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OWP 2009 - 26
D. GORBACHEV, E. LIFLYAND AND S. TIKHONOV
Weighted Fourier Inequalities for Radial Functions
Mathematisches Forschungsinstitut Oberwolfach gGmbH
Oberwolfach Preprints (OWP) ISSN 1864-7596
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WEIGHTED FOURIER INEQUALITIES FOR RADIAL FUNCTIONS
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
Abstract. Weighted Lp (Rn ) → Lq (Rn ) Fourier inequalities are studied. We prove Pitt–Boas type
results on integrability with power weights of the Fourier transform of a radial function.
1. Introduction
Weighted norm inequalities for the Fourier transform provide a natural way to describe the
balance between the relative sizes of a function and its Fourier transform at infinity. What is
more, such inequalities with sharp constants imply the uncertainty principle relations ([2], [3]).
The celebrated Pitt inequality illustrates this idea at the spectral level ([2]):
Z
Z
2
b
Φ(1/|y|)|f (y)| dy ≤ CΦ
Φ(|x|)|f (x)|2 dx,
Rn
Rn
where Φ is an increasing function and fb is the Fourier transform of a function f from the Schwartz
class S(Rn ),
Z
b
(1)
f (y) = Ff (y) =
f (x)eixy dx.
Rn
p
q
In the (L , L ) setting such inequalities have been studied extensively (see for instance [2]–[6], [10],
[11], [12], [18], [23]). In this case Pitt’s inequality is written as follows: for 1 < p ≤ q < ∞,
0 ≤ γ < n/q, 0 ≤ β < n/p0 and n ≥ 1
µZ
¶1/q
µZ
¶1/p
¡ −γ
¢q
¡ β
¢p
b
(2)
|y| |f (y)| dy
≤C
|x| |f (x)| dx
Rn
Rn
with the index constraint
µ
β−γ =n−n
1 1
+
p q
¶
(the primes denote dual exponents, 1/p + 1/p0 = 1).
The restrictions on γ and β can be written as
½
´¾
³1 1
n
+ −1
≤γ< .
(3)
max 0, n
p q
q
It is worth mentioning that inequality (2) contains classical (non-weighted) versions of the
Plancherel theorem, that is, kfbk2 = kf k2 , Hardy–Littlewood’s theorem (1 < p = q ≤ 2, β = 0 or
p = q ≥ 2, γ = 0), and Hausdorff–Young’s theorem (q = p0 ≥ 2, β = γ = 0).
For n = 1, inequality (2) can be found in [4], [16], [17], [21]; for n ≥ 1 see [3], [4]. In [2],
W. Beckner found a sharp constant in (2) for p = q = 2 and used this result to prove a logarithmic
estimate for uncertainty.
In this paper we address the following two problems.
1991 Mathematics Subject Classification. Primary 42B10, 42A38; Secondary 42B35, 46E30.
Key words and phrases: Fourier transforms, Weighted inequalities, General monotone functions, Radial functions.
1
2
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
Problem 1: The range (3) is sharp if f is simply assumed to be in Lpu , u(x) = |x|pβ . Is it
possible to extend this range if additional regularity of f is assumed?
Problem 2: Under which additional assumption on f it is possible to reverse inequality (2) for
p = q?
Let us first recall several known results in dimension 1. Some progress toward extending the range
of γ in (3) was made in [5], [18], and [23], where the authors assumed that the function has vanishing
moments up to certain order.
Another approach, which is related to both Problems 1 and 2, is due to Hardy, Littlewood, and,
later, Boas. The well-known Hardy–Littlewood theorem (see [24, Ch.IV]) states that if 1 < p < ∞
and f is an even non-increasing function that vanishes at infinity, then
µZ
µZ
¶1/p µZ
¶1/p
¶1/p
p
p p−2
p
b
b
(4)
C1
|f (x)| dy
≤
|f (t)| t dx
≤ C2
|f (x)| dy
.
R
R
R+
Boas conjectured in [8] that the weighted version of (4) is also true: under the same conditions
on f and p,
(5)
|x|−γ |fb(x)| ∈ Lp (R)
if and only if
t1+γ−2/p f (t) ∈ Lp (R+ ),
provided −1/p0 = −1 + 1/p < γ < 1/p.
Relation (5) was proved in [19]. Thus, assuming a function to be monotone allows one to extend
the range of γ as well as to reverse inequality (2) for p = q.
In [13], Boas-type results were obtained for the cosine and sine Fourier transforms, separately.
To describe it briefly, we denote
Z ∞
Z ∞
b
b
fc (x) =
f (t) cos xt dt
and
fs (x) =
f (t) sin xt dt.
0
0
We call a function admissible if it is locally of bounded variation on (0, ∞) and vanishes at infinity.
For any admissible non-negative function f satisfying
Z 2t
Z ct
(6)
|dh(u)| ≤ C
u−1 |h(u)| du
t
t/c
for some c > 1, relation (5) holds for f and fbc provided −1/p0 < γ < 1/p, while for f and fbs
provided −1/p0 < γ < 1/p + 1 (note the larger range).
