# Bootstrapping Weighted Inequalities for Hankel Transform

Authors: M. Dziri
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

A method of producing new inequalities for Hankel transform from known ones is given using the theory of positive integral operators. The new inequalities produced depend on six parameters, two real indices, two complex-valued measures, and two positive functions. The method may be iterated using the last inequality generated as input to the next stage.


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## Body

## 1. Introduction

Weighted inequalities for the integral transforms with the general weights are of great importance in many branches of mathematics (functional analysis, integral equation, interpolation theory, etc.). They provide a tool to solve numerous problems related to the estimation of expressions with given integral transform. One of the most important problems is the characterization problem of the operator theory in function spaces such as the criteria of the continuity, compactness, and other qualitative estimations of a classical and nonclassical transformation. Many authors studied some generalizations of the Hardy inequalities and give some applications of these inequalities, [1–7].In this work, we are interested in problems related to weighted inequalities for Hankel transform. More precisely, the main goal of this note is that from a given “input” Hankel inequality, a parametrized collection of “output” Hankel inequalities can be deduced. The idea is to exploit the close relationship of the Hankel transform to the operation of Hankel convolution and then to apply techniques from the theory of positive integral operators.A single application of the theorem will produce new weighted Hankel inequalities from known ones. However, since the output inequalities are of the same form as the input inequality, it becomes possible to “bootstrap” the production of new inequalities by using the output at one stage as the input at the next. The implications of this sort of the iteration are not examined here.Our investigation is inspired by the idea developed by Sinnamon [8], to the classical Fourier transform.Throughout the paper, we adhere to conventions that are more common in the study of positive integral operators than in harmonic analysis generally. When integrals of non negative functions are involved, we will not concern ourselves with convergence; if the integral happens to take the value+∞, then its appearance in formulas is to be interpreted according to arithmetic on [0,∞]. In particular, expressions of the form 0∞,∞/∞,0/0,00 are all taken to be 0, while ∞0=1.


