R𝑛 ).
Decompose 𝑓 = ∑𝑖∈Z𝑛 𝜒𝑄𝑖 𝑓 = ∑𝑖∈Z𝑛 𝑓𝑖 , where {𝑄𝑖 } are
disjoint cubes of common side 10 ⋅ 2𝑗 with 𝑄0 centered at 0.
𝛿,𝑙,𝑗
Since {𝐾𝜆 𝑓𝑖 (𝑥)} have essentially disjoint supports, it suffices
to prove (48) for every 𝑓𝑖 . When 𝑖 = 0, we have proved it. If
𝑖 > 1, then
󵄩2
𝛿 󵄩󵄩𝑓󵄩󵄩󵄩2
𝛼
󵄨2 𝑑𝑥
󵄨
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀(2𝑗 ) ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
|𝑥|𝛼
󵄨2 𝑑𝑥
󵄨
≤ 𝐶2−𝑙 2−𝑗 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
,
|𝑥|𝛼
1
0 < ( + 𝑙) 10 ⋅ 2𝑗
2
where 𝑀 > 𝛼 + 1.
At last, we turn to the case of 𝑖 = 3. Similar to the
aforementioned, we have
−𝛼
< |𝑥|
󵄨󵄨2
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓3 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨
󵄨
󵄩 󵄩2
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀󵄩󵄩󵄩𝑓3 󵄩󵄩󵄩2
1
< ( + 𝑙 + 1) 10 ⋅ 2𝑗
2
(49)
(𝑙 ≥ 0) .
Therefore, we only need to confirm
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀 ∫
1/4≤|𝑥|≤𝐶
󵄨2
󵄨󵄨
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
𝛼
1 −𝛼
󵄨2 1
󵄨󵄨
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀( ) ∫
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 ( ) 𝑑𝑥
4
4
1/4≤|𝑥|≤𝐶
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀4𝛼 ∫
1/4≤|𝑥|≤𝐶
(45)
󵄨󵄨2
󵄨󵄨 𝛿,𝑙,𝑗
󵄨2
󵄨
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥 ≤ 𝐶𝛼 2−𝑙 2−𝑗 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥.
󵄨
󵄨
󵄨2 𝛼
󵄨󵄨
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥| 𝑑𝑥
󵄨󵄨2
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨
󵄨
󵄨2
󵄨
≤ 𝐶2−2𝑀(𝑗+𝑙) 𝛿2𝑀 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
By duality again, we see
(51)
󵄨2
󵄨
≤ 𝐶2−𝑙 2−𝑗 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥.
󵄨󵄨2 𝑑𝑥
󵄨󵄨 𝛿,𝑙,𝑗
∫ 󵄨󵄨󵄨𝐾𝜆 𝑓3 (𝑥)󵄨󵄨󵄨
󵄨 |𝑥|𝛼
󵄨
󵄨2
󵄨
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥
󵄨2 𝑑𝑥
󵄨
𝛿 (2 ) ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
|𝑥|𝛼
−𝑙 −𝑗𝑀 𝑀
(50)
In fact, it follows from Lemma 4 that
󵄨2
󵄨
≤ 𝐶2−𝑙 2−𝑗𝑀𝛿𝑀 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥|𝛼 𝑑𝑥.
≤ 𝐶2 2
(48)
(44)
−𝑙 −𝑗𝑀 𝑀 󵄩
󵄩
≤ 𝐶2 2
(47)
−𝑙 −𝑗
𝑗 𝛼
󵄨2 𝑑𝑥
󵄨
≤ 𝐶2−𝑙 2−𝑗(𝑀−𝛼) 𝛿𝑀 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
|𝑥|𝛼
At present, we complete the proof of Lemma 5.
