Abstract

We define and study Sobolev-type spaces associated with singular second-order differential operator on . Some properties are given; in particular we establish a compactness-type imbedding result which allows a Reillich-type theorem. Next, we introduce a generalized Weierstrass transform and, using the theory of reproducing kernels, some applications are given.

1. Introduction

The Sobolev spaces have served as a very useful tool in the theory of partial differential equations, mostly those related to continuum mechanics or physics. Their uses and the study of their properties were facilitated by the theory of distributions and Fourier analysis. The Sobolev space is defined by the use of the classical Fourier transform as the set of all tempered distribution such that its classical Fourier transform satisfying Generalization of the Sobolev space has been studied by replacing the classical Fourier transform by a generalized one.

In this paper we consider the differential operator on , where is the Chebli-Trimeche function (cf. [1, Section 3.5]) defined on and satisfies the following conditions. (i)There exists a positive even infinitely differentiable function on , with , , such that , .(ii) is increasing on and .(iii) is decreasing on , and .(iv)There exists a constant , such that for all , we have where is on , bounded together with its derivatives.

For , , and , we regain the Bessel operator

For , , , and , we regain the Jacobi operator The purpose of this paper is to introduce and study new Sobolev-type spaces , associated with the singular operator that generalizes the corresponding classical spaces. The Bessel case was treated by Assal and Nessibi [2], while Ben Salem and Dachraoui [3] studied the generalized Soblev spaces in the Jacobi setting theory.

The paper is organized as follows. In Section 2 we recall the main results about the harmonic analysis associated with the operator . In Section 3 Sobolev-type spaces on the dual of the Chébli-Trimèche hypergroup are studied. Some properties including completeness and Sobolev embedding theorems are established. Next, we prove a Reillich-type theorem. Finally, in Section 4, as applications, we give practical real inversion formulas using the theory of reproducing kernels for the generalized Weierstrass transform.

2. Preliminaries

In this section, we collect some harmonic analysis results related to the operator . For details, we refer the reader to [1, 48].

2.1. Eigenfunctions of the Operator

In the following, we denote by the space of even -functions on , the subspace of , consisting of functions rapidly decreasing together with their derivatives,, where is the eigenfunction of the operator associated with the value , the dual topological space of , the dual topological space of , the dual topological space of , the space of even entire functions on which are of exponential type and slowly increasing, the subspace of satisfying We have .

For every , let us denote by the unique solution of the eigenvalue problem:

Remark 1. This function satisfies the following properties. (i), the function is analytic on .(ii) Product formula: where is a measurable positive function on , with support in .(iii)(iv) For , we have where is a positive constant.(v) For , we have (vi) We have the following integral representation of Mehler type, where is an even positive function on with support in .

2.2. Generalized Fourier Transform

For a Borel positive measure on , and , we write for the Lebesgue space equipped with the norm defined by and . When , with a nonnegative function on , we replace the in the norms by .

For , the generalized Fourier transform is defined by

The inverse generalized Fourier transform of a suitable function on is given by where is the spectral measure given by

Remark 2. The function satisfies the following properties. (i)For , we have .(ii)The function is continuous on .(iii)There exist positive constants , , and , such that
If , , ;
If , , ;
If , ;

Proposition 3 (see [7, 9]). (i) The generalized transform and its inverse are topological isomorphisms between the generalized Schwartz space and the Schwartz space .
(ii) The transform is a topological isomorphism from onto . Moreover, for all , we have if and only if .

Next, we give some properties of this transform.(i)For in we have (ii)For in we have

Proposition 4 (see [7, 9]). Plancherel formula for . For all in , we have (ii) Plancherel Theorem. The transform extends uniquely to an isomorphism from onto .

Remark 5. We have for all , but for all .

Proposition 6. Let . The Fourier transform (resp. ) can be extended as a continuous mapping from onto (resp. from onto ) and we have with .

2.3. Generalized Convolution

Definition 7 (see [10]). The translation operator associated with the operator is defined on , by where is the function defined in the relation (9).

Proposition 8 (see [10]). For a suitable function on , we have(i),(ii),(iii),(iv),(v),(vi).

Definition 9 (see [10]). For suitable functions and , we define the convolution product by

Remark 10. It is clear that this convolution product is both commutative and associative:(i). (ii).

