The fractional Hankel transformation of certain tempered distributions and pseudo-differential operators

Transcription

1 Chapter 5 The fractional Hankel transformation of certain tempered distributions and pseudo-differential operators 5.1 Introduction The fractional Hankel transformation is a generalization of the conventional Hankel transformation in the fractional order and is effectively used in the design of lens, analysis of laser cavity study of wave propagation in quadratic refractive index medium when the system is axially symmetric. The earliest work on the fractional Hankel transformation was published by Namias 16 in 198. Recently it is becoming of importance in various applications in optics 34, 39. Kerr 13 has developed a theory of fractional powers of Hankel transformations in the Zemanian spaces. The purpose of the present chapter is to derive certain important properties of the fractional Hankel transformation. Fractional Hankel transformation is studied on Zemanian space H µ I and tempered distributions space H µi. A pseudo-differential operator involving fractional Hankel transformation is introduced and some of its properties including boundedness are investigated in G s µ, I. 57

2 5. Properties of fractional Hankel transformation Proposition 5..1 Let Kµ α x,y be the kernel of fractional Hankel transformation defined in , then r µ,x Kα µ x,y ycscα r Kµ α x,y, r N, where µ,x d + ixcotα d dx dx + 1 4µ + icotα x cot α. 4x Proof: Since Kµ α x,y C α,µ x +y cotα xycscα 1 Jµ xycscα, d dx Kα µ x,y C d α,µ x +y cotα xycscα 1 Jµ xycscα dx C α,µ x +y cotα d dx {xycscα 1 Jµ xycscα} + x +y cotα xycscα 1 Jµ xycscα ixcotα C α,µ x +y cotα ycscα{xycscα µ J µ xycscα µ + 1/xycscα µ 1/ xycsc α µ+1/ xycscα µ J µ+1 xycscα}+ x +y cotα xycscα 1 J µ xycscα ixcotα C α,µ x +y cotα ycscαµ + 1/xycscα 1/ J µ xycscα C α,µ x +y cotα ycscαxycscα 1/ J µ+1 xycscα + C α,µ x +y cotα xycscα 1 Jµ xycscα ixcotα So d dx Kα µ x,y µ + 1/x 1 K α µ x,y ixcotαkα µ x,y C α,µ x +y cotα ycscαxycscα 1/ J µ+1 xycscα, 5.. then from 5.., we have J µ+1 xycscα dx d Kα µ x,y+ixcotαkµ α x,y µ + 1/x 1 Kµ α x,y C α,µ x +y cotα ycscαxycscα 1 58

3 Now again differentiation of 5..1, we get d dx K α µ x,y C α,µ ycscαµ + 1/ d dx J µ xycscα Using 5..3, we have C α,µ ycscα d dx + C α,µ icotα d dx x +y cotα xycscα 1/ x +y cotα xycscα 1/ J µ+1 xycscα x C α,µ ycscαµ + 1/ x +y cotα xycscα 1 Jµ xycscα x +y cotα d dx {xycscα 1/ J µ xycscα}+xycscα 1/ J µ xycscα x +y cotα ix cot α C α,µ ycscα x +y cotα d dx {xycscα1/ J µ+1 xycscα} + xycscα 1/ J µ+1 xycscα x +y cotα ixcotα C α,µ icotα x x +y cotα d dx {xycscα1/ J µ xycscα} + xycscα 1/ J µ xycscα{x x +y cotα ixcotα + x +y cotα } µ 1/4x Kµ α x,y µ + icotαkα µ x,y ycscα K α µ x,y x cot αk α µ x,y+c α,µ ycscαixcotα x +y cotα xycscα 1/ J µ+1 xycscα. d dx Kα µ x,y 1 4µ 4x K α µ x,y icotαk α µ x,y+x cot αk α µ x,y ixcotα d dx Kα µ x,y ycscα K α µ x,y. Now, we have d dx + ixcotα d 1 4µ dx + 4x + icotα x cot α Kµ α x,y 59