## 2. Hankel Convolution Structure and Hankel Transform

In the following, we give some basic definitions and some properties of Hankel transform analogous to those of the classical Fourier transform. For more details, see [9, 10].For fixedα>-1/2, we define(2.1)dμα(x)=x2α+12αΓ(α+1)dx,jα(x)=2αΓ(α+1)x-αJα(x),
where Jα denotes the Bessel function of order α.We denote that byLαp,1≤p<∞, the space of all real-valued, measurable functions f defined on (0,∞) with norm(2.2)‖f‖p,α=(∫0∞|f(x)|pdμα(x))1/p
is finite, whereas Lα∞=L∞ that does not depend on α denotes the space of those measurable functions defined on (0,∞) for which(2.3)‖f‖∞=esssup⁡0<x<∞|f(x)|
is finite.LetΔ(x,y,z) be the area of the triangle with sides x,y,z if such a triangle exists.Set(2.4)Dα(x,y,z)=2(α-1)(Γ(α+1))2Γ(α+1/2)π1/2(xyz)-2αA(x,y,z)2α-1
if Δ exists and zero otherwise. We note that Dα(x,y,z)>0 and that Dα(x,y,z) is symmetric in x,y,z. Further we have the following basic formula:(2.5)jα(xt)jα(yt)=∫0∞Dα(x,y,z)jα(zt)dμα(z),
[11, page 411], from which it follows immediately, on setting t=0, that(2.6)∫0∞Dα(x,y,z)dμα(z)=1.
Using (2.5), we may show that(2.7)|jα(x)|≤1,∀x≥0,
see [10, page 310].For eachf in Lα1, it is clear, by (2.7), that the integral ∫0∞jα(xt)f(t)dμα(t) exists, so that we may define the Hankel transform ℋα(f) of a function f in Lα1 by(2.8)Hα(f)(x)=∫0∞jα(xt)f(t)dμα(t).
For f∈Lα1,ℋα(f) is bounded and continuous for x≥0, see [9, page 336].Proposition 2.1 (Haimo [9, page 338]).
Letf be such that  f and  ℋα(f)∈Lα1. Then
(2.9)f(x)=∫0∞Hα(f)(t)jα(xt)dμα(t)a.e.Proposition 2.2 (Trimèche [12]).
The Hankel transformℋα is an isomorphism from  𝒮*(ℝ) onto itself, and its inverse denoted  ℋα-1=ℋα, where  𝒮*(ℝ) is the space of even infinitely differentiable functions on  ℝ, rapidly decreasing together with all their derivatives equipped with its usual topology.Proposition 2.3 (Trimèche [12]).
The Hankel transform on𝒮*′(ℝ) (the space of even tempered distribution on  ℝ), defined by
(2.10)〈Hα(T),φ〉=〈T,Hα(φ)〉,T∈S*′(R),φ∈S*(R),
is an isomorphism from 𝒮*′(ℝ) onto itself, and ℋα-1=ℋα.Example 2.4.
If we takeT=δ0 where δ0 is the Dirac measure at zero, then
(2.11)Hα(δ0)=1.
Indeed, set φ∈𝒮*(ℝ)(2.12)〈Hα(δ0),φ〉=〈δ0,Hα(φ)〉=Hα(φ)(0)=∫0∞jα(0)φ(t)dμα(t).
Since jα(0)=1, then we get
(2.13)〈Hα(δ0),φ〉=∫0∞φ(t)dμα(t)=〈1,φ〉.
This proves the result.Haimo [9] and Hirschman [10] investigated a convolution operation and translation operation associated to the Hankel transformation. If f,g∈Lα1, the Hankel convolution f*αg of f and g is defined by(2.14)∀x≥0,f*αg(x)=∫0∞Tx(f)(y)g(y)dμα(y),
where the 𝒯x is the Hankel translation given by(2.15)Tx(f)(y)=∫0∞Dα(x,y,z)f(z)dμα(z),forx>0,T0(f)(y)=f(y),∀y≥0.From properties of the kernelDα(x,y,z), we deduce the following properties.(i)
(2.16)Tx(f)(y)=Ty(f)(x).(ii)
Iff∈Lαp,p≥1; then for all x≥0 the function 𝒯xf belongs to Lαp,p≥1 and we have
(2.17)‖Txf‖p,α≤‖f‖p,α.(iii)
Letf∈𝒮*(ℝ); then for all x≥0 the function 𝒯xf belongs to 𝒮*(ℝ) and we have(2.18)Hα(Txf)(y)=jα(xy)Hα(f)(y).Proposition 2.5.
(i) The Hankel convolution*α is commutative and associative. (ii) Assume that p,q,r∈[1,∞] satisfies the Young conditions 1/p+1/q=1+1/r. Then (f,g)→f*αg extends to a continuous map from Lαp×Lαq to Lαr and we have
(2.19)‖f*αg‖r,α≤‖f‖p,α‖g‖q,α.Definition 2.6.
Letf∈𝒮*(ℝ) and T∈𝒮*′(ℝ). The generalized convolution of f and T is defined by the following:
(2.20)T*αf(s)=〈T,Tsf〉,
where 𝒯s is the Hankel translation, which is given by relation (2.15).Proposition 2.7.
Letσ and  ν∈𝒮*′(ℝ); then for all  f∈𝒮*(ℝ), we have
(2.21)Hα(f*ασ)=Hα(f)⋅Hα(σ),Hα(f⋅Hα(ν))=Hα(f)*αν.Proof.
Forσ∈𝒮*′(ℝ) and f∈𝒮*(ℝ), we have that σ*αf belongs to ξ*(ℝ) (the space of even infinitely differentiable function on ℝ) and increase slowly.
Thusσ*αf∈𝒮*′(ℝ), and we have
(2.22)〈Hα(σ*αf),φ〉=〈σ*αf,Hα(φ)〉=∫0∞σ*αf(x)Hα(φ)(x)dμα(x)=∫0∞〈σy,Txf(y)〉Hα(φ)(x)dμα(x)=〈σy,∫0∞Txf(y)Hα(φ)(x)dμα(x)〉=〈σy,∫0∞Tyf(x)Hα(φ)(x)dμα(x)〉=〈σy,∫0∞Hα(Tyf)(x)φ(x)dμα(x)〉=〈σy,∫0∞jα(xy)Hαf(x)φ(x)dμα(x)〉=〈σ,Hα(φHα(f))〉=〈Hα(σ),Hα(f)φ〉=〈Hα(σ)Hα(f),φ〉.
This proves the relation (2.21) on the left.
On the other hand,(2.23)〈Hα(fHα(σ)),φ〉=〈Hα(σ),fHα(φ)〉=〈σ,Hα(fHα(φ))〉=〈σ,Hα(f)*αφ〉.
We complete the proof of the relation (2.21) on the right by the same way as the relation on the left.Remark 2.8.
Since the space𝒮*(ℝ) is a dense subset of Lα1 and it is easy to verify that f→ℋα(f*ασ),f→ℋα(f)ℋα(σ),f→ℋα(fℋα(ν)), and f→ℋα(f)*αν are all continuous maps from Lα1 to L∞, thus, the identities in (2.21) extend to be valid for all f∈Lα1.