Proposition 11 (see [10]). (i) Assume that , satisfy . Then, for every and , we have , and
(ii) If and , then
(iii) If and such that , then where is the conjugate exponent of .
(iv) If and such that , then

Proposition 12. For and , with we have

Proposition 13. Let . Then if and only if belongs to , and in this case we have

Definition 14. The generalized Fourier transform of a distribution in is defined by

Proposition 15. The generalized Fourier transform is a topological isomorphism from onto .

Let be in . We define the distribution , by This distribution satisfy the following property:

3. Sobolev-Type Spaces on the Dual of the Chébli-Trimèche Hypergroup

Definition 16. Let and . We define the Sobolev-type spaces as the set of tempered distributions such that

We provide the space with the norm:

In the sequel, we will give some properties of the space .

Proposition 17. Let . The space is dense in , for

Proof. Firstly, we want to prove that the space is a subset of . Indeed, let . By Proposition 3(i) the function . Thus, using Remark 5, we deduce the claim. Now, we prove the density. Let . Then, from the density of in , we deduce the existence of a sequence in such that On the other hand, according to Proposition 3(i), for all , the function is in and we have Therefore, the result follows by combining (38) and (39).

Proposition 18. (i) Let and let and in such that then
(ii) Let , and let , and be three real numbers: . Then, for all , there exists a nonnegative constant such that for all in

Proof. (i) is clear.
(ii) We consider , with . Moreover it is easy to see Thus, Hence, the proof is completed for .

Proposition 19. (i) Let if and in , if . The space provided with the norm is a Banach space.
(ii) Let and . Then, the operator defined, on , by is an isometric isomorphism from onto . Moreover, for all , , and for all , the function belongs to the space , with .

Proof. (i) Let be a Cauchy sequence in . Then is a Cauchy sequence in . But is complete, so, there exists a function such that and But since when and when , then and consequently . This implies that and from relation (46), we get This achieves the proof of (i).
(ii) Let . By remarking that is an isomorphism from onto and using the fact that we deduce the first part of (ii). Now, let , . Then, for all , the function belongs to the space . Therefore, we obtain the second part of (ii) by using inequality (23).

In the following, we prove a Hardy-Littlewood-Paley type inequality for the transform .

Proposition 20. (1) Let . Then, for and , there exists a positive constant such that for all (2) (i) For , , , and , we have for all
(ii) For , , , and , we have (51) for all .

We start with the following lemma deduced from the hypothesis of the function .

Lemma 21. (i) For any real , there exist positive constants and such that for all ,
(ii) For ,
(iii) For ,

Proof of Proposition 20. (1) Let and . Clearly, the operator defined on , , by is of strong type between the spaces and . Therefore, according to the Marcinkiewicz theorem (cf. [11]), to obtain the result, it suffices to show that is of weak type between the spaces under consideration. Indeed, using assertions (i) and (iii) of Lemma 21 and inequality (23), we obtain for all and and the desired result follows.
(2) Let , , and . According to Proposition 3(ii), for all , it follows that .(i) If and , one can easily see by using Holder inequality, Lemma 21(i), and inequality (23) that for (ii) If , then by virtue of Plancherel Theorem for the transform , we deduce that for all , This completes the proof of the proposition.

Proposition 22. (1) Let . Then for , , and , and we have
(2) (i) Let . Then for , , and , we have for all
(ii) For . Then for , , and , we have for all ,

Proof. (1) The result follows from Proposition 20(1) and the fact that, for all ,
(2) (i) If and using the fact that, for all , it follows, from Holder inequality, that for all and Thus, we deduce the result using Proposition 20(2) and inequality (23).  (ii)By virtue of (64), we obtain the result, for , from Plancherel Theorem and Proposition 20(2).

Proposition 23. Let be a non negative real number. Then, we have (i) For , ,  ; . (ii) If , , ;  , where    is the space of even functions with class    on  .

Proof. (i) Let be in with . It is clear that belongs to .
Thus, from (14) and Proposition 4(ii), we have We identify with the second member, then we deduce that belongs to and the injection of into is continuous.
Now, let be in with such that with . From (12), for all , and such that , we have Using the same method as for and the derivation theorem under the integral sign, we deduce that Then, for all such that belongs to Thus, is in and the injection of into is continuous.
(ii) If and in ; then using assertions (i) and (iii) of Lemma 21 and Holder inequality, we deduce that for all , the function belongs to .
Therefore, (ii) follows from (i).