In the higher-dimensional setting, the situation is expectedly more complex. For radial functions
f (x) = f0 (|x|), x ∈ Rn , the Fourier transform is also radial, i.e. fb(x) = F0 (|x|). One then can
apply the one-dimensional results. For example, in R3 the Fourier transform is given by
Z ∞
−1
b
tf0 (t) sin |x|t dt.
f (x) = 4π|x|
0
So, applying the result for the sine transform fbs to the function tf0 (t), we obtain
(7)
|x|−γ fb(x) ∈ Lp (R3 ) if and only if
t3+γ−4/p f0 (t) ∈ Lp (0, ∞),
provided −2 + 3/p < γ < 3/p. Note that it is enough to assume that f0 itself satisfies (6), since
this implies the same for tf0 (t).
For n 6= 3, we can also apply (5) using fractional integrals. If f0 is such that
Z ∞
(8)
tn−1 (1 + t)(1−n)/2 |f0 (t)| dt < ∞,
0
WEIGHTED FOURIER INEQUALITIES
3
one has the following Leray’s formula (see, e.g., Lemma 25.10 in [20]):
Z ∞
(n−1)/2
b
(9)
f (x) = 2π
I(t) cos |x|t dt,
0
where the fractional integral I is given by
Z ∞
2
I(t) = ¡ n−1 ¢
sf0 (s)(s2 − t2 )(n−3)/2 ds.
Γ 2
t
Then, the one-dimensional Boas’s relation (5) implies that if f0 ≥ 0 satisfies (8), then
|x|−γ fb(x) ∈ Lp (Rn )
if and only if
t1+γ−(n+1)/p I(t) ∈ Lp (0, ∞),
provided −1 + n/p < γ < n/p. However, the condition on I is difficult to verify and so it is
desirable to obtain more direct Boas-type conditions. This is the main goal of the present paper.
Definition. We call an admissible function f0 general monotone, written GM, if for any t > 0
Z ∞
Z ∞
du
(10)
|df0 (u)| ≤ C
|f0 (u)|
u
t
t/c
for some c > 1.
R∞
In the context of our results, we always deal with functions satisfying
|f0 (u)| du/u < ∞. It is
clear that any such function being monotone, or satisfying (6), is general monotone. However, this
class also contains functions with much more complex structure (see, e.g., [14]-[15]).
It is natural in our study that f0 ∈ GM satisfies a less restrictive condition than (8):
Z ∞
Z 1
n−1
t(n−1)/2 |df0 (t)| < ∞.
t |f0 (t)| dt +
(11)
1
0
Our main result reads as follows.
Theorem 1. Let 1 ≤ p < ∞ and n ≥ 1. Then, for any radial function f (x) = f0 (|x|), x ∈ Rn ,
such that f0 ≥ 0, f0 ∈ GM , and satisfying (11),
°
°
°
°
° −γ b °
°β
°
(12)
°|x| f (x)° p n ³ °t f0 (t)° p
L (R )
L (0,∞)
if and only if
β =γ+n−
n+1
p
and
−
n+1 n
n
+ <γ< .
2
p
p
The paper is organized as follows. Section 2 provides some useful facts about the Fourier transform of a radial function. In Sections 3 and 4, we prove auxiliary upper and lower estimates for the
Fourier transform; these estimates are used in Section 6. General (Lp , Lq ) inequalities of Pitt-Boas
type are delivered by Theorem 2 in Section 5. In the case p = q this gives Theorem 1. Section 6
contains the proof of Theorem 2.
Concerning Problem 1, we observe that the upper estimate of fb in Theorem 2 is Pitt’s inequality,
n
n n+1
−
< γ < . Since in
which holds in the case of general monotone functions only when
q
2
p
any case
½
¾
³1 1
´
n n+1
−
< max 0, n
+ −1
,
q
2
p q
we extend the range of γ given by (3). Theorem 1 exhibits a solution of Problem 2. Note that for
n = 1 and n = 3 Theorem 1 gives (5) and (7), correspondingly.
4
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
The notation “ . ” and “ & ” means “ ≤ C ” and “ ≥ C ”, respectively (with C independent of
essential quantities), while “ ³ ” stands for “ . ” and “ & ” to hold simultaneously.
2. The Fourier transform of radial functions
The facts we are going to make use of can be found in [7, 20, 22]. For n ≥ 1, x ∈ Rn , let
f (x) = f0 (|x|) be a radial function. Then
Z
Z ∞
n−1
(13)
f (x) dx = |S |
f0 (t)tn−1 dt,
Rn
0
where |S n−1 | = 2π n/2 /Γ(n/2) is the area of the unit sphere S n−1 = {x ∈ Rn : |x| = 1}.
The Fourier transform (1) of the radial function f is also radial and is given via the Hankel–
Fourier transform [22] as
Z ∞
n−1
b
(14)
f (y) = F0 (|y|) = |S |
f0 (t)jα (|y|t)tn−1 dt.