4 ycscα K α µ x,y. Therefore µ,x Kµ,x α x,y ycscα Kµ α x,y, 5..4 where d µ,x dx + ixcotα d 1 4µ dx + 4x + icotα x cot α Similarly by using 5..4 Continuing in this way, we get µ,x Kα µ x,y µ,x µ,x K α µ x,y µ,x ycscα K α µ x,y ycscα µ,x K α µ x,y ycscα K α µ x,y. r µ,xk α µ x,y ycscα r K α µ x,y, r N Lemma 5..1 For any ϕ H µ I and all non-negative integer k, we have µ,x k k l k k ϕx x l x x n la 1 d n ϕx, dx where µ,x is as 5..5 and a l are constants depending on µ and α only. Proof. At first for k 1, d µ,x ϕx dx + ixcotα d 1 4µ dx + 4x x x d dx + ix3 cotα d 1 4µ dx + 4 x n la l x l x 1 d dx n ϕx. + icotα x cot α ϕx + ix cotα x 4 cot α ϕx 6

5 Similarly, for k, µ,x ϕx µ,x µ,x ϕx d dx + ixcotα d 1 4µ dx + 4x d dx + ixcotα d 1 4µ dx 4x d4 ϕ dx 4 + 4ixcotα d3 ϕ 1 4µ dx 3 + 4x 1 4µ + icotα x+4ixcotα 4x 1 4µ 1 4µ + 6 4x 4 1 4µ + 4x 4ixcotα + icotα x cot α + icotα x cot α + icotα x cot α 4x 3 x 4 n 4 4 l x la l x 1 d n ϕx. dx ϕx d + icotα x cot α +ixcotα ϕ dx dϕ + icotα x cot α dx cot α 4ix cot 3 α ϕx Continuing in this way, we have µ,x k k l k k ϕx x l x x n la 1 d n ϕx. dx We now introduce a new function space H µ,α I. Definition 5..1 Test function space H µ,α I. The space H µ,α I, is defined as follows: ϕ is member of H µ,α I if and only if it is complex valued C function on I and for every choice of m and k of non-negative integers, it satisfies where ϒ α m,k ϕ sup x I x m k µ,xϕx <, 5..7 d µ,x dx + ixcotα d 1 4µ dx + 4x + icotα x cot α, 5..8 and k µ,x is same as the Lemma The topology over H µ,αi is generated by the family {ϒ α m,k }α π m,k N of seminorms. 61

6 Clearly for α π, H µ,αi H µ I, this shown by Pathak. But if α π, the two spaces H µ,α I and H µ I are independent as shown below: It can be easily seen that x m+k k µ,xϕx If x µ+1/ ϕx H µ I, then x m+k k µ,xϕx sup x I k r k l a l sup x I k k l x r la m+l x 1 d r x µ 1/ x ϕx µ+1/. dx x dx xm+l 1 d r x µ 1/ x ϕx µ+1/ <, m, l, r N. So the right-hand side is in H µ I. Therefore x m+k k µ,xϕx <, m, k N, sup x I this implies that ϕx H µ,α I. Proposition 5.. For all ϕ H µ I, we have < r µ,xkµ α x,y,ϕx> <Kµ α x,y, µ,x r ϕx>, where µ,x d ixcotα d dx dx + 1 4µ icotα x cot α. We call 4x µ,x and µ,x as fractional Bessel operator with parameter α. We see that the Bessel operator µ,x µ,x x, for α π. Proof: At first we prove < µ,x K α µ x,y,ϕx> <K α µ x,y, µ,xϕx> Using 5..5, we have < µ,x Kµ α x,y,ϕx> µ,x Kµ α x,y ϕxdx d dx + ixcotα d 1 4µ dx + 4x + icotα x cot α K αµ x,y ϕxdx + d dx Kα µ x,yϕxdx+ ixcotα d dx Kα µ x,yϕxdx 1 4µ 4x Kµ α x,yϕxdx+ icotα x cot αkµ α x,yϕxdx. 6