(46)
Lemma 6. For 𝛿 > 0, 𝑘 ∈ Z, and 0 ≤ 𝛼 < 𝑛, one gets
∫ ∫
R𝑛
2𝑘
2𝑘−1
󵄨2 𝑑𝜆 𝑑𝑥
󵄨󵄨 𝛿
󵄨󵄨2 𝑑𝑥
󵄨󵄨
󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨
󵄨 𝜆 |𝑥|𝛼 ≤ 𝐶𝛼 𝛿 ∫ 󵄨󵄨𝑓 (𝑥)󵄨󵄨 |𝑥|𝛼 .
󵄨
(52)
6
Abstract and Applied Analysis
Proof. Applying Minkowski and Cauchy-Schwartz’s inequalities into the left hand side of (52), we get
󵄨2
󵄨󵄨 2𝑘
󵄨󵄨
𝑑𝜆 󵄨󵄨󵄨󵄨 −𝛼
∫ 󵄨󵄨󵄨∫ 𝐾𝜆𝛿 𝑓 (𝑥)
󵄨 |𝑥| 𝑑𝑥
𝜆 󵄨󵄨󵄨󵄨
R𝑛 󵄨󵄨󵄨 2𝑘−1
2𝑘
1/2
𝑑𝜆
󵄨2
󵄨
(∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥)
)
𝑘−1
𝑛
𝜆
2
R
2
≤ 𝐶(∫
2𝑘
𝑑𝜆
󵄨2
󵄨
≤ 𝐶 ((∫ ∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥 )
𝑘−1
𝑛
𝜆
2
R
× (∫
2𝑘
2𝑘−1
≤ 𝐶∫
2𝑘
2𝑘−1
𝑑𝜆
)
𝜆
1/2
(53)
which satisfies the same estimates as 𝑚𝛿 . Then, by the
fundamental theorem in calculus and Hölder’s inequality, we
have
󵄨
∞ 󵄨󵄨󵄨
𝑑𝐾𝜆𝛿 𝑓 (𝑥) 󵄨󵄨󵄨󵄨
󵄨2
󵄨󵄨 𝛿
󵄨󵄨𝐾∗ 𝑓 (𝑥)󵄨󵄨󵄨 ≤ ∫ 2 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)
󵄨 𝑑𝜆
󵄨
󵄨
󵄨󵄨
𝑑𝜆 󵄨󵄨󵄨󵄨
0
󵄨
󵄨
󵄨 󵄨󵄨 𝛿
󵄨󵄨
∞ 󵄨󵄨𝐾𝛿 𝑓 (𝑥)󵄨󵄨
󵄨 𝜆
󵄨 󵄨󵄨𝑑𝐾𝜆 𝑓 (𝑥) /𝑑𝜆󵄨󵄨󵄨
≤ 2𝛿−1 ∫ 󵄨 1/2 󵄨 𝜆𝛿 󵄨
𝑑𝜆
𝜆
𝜆1/2
0
∞
󵄨
󵄨2 𝑑𝜆 1/2
≤ 2𝛿−1 (∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨
)
𝜆
0
2
1/2
1/2
󵄨󵄨
󵄨2
󵄨󵄨 𝑑𝐾𝜆𝛿 𝑓 (𝑥) 󵄨󵄨󵄨 𝑑𝜆
󵄨󵄨
× (∫ 󵄨󵄨󵄨𝜆𝛿
)
𝑑𝜆 󵄨󵄨󵄨󵄨 𝜆
0 󵄨󵄨󵄨
)
∞
𝑑𝜆
󵄨2
󵄨
∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥 .
𝑛
𝜆
R
= 2𝛿−1 𝐺𝛿 𝑓 (𝑥) 𝐺𝑓 (𝑥) .
