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Harmonic analysis
On estimates for the Fourier transform in the space L 2 ( R n )
Radouan Daher, Mohamed El Hamma
Department of Mathematics, Faculty of Sciences Aïn Chock, Hassan II University, Casablanca, Morocco
a r t i c l e i n f o a b s t r a c t
Article history:
Received 21 September 2013
Accepted after revision 16 December 2013 Available online xxxx
Presented by the Editorial Board
We obtain new inequalities for the Fourier transform in the space L2(Rn), using a generalized spherical mean operator for proving two estimates in certain classes of functions characterized by a generalized continuity modulus.
©2014 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.
1. Introduction and preliminaries
This work is based mainly on Titchmarsh’s theorem ([8], Theorem 84) in the one-dimensional case. In[2], Abilov et al.
proved two useful estimates for the Fourier transform in the space of square integral multivariable functions on certain classes of functions characterized by the generalized continuity modulus, and these estimates are proved by Abilov’s for only two variables, using a translation operator.
In this paper, we prove the analog of Abilov’s results[2]in the Fourier transform for multivariable functions on
R
n. For this purpose, we use a spherical mean operator in the place of the translation operator.Assume thatL2
(R
n)
the space of integrable functions f with the norm:f
2=
Rn
f ( x )
2dx
1/2.
The Fourier transform for the function f
∈
L1( R
n)
is defined by:f (ξ ) = ( 2 π )
−n/2Rn
f ( x ) e
−ix.ξdx .
The inverse Fourier transform is defined by the formula:
f ( x ) = ( 2 π )
−n/2Rn
f (ξ ) e
ix.ξd ξ.
The Plancherel theorem provides an extension of the Fourier transform toL2
( R
n)
, i.e.:Rn
f ( x )
2dx =
Rn
f (ξ )
2d ξ.
Let jp
(
z)
be a normalized Bessel function of the first kind, i.e.:E-mail address:elmohamed77@yahoo.fr(M. El Hamma).
1631-073X/$ – see front matter ©2014 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.
http://dx.doi.org/10.1016/j.crma.2013.12.016
j
p( z ) = 2
pΓ ( p + 1 )
z
pJ
p( z ), ∀ z ∈ R
+, (1)
where Jp
(
z)
is a Bessel function of the first kind.Consider inL2
(R
n)
the spherical mean operator (see[4]):M
hf ( x ) = 1 w
n−1Sn−1
f ( x + hw ) dw ,
where Sn−1 is the unit sphere in
R
n,wn−1 its total surface measure with respect to the usual induced measure dw.The finite differences of the first and higher orders are defined by:
h
f ( x ) = M
hf ( x ) − f ( x ) = ( M
h− I ) f ( x ),
khf ( x ) =
hkh−1
f ( x )
= ( M
h− I )
kf ( x ) =
k
i=0
(− 1 )
k−ii
k
M
hif ( x ),
where M0hf
(
x) =
f(
x)
, Mihf(
x) =
Mh(
Mhi−1f(
x))
fori=
1,
2, . . . ,
kandk=
1,
2, . . .
, I is the identity operator inL2( R
n)
. Thekth order generalized modulus of continuity of function f∈
L2( R
n)
is defined as:Ω
k( f , δ) =
khf ( x )
2
, 0 < h δ.
Let Wr2,,φk
(
D)
denote the class of functions f∈
L2(R
n)
such that Dif∈
L2(R
n)
,i=
1,
2, . . . ,
r(in the sense of Levi, see[6]) andΩ
kD
rf , δ
= O φ
δ
k,
where the operator D
=
n i=1 ∂2∂x2i is the Laplace operator andx
= (
x1,
x2, . . . ,
xn)
D0f=
f, Dif=
D(
Di−1f)
,i=
1,
2, . . . ,
r andφ (
t)
is an arbitrary function defined on[
0, ∞)
i.e.:W
r2,,φk( D ) = f ∈ L
2R
n, D
if ∈ L
2R
n, such that Ω
kD
rf , δ
= O φ
δ
k, i = 1 , 2 , . . . , r .