## 3. Weighted Inequalities for Hankel Transforms

LetLα+ denote the nonnegative, extended real-valued function on the measure space ((0,∞),dμα). We say that a map T:Lα+→Lα+ has a formal adjoint T*:Lα+→Lα+ provided(3.1)∫0∞T(f)(x)g(x)dμα(x)=∫0∞f(x)T*(g)(x)dμα(x)
for all f,g∈Lα+.Let1<q≤p<∞ and a map T:Lα+→Lα+ have formal adjoint. Fix a weight u∈Lα+. For each h∈Lα+, we define(3.2)vh=h1-pT*(u(Th)q-1),C=(∫0∞hpvhdμα)1/q-1/p.
Note that by our convention, Ch=1 when p=q, even if ∫0∞(h(x))pvh(x)dμα(x)=∞.Proposition 3.1.
If0<h<∞μα, almost everywhere, and  0<T(h)<∞uμα, almost everywhere then
(3.3)(∫0∞(T(f)(x))qu(x)dμα(x))1/q≤C(∫0∞(f(x))pvh(x)dμα(x))1/p
for all f∈Lα+.This result is a special case (n=1 and r=1) of Theorem  2.1 in [7].The next result may be deduced from the last by duality argument. It is also a special case of Theorem  3.1 of [6]. Again note that C=1 when p=q.Proposition 3.2.
Suppose1<q≤p<∞,T:Lα+→Lα+ has a formal adjoint  T*, and  v,h∈Lα+ with  0<h<∞. Set
(3.4)u=h(T(v1-p′(T*h)p′-1))1-q,C=(∫0∞hq′u1-q′)1/q-1/p.
Then
(3.5)(∫0∞(T(f)(x))qu(x)dμα(x))1/q≤C(∫0∞(f(x))pv(x)dμα(x))1/p
for all f∈Lα+.Observe that ifp>q, then the formulas for the constants C given in these propositions may take useful alternative forms. In the first,(3.6)∫0∞hpvdμα=∫0∞hT*(u(Th)q-1)dμα=∫0∞(Th)u(Th)q-1=∫0∞(Th)qudμα
so(3.7)C=(∫0∞(Th)qudμα)1/q-1/p.
In the second,(3.8)∫0∞hq′u1-q′=∫0∞hT(v1-p′(T*h)p′-1)dμα=∫0∞(T*h)v1-p′(T*h)p′-1
so(3.9)C=(∫0∞(T*h)p′v1-p′dμα)1/q-1/p.Letσ be finite complex-valued Borel measure on (0,∞) and |σ| denote its absolute value. Then(3.10)∫0∞d|σ|(x)<∞.
The fact that |jα(x)|≤1 allows us to get that the Hankel transform of σ defined by(3.11)Hα(σ)(t)=∫0∞jα(xt)dσ(x)
is continuous for [0,∞[ and that(3.12)‖Hα(σ)‖∞≤|||σ|||,
where(3.13)|||σ|||=∫0∞d|σ|(x)<∞.
Furthermore, the bounded function ℋα(σ) has a Hankel transform in the distributional sense and(3.14)Hα(Hα(σ))=σ.
Moreover, if f is real measurable function on (0,∞) and σ a finite complex-valued Borel measure on (0,∞), we formally set(3.15)f*ασ(x)=∫0∞Txf(y)dσ(y),
where 𝒯x is the Hankel translation given by relation (2.15). If f∈Lα1, then f*ασ is defined in Lα1, and if f∈L∞, the integral (3.15) is defined in L∞.Forσ a finite complex-valued Borel measure on (0,∞), define the positive operator(3.16)Kσ(f)=f*α|σ|
for allf∈Lα+.Thus, iff∈Lα1∪L∞, the convolution operator f*ασ is well defined and we have(3.17)|f*ασ|≤Kσ(|f|).Proposition 3.3.
