Pitt and Boas inequalities for Fourier and Hankel transforms

Accepted Manuscript Pitt and Boas inequalities for Fourier and Hankel transforms L. De Carli, D. Gorbachev, S. Tikhonov PII: DOI: Reference:

S0022-247X(13)00603-3 http://dx.doi.org/10.1016/j.jmaa.2013.06.045 YJMAA 17709

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Journal of Mathematical Analysis and Applications

Received date: 16 March 2013 Please cite this article as: L. De Carli, D. Gorbachev, S. Tikhonov, Pitt and Boas inequalities for Fourier and Hankel transforms, J. Math. Anal. Appl. (2013), http://dx.doi.org/10.1016/j.jmaa.2013.06.045 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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Pitt and Boas inequalities for Fourier and Hankel transforms L. De Carli, D. Gorbachev, and S. Tikhonov Abstract. We prove Pitt and Boas inequalities for products of radial functions and spherical harmonics in Rn . In the process, we obtain upper and lower estimates of the operator norm of the Hankel transform with power weights. Our inequalities are sharp in some specific cases.

1. Introduction Weighted norm inequalities for the Fourier transform provide a natural way to describe the balance between the relative size of a function and its Fourier transform at infinity. A classical example is the Pitt inequality with special radial weights (1.1) k |y|−s Fb kLq (Rn ) ≤ Ck |x|t F kLp (Rn ) , F ∈ S(Rn ). R Here 1 < p ≤ q < ∞, Fb(y) = Rn F (x)e−ixy dx denotes the Fourier transform of F , and   n 1 n 1 (1.2) 0 ≤ s < , 0 ≤ t < ′ , and s = t + n − ′ . q p q p p′ denoting the dual exponent of p, i.e., p1 + p1′ = 1. A proof of this inequality is in [4, 7]. See also [5] and [3]. It is worth mentioning that, when q = 2, the Pitt inequality has been characterized as a Hardy-Rellich inequality in recent literature, with alternative proofs and extensions. See, e.g., [13], [1], and [24]. The range of the parameters in (1.2) cannot be improved in general (see, e.g., [22]), but we can greatly extend this range, and also show that inequality (1.1) can be reversed, i.e, (1.3) Ak |y|−s FbkLq (Rn ) ≤ k |x|t F kLp (Rn ) ≤ Bk |y|−s FbkLq (Rn ) for the case p = q, if we restrict F to certain subspaces of S(Rn ). When n = 1 and 1 < p = q < ∞, and f is even and non increasing in (0, ∞) and vanishes at infinity, the Hardy–Littlewood theorem states that there exist constants A, B > 0 for which 2 (1.4) Akfbkp ≤ k |x|1− p f kp ≤ Bkfbkp ,

see, e.g., [25, Ch. IV]. Here and in the rest of the paper, we let k · kp = k · kLp (0,∞) for the sake of simplicity.

2000 Mathematics Subject Classification. 42B10, 46E30, 26A45. Key words and phrases. Fourier and Hankel transforms, sharp constants, Pitt inequalities, Boas inequalities. The second author was partially supported by the RFFI 10-01-00564. The third author was partially supported by the MTM 2011-27637, 2009 SGR 1303, RFFI 12-01-00169, and NSH-979.2012.1. This work was produced during the special semester in Approximation Theory and Fourier Analysis at the Centre de Recerca Matematica (CRM), Bellaterra, (Spain), Sept. 2011-Febr. 2012. 1

2

L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

Boas conjectured in [6] that weighted versions of (1.4) are also true, i.e., under the same assumptions on f and p, 2

Ak |y|−s fbkp ≤ k |x|s+1− p f kp ≤ Bk |y|−s fbkp .
provided that s ∈ − p1′ , p1 , 1 < p < ∞. This conjecture was proved by Y. Sagher in [20]. More recently, a subclass of the class of functions of bounded variation on [ε, ∞), for any ε > 0, that strictly includes the class of monotonic functions was introduced in [17] (see also [18]). These functions vanish at infinity, and there exist C > 0 and c > 1 so that, for every r > 0, Z ∞ Z ∞ du |f (u)| , |df (u)| ≤ C u r/c r Rb where a g(u) |df (u)| is the Riemann–Stieltjes integral. These functions are called General Monotonic (GM). In [14], the authors proved that inequality (1.5) is valid for GM functions and s ∈
− p1′ , 1p , thus improving Sagher’s theorem. Boas inequalities with general weights (i.e., not necessarily power) are proved in [19]. In this paper we consider functions in Rn that are products of radial functions and spherical harmonics. We prove a sharp version of the Pitt inequality (1.1), and we prove the Boas inequality (1.3) when the radial components are GM functions. The case of the L1 integrability of the Fourier transform for such functions has been considered in Theorem 2 of [16]. See also [11] for other integrability results. (1.5)

1.1. Pitt inequalities and spherical harmonics. Here and throughout the paper, x = (x1 , . . . , xn ), r = |x|, and ω = x/|x|. We let f (|x|) = f (r). Following, e.g., [23], Ch. IV, we say that a solid harmonic of degree k in Rn is a homogeneous harmonic polynomial Hk (x) of degree k. We can write Hk (x) = r k Yk (ω), where Yk is a spherical harmonic of degree k. We denote by Σk the set of the spherical harmonics of degree k. By the Bochner identity, Z ∞ n n (1.6) fd Yk (ρσ) = (2π)n/2 ik Yk (σ)ρ1− 2 r 2 Jn/2+k−1 (rρ)f (r) dr, 0

where ρ = |y|, σ = y/|y| and Js (r) is the standard Bessel function of the first kind. We prove the following