According to[4], we have:
M
hf ( x ) = ( 2 π )
−n/2Rn
f (ξ ) j
n−2 2| ξ | h e
ix.ξd ξ
and
f ( x ) = ( 2 π )
−n/2Rn
f (ξ ) e
ix.ξd ξ,
i.e.:
M
hf ( x ) − f ( x ) = ( 2 π )
−n/2Rn
f (ξ ) j
n−22
| ξ | h
− 1 e
ix.ξd ξ.
By Parseval’s identity, we obtain:
M
hf ( x ) − f ( x )
22
=
Rn
f (ξ )
2j
n−22
|ξ | h
− 1
2d ξ.
It is easy to show that any function f
∈
Wr2,,φk(
D)
implies:kh
D
rf ( x )
22
=
Rn
|ξ |
2r1 − j
n−22
|ξ | h
2kf (ξ )
2d ξ. (2)
From[3], we have:
1 − j
n−22
( r ) min 1 , r
2, (3)
where the symbol
means that the left-hand side is bounded above and below by a positive constant times the right-hand side.2. Main result
Taking into account what was said in Section1for some classes of functions characterized by the generalized modulus of continuity, we can prove two estimates for the integral:
|ξ|N
f (ξ )
2d ξ.
Proposition 2.1.Let r
∈ N ∪ {
0}
and k∈ N
. If f∈
Wr2,,φk(
D)
for some fixedφ
defined on[
0, ∞)
, then|ξ|N
f (ξ )
2d ξ = O N
−2rφ
2N
−kas N
→ ∞
.Proof. In the terms of jp
(
z)
, we have (see[1]):1 − j
p( z ) = O ( 1 ), z 1 , (4)
1 − j
p( z ) = O z
2, 0 z 1 , (5)
√ hz J
p( hz ) = O ( 1 ), hz 0 . (6)
Let f
∈
Wr2,,φm(
D)
. By Hölder inequality, we have:|ξ|N
f (ξ )
2d ξ −
|ξ|N
f (ξ )
2j
n−2 2h |ξ | d ξ =
|ξ|N
1 − j
n−2 2h |ξ | f (ξ )
2d ξ
=
|ξ|N
1 − j
n−2 2h |ξ | f (ξ )
2−1kf (ξ )
1kd ξ
|ξ|N
f (ξ )
2d ξ
2k2k−1|ξ|N
1 − j
n−2 2h |ξ |
2kf (ξ )
2d ξ
2k1|ξ|N
f (ξ )
2d ξ
2k2k−1|ξ|N
1
|ξ |
2r1 − j
n−2 2h |ξ|
2k|ξ|
2rf (ξ )
2d ξ
2k1N
−kr|ξ|N
f (ξ )
2d ξ
2k2k−1|ξ|N
1 − j
n−2 2h |ξ|
2k|ξ |
2rf (ξ )
2d ξ
2k1.
From formula(2), we have the inequality:
|ξ|N
1 − j
n−2 2h |ξ|
2k|ξ|
2rf (ξ )
2d ξ
khD
rf ( x )
22
.
Therefore
|ξ|N
f (ξ )
2d ξ
|ξ|N
f (ξ )
2j
n−2 2h |ξ|
d ξ + N
−kr|ξ|N
f (ξ )
2d ξ
2k2k−1kh
D
rf ( x )
1k2
.
Then
|ξ|N
1 − j
n−2 2h | ξ | f (ξ )
2d ξ = O
N
−kr|ξ|N
f (ξ )
2d ξ
2k2k−1 khD
rf ( x )
21k.
Settingh
=
1N in the last inequality and from inequality(3), we obtain:|ξ|N
f (ξ )
2d ξ = O N
−kr|ξ|N
f (ξ )
2d ξ
2k2k−1φ
1k1
N
k.
Therefore
|ξ|N
f (ξ )
2d ξ = O N
−2rφ
2N
−k.
This completes the proof ofProposition 2.1.
2
Theorem 2.1.Letφ (
t) =
tν. Then|ξ|N
f (ξ )
2d ξ = O N
−r−kν⇐⇒ f ∈ W
r2,,φk( D )
where r
=
0,
1, . . .
;k=
1,
2, . . .
