Inversion of the Dual Dunkl Sonine Transform on R Using Dunkl Wavelets

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1 Symmetry, Integrability and Geometry: Methods and Applications SIGMA ), 071, 12 pages Inversion of the Dual Dunkl Sonine Transform on Using Dunkl Wavelets Mohamed Ali MOUOU Department of Mathematics, Faculty of Sciences, Al-Jouf University, P.O. Box 2014, Al-Jouf, Skaka, Saudi Arabia mohamed ali.mourou@yahoo.fr eceived March 02, 2009, in final form July 04, 2009; Published online July 14, 2009 doi: /sigma Abstract. We prove a Calderón reproducing formula for the Dunkl continuous wavelet transform on. We apply this result to derive new inversion formulas for the dual Dunkl Sonine integral transform. Key words: Dunkl continuous wavelet transform; Calderón reproducing formula; dual Dunkl Sonine integral transform 2000 Mathematics Subject Classification: 42B20; 42C15; 44A15; 44A35 1 Introduction The one-dimensional Dunkl kernel e γ, γ > 1/2, is defined by where e γ z) = j γ iz) + j γ z) = Γγ + 1) z 2γ + 1) j γ+1iz), z C, n=0 1) n z/2) 2n n! Γn + γ + 1) is the normalized spherical Bessel function of index γ. It is well-known see [3]) that the functions e γ λ ), λ C, are solutions of the differential-difference equation where Λ γ u = λu, u0) = 1, Λ γ fx) = f x) + γ + 1 ) fx) f x) 2 x is the Dunkl operator with parameter γ+1/2 associated with the reflection grour Z 2 on. Those operators were introduced and studied by Dunkl [2, 3, 4] in connection with a generalization of the classical theory of spherical harmonics. Besides its mathematical interest, the Dunkl operator Λ α has quantum-mechanical applications; it is naturally involved in the study of onedimensional harmonic oscillators governed by Wigner s commutation rules [6, 11, 16]. It is known, see for example [14, 15], that the Dunkl kernels on possess the following Sonine type integral representation e β λx) = x x K α,β x, y) e α λy) y 2α+1 dy, λ C, x 0, 1.1)

2 2 M.A. Mourou where x 2 y 2) β α 1 a α,β sgnx) x + y) K α,β x, y) := x 2β+1 if y < x, 0 if y x, 1.2) with β > α > 1/2, and a α,β := Γβ + 1) Γα + 1) Γβ α). Define the Dunkl Sonine integral transform X α,β and its dual t X α,β, respectively, by X α,β fx) = t X α,β fy) = x x x y K α,β x, y) fy) y 2α+1 dy, K α,β x, y) fx) x 2β+1 dx. Soltani has showed in [14] that the dual Dunkl Sonine integral transform t X α,β is a transmutation operator between Λ α and Λ β on the Schwartz space S), i.e., it is an automorphism of S) satisfying the intertwining relation t X α,β Λ β f = Λ α t X α,β f, f S). The same author [14] has obtained inversion formulas for the transform t X α,β involving pseudodifferential-difference operators and only valid on a restricted subspace of S). The purpose of this paper is to investigate the use of Dunkl wavelets see [5]) to derive a new inversion of the dual Dunkl Sonine transform on some Lebesgue spaces. For other applications of wavelet type transforms to inverse problems we refer the reader to [7, 8] and the references therein. The content of this article is as follows. In Section 2 we recall some basic harmonic analysis results related to the Dunkl operator. In Section 3 we list some basic properties of the Dunkl Sonine integral trnsform and its dual. In Section 4 we give the definition of the Dunkl continuous wavelet transform and we establish for this transform a Calderón formula. By combining the results of the two previous sections, we obtain in Section 5 two new inversion formulas for the dual Dunkl Sonine integral transform. 2 Preliminaries Note 2.1. Throughout this section assume γ > 1/2. Define L p, x 2γ+1 dx), 1 p, as the class of measurable functions f on for which f p,γ <, where 1/p f p,γ = fx) p x dx) 2γ+1, if p <, and f,γ = f = ess sup x fx). S) stands for the usual Schwartz space. The Dunkl transform of order γ + 1/2 on is defined for a function f in L 1, x 2γ+1 dx) by F γ fλ) = fx) e γ iλx) x 2γ+1 dx, λ. 2.1)

