The Radon transform on hyperbolic space

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The Radon transform on hyperbolic space

Arp´´ ad Kurusa

Abstract. The Radon transform that integrates a function in Hⁿ, the n- dimensional hyperbolic space, over totally geodesic submanifolds with codimension 1 and the dual Radon transform are investigated in this paper. We prove inversion formulas and an inclusion theorem for the range.

0. Introduction

In this paper we investigate the Radon transform and its dual on the hyperbolic space. This problem has been introduced in [7] and has also been considered in [7,14]. The most interesting question concerns the inverse of the transform, and due to [7,14] this is known in odd dimension.

Iff ∈L²(Hⁿ), where Hⁿ is the n-dimensional hyperbolic space, the Radon transform of f is a function Rf defined on the set of hyperplanes, the totally geodesic submanifolds with codimension 1. The value ofRf at a given hyperplane is the integral of f over that hyperplane. The dual Radon transform R^∗F of a function F defined on the set of hyperplanes is a function on Hⁿ. The value of R^∗F at a given pointX is the integral of F over the set of hyperplanes passing throughX by the surface measure of the unit sphere of the tangent space atX (the normals of the hyperplanes atX project the surface measure of this unit sphere to a measure on the set of the hyperplanes throughX).

The points of the hyperplanes nearest to an arbitrarily chosen origin define a hypersurface and the ‘boomerang transform’ integrates a functionf defined on Hⁿ over this hypersurface by the projected measure of the unit sphere in TXHⁿ. This means that the boomerang transform is in principle the dual Radon transform, provided Hⁿ is identified (except at the origin) with the space of its hyperplanes

— each one being represented by its closest point to the origin. The above defined hypersurface is thus the set of the points from which the geodesic segment AMS Subject Classification(1980): 44A05 .

Geom. Dedicata, 40(1991), 325–339. c A. Kurusa^´

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joining the origin and the given point is seen at a right angle. For this reason our boomerang transform is an analog of the spherical means transformation of Cormack and Quinto [3].

The paper is outlined as follows.

We produce explicit formulas for the Radon and boomerang transforms on Hⁿ in Section 1 and invert them in Section 2. These new inversion formulas are obtained in terms of an expansion in spherical harmonics. They give the so-called support theorems as a byproduct. Inversion formulas were known before^∗ only in odd dimensions [8].

Our aim in Section 3 is to prove continuity results and inclusion theorems for the range of the Radon transform and for the null space of the boomerang transform on certain classes of square integrable functions.

In Section 4 we give two new closed inversion formulas valid in odd and even dimension respectively. Our odd dimensional formula exhibit some similarities to Helgason’s formula in [8] and the even dimensional one involves a distribution of the Cauchy principal value type. This shows that the inversion formula is local in odd dimensions and global in even dimensions just as in the Euclidean case.

1. Calculation of the transforms on H

ⁿ

For the calculation we use Poincar´e’s sphere model of Hⁿ, which is the unit ball Bⁿ in Rⁿ(|x| < 1) with Riemannian metric ds² = 4dx²/(1− |x|²)². This metric has constant curvature−1 and the geodesics are circular arcs perpendicular to the boundary ofBⁿ ,Sⁿ⁻¹, and thus the totally geodesic submanifolds are the spheres intersecting Sⁿ⁻¹ orthogonally.

Letµ:Bⁿ→Hⁿbe the parameterization ofHⁿwithBⁿgiven by the Poincar´e model. Since the Euclidean and hyperbolic metrics are conformally related the Euclidean and hyperbolic geodesic polar coordinates onBⁿ differ only in the radial coordinates. With i_B:Sⁿ⁻¹×[0,1)→Bⁿ and i_H:Sⁿ⁻¹×R+ →Hⁿ denoting the respective (polar) coordinatization this means that i⁻¹_H ◦µ◦iB(ω, ρ) = (ω,µ(ρ)),¯ where ¯µ: [0,1)→R⁺ is given as ¯µ(ρ) = ln((1 +ρ)/(1−ρ)), and ¯µ⁻¹(h) = th(h/2).

In what follows we shall frequently refer to the notation of Figure 1.

On the figure, the pointP of the geodesicM N is the nearest to the originO.

The geodesic M N is a piece of the circle centered atO⁰ with radius rinRⁿ. The

∗ We had been informed by the referee that recent results on the matter will appear in the proceedings of the AMS Arcata conference on integral geometry (June 1989). See [1,9].