; 0< ν <
2.Proof. We prove sufficiency by usingProposition 2.1let f
∈
Wr2,,ktν(
D)
, then|ξ|N
f (ξ )
2d ξ
12= O N
−r−kν.
To prove necessity, let:
|ξ|N
f (ξ )
2d ξ = O N
−r−kνi.e.
|ξ|N
f (ξ )
2d ξ = O
N
−2r−2kν.
It is easy to show that there exists a function f
∈
L2( R
n)
such that Drf∈
L2( R
n)
andD
rf ( x ) = ( − 1 )
n( 2 π )
n2Rn
| ξ |
rf (ξ ) e
ix.ξd ξ. (7)
From formula(7)and Parseval’s identity, we have:
khD
rf ( x )
22
=
Rn
1 − j
n−2 2h |ξ |
2k|ξ|
2rf (ξ )
2d ξ.
This integral is split into two:
Rn
=
|ξ|<N
+
|ξ|N
= I
1+ I
2whereN
= [
h−1]
. We estimate them separately from(4); we have:I
2=
|ξ|N
f (ξ )
2| ξ |
2r1 − j
n−22
h | ξ |
2kd ξ
= O
|ξ|N
f (ξ )
2|ξ |
2rd ξ
= O
∞l=0
N+l|ξ|N+l+1
|ξ|
2rf (ξ )
2d ξ
= O
∞l=0
( N + l + 1 )
2rN+l|ξ|N+l+1
f (ξ )
2d ξ
= O
∞l=0
( N + l + 1 )
2r|ξ|N+l
f (ξ )
2d ξ −
|ξ|N+l+1
f (ξ )
2d ξ
= O
∞l=0
( N + l + 1 )
2r|ξ|N+l
f (ξ )
2d ξ −
∞ l=0
( N + l + 1 )
2r|ξ|N+l+1
f (ξ )
2d ξ
= O
N
2r|ξ|N
f (ξ )
2d ξ +
∞ l=0
( N + l + 1 )
2r− ( N + l )
2r|ξ|N+l
f (ξ )
2d ξ
= O
N
2r|ξ|N
f (ξ )
2d ξ +
∞ l=1
( N + l )
2r−1|ξ|N+l
f (ξ )
2d ξ
= O
N
2rN
−2r−2kν+ O
∞l=1
( N + l )
2r−1( N + l )
−2r−2kν= O N
−2kν+ O N
−2kν= O h
2kνi.e.:
I
2= O h
2kν.
Now, we estimate I1, by virtue of(5):
I
1=
|ξ|<N
| ξ |
2r1 − j
n−22
h | ξ |
2kf (ξ )
2d ξ
= O h
4k|ξ|<N
|ξ |
2r|ξ |
4kf (ξ )
2d ξ
= O h
4k|ξ|<N
|ξ |
2r+4kf (ξ )
2d ξ
= O h
4kN−1l=0
l|ξ|<l+1
|ξ|
2r+4kf (ξ )
2d ξ
= O h
4kN−1l=0
( l + 1 )
2r+4kl|ξ|<l+1
f (ξ )
2d ξ
= O h
4kN−1l=0
( l + 1 )
2r+4k|ξ|l
f (ξ )
2d ξ −
|ξ|l+1
f (ξ )
2d ξ
= O h
4k1 +
N−1
l=0
( l + 1 )
2r+4k− l
2r+4k|ξ|l
f (ξ )
2d ξ
= O h
4k1 +
N−1
l=1
l
2r+4k−1|ξ|l
f (ξ )
2d ξ
= O h
4k1 +
N−1
l=1
l
2r+4k−1l
−2r−2kν= O h
4k1 +
N−1
l=0
l
4k−2kν−1= O h
4kO
N
4k−2kν= O h
2kνi.e.:
I
1= O h
2kν.
Combining the estimates for I1and I2 gives:
kh
D
rf ( x )
2
= O h
kν.
The necessity is proved.
2
Theorem 2.1in the caser
=
0 andk=
1 can be found in the works of Platonov[7]and Gioev[5].Acknowledgements
The authors would like to thank the referee for his valuable comments and suggestions.
References
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