0
Here jα (z) is the normed Bessel function
(15)
jα (z) = Γ(α + 1)
³ z ´−α
2
Jα (z) =
∞
Y
Ã
1−
k=1
2
z
ρ2α,k
!
,
where Jα (z) is the classical Bessel function of first kind and order α, and 0 < ρα,1 < ρα,2 < . . . are
the positive zeros of Jα (z). We denote
n
1
α := − 1 ≥ − .
2
2
Let us give several useful properties of the function jα (z), α ≥ −1/2, which follow from the
known properties of Jα (z) (see, e.g., [7, Ch.VII]): j−1/2 (z) = cos z, j1/2 (z) = sinz z ;
(16)
(17)
(18)
(19)
(20)
|jα (z)| ≤ jα (0) = 1,
z ≥ 0;
¢
d ¡ 2α+2
z
jα+1 (z) = (2α + 2)z 2α+1 jα (z);
dz
µ
¶
2α Γ(α + 1)(2/π)1/2
π(α + 1/2)
jα (z) =
cos z −
+ O(z −α−3/2 ),
α+1/2
z
2
|jα (z)| ≤
Mα
,
α+1/2
z
ρα,k = πk + O(1/k),
z → ∞;
z > 0;
k → ∞;
the zeros of the Bessel function are separated:
(21)
0 < ρα,1 < ρα+1,1 < ρα,2 < ρα+1,2 < ρα,3 < . . . .
It follows from (17) and (21) that the function z 2α+2 jα+1 (z) increases when z ∈ [0, ρα,1 ] and
decreases when z ∈ [ρα,1 , ρα+1,1 ]. The function jα+1 (z) decreases on the interval [0, ρα+1,1 ]. This
yields the estimate
(22)
2
z 2α+2 jα+1
(z) ≥ mb > 0,
1/b ≤ z ≤ b,
1 < b = bα < ρα+1,1 .
WEIGHTED FOURIER INEQUALITIES
In what follows we understand integral (14) as improper:
Z A
n−1
f0 (t)jα (st)tn−1 dt,
(23)
F0 (s) = |S | lim
a→0
A→∞
5
s = |y| > 0.
a
Note that for admissible f0 , (16) implies
¯Z A
¯ Z
¯
¯
n−1
¯
¯≤
f
(t)j
(st)t
dt
0
α
¯
¯
a
A
|f0 (t)|tn−1 dt < ∞.
a
Further, for a radial function f (x) = f0 (|x|), by properties (16) and (19), the integral in (14)
converges uniformly for s > 0 in improper sense to the continuous function F0 (s), provided (8)
holds (see [20]). In Lemma 1 below, we prove this fact for F0 (s) via a pointwise estimate of F0 .
Note that for n ≥ 2 condition (11), as well as condition (8), is less restrictive than f ∈ L1 (Rn ).
3. Estimates from above for the Fourier transforms
R 1 n−1
Let
f
(x)
=
f
t |f0 (t)| dt +
0 (|x|) with f0 admissible and satisfying (11), that is,
0
R ∞ (n−1)/2
t
|df0 (t)| < ∞. We observe that (11) implies for t > 1
1
Z ∞
Z ∞
(n−1)/2
(n−1)/2
t
|f0 (t)| ≤ t
|df0 (s)| ≤
s(n−1)/2 |df0 (s)|.
t
t
Therefore
t(n−1)/2 f0 (t) → 0
(24)
as
t → ∞.
Lemma 1. Given f0 as above, for s > 0 the Fourier transform F0 (s) is continuous, and satisfies
Z 1/s
Z ∞
n−1
−(n+1)/2
|F0 (s)| .
t |f0 (t)| dt + s
t(n−1)/2 |df0 (t)|.
0
1/s
Proof. Let for s > 0
Z
∞
F0 (s)
.
|S n−1 |
0
Let ρ > 1 be a zero of the Bessel function Jα+1 (·). Then, by (16),
Z 1/s
Z ρ/s
¯Z ∞
¯
¯
¯
n−1
n−1
n−1
(26)
I≤
|f0 (t)| t
dt +
|f0 (t)| t
dt + ¯
f0 (t)jα (st)t
dt¯ = I1 + I2 + I3 .
(25)
I=
0
1/s
Estimating I2 we obtain
Z ρ/s
³Z
n−1
t
I2 .
Z
(27)
f0 (t)jα (st)tn−1 dt =
1/s
Z
1/s
n
Z
u |df0 (u)| + s
1/s
∞
|df0 (u)| +
t
ρ/s
.
ρ/s
−n
´
|df0 (u)| dt
1/s
∞
|df0 (u)| . s
Z
−(n+1)/2
1/s
∞
t(n−1)/2 |df0 (t)|.