3 Inversion of the Dual Dunkl Sonine Transform on 3 emark 2.2. It is known that the Dunkl transform F γ maps continuously and injectively L 1, x 2γ+1 dx) into the space C 0 ) of continuous functions on vanishing at infinity). Two standard results about the Dunkl transform F γ are as follows. Theorem 2.3 see [1]). i) For every f L 1 L 2, x 2γ+1 dx) we have the Plancherel formula fx) 2 x 2γ+1 dx = m γ F γ fλ) 2 λ 2γ+1 dλ, where m γ = 1 2 2γ+2 Γγ + 1)) ) ii) The Dunkl transform F α extends uniquely to an isometric isomorphism from L 2, x 2γ+1 dx) onto L 2, m γ λ 2γ+1 dλ). The inverse transform is given by Fγ 1 gx) = m γ gλ)e γ iλx) λ 2γ+1 dλ, where the integral converges in L 2, x 2γ+1 dx). Theorem 2.4 see [1]). The Dunkl transform F α is an automorphism of S). An outstanding result about Dunkl kernels on see [12]) is the product formula where T x γ with e γ λx)e γ λy) = T x γ e γ λ )) y), λ C, x, y, stand for the Dunkl translation operators defined by T x γ fy) = 1 2 W γ t) = f x 2 + y 2 2xyt) f x 2 + y 2 2xyt) 1 Γγ + 1) π Γγ + 1/2) 1 + t) 1 t 2) γ 1/2. x y x 2 + y 2 2xyt ) W γ t)dt ) x y W γ t)dt, 2.3) x 2 + y 2 2xyt The Dunkl convolution of two functions f, g on is defined by the relation f γ gx) = Tγ x f y)gy) y 2γ+1 dy. 2.4) Proposition 2.5 see [13]). i) Let p, q, r [1, ] such that 1 p + 1 q 1 = 1 r. If f Lp, x 2γ+1 dx) and g L q, x 2γ+1 dx), then f γ g L r, x 2γ+1 dx) and f γ g r,γ 4 f p,γ g q,γ. 2.5) ii) For f L 1, x 2γ+1 dx) and g L p, x 2γ+1 dx), p = 1 or 2, we have F γ f γ g) = F γ ff γ g. 2.6) For more details about harmonic analysis related to the Dunkl operator on the reader is referred, for example, to [9, 10].

4 4 M.A. Mourou 3 The dual Dunkl Sonine integral transform Throughout this section assume β > α > 1/2. Definition 3.1 see [14]). The dual Dunkl Sonine integral transform t X α,β is defined for smooth functions on by t X α,β fy) := K α,β x, y)fx) x 2β+1 dx, y, 3.1) x y where K α,β is the kernel given by 1.2). emark 3.2. Clearly, if supp f) [ a, a] then supp t X α,β f ) [ a, a]. The next statement provides formulas relating harmonic analysis tools tied to Λ α with those tied to Λ β, and involving the operator t X α,β. Proposition 3.3. i) The dual Dunkl Sonine integral transform t X α,β maps L 1, x 2β+1 dx) continuously into L 1, x 2α+1 dx). ii) For every f L 1, x 2β+1 dx) we have the identity F β f) = F α t X α,β f). 3.2) iii) Let f, g L 1, x 2β+1 dx). Then t X α,β f β g) = t X α,β f α t X α,β g. 3.3) Proof. Let f L 1, x 2β+1 dx). By Fubini s theorem we have But by 1.1), x t X α,β f )y) y 2α+1 dy = x = ) K α,β x, y) fx) x 2β+1 dx y 2α+1 dy x y ) x fx) K α,β x, y) y 2α+1 dy x 2β+1 dx. x K α,β x, y) y 2α+1 dy = e β 0) = ) Hence, t X α,β f is almost everywhere defined on, belongs to L 1, x 2α+1 dx) and t X α,β f 1,α f 1,β, which proves i). Identity 3.2) follows by using 1.1), 2.1), 3.1), and Fubini s theorem. Identity 3.3) follows by applying the Dunkl transform F α to both its sides and by using 2.6), 3.2) and emark 2.2. emark 3.4. From 3.2) and emark 2.2, we deduce that the transform t X α,β maps L 1, x 2β+1 dx) injectively into L 1, x 2α+1 dx). From [14] we have the following result.