(61)
Now, it suffices to prove that
󵄨2
󵄨
∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥
𝑛
R
󵄨2
󵄨
≤ 𝐶𝛼 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 |𝑥|−𝛼 𝑑𝑥,
R𝑛
(54)
Take a Schwartz function 𝜓 such that 𝜓(0) = 0 and 𝜓(𝑡) =
1 if 1/2 ≤ 𝑡 ≤ 2 and 𝜓𝑘 (𝑡) = 𝜓(2𝑘 𝑡). Then, when 2𝑘−1 ≤ 𝜆 ≤
2𝑘 , we have
where 𝐶𝛼 𝛿 is uniform in 2𝑘−1 ≤ 𝜆 ≤ 2𝑘 . For 0 ≤ 𝛼 < 𝑛, it is
equivalent to
󵄨
󵄨2 𝑑𝑥
󵄨󵄨2 𝑑𝑥
󵄨󵄨
.
∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑔 (𝑥)󵄨󵄨󵄨󵄨
𝛼 ≤ 𝐶𝛼 𝛿 ∫ 𝑛 󵄨󵄨𝑔 (𝑥)󵄨󵄨
𝑛
|𝑥|
|𝑥|𝛼
R
R
𝐾𝜆𝛿 𝑓 (𝑥) = ∫ 𝑚𝛿 (𝜆𝑡) 𝜓𝑘 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥) .
(62)
(55)
Using Lemma 6, we get
It is just as the result of Lemma 5.
Now, we come back to the proof of Theorem 2.
∫ ∫
Proof. As in [3], we can decompose
(1 −
R𝑛
𝑡 𝛽 ∞ −𝑘𝛽 2−𝑘 𝑡
) = ∑2 𝑚 ( ) .
𝜆 + 𝑘=0
𝜆
(56)
∞
−𝑘
2𝑘−1
󵄨󵄨 𝛿
󵄨2 𝑑𝜆 𝑑𝑥
󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨
󵄨
󵄨 𝜆 |𝑥|𝛼
󵄨2
󵄨󵄨
󵄨󵄨∫ 𝑚𝛿 (𝜆𝑡) 𝜓 (𝑡) 𝑑𝐸 𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝜆 𝑑𝑥
󵄨󵄨
󵄨
𝑘
𝑡
󵄨
󵄨 𝜆 |𝑥|𝛼
2𝑘−1 󵄨
=∫ ∫
R𝑛
Thus,
𝑆∗𝛽 𝑓 (𝑥) ≤ ∑ 2−𝑘𝛽 𝐾∗2 𝑓 (𝑥) .
2𝑘
󵄨2
󵄨󵄨
󵄨󵄨∫ 𝑚𝛿 (𝜆𝑡) 𝑑𝐸 (𝜓 (𝐻 ) 𝑓) (𝑥)󵄨󵄨󵄨 𝑑𝜆 𝑑𝑥
󵄨󵄨
󵄨
𝑡
𝑘
𝑏
󵄨
󵄨 𝜆 |𝑥|𝛼
2𝑘−1 󵄨
=∫ ∫
(57)
𝑘=0
Consequently, we consider
󵄨2 𝑑𝑥
󵄨
󵄨󵄨2 𝑑𝑥
󵄨󵄨
.
∫ 󵄨󵄨󵄨󵄨𝐾∗𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨
𝛼 ≤ 𝐶𝛼 𝛿 ∫ 𝑛 󵄨󵄨𝑓 (𝑥)󵄨󵄨
𝑛
|𝑥|
|𝑥|𝛼
R
R
(58)
∞
󵄨
󵄨2 𝑑𝜆 1/2
)
𝐺𝛿 𝑓 (𝑥) = (∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨
𝜆
0
(59)
R𝑛
2𝑘
2𝑘
󵄨2 𝑑𝑥
󵄨
≤ 𝐶𝛼 𝛿 ∫ 󵄨󵄨󵄨𝜓𝑘 (𝐻𝑏 ) 𝑓 (𝑥)󵄨󵄨󵄨
|𝑥|𝛼
󵄨󵄨
󵄨󵄨2 𝑑𝑥
= 𝐶𝛼 𝛿 ∫ 󵄨󵄨󵄨󵄨∫ 𝜓𝑘 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)󵄨󵄨󵄨󵄨
.