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Figure 1

Euclidean distance ofO andP (resp. X) is p(resp. x). Here X is a point on the geodesicM N andα(resp. β) is the angle betweenOX andOP (resp. XP).

We start by noting the formulas

(1.1) p= x²+ 1

2xcosα− s

x²+ 1 2xcosα

2

−1 and

(1.2) x=p

r²+ 1 cosα−p

(r²+ 1) cos²α−1,

which derive from the law of cosine applied to the Euclidean trianglesON O⁰ and OXO⁰. Let δ = d(P, X) denote the hyperbolic distance of the points P and X.

Then by the Riemannian metric ds²we have δ=

ln

P M P N/XM

XN

and a straightforward calculation yields

(1.3) δ=1

2 ln

rcosα+ sinα rcosα−sinα

,

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where, as one can easily read off from Figure 1,r= (1/p−p)/2. Furthermore, the law of cosine applied to the triangle O⁰XOgives

(1.4) sinβ= 1−x²

q

x²+1 cosα

²

−4x² .

To get explicit formulas for the Radon and boomerang transforms, we parameter- ize the set of totally geodesic submanifolds of Hⁿ bySⁿ⁻¹×R+, in such a way that ξ(ω, h) denotes the totally geodesic submanifold perpendicular to the geodesic passing through the origin with tangent vector ω at distancehfrom the origin.

Lemma 1.1. Forf ∈L²(Hⁿ), the Radon transform is Rf(¯ω, h) =

Z

S_h,ωⁿ⁻¹

f ω,1

2µ¯ thh hω,ωi¯

(hω,ωi¯ ²cth²h−1)^−n/2

shh dω,

where dω is the surface measure of Sⁿ⁻¹, h., .i is the standard Euclidean scalar product andSⁿ⁻¹_h,¯_ω ={ω∈Sⁿ⁻¹: thh <hω,ωi}.¯

(Here and below we use the shorthandRf(¯ω, h) = (Rf)(ξ(¯ω, h))).

Proof. Let (¯ω, h) be the geodesic polar coordinates ofP ∈Hⁿ(p= th(h/2)) and letX be a point ofξ(¯ω, h) in directionω. (See Figure 1.)

First we assume thatn= 2 and parameterizeS¹by an angleα, with respect to some fixed direction. According to Figure 1, the pointXofξ(¯ω, h) is parameterized byα∈(−arccos_p+r¹ ,arccos_p+r¹ ) (this comes from the triangleOO⁰N) and we can immediately write that

Rf(¯ω, h) =

Z arccos(1/(p+r))

−arccos(1/(p+r))

f(α+ ¯α,µ(x(α)))¯

dδ dα

dα,

where ¯α is the angle of ¯ω with respect to the fixed direction and x(α), δ(α) are given by (1.2), (1.3) for P fixed and X varying withα. It is straightforward from (1.3) that

Rf(¯ω, h) = Z

1/(p+r)<cos(α)

f

α+ ¯α,1

2µ¯(r²+ 1)^−1/2 cosα

r

(r²+ 1) cos²α−1dα.

One can now substitute r = (1/p−p)/2 = (cth(h/2)−th(h/2))/2 = 1/shh to obtain the formula of the lemma.

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The configuration relevant for the case n > 2 can be obtained by rotating Figure 1 around the straight lineOP. It is well known that the surface element of a geodesic sphere with radiusρis shⁿ⁻¹ρdω. The reason for this is that the map ω → i_H(ω, ρ) of Sⁿ⁻¹ onto the geodesic sphere of radius ρin Hⁿ induces a map between their tangent spaces, which is a dilation by shρ. This and the fact that ξ(¯ω, h) is a rotational manifold imply that the surface measure on ξ(¯ω, h) at the point X reads as shⁿ⁻²µ(x(α))|dδ/dαk¯ _cos_α=hω,_ωi_¯ dω. This tells us that

Rf(¯ω, h) = Z

1/(p+r)<hω,ωi¯

f ω,1

2µ¯(r²+ 1)^−1/2 hω,ωi¯

rshⁿ⁻²(¹₂µ¯_(r2+1)^−1/2 hω,ωi¯

(r²+ 1)hω,ωi¯ ²−1 dω.

Substitutingr= 1/shhas above we obtain the statement of the lemma.