Theorem 1.1. For every Yk ∈ Σk , and every radial f ∈ S(Rn ), the Pitt inequality

(1.7)

k |y|−s fd Yk kLq (Rn ) ≤ Ck |x|t f Yk kLp (Rn )

holds with 1 ≤ p ≤ q ≤ ∞, and if and only if       1 1 1 1 1 n 1 − ′ − − , 0 ≤ t < ′ + k. and (n − 1) + max (1.8) s=t+n q p 2 p p′ q p   If p ≤ 2 and q = p′ , and s = t = (n − 1) 12 − p1 , (1.7) holds with (1.9)

n

C = (2π) 2 Ck,p sup

Yk ∈Σk

where (1.10)

Ck,p = 2

1 −1 2 p′

p

(2k+n−1)p+2 4p

(p′ )

Γ

(2k+n−1)p′ +2 4p′

kYk kLq (Sn−1 ) kYk kLp (Sn−1 )



(2k+n−1)p′ +2 4

Γ



 1′

(2k+n−1)p+2 4

The constant C cannot be replaced by any smaller constant.

p

1 . p

PITT AND BOAS INEQUALITIES FOR FOURIER AND HANKEL TRANSFORMS

3

We also show that when C is as in (1.9), equality is attained in (1.7). To the best of our knowledge, the supremum of the ratio of the norms of spherical harmonics in (1.9) is not known. Reverse H¨older inequalities for spherical harmonics have been discussed in [21] and [12]. In [12], it is proved that    ′ k/2 Γ kp+n 1/p Γ n
1/p′ kYk kLp′ (Sn−1 ) 2 2 p ≤ = O(k(n−1)(1/p−1/2) )  ′ 1/p′
kYk kLp (Sn−1 ) p kp +n n 1/p Γ Γ 2 2

whenever 1 < p ≤ 2.

1.2. Best constants in Pitt inequalities. To the best of our knowledge, the sharp constant in the Pitt inequality (1.1) has been evaluated only when p ≤ 2, q = p′ and s = t = 0 (i.e., when it reduces to the Hausdorff–Young inequality) and also when p = q = 2 and s = t < n2 . See [2], [26] and [13]. The best constant in inequality (1.7) is known only in the aforementioned cases, and when p, q, t and s are as in Theorem 1.1. When p = q = 2 and s = t < n2 , it is proved in [26] that (1.11) with

k |y|−t fd Yk kL2 (Rn ) < ck (t)k |x|t f Yk kL2 (Rn )

(1.12)

ck (t) = (2π)

n 2

Γ 12 2−t Γ 12

n 2 n 2



−t+k

. +t+k

Inequality (1.11) is sharp, and equality cannot be attained when t > 0 (see [26, L. 3.8]). The author also observes (see [26, L. 2.1]) that the best constant in inequality (1.1) is c0 (t) = supk ck (t). That means that when p = q = 2, the best constant in (1.1) is attained in the space of radial functions. This fact easily follows from the orthogonality of the spherical harmonics in L2 (Sn−1 ). We use (1.11), the sharp inequality proved in Theorem 1.1, and  a sharp  inequality in [4], 1 1 ′ to estimate the constant in (1.7) when q = p and t > t = (n − 1) 2 − p . Theorem 1.2. Let 1 < p ≤ 2 and let 0 ≤ d <

n 2

+

1 2

− 1p . The Pitt inequality

k |y|−t−d fd Yk kLp′ (Rn ) ≤ Ck |x|t+d f Yk kLp (Rn )

holds with C = BCk,p sup

kYk kLp′ (Sn−1 )

, where Ck,p is as in (1.10), and kYk kLp (Sn−1 )     Γ (−2d+n+1)p−2 2−d Γ (n−1)p+2 n 4p 4p    .  B = (2π) 2 Γ (2d+n−1)p+2 Γ (n+1)p−2 4p 4p

Yk ∈Σk

(1.13)

The constant on the right hand side of(1.13) may not be sharp for all values of p and d, but it is easy to verify that it equals ck (t) = ck (0) in (1.11) when p = 2, and it reduces to Ck,p in Theorem 1.1 when d = 0. 1.3. The Boas conjecture and the Hankel transform. The Hankel transform is a natural generalization of the Fourier transform of radial functions. There are several definitions of the Hankel transform in the literature, which are roughly equivalent to one another (see, e.g., [9] and also [10]).

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L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

In this paper we use the following definition: for every α ≥ − 21 and every k ≥ 0, we let Z ∞ −α e e (1.14) fk (ρ) = fk,α(ρ) = ρ r α+1 Jα+k (ρr)f (r) dr, f ∈ S(0, ∞) 0

be the Fourier–Hankel transform of order k of f . Then, by the Bochner identity (1.6), the Fourier transform of the product of a radial function f (r) and a spherical harmonic Yk (ω) can be written as fd Yk (ρσ) = (2π)n/2 ik Yk (σ)fek, n2 −1 (ρ).

