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Journal of Function Spaces and Applications
Volume 2012, Article ID 315086, 12 pages
http://dx.doi.org/10.1155/2012/315086
Research Article

Bootstrapping Weighted Inequalities for Hankel Transform

Department of Mathematics, Faculty of Sciences of Tunis, Tunisia

Received 30 October 2010; Accepted 12 April 2011

Academic Editor: Lars-Erik Persson

Copyright © 2012 M. Dziri. 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.

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.

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, [17].

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 fixed 𝛼>1/2, we define𝑑𝜇𝛼𝑥(𝑥)=2𝛼+12𝛼𝑗Γ(𝛼+1)𝑑𝑥,𝛼(𝑥)=2𝛼Γ(𝛼+1)𝑥𝛼𝐽𝛼(𝑥),(2.1) where 𝐽𝛼 denotes the Bessel function of order 𝛼.

We denote that by 𝐿𝑝𝛼,1𝑝<, the space of all real-valued, measurable functions 𝑓 defined on (0,) with norm𝑓𝑝,𝛼=0||||𝑓(𝑥)𝑝𝑑𝜇𝛼(𝑥)1/𝑝(2.2) is finite, whereas 𝐿𝛼=𝐿 that does not depend on 𝛼 denotes the space of those measurable functions defined on (0,) for which𝑓=esssup0<𝑥<||||𝑓(𝑥)(2.3) is finite.

Let Δ(𝑥,𝑦,𝑧) be the area of the triangle with sides 𝑥,𝑦,𝑧 if such a triangle exists.

Set𝐷𝛼2(𝑥,𝑦,𝑧)=(𝛼1)(Γ(𝛼+1))2Γ(𝛼+1/2)𝜋1/2(𝑥𝑦𝑧)2𝛼𝐴(𝑥,𝑦,𝑧)2𝛼1(2.4) 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:𝑗𝛼(𝑥𝑡)𝑗𝛼(𝑦𝑡)=0𝐷𝛼(𝑥,𝑦,𝑧)𝑗𝛼(𝑧𝑡)𝑑𝜇𝛼(𝑧),(2.5) [11, page 411], from which it follows immediately, on setting 𝑡=0, that0𝐷𝛼(𝑥,𝑦,𝑧)𝑑𝜇𝛼(𝑧)=1.(2.6) Using (2.5), we may show that||𝑗𝛼||(𝑥)1,𝑥0,(2.7) see [10, page 310].

For each 𝑓 in 𝐿1𝛼, it is clear, by (2.7), that the integral 0𝑗𝛼(𝑥𝑡)𝑓(𝑡)𝑑𝜇𝛼(𝑡) exists, so that we may define the Hankel transform 𝛼(𝑓) of a function 𝑓 in 𝐿1𝛼 by𝛼(𝑓)(𝑥)=0𝑗𝛼(𝑥𝑡)𝑓(𝑡)𝑑𝜇𝛼(𝑡).(2.8) For 𝑓𝐿1𝛼,𝛼(𝑓) is bounded and continuous for 𝑥0, see [9, page 336].

Proposition 2.1 (Haimo [9, page 338]). Let 𝑓 be such that 𝑓 and 𝛼(𝑓)𝐿1𝛼. Then 𝑓(𝑥)=0𝛼(𝑓)(𝑡)𝑗𝛼(𝑥𝑡)𝑑𝜇𝛼(𝑡)𝑎.𝑒.(2.9)

Proposition 2.2 (Trimèche [12]). The Hankel transform 𝛼 is an isomorphism from 𝒮() onto itself, and its inverse denoted 𝛼1=𝛼, where 𝒮() is the space of even infinitely differentiable functions on , rapidly decreasing together with all their derivatives equipped with its usual topology.

Proposition 2.3 (Trimèche [12]). The Hankel transform on 𝒮() (the space of even tempered distribution on ), defined by 𝛼(𝑇),𝜑=𝑇,𝛼(𝜑),𝑇𝒮(),𝜑𝒮(),(2.10) is an isomorphism from 𝒮() onto itself, and 𝛼1=𝛼.