Forσ a finite complex-valued Borel measure on  (0,∞), the positive operator  Kσ defined by relation (3.16) is autoadjoint. That is,  Kσ has a formal adjoint operator  Kσ* and
(3.18)Kσ*=Kσ.Proof.
Iff,g∈Lα+, then by using properties of the kernel Dα and by applying Fubini Tonelli argument we have
(3.19)∫0∞Kσ(f)(x)g(x)dμα(x)=∫0∞(∫0∞Txf(y)d|σ|(y))g(x)dμα(x)=∫0∞(∫0∞g(x)(∫0∞Dα(x,y,z)f(z)dμα(z))dμα(x))d|σ|(y)=∫0∞f(z)(∫0∞Tzg(y)d|σ|(y))dμ(z)=∫0∞f(z)Kσg(z)dμα(z).
This completes the proof.Before introducing any technical details, we give sketch of the argument behind the main following theorem. We suppose that the following Hankel inequality:(3.20)(∫0∞(Hα(f))qudμα)1/q≤(∫0∞|f|pvdμα)1/p
for all f∈Lα1∩Lαp(v) is known to be valid for some fixed p0,q0,u0, and v0. For each appropriate function g, define f=(g*ασ)/ℋα(ν), where σ and ν are finite, complexe-valued Borel measures on [0,∞[.Using Proposition2.7, we get ℋα(g)=(ℋα(f)*αν)/ℋα(σ). For p1≥p0,q1≤q0 and arbitrary functions hσ and hν we apply Propositions 3.1 and 3.2 to give formulas for u1 and v1 so that(3.21)Lαp(v1)⟶Lαp0(v0)⟶Lαq0(u0)⟶Lαq1(u1),g⟼f⟼Hα(f)⟼Hα(g).
The arrow in the middle corresponds to the known “input” Hankel inequality, and the other arrows correspond to the Hankel convolution inequalities for the operators(3.22)g⟶(g*ασ)Hα(ν),Hα(f)⟶(Hα(f)*αν)Hα(σ).
The inequality relating ℋα(g) and g that results from this composition is just the above inequality with new indices p1 and q1 and new weights u1 and v1. This is our “output” inequality.Theorem 3.4.
SupposeC0 is a positive constant,  p0 and  q0 are indices in  (1,∞), and  u0 and  v0 are nonnegative weight functions such that the Hankel inequality
(3.23)(∫0∞(Hα(f)(x))q0u0(x)dμα(x))1/q0≤C0(∫0∞|f(x)|p0v0(x)dμα(x))1/p0
for all f∈Lα1. Let σ and ν be finite complex-valued Borel measure and hσ and hν positive functions on [0,∞[. For p1 and q1 satisfying 1<p0≤p1<∞ and 1<q1≤q0<∞, set
(3.24)wσ=|H(ν)|-p0v0,v1=hσ1-p1Kσ(wσ(Kσhσ)p0-1),wν=hν(Kν(u01-q′(Kνhν)q0′-1))1-q1,u1=|H(σ)|q1wν.
Also set
(3.25)Cσ=(∫0∞hσp1v1dμα)1/p0-1/p1,Cν=(∫0∞hνq1′wν1-q1′dμα)1/q1-1/q0.
If ℋα(ν) is bounded away from zero, then the Hankel inequality
(3.26)(∫0∞(Hα(g)(x))q1u1(x)dμα(x))1/q1≤C1(∫0∞|g(x)|p1v1(x)dμα(x))1/p1
holds for all g∈Lα1. Here C1=CνC0Cσ.Proof.
Letg∈Lα1 and set f=(g*ασ)/ℋα(ν). Since ℋα(ν) is bounded away from zero, 1/ℋα(ν)∈L∞ so f∈Lα1. Taking the Hankel transform of both sides of the equation g*ασ=fℋα(ν) and using identities (2.21) yields