(1.15)

In the polar coordinates,

t

t+(n−1)/p

|x| f Yk p n = kYk k p n−1 r f

L (S ) L (R )

p

n

and the Pitt inequality (1.7), with C = c (2π) 2 sup from (1.16)

Yk ∈Σk





−s+(n−1)/q e fk ≤ c r t+(n−1)/p f ,

ρ

α=

p

q

kYk kLq (Sn−1 ) for some c > 0, follows kYk kLp (Sn−1 ) n − 1, 2

f ∈ S(0, ∞).

Similarly, the Boas inequality (1.3), with F = Yk (ω)f (r), follows from







(1.17) a ρ−s+(n−1)/q fek ≤ r t+(n−1)/p f ≤ b ρ−s+(n−1)/q fek , q

p

q

where p = q and a, b > 0. The proof of Theorem 1.1 (see also Remark 2.1) shows that when s, t and k are as in (1.8), inequality (1.16) holds for every f ∈ S(0, ∞). We show that, when f is a GM function, (1.16) is valid for a wider range of parameters than (1.8), and (1.17) holds. Our results improve a result in [14], where the authors have proved inequalities (1.16) and (1.17) for radial functions in GM (i.e., for k = 0). Let f be a GM function such that Z Z 1 r 2α+k+1 |f (r)| dr + (1.18) I :=

1

0

∞

r α+1/2 |df (r)| < ∞.

Then Lemma 3.1 below implies that its Fourier–Hankel transform is defined in the improper Rb sense (i.e., as lim a ) and fek ∈ C(0, ∞). a→0 b→∞

We prove the following

if

Theorem 1.3. Let 1 < p ≤ q < ∞ and f ∈ GM . The inequality (1.16) holds if and only

(1.19) and (1.20)

(n − 1)

s=t+n





−

1 1 − 2 p



1 1 − q p′



n 1 < t < ′ + k. p p

Clearly, (1.20) is less restrictive than the right-hand-side inequality in (1.8). Note that condition (1.18) follows automatically from the finiteness of the right-hand-side of (1.16) (see Remark 3.1) which is satisfied when p, t and k are as in (1.19) and (1.20). The next theorem reverses Theorem 1.3.

PITT AND BOAS INEQUALITIES FOR FOURIER AND HANKEL TRANSFORMS

5

Theorem 1.4. Let 1 < q ≤ p < ∞, and let f be GM; assume that f ≥ 0 and also that (1.18) is satisfied. If (1.19) and −

n−1 1 − < t < ∞, p p

hold, then the reverse Pitt inequality





−s+(n−1)/q e fk & r t+(n−1)/p f

ρ q

p

holds.

Here and in the sequel the expressions f . g, f & g, f ≍ g mean the inequalities f ≤ Cg, f ≥ Cg, Cg ≤ f ≤ C ′ g, respectively, where C, C ′ , . . . denote positive constants. Taking p = q in Theorems 1.3 and 1.4, we get Corollary 1.1. Let 1 < p < ∞. If f ∈ GM , f ≥ 0 and (1.18) holds, then



−s+(n−1)/p e

fk ≍ r t+(n−1)/p f

ρ p

if and only if

p

n n n+1 < t < ′ + k. − p′ 2 p

Remark 1.1. Theorems 1.3 and 1.4 and Corollary 1.1 hold for any n = 2α + 2 ≥ 1 and k ≥ 0, not necessarily integers. We prove these theorems in Section 3.

2. Proof of Theorems 1.1 and 1.2 In [10], the author considered the Lp (0, ∞)–Lq (0, ∞) mapping properties of operators in the class   Z ∞ 1 L = Lην, µ : Lην, µ f (ρ) = ρµ (rρ)ν Jη (rρ)f (r) dr, η ≥ − and µ, ν ∈ R 2 0 and proved the following

Theorem 2.1. (i) Lην, µ is bounded from Lp (0, ∞) to Lq (0, ∞) p ≤ q ≤ ∞, and if and only if  1 1 1 1 1 (2.1) µ= ′ − and − η − ′ < ν ≤ − max − p q p 2 p′

1 whenever η ≥ − , 1 ≤ 2  1 , 0 . q

(ii) When q = p′ and ν = 21 , the following inequality holds for every 1 < p ≤ 2, η ≥ − 12 and f ∈ S(0, ∞):  1   1 1 p′ 1 p′ η+ 12 + p1 kLη1 , 0 f kLp′ (0, ∞) Γ (η + ) + 2 1 2 2 2 −1 p 2   (2.2) ≤ 2p 2
1 . 1 1 1 p kf kLp (0, ∞) 1 1 p η+ + 2 2 p′ ′ Γ (η + ) + (p ) 2 2 2

The constant on the right-hand side of (2.2) is best possible and is attained by the functions 1 2 fλ (x) = xη+ 2 e−λx , λ > 0.

We show that proving Theorem 1.1 is equivalent to estimating the Lp (0, ∞)–Lq (0, ∞) norm of an operator in the class L.