7 Using integration by parts, we have + Therefore d dx ϕx d dx Kα µ x,ydx d dx ixcotαϕxkα µ x,ydx 1 4µ 4x Kµ α x,yϕxdx+ icotα x cot αkµ α x,yϕxdx Kµ α x,y ixcotα d dx dx ϕx+icotαϕx Kµ α d x,y 1 4µ dx ϕxdx + 4x Kµ α x,yϕxdx+ icotα x cot αkµ α x,yϕxdx d Kµ α x,y dx ixcotα d 1 4µ dx + 4x icotα x cot α ϕxdx <Kµ α x,y, µ,xϕx>. In general, we have < µ,x K α µ x,y,ϕx> <Kα µ x,y, µ,xϕx>. < r µ,x Kα µ x,y,ϕx> <Kα µ x,y, µ,x r ϕx>. Lemma 5.. For any ϕ H µ I and all non-negative integer k, we have r r µ,x r r ϕx x l x n lb l x 1 d n ϕx, dx where µ,x is as Proposition 5.. and b l constants depending on µ and α only. Proof. At first for r1, d µ,xϕx dx ixcotα d 1 4µ dx + 4x x x d dx ix3 cotα d 1 4µ dx + 4 x n lb l x l x 1 d dx n ϕx. icotα x cot α ϕx ix cotα x 4 cot α ϕx 63

8 Similarly, for r, µ,x ϕx µ,x µ,xϕx d dx ixcotα d 1 4µ dx + 4x d dx ixcotα d 1 4µ dx 4x d4 ϕ dx 4 4ixcotα d3 ϕ 1 4µ dx 3 + 4x 1 4µ + icotα x 4ixcotα 4x 1 4µ 1 4µ ixcotα 4x 4 1 4µ + 4x icotα x cot α icotα x cot α icotα x cot α 4x 3 x 4 n 4 4 l x lb l x 1 d n ϕx. dx ϕx d icotα x cot α +ixcotα ϕ dx dϕ icotα x cot α dx cot α + 4ix cot 3 α ϕx Continuing in this way, we have µ,x r r r r ϕx x l x n lb l x 1 d n ϕx. dx Proposition 5..3 Let ϕ H µ I, then i h α µ µ,x r ϕy ycscα r hµϕyfor α every index r, ii r µ,y h α µ ϕ y h α µ xcscα r ϕ yfor every index r, iii h α µ : H µi H µ,α I, is linear and continuous. Proof: i Using Proposition 5.. and 5..1, we have h α µ µ,x r ϕy K α µ x,y µ,x r ϕxdx r µ,xk α µ x,yϕxdx ycscα r Kµ α x,yϕxdx ycscα r h α µϕy. 64

9 ii If ϕ H µ I and for every index r, we have h α µϕ y r µ,ykµ α x,yϕxdx r µ,y xcscα r K α µ x,yϕxdx Kµ α x,y xcscα r ϕxdx h α µ xcscα r ϕ y. iii Linearity of h α µ is obvious. Let m and k be two non- negative integers indices. Then by Proposition 5..3ii, for {ϕ j } j N H µ,α I, we have y m k µ,y hα µϕ j y supy m h α µ xcscα k ϕ j y. sup y I Since ϕ j H µ I, xcscα k ϕ j H µ I. This implies that y I h α µ xcscα k ϕ j y Hµ I, hence ϒm,k α ϕ j supy m k µ,y hα µϕ j y, y I if ϕ j in H µ,α I as j. This implies continuity of h α µ. Proposition 5..4 Parseval s identity for fractional Hankel transformation. If ϕ, ψ H µ I, we have the equalities ϕxψxdx sinα and ϕx dx sinα h α µϕyh α µψydy, hµϕy α dy. Proof: By using and 1.11., we have ϕxψxdx ϕx Kµ α x,yhµψydy α dx. Apply change of order of integration and Fubini s theorem in the above, we have ϕxψxdx sin α hµψy α Kµ α x,yϕxdx dy sinα h α µϕyh α µψydy. 65