Example 2.4. If we take 𝑇=𝛿0 where 𝛿0 is the Dirac measure at zero, then 𝛼𝛿0=1.(2.11) Indeed, set 𝜑𝒮()𝛼𝛿0,𝜑=𝛿0,𝛼(𝜑)=𝛼(𝜑)(0)=0𝑗𝛼(0)𝜑(𝑡)𝑑𝜇𝛼(𝑡).(2.12) Since 𝑗𝛼(0)=1, then we get 𝛼𝛿0=,𝜑0𝜑(𝑡)𝑑𝜇𝛼(𝑡)=1,𝜑.(2.13) This proves the result.

Haimo [9] and Hirschman [10] investigated a convolution operation and translation operation associated to the Hankel transformation. If 𝑓,𝑔𝐿1𝛼, the Hankel convolution 𝑓𝛼𝑔 of 𝑓 and 𝑔 is defined by𝑥0,𝑓𝛼𝑔(𝑥)=0𝒯𝑥(𝑓)(𝑦)𝑔(𝑦)𝑑𝜇𝛼(𝑦),(2.14) where the 𝒯𝑥 is the Hankel translation given by𝒯𝑥(𝑓)(𝑦)=0𝐷𝛼(𝑥,𝑦,𝑧)𝑓(𝑧)𝑑𝜇𝛼𝒯(𝑧),for𝑥>0,0(𝑓)(𝑦)=𝑓(𝑦),𝑦0.(2.15)

From properties of the kernel 𝐷𝛼(𝑥,𝑦,𝑧), we deduce the following properties.(i)𝒯𝑥(𝑓)(𝑦)=𝒯𝑦(𝑓)(𝑥).(2.16)(ii)If 𝑓𝐿𝑝𝛼,𝑝1; then for all 𝑥0 the function 𝒯𝑥𝑓 belongs to 𝐿𝑝𝛼,𝑝1 and we have 𝒯𝑥𝑓𝑝,𝛼𝑓𝑝,𝛼.(2.17)(iii) Let 𝑓𝒮(); then for all 𝑥0 the function 𝒯𝑥𝑓 belongs to 𝒮() and we have𝛼𝒯𝑥𝑓(𝑦)=𝑗𝛼(𝑥𝑦)𝛼(𝑓)(𝑦).(2.18)

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)𝑓𝛼𝑔 extends to a continuous map from 𝐿𝑝𝛼×𝐿𝑞𝛼 to 𝐿𝑟𝛼 and we have 𝑓𝛼𝑔𝑟,𝛼𝑓𝑝,𝛼𝑔𝑞,𝛼.(2.19)

Definition 2.6. Let 𝑓𝒮() and 𝑇𝒮(). The generalized convolution of 𝑓 and 𝑇 is defined by the following: 𝑇𝛼𝑓(𝑠)=𝑇,𝒯𝑠𝑓,(2.20) where 𝒯𝑠 is the Hankel translation, which is given by relation (2.15).

Proposition 2.7. Let 𝜎 and 𝜈𝒮(); then for all 𝑓𝒮(), we have 𝛼𝑓𝛼𝜎=𝛼(𝑓)𝛼(𝜎),𝛼𝑓𝛼(𝜈)=𝛼(𝑓)𝛼𝜈.(2.21)