󵄨
󵄨 |𝑥|𝛼
(63)
Let
𝛿
and 𝐺 be defined in the same way but using instead of 𝑚𝛿
the function
𝑑𝑚𝛿 (𝜆)
,
𝑚𝛿 (𝜆) = 𝛿𝜆
𝜆
(60)
From Theorem 3.1 in page 411 of [10] and the density of
𝑝
𝐿∞
𝑐 in 𝐿 (𝑤), we induce that
1/2
󵄨󵄨2
󵄨󵄨
( ∑ 󵄨󵄨󵄨󵄨∫ 𝜓𝑘 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)󵄨󵄨󵄨󵄨 )
󵄨
󵄨
∞
𝑘=−∞
(64)
Abstract and Applied Analysis
7
is bounded in 𝐿2 (𝑑𝑥/|𝑥|𝛼 ). As a result,
∫ (𝐺𝛿 𝑓 (𝑥))
2
[8] C. Fefferman, “The multiplier problem for the ball,” Annals of
Mathematics, vol. 94, pp. 330–336, 1971.
[9] A. Carbery, J. L. Rubio de Francia, and L. Vega, “Almost
everywhere summability of Fourier integrals,” Journal of the
London Mathematical Society, vol. 38, no. 3, pp. 513–524, 1988.
[10] J. Garcia-Cuerva and J. L. Rubio De Francia, Weighted Norm
Inequalities and Related Topics, North-Holland, Amsterdam,
The Netherlands, 1985.
∞
𝑑𝑥
󵄨2 𝑑𝜆 𝑑𝑥
󵄨
= ∫ ∫ 󵄨󵄨󵄨󵄨𝐾𝜆𝛿 𝑓 (𝑥)󵄨󵄨󵄨󵄨
𝜆 |𝑥|𝛼
|𝑥|𝛼
0
∞
2𝑘
𝑘=−∞
2𝑘−1
=∫ ∑ ∫
∞
2𝑘
𝑘=−∞
2𝑘−1
= ∑ ∫∫
󵄨󵄨 𝛿
󵄨2 𝑑𝜆 𝑑𝑥
󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨
󵄨
󵄨 𝜆 |𝑥|𝛼
󵄨󵄨 𝛿
󵄨2 𝑑𝜆 𝑑𝑥
󵄨󵄨𝐾𝜆 𝑓 (𝑥)󵄨󵄨󵄨
󵄨
󵄨 𝜆 |𝑥|𝛼
∞
󵄨󵄨
󵄨󵄨2 𝑑𝑥
≤ ∑ 𝐶𝛼 𝛿 ∫ 󵄨󵄨󵄨󵄨∫ 𝜓𝑘 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)󵄨󵄨󵄨󵄨
󵄨
󵄨 |𝑥|𝛼
𝑘=−∞
∞ 󵄨
󵄨
󵄨󵄨2 𝑑𝑥
≤ 𝐶𝛼 𝛿 ∫ ∑ 󵄨󵄨󵄨󵄨∫ 𝜓𝑘 (𝑡) 𝑑𝐸𝑡 𝑓 (𝑥)󵄨󵄨󵄨󵄨
󵄨
󵄨 |𝑥|𝛼
𝑘=−∞
󵄨2 𝑑𝑥
󵄨
≤ 𝐶𝛼 𝛿 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
|𝑥|𝛼
(65)
At last, with (61) and Hölder’s inequality, we come to the result
that
2 𝑑𝑥
󵄨2 𝑑𝑥
󵄨
≤ 𝐶𝛼 ∫ 󵄨󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
.
∫ (𝐾∗𝛿 𝑓 (𝑥))
(66)
|𝑥|𝛼
|𝑥|𝛼
Acknowledgments
The authors deeply thank the referee for reviewing their
paper. The first author was supported by the NSF of China
(71201051, 71031004, and 71073047), NSF of Hunan Province
(2012RS4028) and Postdoctoral Science Foundation of China
(2012M521513). The second author was supported by the NSF
of China (10771054) and NSF of Hunan Province (09JJ5002).
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