Lemma 1.2. Forf ∈L²(Hⁿ), the boomerang transform is Bf(¯ω, h) =

Z

Sⁿ⁻¹₀ ,¯ω

f ω,1

2µ(hω,¯ ωi¯ thh)(1− hω,ωi¯ ²th²h)⁻¹

chh dω.

Proof. Let (¯ω, h) be the geodesic polar coordinates ofX∈Hⁿ(x= th(h/2)). (See Figure 1)

First we assume thatn= 2 and parameterizeS¹by an angleα, with respect to some fixed direction. According to Figure 1, the point P is parameterized by α∈[−π/2, π/2] and we can write immediately that

Bf(¯ω, h) = Z π/2

−π/2

f(α+ ¯α,µ(p(α)))¯

dβ dα

dα,

where ¯αis the angle of ¯ω with respect to the fixed direction and p(α), β(α) are given by (1.1), (1.4) forX fixed andP varying withα. It is immediate from (1.4) that

Bf(¯ω, h) = Z π/2

−π/2

f

α+ ¯α,1

2µ¯ 2x

x²+ 1cosα 1−x⁴

(x²+ 1)²−4x²cos²αdα.

One can now substitutex= ¯µ⁻¹(h) = th(h/2) to obtain the formula of the lemma.

The configuration relevant for the case n > 2 can be obtained by rotating Figure 1 around the straight lineOX. Thus the conformality of the model implies the lemma.

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2. Inversion formulas and support theorems

We need the following two technical lemmas that can be easily proven from the formulas given in [2] and [4].

Lemma 2.1. If m∈ZthenI=π/2, where I=

Z q t

cos(marccos(thh/thq)) q

1−th²h/th²q

×ch(marcch(thh/tht)) q

th²h/th²t−1

dh shhchh. Lemma 2.2. If m ∈ Z, n >2, λ = (n−2)/2 and C_m^λ denotes the Gegenbauer polynomials of the first kind, then

M

sh(q−t) shq sht

n−2

= Z q

t

cthⁿ⁻³h C_m^λ thh

tht

C_m^λ thh

thq

×

× th²h

th²t −1

n−3

2

1−th²h th²q

n−3

2 dh

sh²h, where

M =π2³⁻ⁿ

Γ(m+n−2) Γ(m+ 1)Γ(λ)

² 1 Γ(n−1).

Now we present two propositions that describe our transformations in terms of spherical harmonics. For this purpose we recall the following facts.

A complete orthonormal system in the Hilbert spaceL²(Sⁿ⁻¹) can be chosen consisting of spherical harmonics Yl,m, where Yl,m is of degree m. If Yl,m is a member of such a system,f ∈C^∞(Sⁿ⁻¹×R+) and p∈R+ let the corresponding coefficients of the series in this system for f(ω, p) bef_l,m(p). Then the series

∞

X

l,m

fl,m(p)Yl,m(ω)

converges uniformly absolutely on compact subsets of Sⁿ⁻¹×R+ to f(ω, p) [13].

Below we use the expansions f(ϕ, q) =

∞

X

m=−∞

f_m(q) exp(imϕ) and f(ω, q) =

∞

X

l,m

f_l,m(q)Y_l,m(ω) in dimension 2 and in higher dimensions, respectively. These expansions will be used for the Radon and boomerang transforms Rf andBf as well.

Geom. Dedicata, 40 (1991), 325–339. c A. Kurusa^´

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Proposition 2.3. i)If f(ϕ, p)∈L²(H²) then (R) (Rf)m(h) = 2

Z ∞ h

fm(q)cos(marccos(thh/thq)) chh

q

1−th²h/th²q dq.

ii) If n >2,f(ω, p)∈L²(Hⁿ)andλ= (n−2)/2 then (RN) (Rf)_l,m(h) =|Sⁿ⁻²|

C_m^λ(1) Z ∞

h

f_l,m(q)C_m^λ thh

thq 1−th²h th²q

n−3

2 shⁿ⁻²q chh dq, where |S^k| is the area ofS^k andC_m^λ is the Gegenbauer polynomial.

Proof. Ifn= 2 substituting the expansions off andRfinto the formula of Lemma 1.1 we get

(Rf)_m(h) = Z

thh<cosα

f_m1

2µ¯thh cosα

1/shh

cth²h cos²α−1exp(imα)dα.

The factor exp(imα) can be replaced by its real part, because the domain of inte- gration is symmetric and the remainder of the integrand is an even function ofα.