6

L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

2.1. Proof of Theorem 1.1. We have observed in Section 1.3 that when F = f (r)Yk (ω), where f is radial and Yk is a spherical harmonic of degree k, the Pitt inequality kYk kLq (Sn−1 ) n , follows if we prove (1.1), with C = c (2π) 2 sup Yk ∈Σk kYk kLp (Sn−1 )





−s+(n−1)/q e fk ≤ c y t+(n−1)/p f , f ∈ S(0, ∞),

x p

q

where fek is given by (1.14) and α = n2 − 1. We let r t+(n−1)/p f = g; thus, kr t+(n−1)/p f kp = kgkp , and Z ∞ q Z ∞



−s+(n−1)/q e q r α+1 Jα+k (ρr)f (r) dr dρ ρq(−s+(n−1)/q−α) fk =

ρ q

=

0

0

∞

0

0

Z q(−s+(n−1)/q−α) ρ

∞

−t−(n−1)/p+α+1

q g(r) dr dρ

Jα+k (ρr)r q Z ∞ ∞ ρq(−s+(n−1)/q−2α+t+(n−1)/p−1) Jα+k (ρr)(ρr)−t−(n−1)/p+α+1 g(r) dr dρ =

Z

Z

and recalling that α = Z

−s+(n−1)/q e q fk =

ρ q

0

n 2 − ∞

ρ

0

0

1, we obtain

Z

q(t−s+(n−1)(1/q−1/p′ )

∞

−t−(n−1)/p+α+1

Jα+k (ρr)(ρr)

0

q = kLα+k ν,µ gkq

q g(r) dr dρ

n n−1 where µ = t − s + (n − 1) (1/q − 1/p′ ) and ν = α + 1 − t − n−1 p = 2 −t− p . Let us show that condition (1.8) is equivalent to condition (2.1).   First, note that by (2.1), µ = p1′ − 1q , which implies t − s = n p1′ − 1q as in (1.2). Also,

the condition −k − α −

< ν = α + 1 − t − n−1 t < pn′ + k. p yields that n n o o We also need ν ≤ 12 − max p1′ − 1q , 0 ; assume that max p1′ − 1q , 0 = 0, since the other 1 p′

case is similar. So, ν = α + 1 − t −

n−1 p

=

n 2

− t − n−1 p  1 − t ≥ t = (n − 1) 2

This concludes the first part of the theorem.

≤ 12 implies  1 . p

By Theorem 2.1, the Lp –Lq norm of Lα+k ν,µ can be explicitly evaluated, and is as in (2.2), if 1 < p ≤ 2, q = p′ , ν = 21 and µ = 0. In particular we can take, s = t = t. Then Ck,p in  (1.9) equals the right-hand side of (2.2) with η = k + n2 − 1, which is (1.10). The proof of Theorem 1.1 yields the following

Remark 2.1. Under the assumptions of Theorem 1.1, and with the notation of Theorem 2.1, the Pitt inequality (1.16) holds whenever s, t and k are as in (2.1), with the sharp constant c = kLα+k ν,µ kLp →Lq ,   1 1 where µ = t − s + (n − 1) q − p′ and ν = n2 − t − n−1 p . Remark 2.2. In [10] it is conjectured that the Lp –Lq norm of Lην,µ is
1 1 (1+η−ν+ p1 ) 2 q + 12 q Γ η+ν+mu ν− q1 p η 2 C = Cν, p, q = 2 1  1 (η+ν+ p1′ ) p q2 p + 12 Γ 1+η−ν 2

PITT AND BOAS INEQUALITIES FOR FOURIER AND HANKEL TRANSFORMS

7

for all admissible values of the parameters, but unfortunately the conjecture does not hold in general. We have observed in Section 1.2 that when q = p = 2 and t = s < n2 , the best constant in the Pitt inequality (1.7) is ck (t) in (1.12); by Remark 2.1 and the discussion in k+ n −1

n

2 ||L2 →L2 . If the conjecture was true, ck (t) would then Section 1.3, ck (t) = (2π) 2 ||L 1 −t,0 n 2

−1 k+ n 2

equal (2π) C 1 −t, 2,2 = (2π) 2

n 2

q2 Γ(k+ n −t) 2 q , Γ(k+ n +t) 2

but it is not too difficult to see that when k, t > 0,

this constant is strictly smaller than the one in (1.12). 2.2. Proof of Theorem 1.2. We use the well-known formula −z

n+z

2 π 2 dz n+z π

−n−z , n+z |x| = (2π) −z |x| Γ 2 Γ 2

−n < z < 0,

and, by letting z = −n+d, we can see at once that |x|−d is the Fourier transform of cd |x|−n+d , with
Γ n−d −d − n 2
. (2.3) cd = 2 π 2 Γ d2   We recall that t = (n − 1) 21 − 1p . Thus, if we let f = g|x|−d and we denote with F the Fourier transform for ease of notation, we obtain k |x|−t−d F(Yk g|x|−d )kLp′ (Rn ) = cd k |x|−t F(|x|−n+d )F(Yk g|x|−d )kLp′ (Rn )   = cd k |x|−t F |x|−n+d ∗ Yk g|x|−d kLp′ (Rn ) .