1/s
It follows from (17) that
(28)
d n
(t jα+1 (st)) = ntn−1 jα (st).
dt
Integrating by parts, we obtain
Z
Z
¯∞
1 ∞ n
1 ∞
1
¯
n
n
I3 =
f0 (t) d(t jα+1 (st)) = f0 (t)t jα+1 (st)¯ −
t jα+1 (st) df0 (t).
n ρ/s
n
n ρ/s
ρ/s
6
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
Then (19) and (24) yield
|f0 (t)tn jα+1 (st)| . |f0 (t)|tn (st)−(n+1)/2 . |f0 (t)|t(n−1)/2 → 0 as t → ∞,
and hence
Z
(29)
Z
∞
I3 .
n
−(n+1)/2
t (st)
∞
−(n+1)/2
|df0 (t)| . s
ρ/s
t(n−1)/2 |df0 (t)|.
1/s
Combining (27) and (29), we finish the proof of the lemma.
¤
We will also use similar estimates of the Fourier transform in terms of the following functions:
Z 2t
Z ∞
Z ∞
∗
Φ (t) =
|df0 (u)|, Φ(t) =
|df0 (u)|, Ψ(t) =
s(n−1)/2 |df0 (s)|.
t
t
t
∗
These functions are continuous for t > 0, and Φ (t) ≤ Φ(t).
Corollary 1. The estimate holds for s > 0
Z 1/s
Z
n−1 ∗
−(n+1)/2
|F0 (s)| .
t Φ (t) dt + s
Z
0
Z
1/s
.
t
n−1
−(n+1)/2
∞
1/s
∞
Φ(t) dt + s
0
t(n−3)/2 Φ∗ (t) dt
t(n−3)/2 Φ(t) dt.
1/s
Proof. Similar to (27), we first get
Z 1/s
Z
n−1
(30)
t |f0 (t)| dt .
0
Z
1/s
n
−(n+1)/2
∞
t |df0 (t)| + s
0
t(n−1)/2 |df0 (t)|.
1/s
Then the required estimates follows from Lemma 1 and inequalities
Z 2t
Z B
Z B
−1
|ψ(u)| du dt,
t
|ψ(u)| du ≤
(31)
ln 2
Z
(32)
t
0
0
Z
∞
ln 2
|ψ(u)| du ≤
2A
Z
∞
t
2t
−1
|ψ(u)| du dt,
A
t
valid for any integrable ψ.
¤
Corollary 2. The estimate holds for s > 0
(33)
Z
1/s
|F0 (s)| .
t(n−1)/2 Ψ(t) dt.
0
Proof. Indeed, by Lemma 1 and (30),
Z
Z 1/s
n
−(n+1)/2
|F0 (s)| .
t |df0 (t)| + s
We have
Z
I2 = s
Z
1/s
(n−1)/2
Ψ(1/s) ³ Ψ(1/s)
t
0
2t
¶
Z
t
t
µZ
t
0
Z
(n−1)/2
s
t
(n−1)/2
t(n−1)/2 Ψ(t) dt.
0
¶
2t
(n−1)/2
1/s
Ψ(t) dt ≤
1/(2s)
1/s
|df0 (s)| dt ³
1/s
dt ≤
1/(2s)
Using (31), we get
µZ
Z 1/s
n−1
I1 .
t
t(n−1)/2 |df0 (t)| = I1 + I2 .
1/s
0
−(n+1)/2
∞
Z
1/s
|df0 (s)| dt ≤
0
t(n−1)/2 Ψ(t) dt.
WEIGHTED FOURIER INEQUALITIES
7
The obtained bounds for I1 and I2 give (33).
¤
Note that in this section we have no assumption on positivity of f0 so far. This will come into
play in the next section.
4. Estimates from below for the Fourier transforms
Let us consider a radial function f (x) = f0 (|x|) such that f0 is admissible and f0 (t) ≥ 0 when
t > 0. We assume that f0 satisfies condition (11). Then, by Lemma 1, the integral in (23) converges
uniformly on any compact set away from zero and F0 (s) is continuous for s > 0. Suppose also that
Z 1
(34)
|F0 (s)|s(n−1)/2 ds < ∞.
0
In particular, this implies that fb is integrable in a neighborhood of zero. We will need the following
Lemma 2. For u > 0 and 1 < b < ρα+1,1 , the inequality holds
Z 2/u
Z bu
f0 (t)
(1−n)/2
(n−1)/2
u
s
|F0 (s)| ds &
dt.
t
0
u/b
Proof. We denote by B n = {x ∈ Rn : |x| ≤ 1} the unit ball, |B n | = |S n−1 |/n is the volume of this
ball.
Let us consider the following well-known compactly supported function
k(y) = |B n |−1 (χ ∗ χ)(y),
where χ is the indicator function of the unit ball B n . For n = 1, it is the Fejér kernel (1 − |y|/2)+ .
The kernel k is radial k(y) = k0 (|y|) and possesses the following properties:
(35)
0 ≤ k0 (s) ≤ k0 (0) = 1,
0 ≤ s ≤ 2;
k0 (s) = 0,
s ≥ 2;
and the Fourier transform of k is
b
k(x) = K0 (|x|) = |B n |−1 (b
χ(x))2 ≥ 0.