10 If ψ ϕ H µ I, then ϕx dx sinα Proposition 5..5 Let ϕ H µ I, then h α µ ϕy dy. i hµϕcxy α C ycscα α,µ c y cotα h µ z c cotα ϕ,c >, c ii hµx α 1 ϕxy C α,µ µ ycscα 1 hµ 1 α C ϕy+ 1 hµ+1 α α,µ 1 C ϕy, α,µ+1 d iii h α µ dx ϕx y C α,µ 4µ ycscα 1 µ 1hµ+1 α C ϕy α,µ+1 1 µ + 1hµ 1 α C ϕy + icotαhµxϕy, α α,µ 1 iv h α µ x µ+1/ x cotα Hc x where Hc x Heaviside s unit step function, v h α µ δx cy Kα µ c,y,c >. Proof: i Using , we have h α µϕcxy C α,µ C α,µ c ii Using , we have C α,µ c y cotα y C α,µ y cotα ycscα 1/ c µ+1 J µ+1 cycscα, x +y cotα xycscα 1 Jµ xycscαϕcxdx, z c +y cotα z z ycscα c C α,µ c y cotα h µ z c cotα ϕ h α µ x 1 ϕxy C α,µ C α,µ 1 c ycscα z J µ ϕzdz c ycscα 1 J µ z ycscα z c c cotα ϕzdz ycscα. c x +y cotα xycscα 1 Jµ xycscαx 1 ϕxdx x +y cotα x 1 ycscα 1 Jµ xycscαϕxdx. 66

11 Using the recurrence relation J µ x µ x Jµ 1 x+j µ+1 x, we have h α µx 1 ϕxy C α,µ ycscαc α,µ x +y cotα x 1 1 xycscα ycscα µ x +y cotα Jµ 1 xycscα+j µ+1 xycscα ϕxdx xycscα 1 Jµ 1 xycscα+j µ+1 xycscα ϕxdx, µ so h α µ x 1 ϕxy C α,µ µ ycscα 1 hµ 1 α C ϕy+ 1 hµ+1 α α,µ 1 C ϕy. α,µ+1 iii Using , we have d h α µ dx ϕx y C α,µ d C α,µ dx C α,µ x +y cotα xycscα 1 Jµ xycscα d dx ϕxdx. x +y cotα xycscα 1 Jµ xycscα ϕxdx x +y cotα xycscα 1 ycscαj µ xycscα+j µ xycscα x +y cotα ycscα 1 Using the recurrence relation J µ x 1 Jµ 1 x J µ+1 x, we have d h α µ dx ϕx y C α,µ x +y cotα xycscα 1 1 ycscα + J µ xycscα ycscα x +y cotα ycscα 1 1 x 1 1 x 1 Cα,µ C α,µ+1 h α µ+1 ϕy C α,µ C α,µ 1 h α µ 1 ϕy + x +y cotα xycscα 1 ixcotα ϕxdx. Jµ 1 xycscα J µ+1 xycscα + x +y cotα xycscα 1 ixcotα ϕxdx 1 hα µ x 1 ϕy+icotαh α µ xϕy. Now using Proposition 5..5ii, we have d h α µ dx ϕx y C α,µ 4µ ycscα 1 µ 1hµ+1 α C ϕy 1 µ + 1hµ 1 α α,µ+1 C ϕy α,µ 1 + icotαh α µ xϕy. 67

12 iv Using , we have h α µ x µ+1/ x cotα Hc x y C α,µ Using and 5..1, we have h α µ x µ+1/ x cotα Hc x v Using , we have h α µ δx cy C α,µ x +y cotα xycscα 1 Jµ xycscα xµ+1/ C α,µ y cotα ycscα 1 c c y C α,µ y cotα ycscα 1 x cotα Hc xdx x µ+1 J µ xycscαdx ycscα D x µ+1 J µ+1 xycscα dx C α,µ y cotα ycscα 1/ c µ+1 J µ+1 cycscα. x +y cotα xycscα 1 Jµ xycscαδx cdx C α,µ c +y cotα cycscα 1 Jµ cycscα K α µ c,y,c >. 5.3 The fractional Hankel transformation of tempered distributions Theorem The fractional Hankel transformation is a continuous linear map of H µ I onto itself. Proof: Let ϕx H µ I L 1 I; then using Zemanian s technique 38 and 1.1.1, we have y 1 D β y µ 1/ ϕ α µ y y 1 D β y µ 1/ C α,µ y cotα Φ α µ y C α,µ β β C α,µ β β β y 1 β D β y cotα y 1 D β β y µ 1/ Φ α µ y β β icotα β y cotα y 1 D β β y µ 1/ Φ α µ y. 68