5 Inversion of the Dual Dunkl Sonine Transform on 5 Theorem 3.5. The dual Dunkl Sonine integral transform t X α,β is an automorphism of S) satisfying the intertwining relation t X α,β Λ β f = Λ α t X α,β f, f S). Moreover t X α,β admits the factorization t X α,β f = t V 1 α t V β f for all f S), where for γ > 1/2, t V γ denotes the dual Dunkl intertwining operator given by t Γγ + 1) V γ fy) = sgnx) x + y) x 2 y 2) γ 1/2 fx) dx. π Γγ + 1/2) x y Definition 3.6 see [14]). The Dunkl Sonine integral transform X α,β is defined for a locally bounded function f on by x K α,β x, y) fy) y 2α+1 dy if x 0, X α,β fx) = x 3.5) f0) if x = 0. emark 3.7. i) Notice that by 3.4), X α,β f f if f L ). ii) It follows from 1.1) that e β λx) = X α,β e α λ )x) 3.6) for all λ C and x. Proposition 3.8. i) For any f L ) and g L 1, x 2β+1 dx) we have the duality relation X α,β fx)gx) x 2β+1 dx = fy) t X α,β gy) y 2α+1 dy. 3.7) ii) Let f L 1, x 2β+1 dx) and g L ). Then X α,β t X α,β f α g ) = f β X α,β g. 3.8) Proof. Identity 3.7) follows by using 3.1), 3.5) and Fubini s theorem. Let us check 3.8). Let ψ S). By using 3.3), 3.7) and Fubini s theorem, we have f β X α,β gx)ψx) x 2β+1 dx = X α,β gx)ψ β f x) x 2β+1 dx = gy) t X α,β ψ β f )y) y 2α+1 dy = gy) t X α,β ψ t α X α,β f ) y) y 2α+1 dy, where f x) = f x), x. But an easy computation shows that t X α,β f = t X α,β f ). Hence, f β X α,β gx)ψx) x 2β+1 dx = gy) t X α,β ψ t α X α,β f ) y) y 2α+1 dy = t X α,β f α gy) t X α,β ψy) y 2α+1 dy = X t α,β X α,β f α g ) x)ψx) x 2β+1 dx. This clearly yields the result.

6 6 M.A. Mourou 4 Calderón s formula for the Dunkl continuous wavelet transform Throughout this section assume γ > 1/2. Definition 4.1. We say that a function g L 2, x 2γ+1 dx) is a Dunkl wavelet of order γ, if it satisfies the admissibility condition 0 < C γ g := emark F γ gλ) 2 dλ λ = F γ g λ) 2 dλ λ i) If g is real-valued we have F γ g λ) = F γ gλ), so 4.1) reduces to 0 < C γ g := 0 F γ gλ) 2 dλ λ <. ii) If 0 g L 2, x 2γ+1 dx) is real-valued and satisfies 0 η > 0 such that F γ gλ) F γ g0) = Oλ η ) as λ 0 + then 4.1) is equivalent to F γ g0) = 0. <. 4.1) Note 4.3. For a function g in L 2, x 2γ+1 dx) and for a, b) 0, ) we write g γ a,b 1 x) := a b T 2γ+2 γ g a x), where T b γ are the generalized translation operators given by 2.3), and g a x) := gx/a), x. emark 4.4. Let g L 2, x 2γ+1 dx) and a > 0. Then it is easily checked that g a L 2, x 2γ+1 dx), g a 2,γ = a γ+1 g 2,γ, and F γ g a )λ) = a 2γ+2 F γ g)aλ). Definition 4.5. Let g L 2, x 2γ+1 dx) be a Dunkl wavelet of order γ. We define for regular functions f on, the Dunkl continuous wavelet transform by Φ γ gf)a, b) := fx)g γ a,b x) x 2γ+1 dx which can also be written in the form Φ γ gf)a, b) = 1 a 2γ+2 f γ g a b), where γ is the generalized convolution product given by 2.4), and g a x) := g x/a), x. The Dunkl continuous wavelet transform has been investigated in depth in [5] in which precise definitions, examples, and a more complete discussion of its properties can be found. We look here for a Calderón formula for this transform. We start with some technical lemmas. Lemma 4.6. For all f, g L 2, x 2γ+1 dx) and all ψ S) we have the identity f γ gx)fγ 1 ψx) x 2γ+1 dx = m γ F γ fλ)f γ gλ)ψ λ) λ 2γ+1 dλ, where m γ is given by 2.2).