Putting now q= ¹₂µ(¯ _cos^th^h_α) (cosα= thh/thq) this becomes (R).

Forn >2 write the expansions of f and Rf into the formula of Lemma 1.1 and use the Funk-Hecke theorem [13] to obtain

(Rf)l,m(h) = |Sⁿ⁻²| C_m^λ(1)

Z 1 thh

fl,m

1 2µ¯thh

t ×

×C_m^λ(t) 1−t²ⁿ⁻³₂ (t²cth²h−1)^−n/2

shh dt.

Puttingq=¹₂µ(¯ ^th_t^h) (t= thh/thq) this becomes (RN) which was to be proved.

Proposition 2.4. i)If f(ϕ, p)∈L²(H²) then (B) (Bf)m(h) = 2

Z h 0

fm(q)cos(marccos(thq/thh)) shh

q

1−th²q/th²h dq.

ii) If n >2,f(ω, p)∈L²(Hⁿ)andλ= (n−2)/2 then (BN) (Bf)_l,m(h) = |Sⁿ⁻²|

C_m^λ(1) Z h

0

f_l,m(q)C_m^λ thq

thh 1−th²q th²h

n−3

2 1

shhdq.

This proposition can be proven from Lemma 1.2 in the same way as the previous one so we leave the proof to the reader. Our following theorems give the inversion formulas. Since their proofs are very similar we only give the first proof.

Geom. Dedicata, 40 (1991), 325–339. c A. Kurusa^´

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Theorem 2.5. i)If f ∈C_c^∞(H²)⊂C_c^∞(S¹×R+)then (RI) fm(t) = −1

π d dt

Z ∞ t

(Rf)m(h)ch(marcch(thh/tht)) shh

q

th²h/th²t−1 dh.

ii) If n >2,f ∈C_c^∞(Hⁿ)⊂C_c^∞(Sⁿ⁻¹×R+) then (RNI) fl,m(t) = (−1)ⁿ⁻¹ Γ(m+ 1)Γ(λ)

2π^n/2Γ(m+n−2) _d

dtδ2δ4. . . δ_n−2F(t) if neven δ₁δ₃δ₅. . . δ_n−2F(t) if nodd, where δk =_dt^d²2 −k² (k∈N) and

F(t) = Z ∞

t

(Rf)l,m(h)C_m^λ thh

tht

th²h th²t −1

n−3

2 shⁿ⁻²t

shh cthⁿ⁻²h, dh Theorem 2.6. i)If f ∈C^∞(H²)⊂C^∞(S¹×R+)then

(BI) fm(t) = 1 π

d dt

Z t 0

(Bf)m(h)ch(marcch(tht/thh)) chh

q

th²t/th²h−1 dh.

ii) If n >2,f ∈C^∞(Hⁿ)⊂C^∞(Sⁿ⁻¹×R⁺) then (BNI) fl,m(t) = Γ(m+ 1)Γ(λ)

2π^n/2Γ(m+n−2)chⁿ⁻²t d

dtδ₂δ₄. . . δ_n−2F(t) if neven δ1δ3δ5. . . δn−2F(t) if nodd, where

F(t) = Z t

0

(Bf)l,m(h)C_m^λ tht

thh

th²t th²h−1

n−3

2 chⁿ⁻²t

chh thⁿ⁻²h dh.

Proof. To prove (RI) one multiplies (R) by ch(marcch(thh/tht))/ shh

q

th²h/th²t−1

and integrates fromtto infinity. Then one simplifies the integral by using Lemma 2.1 and finally differentiates with respect tot.

To prove (RNI) one multiplies (RN) by

C_m^λ thh

tht

th²h th²t −1

n−3

2 shⁿ⁻²t

shh cthⁿ⁻²h

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and integrates fromt to infinity again. Then Lemma 2.2 leads to F(t) =M|Sⁿ⁻²|

C_m^λ(1) Z ∞

t

fl,m(q) shⁿ⁻²(q−t)dq.

To finish the proof it is enough to observe that d²

dt²sh^k(q−t) =k²sh^k(q−t) +k(k−1) sh^k−2(q−t).

The following two corollaries are direct consequences of the above theorems.

The first one has a stronger version in [8,Theorem III.1.2].

Corollary 2.7. If f ∈ C_c^∞(Hⁿ) andA > 0 then the values of Rf(ω, p) forp≥A determinef(ω, p)forp≥A. IfRf(ω, p) = 0on this domain, thenf(ω, p) = 0too.