The function |x|−n+d ∗ (Yk g|x|−d ) is a product of a radial function times Yk . By Theorem 1.1,   k |x|−t−d F(Yk g|x|−d )kLp′ (Rn ) = cd k |x|−t F |x|−n+d ∗ Yk g|x|−d kLp′ (Rn )   n ≤ (2π) 2 Ck,p cd k |x|t |x|−n+d ∗ (Yk g|x|−d ) kLp (Rn ) . where Ck,p is as in (1.9). To conclude the proof of the Theorem, we apply the following Lemma (see Theorem 2 in [4]). Lemma 2.1. For every a < n/p and b < n/p′ , a + b > 0, and for c = n − a − b, the following sharp inequality holds   (2.4) k |x|−a |x|−c ∗ (|x|−b h) kLp (Rn ) ≤ Da,b khkLp (Rn )

with

  
n b n Γ 2p − a2 Γ 2p ′ − 2   . =
 n n Γ 12 (−a − b + n) Γ a2 + 2p Γ 2b + 2p ′ π n/2 Γ

Da,b

So, the inequality (2.5)

a+b 2

  k |x|t |x|−n+d ∗ (g|x|−d Yk ) kLp (Rn ) ≤ Bk |x|t gYk kLp (Rn )

is equivalent to (2.4) if h = g|x|t Yk , a = −t, b = d + t and c = n − d.

8

L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

For these values of the parameters, the constant in (2.5) is then   
 (−2d+n+1)p−2 π n/2 Γ d2 Γ (n−1)p+2 Γ 4p 4p D−t,d+t =
 (n+1)p−2   (2d+n−1)p+2  n−d Γ 2 Γ Γ 4p 4p n

and the Pitt inequality in Theorem 1.2 holds with C = (2π) 2 D−t,d+t cd Ck,p . But     2−d Γ (n−1)p+2 Γ (−2d+n+1)p−2 4p 4p     D−t,d+t cd = Γ (n+1)p−2 Γ (2d+n−1)p+2 4p 4p



which is the same as B in (1.13).

3. Proof of Theorems 1.3 and 1.4 We call a function admissible if it is of bounded variation on (ε, ∞), for every ε > 0, and vanishes at infinity. Before proving the theorems, we need to prove two technical lemmas. Lemma 3.1. For an admissible function f such that (1.18) holds, we have fek ∈ C(0, ∞) and for ρ > 0 Z 1/ρ Z ∞ (3.1) |fek (ρ)| . ρk r 2α+k+1 |f (r)| dr + ρ−α−3/2 r α+1/2 |df (r)|. 0

1/ρ

Proof. Define (3.2)

ψ(t) =

Z

t

uα+1 Jα+k (u) du.

0

Let us estimate |ψ(t)| for t > 1, α ≥ − 12 and k ≥ 0. We have ψ(t) = Js (u) ≍ uα+k for small values of u, then Z 1 Z 1 u2α+k+1 du ≍ 1. ≍ 0

R1 0

+

Rt

1.

Since

0

It is well known that u1/2 Jα+k (u) = C1 cos(u − c) + C2 sin(u−c) + O(u−2 ) for u > 1. Therefore, u   Z t Z t sin(u − c) α+1/2 −2 u = + O(u ) du. C1 cos(u − c) + C2 u 1 1 Using second mean value theorem, for some τ ∈ (1, t) we have  Z τ Z t sin(u − c) = C1 cos(u − c) + C2 du u 1 1   Z t Z t sin(u − c) u−2 du . + tα+1/2 du + tα+1/2 O C1 cos(u − c) + C1 u 1 τ Here all integrals are bounded and therefore,

|ψ(t)| . tα+1/2 ,

t > 1.

To get (3.1), we first use (1.14): Z Z 1/ρ r α+1 Jα+k (ρr)f (r) dr + fek (ρ) = ρ−α 0

∞

1/ρ

r

α+1

Jα+k (ρr)f (r) dr

!

=: I1 + I2 .

PITT AND BOAS INEQUALITIES FOR FOURIER AND HANKEL TRANSFORMS

9

Let us estimate I1 . Since Jα+k (t) ≍ tα+k when 0 < t < 1, then Z Z 1/ρ 1/ρ α+1 −α r 2α+k+1 |f (r)| dr. r Jα+k (ρr)f (r) dr . ρk ρ 0 0

which is the first integral in (3.1). Estimating I2 is more complicated. Using (3.2) we have

[ψ(ρr)]′r = ρ(ρr)α+1 Jα+k (ρr), and hence I2 = ρ−2α−2

Z

∞

1/ρ

Integrating by parts, I2 = ρ

−2α−2

(3.3) −2α−2

=ρ

[ψ(ρr)]′r f (r) dr.

Z ∞ ψ(ρr)f (r) − 1/ρ

lim ψ(ρr)f (r) − ψ(1)f (1/ρ) − ρ

α+ 21

r→∞

∞

!

ψ(ρr) df (r)

1/ρ

Z

∞h

−α− 12

(ρr)

1/ρ

! i α+ 12 ψ(ρr) r df (r) .

1

1

We have shown that |ψ(t)| . tα+ 2 . So, (ρr)−α− 2 |ψ(ρr)| . 1, and the integral in (3.3) is Z Z ∞ ∞h i 1 1 −α− 12 α+ 12 α+ 12 (ρr) (3.4) ρ ψ(ρr) r df (r) . ρα+ 2 r α+ 2 |df (r)|. 1/ρ 1/ρ

Also, when r is much larger that ρ1 ,

1

1

|ψ(ρr)f (r)| . (ρr)α+ 2 |f (r)| . r α+ 2 |f (r)|. 1

1

By (1.18), r α+ 2 |f (r)| is integrable in [1, ∞), and so lim r α+ 2 f (r) = 0. r→∞

Clearly, also

lim f (r) = 0. On the other hand, r→∞ Z ∞ Z ∞ 1 1 df (t) . ρα+ 2 (3.5) |ψ(1)f (1/ρ)| = ψ(1) tα+ 2 |df (t)|. 1/ρ 1/ρ By (3.3), (3.4), and (3.5),

3

|I2 | . ρ−α− 2

Z

∞

1/ρ

1

tα+ 2 |df (t)|

as required. This concludes the proof of the Lemma.