Proof. For 𝜎𝒮() and 𝑓𝒮(), we have that 𝜎𝛼𝑓 belongs to 𝜉() (the space of even infinitely differentiable function on ) and increase slowly.
Thus 𝜎𝛼𝑓𝒮(), and we have 𝛼𝜎𝛼𝑓,𝜑=𝜎𝛼𝑓,𝛼(𝜑)=0𝜎𝛼𝑓(𝑥)𝛼(𝜑)(𝑥)𝑑𝜇𝛼=(𝑥)0𝜎𝑦,𝒯𝑥𝑓(𝑦)𝛼(𝜑)(𝑥)𝑑𝜇𝛼=𝜎(𝑥)𝑦,0𝒯𝑥𝑓(𝑦)𝛼(𝜑)(𝑥)𝑑𝜇𝛼=𝜎(𝑥)𝑦,0𝒯𝑦𝑓(𝑥)𝛼(𝜑)(𝑥)𝑑𝜇𝛼=𝜎(𝑥)𝑦,0𝛼𝒯𝑦𝑓(𝑥)𝜑(𝑥)𝑑𝜇𝛼=𝜎(𝑥)𝑦,0𝑗𝛼(𝑥𝑦)𝛼𝑓(𝑥)𝜑(𝑥)𝑑𝜇𝛼=(𝑥)𝜎,𝛼𝜑𝛼(𝑓)=𝛼(𝜎),𝛼(𝑓)𝜑=𝛼(𝜎)𝛼(𝑓),𝜑.(2.22) This proves the relation (2.21) on the left.
On the other hand, 𝛼𝑓𝛼(𝜎),𝜑=𝛼(𝜎),𝑓𝛼(𝜑)=𝜎,𝛼𝑓𝛼(𝜑)=𝜎,𝛼(𝑓)𝛼𝜑.(2.23) 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 space 𝒮() is a dense subset of 𝐿1𝛼 and it is easy to verify that 𝑓𝛼(𝑓𝛼𝜎),𝑓𝛼(𝑓)𝛼(𝜎),𝑓𝛼(𝑓𝛼(𝜈)), and 𝑓𝛼(𝑓)𝛼𝜈 are all continuous maps from 𝐿1𝛼 to 𝐿, thus, the identities in (2.21) extend to be valid for all 𝑓𝐿1𝛼.

3. Weighted Inequalities for Hankel Transforms

Let 𝐿+𝛼 denote the nonnegative, extended real-valued function on the measure space ((0,),𝑑𝜇𝛼). We say that a map 𝑇𝐿+𝛼𝐿+𝛼 has a formal adjoint 𝑇𝐿+𝛼𝐿+𝛼 provided0𝑇(𝑓)(𝑥)𝑔(𝑥)𝑑𝜇𝛼(𝑥)=0𝑓(𝑥)𝑇(𝑔)(𝑥)𝑑𝜇𝛼(𝑥)(3.1) for all 𝑓,𝑔𝐿+𝛼.

Let 1<𝑞𝑝< and a map 𝑇𝐿+𝛼𝐿+𝛼 have formal adjoint. Fix a weight 𝑢𝐿+𝛼. For each 𝐿+𝛼, we define𝑣=1𝑝𝑇𝑢(𝑇)𝑞1,𝐶=0𝑝𝑣𝑑𝜇𝛼1/𝑞1/𝑝.(3.2) Note that by our convention, 𝐶=1 when 𝑝=𝑞, even if 0((𝑥))𝑝𝑣(𝑥)𝑑𝜇𝛼(𝑥)=.

Proposition 3.1. If 0<<𝜇𝛼, almost everywhere, and 0<𝑇()<𝑢𝜇𝛼, almost everywhere then 0(𝑇(𝑓)(𝑥))𝑞𝑢(𝑥)𝑑𝜇𝛼(𝑥)1/𝑞𝐶0(𝑓(𝑥))𝑝𝑣(𝑥)𝑑𝜇𝛼(𝑥)1/𝑝(3.3) for all 𝑓𝐿+𝛼.

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. Suppose 1<𝑞𝑝<,𝑇𝐿+𝛼𝐿+𝛼 has a formal adjoint 𝑇, and 𝑣,𝐿+𝛼 with 0<<. Set 𝑇𝑣𝑢=1𝑝𝑇𝑝11𝑞,𝐶=0𝑞𝑢1𝑞1/𝑞1/𝑝.(3.4) Then 0(𝑇(𝑓)(𝑥))𝑞𝑢(𝑥)𝑑𝜇𝛼(𝑥)1/𝑞𝐶0(𝑓(𝑥))𝑝𝑣(𝑥)𝑑𝜇𝛼(𝑥)1/𝑝(3.5) for all 𝑓𝐿+𝛼.