7 Inversion of the Dual Dunkl Sonine Transform on 7 Proof. Fix g L 2, x 2γ+1 dx) and ψ S). For f L 2, x 2γ+1 dx) put S 1 f) := f γ gx)fγ 1 ψx) x 2γ+1 dx and S 2 f) := m γ F γ fλ)f γ gλ)ψ λ) λ 2γ+1 dλ. By 2.5), 2.6) and Theorem 2.3, we see that S 1 f) = S 2 f) for each f L 1 L 2, x 2γ+1 dx). Moreover, by using 2.5), Hölder s inequality and Theorem 2.3 we have and S 1 f) f γ g Fγ 1 ψ 1,γ 4 f 2,γ g 2,γ Fγ 1 ψ 1,γ S 2 f) m γ F γ ff γ g 1,γ ψ m γ F γ f 2,γ F γ g 2,γ ψ = f 2,γ g 2,γ ψ, which shows that the linear functionals S 1 and S 2 are bounded on L 2, x 2γ+1 dx). Therefore S 1 S 2, and the lemma is proved. Lemma 4.7. Let f 1, f 2 L 2, x 2γ+1 dx). Then f 1 γ f 2 L 2, x 2γ+1 dx) if and only if F γ f 1 F γ f 2 L 2, x 2γ+1 dx) and we have F γ f 1 γ f 2 ) = F γ f 1 F γ f 2 in the L 2 -case. Proof. Suppose f 1 γ f 2 L 2, x 2γ+1 dx). By Lemma 4.6 and Theorem 2.3, we have for any ψ S), m γ F γ f 1 λ)f γ f 2 λ)ψλ) λ 2γ+1 dλ = f 1 γ f 2 x)fγ 1 ψ x) x 2γ+1 dx = f 1 γ f 2 x)fγ 1 ψx) x 2γ+1 dx = m γ F γ f 1 γ f 2 )λ)ψλ) λ 2γ+1 dλ, which shows that F γ f 1 F γ f 2 = F γ f 1 γ f 2 ). Conversely, if F γ f 1 F γ f 2 L 2, x 2γ+1 dx), then by Lemma 4.6 and Theorem 2.3, we have for any ψ S), f 1 γ f 2 x)fγ 1 ψx) x 2γ+1 dx = m γ F γ f 1 λ)f γ f 2 λ) ψλ) λ 2γ+1 dλ = F 1 γ F γ f 1 F γ f 2 )x)fγ 1 ψx) x 2γ+1 dx, which shows, in view of Theorem 2.4, that f 1 γ f 2 = Fγ 1 F γ f 1 F γ f 2 ). This achieves the proof of Lemma 4.7. A combination of Lemma 4.7 and Theorem 2.3 gives us the following. Lemma 4.8. Let f 1, f 2 L 2, x 2γ+1 dx). Then f 1 γ f 2 x) 2 x 2γ+1 dx = m γ F γ f 1 λ) 2 F γ f 2 λ) 2 λ 2γ+1 dλ, where both sides are finite or infinite.