By (28), for t = |x|
Z
(36)
χ
b(x) = |S
n−1
1
|
jα (ts)sn−1 ds =
0
Therefore,
(37)
Z
K0 (t) = |S
n−1
2
|
0
|S n−1 |
jα+1 (t) = |B n |jα+1 (t).
n
2
(t).
k0 (s)jα (ts)sn−1 ds = |B n |jα+1
Let ε be small enough. Denoting
Z 2/u
Z
n−1
−n
Jε :=
F0 (s)k0 (us)s
ds = u
ε/u
We have, by (34) and (35),
Z
(38)
|Jε | ≤
F0 (s/u)k0 (s)sn−1 ds.
ε
2/u
Z
n−1
|F0 (s)|s
0
2
ds . u
2/u
(1−n)/2
0
s(n−1)/2 |F0 (s)| ds.
8
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
The uniform convergence of integral (23) implies
¶
Z 2µ
Z ∞
−n
n−1
n−1
Jε = u
|S |
f0 (t)jα (st/u)t
dt k0 (s)sn−1 ds
ε
0
µ
¶
Z ∞
Z 2
−n
n−1
n−1
=u
f0 (t) |S |
k0 (s)jα (st/u)s
ds tn−1 dt.
0
Using (37), we get
ε
Z
|S
n−1
2
|
k0 (s)jα (st/u)sn−1 ds = K0 (t/u) − λε (t),
ε
where
Z
λε (t) = |S
n−1
ε
k0 (s)jα (st/u)sn−1 ds.
|
0
Taking into account (22) and (37), we have (t/u)n K0 (t/u) & 1 for u/b ≤ t ≤ bu. Therefore,
Z ∞
Z bu
f0 (t)
−n
0
0
(39)
Jε &
dt − Jε ,
Jε = u
f0 (t)λε (t)tn−1 dt.
t
u/b
0
We are going to prove that Jε0 → 0 as ε → 0. Take A > 1. It follows from (35) and (16) that
Z ε
n−1
(40)
|λε (t)| ≤ |S |
sn−1 ds . εn ,
0
and hence
¯
Z
¯ −n
¯u
¯
(41)
Let t ≥ A. Define
Z
f0 (t)λε (t)t
0
λε (v)v
n−1
dv = |S
n−1
ε
|
0
|S n−1 |tn
Λε (t) =
n
Z
A
|f0 (t)|tn−1 dt.
0
µZ
ε
¶
t
n−1
jα (sv/u)v
k0 (s)s
0
Making use of (28), we obtain
(42)
¯
Z
¯
n
dt¯¯ . ε
n−1
Z
t
Λε (t) =
A
n−1
dv
ds.
0
k0 (s)jα+1 (st/u)sn−1 ds.
0
For n = 1,
¯
¯ Z ε
¯ ¯¯ Z εt/u
¯
2
¯
¯ ¯
sin
s
sin(st/u)
u
(1
−
cos(εt/u))
¯
(1 − s/2)
|Λε (t)| = ¯¯2t
ds¯¯ = ¯2u
ds −
¯.
¯
¯
st/u
s
t
0
0
¯ Z π
¯Z v
¯
sin s ¯¯
sin s
¯
ds¯ ≤
ds for v > 0, and |Λε (t)| . 1 . t(n−1)/2 .
It is well-known that ¯
s
s
0
0
Let now n ≥ 2. We have
ÃZ
Z ε!
ε/t
|S n−1 |tn
+
k0 (s)jα+1 (st/u)sn−1 ds.
Λε (t) =
n
0
ε/t
As above
¯
¯
Z ε/t
¯ |S n−1 |tn Z ε/t
¯
¯
¯
n
n−1
sn−1 ds . εn . 1 . t(n−1)/2 .
k0 (s)jα+1 (st/u)s
ds¯ . t
¯
¯
¯
n
0
0
WEIGHTED FOURIER INEQUALITIES
9
Applying (19), we get
¯ n−1 n Z ε
¯
Z ε
¯ |S |t
¯
n−1
n
¯
¯
k0 (s)jα+1 (st/u)s
ds¯ . t
|jα+1 (st/u)|sn−1 ds
¯
n
ε/t
ε/t
Z ε
. tn (t/u)−(n+1)/2
s(n−1)/2−1 ds . t(n−1)/2 ε(n−1)/2 . t(n−1)/2 .
ε/t
Therefore, |Λε (t)| . t(n−1)/2 for t ≥ A and n ≥ 1.
Integrating by parts yields
Z
Z ∞
Z ∞
¯∞
n−1
f0 (t)λε (t)t
dt =
f0 (t) dΛε (t) = f0 (t)Λε (t)¯A −
A
A
∞
Λε (t) df0 (t).
A
It follows from (24) and |Λε (t)| . t(n−1)/2 that f0 (t)Λε (t) → 0 as t → ∞. Since (40) and (42) imply
|Λε (A)| . εn An ,
¯Z ∞
¯
Z ∞
¯
¯
n
n
n−1
¯
¯
t(n−1)/2 |df0 (t)|.