Observe that if 𝑝>𝑞, then the formulas for the constants 𝐶 given in these propositions may take useful alternative forms. In the first,0𝑝𝑣𝑑𝜇𝛼=0𝑇𝑢(𝑇)𝑞1𝑑𝜇𝛼=0(𝑇)𝑢(𝑇)𝑞1=0(𝑇)𝑞𝑢𝑑𝜇𝛼(3.6) so𝐶=0(𝑇)𝑞𝑢𝑑𝜇𝛼1/𝑞1/𝑝.(3.7) In the second,0𝑞𝑢1𝑞=0𝑣𝑇1𝑝𝑇𝑝1𝑑𝜇𝛼=0𝑇𝑣1𝑝𝑇𝑝1(3.8) so𝐶=0𝑇𝑝𝑣1𝑝𝑑𝜇𝛼1/𝑞1/𝑝.(3.9)

Let 𝜎 be finite complex-valued Borel measure on (0,) and |𝜎| denote its absolute value. Then0𝑑|𝜎|(𝑥)<.(3.10) The fact that |𝑗𝛼(𝑥)|1 allows us to get that the Hankel transform of 𝜎 defined by𝛼(𝜎)(𝑡)=0𝑗𝛼(𝑥𝑡)𝑑𝜎(𝑥)(3.11) is continuous for [0,[ and that𝛼(𝜎)|||||||||𝜎|,(3.12) where||||||||=|𝜎|0𝑑|𝜎|(𝑥)<.(3.13) Furthermore, the bounded function 𝛼(𝜎) has a Hankel transform in the distributional sense and𝛼𝛼(𝜎)=𝜎.(3.14) Moreover, if 𝑓 is real measurable function on (0,) and 𝜎 a finite complex-valued Borel measure on (0,), we formally set𝑓𝛼𝜎(𝑥)=0𝒯𝑥𝑓(𝑦)𝑑𝜎(𝑦),(3.15) where 𝒯𝑥 is the Hankel translation given by relation (2.15). If 𝑓𝐿1𝛼, then 𝑓𝛼𝜎 is defined in 𝐿1𝛼, and if 𝑓𝐿, the integral (3.15) is defined in 𝐿.

For 𝜎 a finite complex-valued Borel measure on (0,), define the positive operator𝐾𝜎(𝑓)=𝑓𝛼|𝜎|(3.16) for all𝑓𝐿+𝛼.

Thus, if 𝑓𝐿1𝛼𝐿, the convolution operator 𝑓𝛼𝜎 is well defined and we have||𝑓𝛼𝜎||𝐾𝜎||𝑓||.(3.17)

Proposition 3.3. For 𝜎 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 𝐾𝜎=𝐾𝜎.(3.18)

Proof. If 𝑓,𝑔𝐿+𝛼, then by using properties of the kernel 𝐷𝛼 and by applying Fubini Tonelli argument we have 0𝐾𝜎(𝑓)(𝑥)𝑔(𝑥)𝑑𝜇𝛼(𝑥)=00𝒯𝑥𝑓(𝑦)𝑑|𝜎|(𝑦)𝑔(𝑥)𝑑𝜇𝛼=(𝑥)00𝑔(𝑥)0𝐷𝛼(𝑥,𝑦,𝑧)𝑓(𝑧)𝑑𝜇𝛼(𝑧)𝑑𝜇𝛼=(𝑥)𝑑|𝜎|(𝑦)0𝑓(𝑧)0𝒯𝑧𝑔(𝑦)𝑑|𝜎|(𝑦)𝑑𝜇(𝑧)=0𝑓(𝑧)𝒦𝜎𝑔(𝑧)𝑑𝜇𝛼(𝑧).(3.19) 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:0𝛼(𝑓)𝑞𝑢𝑑𝜇𝛼1/𝑞0||𝑓||𝑝𝑣𝑑𝜇𝛼1/𝑝(3.20) for all 𝑓𝐿1𝛼𝐿𝑝𝛼(𝑣) is known to be valid for some fixed 𝑝0,𝑞0,𝑢0, and 𝑣0. For each appropriate function 𝑔, define 𝑓=(𝑔𝛼𝜎)/𝛼(𝜈), where 𝜎 and 𝜈 are finite, complexe-valued Borel measures on [0,[.