8 8 M.A. Mourou Lemma 4.9. Let g L 2, x 2γ+1 dx) be a Dunkl wavelet of order γ such that F γ g L ). For 0 < < δ < define and Then and G,δ x) := 1 C γ g K,δ λ) := 1 C γ g δ δ da g a γ g a x) a 4γ+5 4.2) F γ gaλ) 2 da a. 4.3) G,δ L 2, x 2γ+1 dx), K,δ L 1 L 2), x 2γ+1 dx), 4.4) F γ G,δ ) = K,δ. Proof. Using Schwarz inequality for the measure so da we obtain a4γ+5 G,δ x) 2 1 δ ) da δ Cg γ ) 2 a 4γ+5 g a γ g a x) 2 da a 4γ+5, G,δ x) 2 x 2γ+1 dx 1 δ ) da δ Cg γ ) 2 a 4γ+5 g a γ g a x) 2 x 2γ+1 dx da a 4γ+5. By Theorem 2.3, Lemma 4.8, and emark 4.4, we have g a γ g a x) 2 x 2γ+1 dx = m γ F γ g a )λ) 4 λ 2γ+1 dλ m γ F γ g a ) 2 F γ g a )λ) 2 λ 2γ+1 dλ Hence = F γ g a ) 2 g a 2 2,γ = a6γ+6 F γ g 2 g 2 2,γ. G,δ x) 2 x 2γ+1 dx F γg 2 g 2 2,γ C γ g ) 2 δ ) δ ) a 2γ+1 da da a 4γ+5 <. The second assertion in 4.4) is easily checked. Let us calculate F γ G,δ ). Fix x. From Theorem 2.3 and Lemma 4.7 we get g a γ g a x) = m γ F γ g a )λ) 2 e γ iλx) λ 2γ+1 dλ, so G,δ x) = m γ C γ g δ ) da F γ g a )λ) 2 e γ iλx) λ 2γ+1 dλ a 4γ+5. As e γ iz) 1 for all z see [12]), we deduce by Theorem 2.3 that δ m γ F γ g a )λ) 2 e γ iλx) λ 2γ+1 dλ da a 4γ+5

9 Inversion of the Dual Dunkl Sonine Transform on 9 δ g a 2 da 2,γ a 4γ+5 = g 2 2,γ δ Hence, applying Fubini s theorem, we find that G,δ x) = m γ = m γ 1 C γ g δ F γ gaλ) 2 da a K,δ λ)e γ iλx) λ 2γ+1 dλ da <. a2γ+3 ) e γ iλx) λ 2γ+1 dλ which completes the proof. We can now state the main result of this section. Theorem 4.10 Calderón s formula). Let g L 2, x 2γ+1 dx) be a Dunkl wavelet of order γ such that F γ g L ). Then for f L 2, x 2γ+1 dx) and 0 < < δ <, the function f,δ x) := 1 C γ g δ Φ γ gf)a, b)g a,b x) b 2γ+1 db da a belongs to L 2, x 2γ+1 dx) and satisfies lim 0, δ f,δ f 2,γ = ) Proof. It is easily seen that f,δ = f γ G,δ, where G,δ is given by 4.2). It follows by Lemmas 4.7 and 4.9 that f,δ L 2, x 2γ+1 dx) and F γ f,δ ) = F γ f) K,δ, where K,δ is as in 4.3). From this and Theorem 2.3 we obtain f,δ f 2 2,γ = m γ F γ f,δ f)λ) 2 λ 2γ+1 dλ = m γ F γ fλ) 2 1 K,δ λ)) 2 λ 2γ+1 dλ. But by 4.1) we have lim K,δλ) = 1, for almost all λ. 0, δ So 4.5) follows from the dominated convergence theorem. Another pointwise inversion formula for the Dunkl wavelet transform, proved in [5], is as follows. Theorem Let g L 2, x 2γ+1 dx) be a Dunkl wavelet of order γ. If both f and F γ f are in L 1, x 2γ+1 dx) then we have fx) = 1 C γ g 0 ) Φ γ gf)a, b)g γ da a,b x) b 2γ+1 db a, where, for each x, both the inner integral and the outer integral are absolutely convergent, but possibly not the double integral. a.e.,