13 Therefore, as per Koh 14, we have sup y I y m y 1 D β y µ 1/ ϕ µ α y Cα,µ β β β β cotα β supy m y 1 D β β y µ 1/ Φ α µ y y I < Hence Φ α µy H µ I. Thus ϕ α µ y h α µϕy H µ I. Also from and 1.11., we see that for all ϕ H µ I, h α µ 1 h α µϕ ϕ h α µ hα µ 1 ϕ. It follows that h α µ is a one-one function of H µi onto itself. h α µ is clearly a linear map of H µ I onto itself. To show that it is continuous, assume that the sequence {ϕ j } j N converges in H µ I to zero, then from the continuity of the fractional Hankel transformation follows. Definition : The generalized fractional Hankel transformation h α µ f of f H µ I is defined by where ϕ H µ I. <h α µ f,ϕ> < f,hα µ ϕ>, 5.3. Theorem 5.3. The generalized fractional Hankel transformation h α µ is a continuous linear map of H µ I onto itself. Proof: Since by Theorem 5.3.1, h α µϕ H µ I whenever ϕ H µ I, the right-hand side of 5.3. is well defined. Also, if {ϕ j } j N converges in H µ I to zero, then by continuity of the fractional Hankel transformation {h α µϕ j }, and so the right-hand side of 5.3. converges to zero, which in turn implies that{<h α µ f,ϕ j>} as j. Thus h α µ f is continuous on H µ I. Clearly hα µ f is linear on H µ I. Similarly the inverse fractional Hankel transformation h α µ 1 f of f H µ I is defined by <h α µ 1 f,ϕ> < f,h α µ 1 ϕ>, ϕ H µ I

14 It can be shown, as in the above, that h α µ 1 f H µi. Thus from 5.3. and it follows that for f H µ I and ϕ H µi, So that Similarly < f,ϕ> <h α µ f,h α µ 1 ϕ> < f,h α µh α µ 1 ϕ>. h α µ 1 h α µ f f h α µh α µ 1 f f. Thus h α µ and hα µ 1 are one-one map of H µ I onto itself. 5.4 Pseudo-differential operator involving fractional Hankel transformation A linear partial differential operator Ax, µ,x on I is given by Ax, µ,x m a r x µ,x r, r where the cofficient a r x are functions defined on I and µ,x is the same as in Proposition 5... If we replace µ,x r in by monomial ycscα r in I, then we obtain the so called symbol Ax,y m a r x ycscα r r In order to get another representation of the operator Ax, µ,x, let us take any function ϕ H µ I, then by using and Proposition 5..3i, we have Ax, µ,x ϕ x m r m r a r x µ,x r ϕx a r xh α µ 1 h α µ µ,x r ϕx m a r xh α µ 1 ycscα r hµϕ α x r 7

15 K α µ x,y m ra r x ycscα r K α µ x,yax,y ϕ α µ ydy, ϕ α µ ydy where K α µ x,y is the same as in If we replace the symbol Ax,y by more general symbol ax, y which is no longer polynomial in y, we get the pseudo-differential operator h α µ,a defined below. For pseudo-differential operator involving Fourier transformation and fractional Fourier transformation we may refer to 8, 36 and 31 respectively. Definition Let ax,y be a symbol. Then the pseudo-differential operator A µ,a associated to ax,y is defined by where ϕ µ α y as Aµ,a ϕ x Kµ α x,yax,y ϕ µ α ydy, Definition 5.4. Let l R. Then we define H l 1 to the set of all functions ax,y C I I such that for any two non-negative integers ν and β, there exists a positive constant C l,ν,β such that where x,y I. x 1 d ν y 1 d β ax,y dx dy C l,ν,β1+y l β, Zemanian 38, p.141 and Pathak, p.95, have given the following result 1 m+n x m x 1 d dx n x µ 1/ h µ ϕx y µ+m+n+1 y dy 1 d m y µ 1/ ϕy xy µ+n J µ+m+n xydy, m,n N Theorem The pseudo-differential operator A µ,a is a continuous linear mapping of H µ I into itself. 71