Using Proposition 2.7, we get 𝛼(𝑔)=(𝛼(𝑓)𝛼𝜈)/𝛼(𝜎). For 𝑝1𝑝0,𝑞1𝑞0 and arbitrary functions 𝜎 and 𝜈 we apply Propositions 3.1 and 3.2 to give formulas for 𝑢1 and 𝑣1 so that𝐿𝑝𝛼𝑣1𝐿𝑝0𝛼𝑣0𝐿𝑞0𝛼𝑢0𝐿𝑞1𝛼𝑢1,𝑔𝑓𝛼(𝑓)𝛼(𝑔).(3.21) The arrow in the middle corresponds to the known “input” Hankel inequality, and the other arrows correspond to the Hankel convolution inequalities for the operators𝑔𝑔𝛼𝜎𝛼(𝜈),𝛼(𝑓)𝛼(𝑓)𝛼𝜈𝛼(𝜎).(3.22) The inequality relating 𝛼(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. Suppose 𝐶0 is a positive constant, 𝑝0 and 𝑞0 are indices in (1,), and 𝑢0 and 𝑣0 are nonnegative weight functions such that the Hankel inequality 0𝛼(𝑓)(𝑥)𝑞0𝑢0(𝑥)𝑑𝜇𝛼(𝑥)1/𝑞0𝐶00||||𝑓(𝑥)𝑝0𝑣0(𝑥)𝑑𝜇𝛼(𝑥)1/𝑝0(3.23) for all 𝑓𝐿1𝛼. Let 𝜎 and 𝜈 be finite complex-valued Borel measure and 𝜎 and 𝜈 positive functions on [0,[. For 𝑝1 and 𝑞1 satisfying 1<𝑝0𝑝1< and 1<𝑞1𝑞0<, set 𝑤𝜎=||||(𝜈)𝑝0𝑣0,𝑣1=1𝑝1𝜎𝐾𝜎𝑤𝜎𝐾𝜎𝜎𝑝01,𝑤𝜈=𝜈𝐾𝜈𝑢1𝑞0𝐾𝜈𝜈𝑞011𝑞1,𝑢1=||||(𝜎)𝑞1𝑤𝜈.(3.24) Also set 𝐶𝜎=0𝑝1𝜎𝑣1𝑑𝜇𝛼1/𝑝01/𝑝1,𝐶𝜈=0𝑞1𝜈𝑤1𝑞1𝜈𝑑𝜇𝛼1/𝑞11/𝑞0.(3.25) If 𝛼(𝜈) is bounded away from zero, then the Hankel inequality 0𝛼(𝑔)(𝑥)𝑞1𝑢1(𝑥)𝑑𝜇𝛼(𝑥)1/𝑞1𝐶10||||𝑔(𝑥)𝑝1𝑣1(𝑥)𝑑𝜇𝛼(𝑥)1/𝑝1(3.26) holds for all 𝑔𝐿1𝛼. Here 𝐶1=𝐶𝜈𝐶0𝐶𝜎.

Proof. Let 𝑔𝐿1𝛼 and set 𝑓=(𝑔𝛼𝜎)/𝛼(𝜈). Since 𝛼(𝜈) is bounded away from zero, 1/𝛼(𝜈)𝐿 so 𝑓𝐿1𝛼. Taking the Hankel transform of both sides of the equation 𝑔𝛼𝜎=𝑓𝛼(𝜈) and using identities (2.21) yields 𝛼(𝑔)𝛼(𝜎)=𝛼(𝑓)𝛼𝜈.(3.27) Proposition 3.2 shows that 0𝒦𝜈𝛼(𝑓)𝑞1𝑤𝜈𝑑𝜇𝛼1/𝑞1𝐶𝜈0||𝛼||(𝑓)𝑞0𝑢0𝑑𝜇𝛼1/𝑞0(3.28) and the estimate ||𝛼(𝑔)𝛼||=||(𝜎)𝛼(𝑓)𝛼𝜈||𝒦𝜈||𝛼||(𝑓)(3.29) gives 0||𝛼||(𝑔)𝑞1𝑢1𝑑𝜇𝛼1/𝑞1𝐶𝜈0||𝛼||(𝑓)𝑞0𝑢0𝑑𝜇𝛼1/𝑞0.(3.30) Proposition 3.1 shows that 0𝒦𝜎||𝑔||𝑝0𝑤𝜎𝑑𝜇𝛼1/𝑝0𝐶𝜎0||𝑔||𝑝1𝑣1𝑑𝜇𝛼1/𝑝1(3.31) and the trivial estimates ||𝑓𝛼||=||(𝜈)𝑔𝛼𝜎||𝒦𝜎||𝑔||(3.32) give 0||𝑓||𝑝0𝑣0𝑑𝜇𝛼1/𝑝0𝐶𝜎0||𝑔||𝑝1𝑣1𝑑𝜇𝛼1/𝑝1.(3.33) 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. Suppose 𝐶 is a positive constant and 𝑢0 and 𝑣0 are nonnegative weight functions such that the Hankel inequality 0𝛼(𝑓)(𝑥)2𝑢0(𝑥)𝑑𝜇𝛼(𝑥)𝐶0||||𝑓(𝑥)2𝑣0(𝑥)𝑑𝜇𝛼(𝑥)(3.34) holds for all 𝑓𝐿1𝛼. Let 𝜎 and 𝜈 be finite complex-valued Borel measures and 𝜎 and 𝜈 positive function on [0,[. Set 𝑣1=1𝜎𝐾𝜎𝑣0𝐾𝜎𝜎||𝛼||(𝜈)2,𝑢1=||𝛼||(𝜎)2𝜈𝐾𝜈𝐾𝜈𝜈/𝑢0.(3.35) If 𝛼(𝜈) is bounded away from zero, then the Hankel inequality 0𝛼(𝑔)(𝑥)2𝑢1(𝑥)𝑑𝜇𝛼(𝑥)𝐶0||||𝑔(𝑥)2𝑣1(𝑥)𝑑𝜇𝛼(𝑥)(3.36) holds for all 𝑔𝐿1𝛼.