10 10 M.A. Mourou 5 Inversion of the dual Dunkl Sonine transform using Dunkl wavelets From now on assume β > α > 1/2. In order to invert the dual Dunkl Sonine transform, we need the following two technical lemmas. Lemma 5.1. Let 0 g L 1 L 2, x 2α+1 dx) such that F α g L 1, x 2α+1 dx) and satisfying η > β 2α 1 such that F α gλ) = O λ η ) as λ ) Then X α,β g L 2, x 2β+1 dx) and F β X α,β g)λ) = m α F α gλ) λ 2β α). Proof. By Theorem 2.3 we have gx) = m α F α gλ)e α iλx) λ 2α+1 dλ, a.e. So using 3.6), we find that with X α,β gx) = h α,β λ) := m α h α,β λ)e β iλx) λ 2β+1 dλ, a.e. 5.2) F α gλ) λ 2β α). Clearly, h α,β L 1, x 2β+1 dx). So it suffices, in view of 5.2) and Theorem 2.3, to prove that h α,β belongs to L 2, x 2β+1 dx). We have h α,β λ) 2 λ 2β+1 dλ = = mα ) 2 + λ 1 mα λ 1 ) 2 λ 4α 2β+1 F α gλ) 2 dλ ) By 5.1) there is a positive constant k such that I 1 k λ 2η+4α 2β+1 dλ = λ 1 From Theorem 2.3, it follows that I 2 = mα mα ) 2 ) 2 λ 4α 2β+1 F α gλ) 2 dλ := I 1 + I 2. k η + 2α β + 1 <. λ 2α β) F α gλ) 2 λ 2α+1 dλ λ 1 F α gλ) 2 λ 2α+1 dλ λ 1 mα ) 2 F α g 2 2,α = m α ) 2 g 2 2,α < which ends the proof.

11 Inversion of the Dual Dunkl Sonine Transform on 11 Lemma 5.2. Let 0 g L 1 L 2, x 2α+1 dx) be real-valued such that F α g L 1, x 2α+1 dx) and satisfying η > 2β α) such that F α gλ) = Oλ η ) as λ ) Then X α,β g L 2, x 2β+1 dx) is a Dunkl wavelet of order β and F β X α,β g) L ). Proof. By combining 5.3) and Lemma 5.1 we see that X α,β g L 2, x 2β+1 dx), F β X α,β g) is bounded and F β X α,β g)λ) = O λ η 2β α)) as λ 0 +. Thus, in view of emark 4.2, X α,β g satisfies the admissibility condition 4.1) for γ = β. emark 5.3. In view of emark 4.2, each function satisfying the conditions of Lemma 5.1 is a Dunkl wavelet of order α. Lemma 5.4. Let g be as in Lemma 5.2. Then for all f L 1, x 2β+1 dx) we have Φ β X α,β g f)a, b) = 1 a 2β α) X [ α,β Φ α t g X α,β f ) a, ) ] b). Proof. By Definition 4.5 we have But Φ β 1 X α,β g f)a, b) = a 2β+2 f β X α,β g) a b). Xα,β g) a = X α,β g a ) by virtue of 1.2) and 3.5). So using 3.8) we find that Φ β 1 X α,β g f)a, b) = which gives the desired result. a 2β+2 f β [X α,β g a )] b) = 1 a 2β+2 X α,β [ t X α,β f α g a ] b) = 1 a 2β α) X α,β [ Φ α g t X α,β f ) a, ) ] b), Combining Theorems 4.10, 4.11 with Lemmas 5.2, 5.4 we get Theorem 5.5. Let g be as in Lemma 5.2. Then we have the following inversion formulas for the dual Dunkl Sonine transform: i) If both f and F β f are in L 1, x 2β+1 dx) then for almost all x we have fx) = 1 C β X α,β g 0 [ X α,β Φ α t g X α,β f ) a, ) ] b) X α,β g ) ) β da a,b x) b 2β+1 db a 2β α)+1. ii) For f L 1 L 2, x 2β+1 dx) and 0 < < δ <, the function f,δ x) := 1 δ [ C β X α,β Φ α t g X α,β f ) a, ) ] b) X α,β g ) β da a,b x) b 2β+1 db X α,β g a 2β α)+1 satisfies lim 0, δ Acknowledgements f,δ f 2,β = 0. The author is grateful to the referees and editors for careful reading and useful comments.

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