Corollary 2.8. Iff ∈C^∞(Hⁿ)andA >0then the values ofBf(ω, p)for0≤p≤A determinef(ω, p)for0≤p≤A. IfBf(ω, p) = 0 on this domain, thenf(ω, p) = 0 too.

3. Null spaces and ranges

Our first proposition in this section establishes the continuity of the Radon and boomerang transforms. Let ψ(ω, p) denote the hypersurface from the points of which the geodesic segment joining O andi_H(ω, p) is seen at a right angle. To calculate the boomerang transform at the point i_H(ω, p) one needs only integrate onψ(ω, p).

Proposition 3.1. Let S be a measurable set in Hⁿ and n ≥ 3. The maps R:L²(S,shⁿ⁻¹δxdx) → L²(S^R) and B:L²(S,sh¹⁻ⁿδxdx) → L²(S^B) are contin- uous, where

S^R={(ω, p)∈Sⁿ⁻¹×R+:ξ(ω, p)∩S6=∅}, S^B ={(ω, p)∈Sⁿ⁻¹×R+:ψ(ω, p)∩S6=∅},

δ_x is the distance ofxfrom the origin anddx is the Lebesgue measure onHⁿ.

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Proof. We can assumeS=Hⁿ without loss of generality. By the orthogonality of the spherical harmonics Yl,m it is enough to prove that for f(ω, p) =g(p)Yl,m(ω)

kRfk²_L2(Sⁿ⁻¹×R+)≤16|Sⁿ⁻²|²kfk²_L2(Hⁿ,shⁿ⁻¹δxdx). Using (RN) we see that

kRfk²_L2(Sⁿ⁻¹×R+)= 4k(Rf)l,mk²_L2(R+)

= 4|Sⁿ⁻²|² (C_m^λ(1))² ×

Z ∞ 0

1 ch²h

Z ∞ h

g(q)C_m^λthh thq

1−th²h th²q

ⁿ⁻³₂

shⁿ⁻²q dq2

dh.

Since |C_m^λ(t)| ≤ |C_m^λ(1)|for t∈[0,1], ch²h≥1 and |1−th²h/th²q| ≤1 we get kRfk²_L2(Sⁿ⁻¹×R+)≤4|Sⁿ⁻²|²

Z ∞ 0

Z ∞ h

|g(q)|shⁿ⁻²q dq 2

dh

= 4|Sⁿ⁻²|² Z ∞

0

1 h

Z h 0

g

1 q

shⁿ⁻²

1 q

dq q²

!2

dh.

At the same time Hardy’s inequality,

1 v

Z v 0

k(u)du _L₂₍

R+)≤2kkkL²(R+), gives

kRfk²_L2(Sⁿ⁻¹×R+)≤16|Sⁿ⁻²|² g

1 q

shⁿ⁻²

1 q

q⁻²

2

L²(R+).

Puttingp= 1/q, then usingq≤shqwe find the theorem forR. The proof for the boomerang transform is very similar and is left to the reader.

Theorem 3.2. Let S be a measurable set in Hⁿ (may be Hⁿ), n ≥ 3 and g_j,l,m(ω, h) = ^th_chh^j^hY_l,m(ω) for j, l, m ∈ N, where 0 ≤ j < m and (m−j) is even. The Radon transform is an injection of L²(S,shⁿ⁻¹δxdx)into

A= (Cl Sp{gj,l,m(ω, h)})^⊥∩L²(S^R),

where Cl Spmeans the closure of the span of the set of functions indicated.

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Proof. g_j,l,m∈L²(Sⁿ⁻¹×R+) is proved by the fact that kgj,l,mk_L2(Sⁿ⁻¹×R+)= 2/p

2j+ 1.

Now let

f(ω, q) =v(q)Yi,k(ω)∈L²(S,shⁿ⁻¹δxdx) and

g^∗(ω, h) =g(thh)Yl,m(ω)/chh.

We get immediately from (RN) by changing the order of integrations that hg^∗,Rfi_L2(Sⁿ⁻¹×R+)

=δl,iδm,k

4|Sⁿ⁻²| C_m^λ(1)

Z ∞ 0

v(q)shⁿ⁻¹q chq

Z 1 0

g(xthq)C_m^λ(x) 1−x²ⁿ⁻³₂ dx dq, where δm,k is the Kronecker delta. This means thatRf is orthogonal tog^∗ for all f ∈L²(S,shⁿ⁻¹δ_xdx) if and only if

Z 1 0

g(xy)C_m^λ(x) 1−x²ⁿ⁻³2 dx≡0.