16 Proof: Let ϕ H µ I. Then for any two non-negative integers m and k, we need only prove that sup x I x m x 1 D k x µ 1/ A µ,a ϕx <. Following the technique of proof of Theorem and we have x m x 1 D k x µ 1/ A µ,a ϕx Now x m x 1 D k x µ 1/ Kµ α x,yax,y ϕ µ α ydy x m x 1 D k x µ 1/ Cα,µe i x cotα h µ e i y cotα ax,y ϕ µ α y xcscα x m C k k α,µ x 1 D k ν e i x cotα ν ν x 1 D ν x µ 1/ h µ e i y cotα ax,y ϕ µ α y xcscα. x m x 1 D k x µ 1/ A µ,a ϕx 1 m+ν C α,µ k ν k icotα k ν e i x cotα 1 m+ν x m ν x 1 D ν x µ 1/ h µ e i y cotα ax,y ϕ α µ y xcscα. Applying the Zemanian s result from 5.4.5, we have C k k α,µ icotα k ν e i x cotα sinα m ν µ 1/ ν ν y µ+m+ν+1 y 1 D m y µ 1/ e i y cotα ax,y ϕ µ α y xycscα µ+ν J µ+m+ν xycscαdy C k k α,µ icotα k ν e i x cotα sinα m ν µ 1/ ν ν y µ+m+ν+1 m m s m m s s t s t y 1 D m s t e i y cotα y 1 D t ax,yy 1 D s y µ 1/ ϕ α µ y xycscα µ+ν J µ+m+ν xycscαdy. 7

17 Suppose N is an integer no less than µ + m+ν + 1 and using 5.4.4, then we have x sup m x I x 1 D k x µ 1/ A µ,a ϕx k ν k cotα k ν sinα m ν µ 1/ m ν m s s t xycscα sup µ+ν x,y I J µ+m+ν xycscα m s m s t cotα m s t sup y I 1+y N+ y 1 D s y µ 1/ ϕ µ α y y 1 D t dy ax,y 1+y k k cotα k ν sinα m ν µ 1/ m m s m m s ν ν cotα m s t s t s t 1+y A µ,α C l,t sup N++l t y I y 1 D s y µ 1/ ϕ µ α, y xycscα where < sup µ+ν x,y I J µ+m+ν xycscα B µ,α. Using binomial formula and Theorem 5.3.1, we have x m x 1 D k x µ 1/ A µ,a ϕx <. sup x I 5.5 Boundedness of pseudo-differential operator In this section we assume that the symbol bx,y ax,yxycscα 1/ J µ xycscα satisfies the following condition instead of For given l R, assume that 1+x q x 1 d ν y 1 d β bx,y dx dy D l,ν,β,q,α1+y l β, where x,y I,D l,ν,β,q,α > and q,ν,β N. The class of all such symbols is denoted by H l. Definition A tempered distributions ϕ belongs to the Sobolev- type space G s µ,, s, µ R, if its fractional Hankel transformation h α µϕ corresponds to a locally integrable function on I, and 1+y s/ h α µϕy L I. A norm in this space is defined by ϕ G s µ, I 1+y s/ hµϕy α dy

18 Lemma For any symbol bx,y H l,l R and k N, there exists a positive constant C l such that Proof: Since so that, b α µy,ξ C l 1+ξ l 1+y csc α k/ b α µy,ξ 1+y csc α k bα µ y,ξ y,ξ I,k N. Then we have 1 µ,x k bx,ξ k r k r k r K α µ x,ybx,ξdx, K α µ x,y1 µ,x k bx,ξdx, r µ,x r bx,ξ k r k 1 r x r k 1 r x r r r n n r r r n bx,ξ k k x x km 1 d dx r m k x D k l,n,q,α 1+ξ l 1+x q. k So 1 µ,x k bx,ξ k r k r r r m k D l,n,q,α 1+ξ l n k 1+x k r q Hence by and 5.5.5, we get 1+y csc α k b α µy,ξ C α,µ k r k r r r m k D l,n,q,α 1+ξ l n k 1+x k r q dx. Since x- integral is convergent for sufficient large value of q, there exists a constant C l > such that b α µy,ξ C l 1+ξ l 1+y csc α k/. 74