(43)
f0 (t)λε (t)t
dt¯ ≤ ε |f0 (A)|A +
¯
A
A
Combining (41) and (43), we get
¶ Z
µZ A
n−1
n
0
n
|f0 (t)|t
dt + |f0 (A)|A +
|Jε | . ε
0
∞
t(n−1)/2 |df0 (t)|.
A
Letting first ε → 0 and then A → ∞, we obtain Jε0 → 0. Using this, (38), and (39), we arrive at
the assertion of the lemma.
¤
5. Pitt-type theorem: Lp –Lq Fourier inequalities with power weights
The following result captures the part “if” of Theorem 1. To show this, take p = q.
Theorem 2. Let 1 ≤ p, q < ∞ and n ∈ N. Let f be radial on Rn such that f0 is a general monotone
function on R+ .
¡ ¢
A
If p ≤ q and
n n+1
n
−
<γ< ,
q
2
q
(44)
then
¡ ¢
B
tn+γ−n/q−1/p f0 (t) ∈ Lp (0, ∞)
implies
|x|−γ fb(x) ∈ Lq (Rn );
Let a non-negative function f0 satisfy (11). If q ≤ p and
n n+1
−
< γ,
q
2
(45)
then
|x|−γ fb(x) ∈ Lq (Rn )
implies
tn+γ−n/q−1/p f0 (t) ∈ Lp (0, ∞).
10
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
5.1. Discussion. Let us discuss the conditions for a function and its Fourier transform in Theorem
2. To this end, we make sure that (A) and (B) imply corresponding assumptions of Lemmas 1 and
2.
n 1
(A) If tβ f0 (t) ∈ Lp (0, ∞), β = n + γ − − , then Hölder’s inequality implies
q
p
Z ∞
tn−1 (1 + t)−(n+1)/2 |f0 (t)| dt ≤ ktn−1−β (1 + t)−(n+1)/2 kLp0 (0,∞) ktβ f0 (t)kLp (0,∞) = I1 I2 . I1 .
0
Since (44) is equivalent to
and
n−1 1
1
− < β < n − , we get
2
p
p
µ
¶
1
0
(n − 1 − β)p > n − 1 + − n p0 = −1
p
µ
¶
µ
¶
n+1 0
1 n−1 n+1 0
n−1−β−
p < n−1+ −
−
p = −1.
2
p
2
2
This guarantees that the integral I1 converges. Therefore,
Z
∞
(46)
tn−1 (1 + t)−(n+1)/2 |f0 (t)| dt < ∞.
0
Since for any GM function f0 we have
Z ∞
Z
σ
t |df0 (t)| .
1
∞
tσ−1 |f0 (t)| dt,
σ > 0,
1
provided tσ−1 f0 (t) ∈ L1 (0, ∞), inequality (46) implies condition (11) of Lemma 1. Thus, Theorem 2
(A) states that condition tn+γ−n/q−1/p f0 (t) ∈ Lp (0, ∞) ensures the existence of the Fourier transform
in the improper sense and that |x|−γ fb(x) ∈ Lq (Rn ).
n−1
Let us now proceed to (B). Assume |y|−γ fb(y) ∈ Lq (Rn ), or equivalently, s q −γ F0 (s) ∈ Lq (0, ∞).
Applying Hölder’s inequality, we obtain
Z 1
s(n−1)/2 |F0 (s)| ds ≤ ks(n−1)/2−(n−1)/q+γ kLq0 (0,1) ks(n−1)/q−γ F0 (s)kLq (0,1) = I1 I2 . I1 .
0
³
´
³
´
n−1
n−1
n+1
1
0
Condition (45) yields n−1
−
+
γ
q
>
−
+
q 0 = −1. Therefore, I1 . 1 and
2
q
2
2
q
condition (34) of Lemma 2 is fulfilled. Hence, part (B) of Theorem 2 asserts that condition (11)
and |y|−γ fb(y) ∈ Lq (Rn ) imply that F0 is the Fourier transform (23), continuous for s > 0, and
tn+γ−n/q−1/p f0 (t) ∈ Lp (0, ∞).
5.2. Sharpness of conditions on γ. Let us rewrite part (A) of Theorem 2 in the following way.
Theorem 20 . Let 1 ≤ p ≤ q < ∞ and n ∈ N. Let f be radial on Rn such that f0 is a general
monotone function on R+ . Then
°
°
°
°
°
°
° −γ b °
(47)
°|x| f (x)° q n . °tβ f0 (t)° p
L (0,∞)
L (R )
if and only if
(48)
β =γ+n−
n 1
−
q
p
and
n n+1
n
−
<γ< .
q
2
q
WEIGHTED FOURIER INEQUALITIES
11
We restrict ourselves to the “only if” direction in Theorem 20 so far. This captures the corresponding part in Theorem 1 when p = q. The proof of the “if” part will be given in Section
6.
Proof. Consider f (x) = χ(x), then f0 (t) = χ[0,1] (t) ∈ GM . Then we have
µZ 1
¶1/p
pn+pγ−pn/q−1
n+γ−n/q−1/p
kt
f0 (t)kLp (0,∞) =
t
dt
.