Remark 3.1. First let us recall the following inequality: for any f ∈ GM there is c > 1 such that Z ∞ Z ∞ (3.6) uσ |df (u)| . uσ−1 |f (u)| du, σ ≥ 0, r

r/c

which follows from the definition of GM functions, see also [14, p. 111]. Let us show that I

defined by (1.18) is bounded by r t+(2α+1)/p f p under conditions of Theorem 1.3. Indeed, using (3.6), Z 1 Z ∞ Z 1 r 2α+k+1 r α−1/2 |f (r)| dr . r 2α+k+1 |f (r)| dr + I. |f (r)| dr. α+k+3/2 0 (1 + r) 1 0

10

L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

Then using the H¨older inequality, we estimate



r 2α+k+1

−t−(2α+1)/p t+(2α+1)/p

. I . r t+(2α+1)/p f r f

r

, p p (1 + r)α+k+3/2 p′   provided that (n − 1) 12 − p1 − p1 < t < pn′ + k, i.e., (1.20).



Lemma 3.2. For an admissible non-negative function f such that (1.18) holds and Z 1 |fek (ρ)|ρ2α+k+1 dρ < ∞; 0

we have for any z > 0 and any 1 < b < 2π Z Z 1/z (3.7) zk |fek (ρ)|ρ2α+k+1 dρ & 0

zb z/b

f (r) dr. r

Proof. The proof is in two steps: first, we construct a non-negative kernel with finite support similar to K from [14]. Then, we use its properties. To construct our kernel, we apply the Bessel–Hankel transform and the generalized Bessel convolution; see, e.g., [15]. Let jν (t) = 2ν Γ(ν + 1)t−ν Jν (t) be the normalized Bessel function, with ν ≥ −1/2. Recall that jν (0) = 1. We let Z ∞ 1 −1 f (t)jν (st)t2ν+1 dt Bν f (s) = Bν f (s) = ν 2 Γ(ν + 1) 0 be the Bessel–Hankel transform. The Bessel generalized translation operator is given by f (t + s) + f (|t − s|) , ν = −1/2, Ts f (t) = 2 Z π p 1 Ts f (t) = R π 2ν f ( t2 + s2 − 2ts cos ξ) sin2ν ξ dξ, sin ξ dξ 0 0 Using this, we define the generalized Bessel convolution as follows Z ∞ 1 Ts f (t)g(t)t2ν+1 dt. (f ∗ g)ν (s) = ν 2 Γ(ν + 1) 0

ν > −1/2.

Note that the Fourier transform and convolution of radial functions are related to the Bessel translation and convolution operators. We recall some properties of the generalized Bessel convolution. Let Then

F (u) = (f ∗ f )ν (s),

f ∈ L1loc (R+ ),

supp f ⊂ [0, a],

a > 0.

(i) F ∈ C(R+ ) and supp F ⊂ [0, 2a]; (ii) for every s > 0, 0 ≤ F (s) ≤ F (0) = (Bν f 2 )(0); (iii) Bν F = (Bν f )2 . Let χa = χ[0,a] . Then Z Z a aν+2 1 Jν (st) 2ν+1 Jν (ast)tν+1 dt. t dt = Bν χa (s) = ν sν 0 0 (st)

Taking into account that

Z

0

1

Jν (ut)tν+1 dt = u−1 Jν+1 (u),

PITT AND BOAS INEQUALITIES FOR FOURIER AND HANKEL TRANSFORMS

11

we get a2ν+2 . + 2) Define K(ρ) = Ca−1 (χa ∗ χa )ν (ρ). Using properties of convolution, we get the following properties of K: (i) K ∈ C(R+ ) and supp K ⊂ [0, 2a]; (ii) 0 ≤ K(ρ) ≤ K(0), ρ > 0, where Z a C −1 K(0) = Ca−1 (χa ∗ χa )ν (0) = ν a t2ν+1 dt = 1; 2 Γ(ν + 1) 0 Bν χa (s) = Ca jν+1 (as),

with Ca =

2ν+1 Γ(ν

2 (ar). (iii) Bν K(r) = Ca−1 (Bν χa (r))2 = Ca jν+1

We now let f be a non-negative function satisfying the assumptions of the lemma. We recall that Z ∞ fek (ρ) = ρ−α r α+1 Jα+k (ρr)f (r) dr, ρ > 0. 0

Multiplying both sides by K(ρ)ρ2α+k+1 and integrating we get, on the left-hand side Z 2a Z 2a |fek (ρ)|ρ2α+k+1 dρ fek (ρ)K(ρ)ρ2α+k+1 dρ ≤ (3.8) I= 0

0

by (ii). The left-hand side is Z Z 2a K(ρ)ρα+k+1 I= 0

0

∞

 r α+1 Jα+k (ρr)f (r) dr dρ.