(3.27)Hα(g)Hα(σ)=Hα(f)*αν.
Proposition 3.2 shows that
(3.28)(∫0∞Kν(Hα(f))q1wνdμα)1/q1≤Cν(∫0∞|Hα(f)|q0u0dμα)1/q0
and the estimate
(3.29)|Hα(g)Hα(σ)|=|Hα(f)*αν|≤Kν|Hα(f)|
gives
(3.30)(∫0∞|Hα(g)|q1u1dμα)1/q1≤Cν(∫0∞|Hα(f)|q0u0dμα)1/q0.
Proposition 3.1 shows that
(3.31)(∫0∞(Kσ|g|)p0wσdμα)1/p0≤Cσ(∫0∞|g|p1v1dμα)1/p1
and the trivial estimates
(3.32)|fHα(ν)|=|g*ασ|≤Kσ|g|
give
(3.33)(∫0∞|f|p0v0dμα)1/p0≤Cσ(∫0∞|g|p1v1dμα)1/p1.
The three inequalities (3.30), (3.23), and (3.33) combine to yield (3.26) as required. This completes the proof.When all indices are taken to 2 the theorem simplifies substantially.Corollary 3.5.
SupposeC is a positive constant and  u0 and  v0 are nonnegative weight functions such that the Hankel inequality
(3.34)∫0∞(Hα(f)(x))2u0(x)dμα(x)≤C∫0∞|f(x)|2v0(x)dμα(x)
holds for all f∈Lα1. Let σ and ν be finite complex-valued Borel measures and hσ and hν positive function on [0,∞[. Set
(3.35)v1=1hσKσ(v0(Kσhσ)|Hα(ν)|2),u1=|Hα(σ)|2hνKν(Kνhν/u0).
If ℋα(ν) is bounded away from zero, then the Hankel inequality
(3.36)∫0∞(Hα(g)(x))2u1(x)dμα(x)≤C∫0∞|g(x)|2v1(x)dμα(x)
holds for all g∈Lα1.A further simplification yields a following new weighted Hankel inequality.Corollary 3.6.
Ifσ is finite positive Borel measure and  h is a positive function on  [0,∞[then for  α>-1/2(3.37)∫0∞|Hα(g)(x)|2|Hα(σ)(x)|2dμα(x)≤C∫0∞|g(x)|2(σ*α(σ*αh))(x)h(x)dμα(x)
holds for all g∈Lα1.Proof.
It is well known that the Hankel transform is continuous fromLαp into Lαp′, where 1/p+1/p′=1,1≤p≤2. This shows that the inequality (3.33) holds with p=2 and u0=1 and v0=1. Thus if we take hν=1,hσ=h, and take ν to be the Dirac measure at zero. Then 𝒦ν=f*αδ0=f, which proves that 𝒦ν reduce to the identity. Furthermore, since ℋα(ν)=1, then we get
(3.38)u1=|Hα(σ)|2hνKν((Kνhν)/u0)=|Hα(σ)|2hνhν/u0=|Hα(σ)|2.
From the fact that Kσ is just convolution by σ and ℋα(ν)=1, it follows
(3.39)v1=1hσKσ(v0(Kσhσ)|Hα(ν)|2)=1hσKσ(v0(Kσhσ)).
Then if we take hσ=h, we get the result.If we takeσ and h as Gaussian functions, that is, σt=C1e-tx2 and hs=C2e-sx2, then we obtain the following new particular case of weighted Hardy-type inequality of Hankel transform.Corollary 3.7.
Forα>-1/2 and  s,t>0, we have
(3.40)∫0∞|Hα(g)(x)|2e-(1/2t)x2dμα(x)≤Cα,t,s∫0∞|g(x)|2e(2s2/(2s+t))x2dμα(x).Proof.
The result is obtained by using Propositions2.1 and 2.7 and the fact that
(3.41)Hα(σt)(x)=C1e-x2/4t(2t)α+1,Hα(hs)(x)=C2e-x2/4s(2s)α+1.



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