By Lemma 5.1 and Theorem 5.2 of [11] and Theorem 3.2 of [10] this is equivalent to g being in the closure of the span of the functions x^j, where 0 ≤ j < mand m−j is even. This proves that the range of the Radon transform is inA.

Now we prove the injectivity ofR. We are looking for a function f(ω, q) = v(q)Yi,k(ω)∈L²(S,shⁿ⁻¹δxdx), which has zero Radon transform. Thus we should find a functionv∈L²(R+,sh²ⁿ⁻²q dq) which satisfies the equation

0≡ Z ∞

h

v(q)C_m^λthh thq

1−th²h th²q

ⁿ⁻³₂

shⁿ⁻²q dq.

Assuming v(q) = w(cthq) sh¹⁻ⁿq/chq and changing the variables, s = thq and t= thh, we get the integral equation

0≡ Z 1

t

1 sw

1 s

C_m^λ

t

s 1−t² s²

ⁿ⁻³2

ds,

which has to be satisfied by w for all t ∈ [0,1]. Since w(1/s) ∈ L²([0,1]) this Volterra integral equation is of the type (3.7) of [12], so its only solution is the zero function, which completes the proof.

A similar method can be used for getting the corresponding result for the boomerang transform.

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Theorem 3.3. Let S be a measurable set in Hⁿ (may be Hⁿ), n ≥ 3 and gj,l,m(ω, h) = ^th_ch^j2^hhYl,m(ω) forj, l, m∈N, where 0≤j < mand(m−j)is even.

The kernel of the boomerang transform inL²(S,sh¹⁻ⁿδ_xdx)isCl Sp{g_j,l,m(ω, h)}.

4. Closed inversion formula

Theorem 4.1. Let n≥2,λ= (n−2)/2andf ∈C_c^∞(Hⁿ). If nis odd then f(¯ω, t) = (−1)ⁿ⁻¹² 2¹⁻ⁿ

πⁿ⁻¹δ₁δ₃. . . δ_n−2 B

Rf(ω, h)cth^2λh shh

(¯ω, t) shⁿ⁻¹t .

If nis even then

f(¯ω, t) = (−1)ⁿ² 2¹⁻ⁿ πⁿ

d

dtδ2δ4. . . δ_n−2 B

HD

Rf(ω, h)cth^2λh shh

E

(¯ω, t) shⁿ⁻¹t ,

where the Hdistribution is Hf(ω, h) = 1

ch²h Z ∞

−∞

f(ω, r) 1

thr−thhdr.

(We use here the natural identification f(ω, r) =f(−ω,−r) for r <0.) Proof. We start with the odd-dimensional case, where Theorem 2.5 tells us that

fl,m(t) =CD Z ∞

t

tht

th²h th²t −1

n−3

2 shⁿ⁻²t

shh cthⁿ⁻²h dh, whereC=_2π^{Γ(m+1)Γ(λ)}_n/2_Γ(m+n−2) and D=δ1δ3. . . δn−2. The integralR∞

t can be modified by making use of R∞

t =R∞ 0 −Rt

0 which yields (∗)

fl,m(t) =I+C(−1)^n−1/2D Z t

0

tht

×

1−th²h th²t

n−3

2 shⁿ⁻²t

shh cthⁿ⁻²h dh, where

I=CD Z ∞

0

tht

th²h th²t −1

n−3

2 shⁿ⁻²t

shh cthⁿ⁻²h dh.

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Using (RN) and reversing the order of integrations it turns out thatIis proportional to

Z ∞ 0

f_l,m(q) sh^2λq Z

Ψ^q₀C_m^λ thh

thq 1−th²h th²q

n−3 2

×

× D



C_m^λ thh

tht

th²h th²t −1

n−3 2

sh^2λt





cthⁿ⁻¹h ch²h dh dq.

Substitutingx= thh/thqin J, the integral with respect to h, and using that the integrand is an even function we obtain

J = cth^2λq 2

Z 1

−1

C_m^λ(x) 1−x²ⁿ⁻³₂

×

× D



C_m^λ thq

thtx th²q th²tx²−1

n−3 2

sh^2λt



x¹⁻ⁿdx.