Changing the order of integration (we can justify this using (1.18) and the properties of K, as in [14]), we obtain  Z 2a Z ∞ K(ρ)ρα+k+1 Jα+k (ρr) dρ dr. I= r α+1 f (r) 0

0

Here

Z

0

2a

2 K(ρ)ρα+k+1 Jα+k (ρr) dρ = r α+k Bα+k K(r) = Ca r α+k jν+1 (ar),

Thus,

I = Ca

Z

0

∞

ν = α + k.

2 r 2α+k+1 f (r)jν+1 (ar) dr.

2 (t) & 1 for (2b)−1 < t < b/2, 1 < b < 2π. Indeed, Let us show that t2α+k+2 jν+1 2α+k+2 2α + k + 2 > 0, therefore t & 1 for t > (2b)−1 . Moreover, the Bessel function jν+1 (t) is decreasing on [0, qν+1 ], where qν+1 is its first positive zero. Then jν+1 (t) & 1 for 0 < t < b/2, where 1 < b < 2 inf qν+1 = 2π. ν+1≥1/2

Then using non-negativity of f and Ca ≍ a2ν+2 , the integral I can be rewritten as follows: Z Z b/(2a) f (r) a2(α+k)+2 ∞ f (r) 2α+k+2 2 k (ar) jν+1 (ar) dr & a dr. (3.9) I ≍ 2α+k+2 a r 0 1/(2ab) r

Combining estimates (3.8) and (3.9) and putting 2a = 1/z, we complete the proof of the Lemma.  We will also need the following weighted Hardy inequality (see [8]).

12

L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

Lemma 3.3. Let u, v be non-negative measurable functions which satisfy  1′  1 Z a Z ∞ q p −p′ q <∞ v(t) dt u(t) dt (3.10) sup a>0

0

a

for some 1 ≤ p ≤ q ≤ ∞. Then, there exists a constant C > 0 such that for every nonnegative measurable F , q  q1 Z ∞  1 Z ∞ Z t p
p . F (s) ds dt (3.11) u(t) v(t)F (t) dt ≤C 0

0

0

3.1. Proof of Theorem 1.3. We recall that inequality (1.16) is equivalent to

2α+1

−s+ 2α+1

t+ p ek q f f

ρ

. r q

where fek is defined by (1.14), α =

(3.12)

−1≥

n 2

p

− 12 ,

s = t + (2α + 2)

and (3.13)

(2α + 1)



1 1 − 2 p





1 1 − ′ q p



,

2α + 2 1
−

By Lemma 3.1 and Remark 3.1, Z 1/ρ Z |fek (ρ)| . ρk r 2α+k+1 |f (r)| dr + ρ−α−3/2 (3.14)

. ρk

Z

0

1/ρ

0

r 2α+k+1 |f (r)| dr + ρ−α−3/2

Z

∞

1/ρ

∞

1/ρ

r α+1/2 |df (r)|

r α−1/2 |f (r)|dr =: I1 + I2 .

By (3.14), we can see at once that





−s+ 2α+1

−s+ 2α+1

−s+ 2α+1 ek q f q I + r q I = K + K .

ρ

≤ r 1 2 1 2 q

q

So, with the change of variables r → K1 =

Z

∞

r

r −1 ,

and the substitution s = t + (2α + 2)

−sq+kq+2α+1

0

=

Z

0

=

Z

0

q

Z

1 r

z

2α+k+1

0

∞

r sq−kq−2α−3

∞

Z

r

0

r

t− 1q −k− 2α+2 p′

Z

|f (z)| dz

z 2α+k+1 |f (z)| dz

r

z

2α+k+1

0

q

|f (z)| dz

We apply the Hardy inequality (Lemma 3.3) with u(r) = r

!q

!1 q

dr



1 q

−

1 p′



1 q

dr

q

t− 1q −k− 2α+2 p′

− 2α+1 −k+t p′

1 q

dr

.

, F (z) = z 2α+k+1 |f (z)|,

< − 1q and and v(r) = r . By the assumptions on t in (3.13), t − 1q − k − 2α+2 p′ 2α+1 1 − p′ − k + t < p′ , so that both integrals in (3.10) are finite, and  1Z a  1′ Z ∞ Z ∞ q  q1Z a 2α+1 −p′  p1′ q p ′ − t− 1 −k− 2α+2 −k+t p′ r p′ u(r)q dr r q dr v(r)−p dr dr = a

0

.a

a t−k− 2α+2 p′

0

a

2α+2 +k−t p′

= 1.

PITT AND BOAS INEQUALITIES FOR FOURIER AND HANKEL TRANSFORMS

So, by (3.11), K1 .

Z

∞ 0

r tp+2α+1 |f (r)|p dr

1

p

13

.

−1 The norm K2 can be estimated similarly. We use again the change of variables r → r ,  1 1 and the substitution s = t + (2α + 2) q − p′ :

K2 = =

Z

Z

∞

r

−sq+2α+1−αq− 32 q

0

0

∞

r

t+ 32 +α− 2α+2 − 1q p′

Z

Z

r

∞

1 r

∞

z

α− 12

|f (z)| dz

1

z α− 2 |f (z)| dz

!q

q

!1 q

dr

1 q

dr

.