0
This integral converges if pn + pγ − pn/q > 0, or equivalently γ > nq − n.
Let us figure out when |y|−γ χ
b(y) ∈ Lq (Rn ). By (36), the Fourier transform of f is χ
b(y) =
n
|B |jα+1 (|y|) = F0 (s). Therefore, we obtain
µZ ∞
¶1/q µZ ∞
¶1/q
°
°
¡ −γ
¢q n−1
°
° −γ
n−qγ−1
q
³
s |F0 (s)| s
ds
³
(49)
b(y)°
s
|jα+1 (s)| ds
.
°|y| χ
Lq (Rn )
0
0
There holds jα+1 (s) ³ 1 in a neighborhood of zero, hence the integral in (49) converges if n−qγ > 0,
that is, when γ < nq . The upper bound is established.
n n+1
There holds for s large, jα+1 (s) . s−(n+1)/2 , therefore the integral in (49) converges if −
<
q
2
γ. We will now show that if this condition does not hold, then the integral in (49) diverges. It
follows from (20) that for an integer number k0 large enough
ρα+1,k ³ k,
ρα+1,k+1 − ρα+1,k ³ 1,
k ≥ k0 ,
and there is a small ε > 0, independent of k, such that
|jα+1 (s)| & s−(n+1)/2 ,
Therefore,
Z ∞
n−qγ−1
s
q
|jα+1 (s)| ds &
0
&
s ∈ [ρα+1,k + ε, ρα+1,k+1 − ε],
∞ Z
X
k=k0
∞
X
ρα+1,k+1 −ε
k ≥ k0 .
sn−qγ−1 s−q(n+1)/2 ds
ρα+1,k +ε
n−qγ−1−q(n+1)/2
(ρα+1,k+1 − ε)
k=k0
&
∞
X
k n−qγ−1−q(n+1)/2 .
k=k0
n n+1
−
.
q
2
Let us verify that β and γ should be related by β = γ + n − n/q − 1/p. Let u > 0 and
g(x) = f0 (|x|/u) = χ(x/u). Then for t = |y| with 0 < t < 1/u
The last series diverges provided γ ≤
gb(y) = G0 (|y|) = un F0 (u|y|) = |B n |un jα+1 (ut) ³ un .
We then have
µZ
u
β
kt g0 (t)kLp (0,∞) ³
βp+1
t
0
and
° −γ
°
°|x| gb(x)°
µZ
u
dt
|G0 (t)|
t
dt
t
¶1/p
³ uβ+1/p ,
¶1/q
µZ
u
dt
&
t
&
u
t
q
n
L (R )
t
0
0
n 1
These yield uβ+1/p & uγ+n−n/q for any u > 0, that is, β = γ + n − − .
q
p
−γq+n
q
n
−γq+n
¶1/q
³ uγ+n−n/q .
¤
12
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
6. Proof of Theorem 2
°
°
We begin with the upper estimate of °fb(x)|x|−γ °Lq . First, by Corollary 1,
ÃZ
!1/q
·Z
¸1/q
|fb(x)|q
q n−qγ−1
dx
.
|F0 (t)| t
dt
qγ
Rn |x|
R+
ÃZ
"Z
!q #1/q
t
.
1/t
n−qγ−1
µZ
∞
n−qγ−1−nq/2−q/2
+
dt
0
R+
·Z
sn−1 Φ(s) ds
t
s
R+
n/2−3/2
¶q ¸1/q
Φ(s) ds dt
1/t
=: K1 + K2 .
We will use the (p, q) version of Hardy’s inequalities ([9]) with general weights u, v ≥ 0: for
1 ≤ α ≤ β < ∞,
"Z
µZ t
¶β #1/β
·Z ∞
¸1/α
∞
α
u(t)
ψ(s) ds dt
≤C
v(t)ψ(t) dt
(50)
0
0
0
holds for every ψ ≥ 0 if and only if
¶1/α0
¶1/β µZ r
µZ ∞
1−α0
v(t)
dt
< ∞,
u(t) dt
sup
r>0
and
"Z
µZ
∞
(51)
0
r
¶β
∞
ψ(s) ds
u(t)
#1/β
dt
·Z
α
v(t)ψ(t) dt
≤C
0
t
0
¸1/α
∞
if and only if
µZ
sup
r>0
¶1/β µZ
r
∞
u(t) dt
1−α0
v(t)
r
0
Here we consider the usual modification of the integral
¶1/α0
dt
hR
< ∞.
v(t)θ dt
i1/θ
when θ = ∞.
Remark 1. In particular, (50) holds with u(t) = tε−1 and v(t) = tδ−1 if and only if ε < 0 and
δ = εα/β + α.
To estimate K1 , substitution 1/t → t yields
·Z
µZ t
¶q ¸1/q
qγ−n−1
n−1
K1 .
t
s Φ(s) ds dt
.