Proposition 24. For and , the space is separable.

Proof. Let . It is well known that is separable. More precisely, the set is countable and dense in . Thus, for all , , there exists a sequence in such that On the other hand, for all , , and so . Therefore, for all , there exists such that . Hence, from (70), we obtain This implies that is countable and dense in and the proposition is proved.

3.1. Reillich-Type Theorem

In this subsection, using Hahn Banach’s and Riesz’s theorems [12, 13], we describe the dual space of . We prove also that a compact imbedding theorem and a Reillich-type theorem are established. We need firstly the following lemmas.

Lemma 25. Let and . For all positive continuous function , we have

Proof. The result follows by using the following classical Peetre’s inequality: and the fact that the kernel is positive with support in .

Lemma 26. For all and , we have where, for in and in , the function is the generalized convolution product of the distribution and the function defined by

Proof. For all and , the function belongs to (cf. [4]) and we have Therefore, the result follows by using the fact that is an isomorphism from (resp. ) onto (resp. ).

Theorem 27. Let , , and .(i)If , then the mapping is continuous.(ii) If . We have the same result as in (i) if .(iii) If and , then the mapping is continuous.(iv)Let , , and . Then, the mapping is continuous.

Proof. (1)  (i) According to Lemmas 25 and 26, we have, for and , On the other hand, from the hypothesis on and , there exists Thus, using Proposition 11(i), we obtain the result.
(ii) As , it is easy to see that for . Moreover, we proceed as above and using Proposition 11(i), we obtain for Hence, using the fact that the imbedding is continuous, the desired result follows.(iii)Let . Then . Therefore, from (80) and using Proposition 11(iii), we deduce the result.(iv)Using (23) and Holder’s inequality, we obtain, for all and , where and are,respectively, the conjugates of and . Therefore, we deduce the result by remarking that the embedding is continuous. This achieves the proof of theorem.

Now, we shall characterize the dual space of .

Theorem 28. Let when and when . The dual space of can be identified with , where is the conjugate of .

Proof. Let when and when . Then, for , we have, for all , This proves, from the density of in , that admits a unique continuous extension to .
Conversely, suppose that . Then the mapping is continuous, where is the isometric isomorphism from into , defined by with inverse Thus, and using the fact that , there exists such that Hence, from the density of in and using Riez’s theorem, we deduce that belongs to . This completes the proof of Theorem 28.

Proposition 29. Let when and when . Let be in and . Then for all such that if and if , the mapping is compact.

Proof. Let be a sequence in such that , for all . Then, from Theorem 28, we deduce that can be regarded as a sequence in with , and using Holder inequality, we obtain for all Therefore, by virtue of Propositions 24 and 19(i), is a separable Banach space. Which implies, from Alaoglu theorem (cf. [14]), that there exists a subsequence weakly converging in . We denote by its weak limit. Using Lemma 26, for all , we have But is weakly converging to zero in , then it follows that for all .
Now, according to Theorem 27, for all , and using (82) we get Hence, by Lebesgue’s theorem, we deduce that for all , On the other hand, for with , , and , it follows from Theorem 27(i) that for and satisfying , we have And for with , , and , we obtain from Theorem 27(iv) Which implies that these last integrals, (95) and (96), tend to zero when tends to uniformly with respect to and so, by virtue of (94), we conclude that is strongly converging in ; that is, . This achieves the proof of Proposition 29.

Notation. Let be a compact contained in . We denote by the subspace of defined by As a consequence of Proposition 29, with supported on a bounded set containing the compact and satisfying on , we deduce the main result of this section.

Theorem 30 (Reillich-type theorem). Let when and when . Let be a compact contained in and let . Then for all such that if and if , the canonical imbedding is compact.

Using Reillich-type theorem, we prove the following inequalities.

Corollary 31. Let be a compact contained in and . Then, for when and when , there exists such that, for all , we have where .

Proof. It is clear that, for all , and we have Now, let us prove the left hand side inequality. Suppose that for all positive integer , there exists such that Without loss of generality, one can suppose that . Then we have and, by using Reillich-type theorem, we deduce that there exists a subsequence of strongly converging in , . Let be its strong limit. Therefore, by (101) and the fact that for it follows that . This implies that and so . On the other hand, there exists a positive constant such that Hence, by tending to infinity and from (101), it follows that which is impossible. Thus, the required inequality is satisfied. Combining the left hand side inequality of (98) and (99), we obtain This completes the proof.