19 Proposition For any symbol bx,y ax,yxycscα 1 J µ xycscα H l,l R, the associated operator A µ,a ϕ admits the representation A µ,a ϕ e i x cotα Cα,µ 1 Kµ α x,y e i ξ cotα bα µ y,ξ ϕ µ α ξdξ dy, where ϕ H µ I and all involved integrals are convergent. Proof: We first note that and so that, from and 5.5.8, we have where D l max1, l/. Hence from ξ l l/ 1+ξ l/ if l, ξ l 1+ξ l/ if l <, ξ l D l 1+ξ l/,ξ I, b α µy,ξ C l D l 1+ξ l/ 1+y csc α k/ Let g α y e i ξ cotα bα µ y,ξ ϕ α µ ξdξ, is in L 1 I. We now compute its inverse fractional Hankel transformation Kµ α x,yg α ydy Kµ α x,y e i ξ cotα bα µ y,ξ ϕ µ α ξdξ dy C α,µ e i x +y cotα xycscα 1 Jµ xycscα e i ξ cotα bα µ y,ξ ϕ µ α ξdξ dy x cotα {C α,µ C α,µ C α,µ e i x +y cotα xycscα 1 Jµ xycscα e i x +ξ cotα bα µ y,ξ ϕ µ α ξdξ }dy 75

20 x cotα {C α,µ C α,µ x cotα C α,µ x cotα C α,µ C α,µ e i x +ξ cotα ϕα µ ξ e i x +y cotα xycscα 1 Jµ xycscα b α µy,ξdy}dξ C α,µ e i x +ξ cotα bx,ξ ϕ µ α ξdξ C α,µ e i x +ξ cotα xξ cscα 1 Jµ xξ cscα bx,ξ ϕ xξ cscα 1 µ α ξdξ J µ xξ cscα x cotα C α,µ e i x +ξ cotα xξ cscα 1 Jµ xξ cscαax,ξ ϕ µ α ξdξ C α,µ x cotα C α,µ A µ,a ϕx Kµ α x,ξax,ξ ϕ µ α ξdξ x cotα C α,µ Theorem Let ax, y. bx,y xycscα 1 J µ xycsc α H l and A µ,a ϕ be associated operator. Then the following estimate holds true: A µ,a ϕ G µ, I Cα ϕ G l µ, I, where ϕ H µ I, for a certain constant C Cα Proof: We have seen the tempered distributions x cotα A µ,a ϕ has fractional Hankel transformation equal to Let us set C µ,α 1 h α i µ e x cotα A µ,a ϕ g α y e i ξ cotα bα µ y,ξ ϕ µ α ξdξ. U α y U α y e i ξ cotα bα µ y,ξ ϕ α µ ξdξ b α µy,ξ ϕ α µ ξ dξ. 76

21 Using 5.5.1, we have U α y C l D l 1+ξ l/ 1+y csc α k/ ϕ µ α ξ dξ C l D l 1+y csc α k/ 1+ξ l/ ϕ µ α ξ dξ. Let f α y 1+y csc α k/ and θ α ξ 1+ξ l/ ϕ µ α ξ. If k is sufficient large, f α L 1 I as ϕ µ α H µ I,θ α L I. Then Therefore, using 5.5.6, we have U α y L I Cα ϕ G l µ, I. C µ,α 1 x cotα A µ,a ϕ G µ, I C µ,α 1 h α i µ e x cotα A µ,a ϕ L I Hence this implies that U α y L I Cα ϕ G l µ, I. This completes the proof theorem. A µ,a ϕ G µ, I Cα ϕ G l µ, I. ***** 77