A further simplification yields a following new weighted Hankel inequality.

Corollary 3.6. If 𝜎 is finite positive Borel measure and is a positive function on [0,[then for 𝛼>1/20||𝛼||(𝑔)(𝑥)2||𝛼||(𝜎)(𝑥)2𝑑𝜇𝛼(𝑥)𝐶0||||𝑔(𝑥)2𝜎𝛼𝜎𝛼(𝑥)(𝑥)𝑑𝜇𝛼(𝑥)(3.37) holds for all 𝑔𝐿1𝛼.

Proof. It is well known that the Hankel transform is continuous from 𝐿𝑝𝛼 into 𝐿𝛼𝑝, where 1/𝑝+1/𝑝=1,1𝑝2. This shows that the inequality (3.33) holds with 𝑝=2 and 𝑢0=1 and 𝑣0=1. Thus if we take 𝜈=1,𝜎=, and take 𝜈 to be the Dirac measure at zero. Then 𝒦𝜈=𝑓𝛼𝛿0=𝑓, which proves that 𝒦𝜈 reduce to the identity. Furthermore, since 𝛼(𝜈)=1, then we get 𝑢1=||𝛼||(𝜎)2𝜈𝐾𝜈𝐾𝜈𝜈/𝑢0=||𝛼||(𝜎)2𝜈𝜈/𝑢0=||𝛼||(𝜎)2.(3.38) From the fact that 𝐾𝜎 is just convolution by 𝜎 and 𝛼(𝜈)=1, it follows 𝑣1=1𝜎𝐾𝜎𝑣0𝐾𝜎𝜎||𝛼||(𝜈)2=1𝜎𝒦𝜎𝑣0𝒦𝜎𝜎.(3.39) Then if we take 𝜎=, we get the result.

If we take 𝜎 and as Gaussian functions, that is, 𝜎𝑡=𝐶1𝑒𝑡𝑥2 and 𝑠=𝐶2𝑒𝑠𝑥2, then we obtain the following new particular case of weighted Hardy-type inequality of Hankel transform.

Corollary 3.7. For 𝛼>1/2 and 𝑠,𝑡>0, we have 0||𝛼||(𝑔)(𝑥)2𝑒(1/2𝑡)𝑥2𝑑𝜇𝛼(𝑥)𝐶𝛼,𝑡,𝑠0||||𝑔(𝑥)2𝑒(2𝑠2/(2𝑠+𝑡))𝑥2𝑑𝜇𝛼(𝑥).(3.40)

Proof. The result is obtained by using Propositions 2.1 and 2.7 and the fact that 𝛼𝜎𝑡(𝑥)=𝐶1𝑒𝑥2/4𝑡(2𝑡)𝛼+1,𝛼𝑠(𝑥)=𝐶2𝑒𝑥2/4𝑠(2𝑠)𝛼+1.(3.41)

Acknowledgment

The author is very grateful to Professor Gord Sinnamon for his help and critical comments.

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