4. Applications

4.1. Weierstrass Transform on the Dual of the Chébli-Trimèche Hypergroup

This subsection is devoted to define and establish some properties for the Weierstrass transform , on the dual of the Chébli-Trimèche hypergroup, which we need later.

Definition 32. Let . We define the generalized Weierstrass transform of order on as follows:

For , we denote by The norm in is given by

Remark 33. Let . For all , we have where , , are the heat functions on the Chébli-Trimèche hypergroup (). In particular, is the Gaussian kernel on (), see [15].
In the case of the Bessel-Kingman hypergroup (when the function is of the form and ), the Weierstrass transform associated with the Hankel transform is studied in [16]. For the classical Weierstrass transform, one can see [1719].

In the following, we show some properties for and .

Proposition 34. (i) Let . For all , (ii) For all , we have (iii) Let and . Then, for all , we have (iv) Let . Then, . Moreover, for all , we have

Proof. (i) For all , we have and .
(ii) Clear.
(iii) Let . By (i) and Definition 32, we obtain
(iv) We deduce the result by using Proposition 4(ii).

Now, under a sufficient condition on and , we shall prove that is a bounded operator from into .

Proposition 35. Let and . Then, for all and , there exists a positive constant such that for all , we have

Proof. According to Lemma 21 and Definition 32, we obtain the result by using Proposition 6 and applying Holder inequality.

4.2. Kernel Reproducing

Let . The space provided with the inner product, and the norm , is a Hilbert space.

Proposition 36. For when and when , the Hilbert space admits the following reproducing kernel: that is,(i)for all  , the function    belongs to  .(ii) The reproducing property: for all    and  ,

Proof. (i) It is clear from Lemma 21 and relations (9) and (10) that, for all , the function belongs to when when and when . Thus, the function is well defined and we can write Moreover, from Proposition 4, we can see that the function belongs to , and we have Therefore, according to Lemma 21 and using relations (9) and (10), we deduce that This proves that for all , the function belongs to .
(ii) Let be in and . Then by (121), we get and from inversion formula, we obtain the reproducing property
This completes the proof of the theorem.

Definition 37. For all positive real numbers , , and , we define the Hilbert space as the subspace of with the inner product: The norm associated to the inner product is defined by

Proposition 38. Let when and when . For all , the Hilbert space admits the following reproducing kernel:

Proof. As in Proposition 36, we can deduce that for all , there exists a function in such that we have This proves that for all the function belongs to .
On the other hand, for in and , we have where But from (116) and (128), we have and from (128), it follows, by using Parseval formula for the transform , that Thus, by virtue of (129), and combining (131) and (132), we deduce that

4.3. Extremal Function for Generalized Weierstrass Transform

In this subsection, we show the existence and unicity of the extremal function related to the generalized Weierstrass transform . We start with the following fundamental theorem (cf. [20]).

Theorem 39. Let be a Hilbert space admitting the reproducing kernel on a set and let be a Hilbert space. Let be a bounded linear operator on into . For , we introduce the inner product in and we call it as Then,(i) is a Hilbert space with the reproducing kernel on and satisfying the equation where is the adjoint operator of .(ii) For any and for any in , the infinitum is attained by a unique function in and this extremal function is given by

We can now state the main result of this paragraph.

Theorem 40. Let when and when .(i)For any and for any , the best approximate function in the sense exists uniquely and is represented by where  (ii)Let . Then, If we take , we have  (iii)Let and let and satisfy . Then

Proof. (i) By Proposition 38 and Theorem 39(ii), the infinitum given by (138) is attained by a unique function , and the extremal function is represented by where is the kernel given by Proposition 38. Hence, by (128), we obtain the expression (140) of .
(ii) From Proposition 35, the function belongs to . According to Lemma 21 and relations (9) and (10), it can be observed, using Cauchy-Schwartz inequality, that for all the function belongs to , which implies, from inversion formula for the transform , that Therefore, by (132), it follows that Hence, by dominated convergence theorem we deduce the result.
(iii) It is clear, from (140), that then, using Parseval formula for the Fourier transform , it follows, by (139), that and so Hence, Using the inequality , we obtain Thus, and from Proposition 4(ii), we obtain which gives the desired result.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.