R+
0
Using Remark 1 with ε = qγ − n and α = p, β = q, we obtain
¸1/p
¸1/p ·Z
·Z
¡ n+γ−n/q−1/p
¢p
¡ n−1
¢p
γp−np/q+p−1
,
=
t
Φ(t) dt
K1 .
t
t Φ(t) dt
R+
R+
which holds for γ < n/q.
Further, to estimate K2 , we change variables 1/s → s in the inner integral:
WEIGHTED FOURIER INEQUALITIES
·Z
µZ
K2 .
t
·Z
µZ
−n/2−1/2
s
−γp−np/2+p/2+np/q−1
¡
t
−n/2−1/2
R+
¡
=
dt
.
¶q ¸1/q
Φ(1/s) ds dt
0
t
·Z
t
n−qγ−1−nq/2−q/2
R+
.
Φ(1/s) ds
¸1/q
0
·Z
t
−n/2−1/2
s
R+
We wish to have
¶q
t
n−qγ−1−nq/2−q/2
13
γ−n/q+n−1/p
t
Φ(t)
¢p
Φ(1/t)
¸1/p
¢p
dt
¸1/p
dt
.
R+
This estimate holds, by Hardy’ inequality (see Remark 1), with ε = n − qγ − nq/2 − q/2, α = p,
and β = q under assumption
n n+1
ε < 0 ⇐⇒ γ > −
.
q
2
Combining estimates for K1 and K2 , and using the definition of the GM class, we get
ÃZ
Rn
|fb(x)|q
dx
|x|qγ
!1/q
·Z
¡
≤ C
·Z
R+
¢p
¸1/p
dt
¶p
∞
γp−np/q−1
−1
s |f0 (s)| ds
t/c
R+
t
≤ C
Φ(t)
µZ
t
≤ C
·Z
t
γ−n/q+n−1/p
¯
γp−np/q+np−1 ¯
¯p
f0 (t)¯ dt
¸1/p
dt
¸1/p
.
R+
In the second estimate we have used, after appropriate changes of variables, inequality (51) with
α = β = p under condition γ > n/q − n. This proves the first part of the theorem.
To prove the part (B), we first note that for any f0 ∈ GM there holds
Z
(52)
Z
∞
|f0 (x)| ≤
f0 (t)
x/c
x
Secondly, by (32) and Lemma 2, we have
Z ∞
Z
|f0 (x)| ≤
|df0 (t)| .
Z
x
∞
.
Z
(53)
∞
t
−1
bt
0
2bc/x
t
0
³Z
dt
.
t
f0 (s) ´
ds dt
s
t/b
x/bc
³ Z 2/t
´
(1−n)/2−1
(n−1)/2
t
z
|F0 (z)|dz dt
x/bc
.
∞
|df0 (t)| .
(n−1)/2−1
³Z
t
z
(n−1)/2
´
|F0 (z)|dz dt.
0
Let us assume p ≥ q and obtain appropriate upper estimates for
·Z
¸1/p
γp−np/q+np−1
p
.
(54)
J :=
s
|f0 (s)| ds
R+
14
D. GORBACHEV, E. LIFLYAND, AND S. TIKHONOV
By (53), we have
·Z
µZ
s
−γp+np/q−np−1
s
J .
(n−1)/2−1
³Z
t
t
z
0
R+
(n−1)/2
¶p
´
|F0 (z)|dz dt
¸1/p
ds
.
0
Using Remark 1 with α = q, β = p, and
ε = −γp + np/q − np,
δ = −γq + n − nq + q,
γ > n/q − n,
we get
·Z
µZ
J .
t
R+
¶q
t
−γq+n−q(n+1)/2−1
z
(n−1)/2
|F0 (z)|dz
¸1/q
dt
.
0
Applying again Remark 1, now with α = β = q, and
ε = −γq + n − q(n + 1)/2,
δ = −γq + n − q(n − 1)/2,
we obtain
·Z
¯
¯q ¸1/q
¯
n−qγ−1 ¯
t
=
¯F0 (t)¯ dt
J .
ÃZ
Rn
R+
γ > n/q − (n + 1)/2,
|fb(x)|q
dx
|x|qγ
!1/q
,
¤
the required bound.
7. Acknowledgements
A part of the present work was done at the Mathematisches Forschungsinstitut Oberwolfach
during a stay within the Research in Pairs Programme from October 11 to October 24, 2009. The
authors appreciate the hospitality and creative atmosphere of the Institute.
This research was also supported by the Centre de Recerca Matemàtica (Barcelona), the RFFI
09-01-00175, NSH-2787.2008.1, MTM 2008-05561-C02-02, and 2009 SGR 1303.
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Tula State University, Department of Mechanics and Mathematics, 300600 Tula, Russia
E-mail address: [email protected]
Department of Mathematics, Bar-Ilan University, 52900 Ramat-Gan, Israel
E-mail address: [email protected]
S. Tikhonov, ICREA and Centre de Recerca Matemàtica, Apartat 50, Bellaterra 08193
Barcelona
E-mail address: [email protected]