To estimate the latter integral, we use the dual version of the Hardy inequality (3.11) which is as follows: if  1′  1 Z ∞ Z a q p −p′ q <∞ v(t) dt u(t) dt (3.15) sup a>0

a

0

for some 1 ≤ p ≤ q ≤ ∞, then q  1 Z ∞  Z Z ∞ q F (s) ds dt (3.16) u(t) ≤C 0

t

∞

0

t− 1 + 3 +α− 2α+2

1


p v(t)F (t) dt

1

p

.

p′ , F (z) = z α− 2 |f (z)|, and v(r) = r We apply this with u(r) = r q 2   Then condition (3.15) holds provided (2α + 1) 12 − 1p − p1 < t and we get

K2 ≤

Z

∞

0

r tp+2α+1 |f (r)| dr

1

p

p

and since 2α + 1 = n − 1, Theorem 1.3 is proved.

3.2. Proof of Theorem 1.4. Let s = t + (2α + 2) (3.17)

t>−

.

.

To summarize, we have proved that

2α+1

t+ p

−s+ 2α+1 ek q f f

. r

ρ q

− 2α+1 +α+ 32 +t p′

2α + 2 . p



1 q

−

1 p′



and

Note that this assumption is less restrictive than that of Theorem 1.3. Let us use (3.7) in Lemma 3.2 Z 1/z Z zb f (r) dr . z k |fek (ρ)|ρ2α+k+1 dρ, z > 0, 0 z/b r

and the following inequality, which is valid for a > 0 and b > 1 ! Z ∞ Z bz Z ∞ dz |ψ(x)| dx |ψ(x)| dx . (3.18) . z z/b a/b a Since f is GM and is non-negative, by (3.18) Z ∞ Z ∞ Z ∞ 1 f (x) dx . |df (x)| . f (r) . r x z r/c r bc

Z

bz z/b

! f (x) dx dz. x

14

L. DE CARLI, D. GORBACHEV, AND S. TIKHONOV

By (3.7) in Lemma 3.2 and with the substitution z −1 → z, ! Z Z bc Z 1/z Z ∞ r 2α+k+1 k−1 −k−1 e |fk (ρ)|ρ dρ dz = f (r) . z z r bc

0

0

0

z

 2α+k+1 e |fk (ρ)|ρ dρ dz.

Raising the left-hand side and the right-hand side of the last inequality to the p-th power, we multiply by r tp+2α+1 and integrate to obtain  !p ! p1 Z z 1 Z ∞ Z ∞ Z bc p r −k−1 2α+k+1 tp+2α+1 tp+2α+1 p z |fek (ρ)|ρ dρ dz dr r r |f (r)| dr . 0

0

0

and after the change of variables r = Z

0

∞

r tp+2α+1 |f (r)|p dr

1

p

.

Z

∞

0

bc r′ ,

−t− 2α+3 p

(r ′ )

0

Z

r′

z −k−1

0

Z

0

z

 !p ! p1 . |fek (ρ)|ρ2α+k+1 dρ dz dr ′ −t− 2α+3

p , We use the Hardy inequality in Lemma 3.3 with u(r) = r F (z) = Rz 2α+2 1 +1− −t− −k−1 2α+k+1 ek (ρ)|ρ p q. z | f dρ, and v(r) = r 0 Let us verify that u and v satisfy condition (3.10) in Lemma 3.3. Indeed, because of the definition of t, both integrals in (3.11) are finite, and  1 Z a  1′ Z ∞ p q ′ u(r)p dr v(r)−q dr

a

=

Z

∞

r

a

−t−

0

 2α+3 p p

dr

+ p1 −t− 2α+3 p

.a provided that condition (3.17) holds. We get  1 Z Z ∞ p r tp+2α+1 |f (r)|p dr . 0

 1 Z

0

p

a

r

−t− 2α+2 +1− q1 p

0 t+ 2α+2 −1+ 1q + q1′ p

a

∞

r −t−

2α+2 1 − q −k p

Using again the Hardy inequality (3.11) with q = p, −t− 2α+2 − 1q −k p

provided that t >

Z

r 0

∞

ρ−s+

2α+1 q

0

− k. This concludes the proof. 

dr

 1′ q

|fek (ρ)|ρ2α+k+1 dρ

  ′ −t−(2α+2) q1 − p1 −k− 2α+1 q′

u(r) = r , v(r) = r 2α+k+1 e and F (ρ) = |fk (ρ)|ρ , we obtain Z 1 Z ∞ p . r tp+2α+1 |f (r)|p dr 0 − 2α+2 p

= 1,

−q′

=r

q  1

−s−k− 2α+1 q′

|fek (ρ)|q dρ

q

.

,

1 q

References [1] Barbatis G. and Tertikas A. On a class of Rellich inequalities. J. Comp. Appl. Math. 194 (2006), 156–172. [2] Beckner W. Inequalities in Fourier Analysis. Ann. Math. 102 (1975) 159–182. [3] Beckner W. Weighted inequalities and Stein–Weiss potentials. Forum Math. 20 (2008), no. 4, 587–606. [4] Beckner W. Pitt’s inequality with sharp convolution estimates. Proc. Amer. Math. Soc. 136 (2008), no. 5, 1871–1885. [5] Beckner W. Pitt’s inequality and the uncertainty principle. Proc. Amer. Math. Soc. 123 (1995), no. 6, 1897–1905. [6] Boas R. P. The integrability class of the sine transform of a monotonic function. Studia Math. 44 (1972), 365–369.

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