Let us recall thatC_m^λ(x) 1−x²ⁿ⁻³₂

is a polynomial of degreem+n−3 and that {C_m^λ(x)} is an orthogonal system on [-1,1] with weight-function 1−x²ⁿ⁻³₂

[6].

Now we prove that the polynomialD

C_m^λ(x^th_th^q_t)_th2q

th²tx²−1ⁿ⁻³₂ sh^2λt

is divis- ible by xⁿ⁻¹ which impliesJ = 0 andI= 0 by the above facts. The coefficient of x^k, for 0≤k≤n−2, vanishes if

δ₁δ₃. . . δ_n−2

shⁿ⁻²t cth^kt

= 0, which can be verified by induction after establishing that

δk(ch^kt) =−k(k−1) ch^k−2t and δk(sh^kt) =k(k−1)sh^k−2t.

Now equation (∗) gives just the expansion of the closed inversion formula stated for odd dimension.

Let us consider now the even-dimensional case, when λ is integer. Let D denote the differential operator _dt^dδ2δ4. . . δ_n−2. Theorem 2.5 says that

(∗∗)

fl,m(t) =−CD Z ∞

t

tht

th²h th²t −1

n−3

2 shⁿ⁻²t

shh cthⁿ⁻²h dh.

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For further use, we recall that D_m^λ is the Gegenbauer function of the second kind and the polynomialE^λ_m+2λ−1is given by (A.14) and (A.4) of [5] as

E_m+2λ−1^λ =

(−1)^λ+12^λ−1Γ(λ)(1−x²)^λ−1/2D^λ_m(x) if 0< x <1 2^λ−1Γ(λ)(x²−1)^λ−1/2(C_m^λ(x)−2D^λ_m(x)) if 1< x.

The function I_m^λ is defined by (22) of [5] as I_m^λ(x) =

Z 1

−1

C_m^λ(t) 1−t²^λ−1/2

(x−t)⁻¹dt and in (24) and (25) of [5] it is proved that

I_m^λ(x) =

π(1−x²)^λ−1/2D^λ_m(x) if 0< x <1 2πe^−iπλ(x²−1)^λ−1/2D^λ_m(x) if 1< x.

As in the odd-dimensional case, one can easily see that 0 =CD

Z ∞ 0

(Rf)l,m(h)E_m+2λ−1^λ thh tht

shⁿ⁻²t

shh cthⁿ⁻²h dh,

because E_m+2λ−1^λ is a polynomial of degree m+n−3 [5]. To proceed one has to rewrite this integral as Rt

0+R∞

t and substitute the appropriate expressions of E^λ_m+2λ−1 into Rt

0 and R∞

t respectively. Adding the result to equation (∗∗) one obtains

fl,m(t) =−CDhZ ∞ t

(Rf)l,m(h)2D^λ_mthh tht

th²h

th²t −1ⁿ⁻³₂ shⁿ⁻²t

shh cthⁿ⁻²h dh+

+ Z t

0

(Rf)l,m(h)(−1)^λD^λ_mthh tht

1−th²h th²t

ⁿ⁻³₂ shⁿ⁻²t

shh cthⁿ⁻²h dhi that can be written in the form

fl,m(t) =−C

π(−1)^λD Z ∞

0

(Rf)l,m(h)I_m^λthh tht

shⁿ⁻²t

shh cthⁿ⁻²h dh, where

I_m^λthh tht

= Z 1

−1

C_m^λ(x)(1−x²)ⁿ⁻³² thh

tht −x−1

dx.

Now the substitutionx= thr/thtand a change in the order of integrations result in

f_l,m(t) =−C(−1)^λ

π D

Z t

−t

C_m^λ thr

tht 1−th²r th²t

n−3 2

×

×shⁿ⁻¹t sht

1 ch²r

Z ∞ 0

(Rf)l,m(h)cthⁿ⁻²h/shh thh−thr dh dr.

This equation is just the expansion of the closed inversion formula stated for even dimensions.

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The author would like to thank L.Feh´er and the referee for making valuable suggestions on the form and contents of this paper.

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[11] E. T. QUINTO, Null spaces and ranges for the classical and spherical Radon transforms, J. Math. Anal. Appl., 90(1982), 408–420.

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A. K´ URUSA, Bolyai Institute, Aradi v´ertan´uk tere 1. 6720 Szeged, Hungary; e-mail:

kurusa@math.u-szeged.hu

Geom. Dedicata, 40(1991), 325–339. c A. Kurusa´