Identifying rotational Radon transforms

Identifying rotational Radon transforms

Arp´´ ad Kurusa^∗

Abstract. We show classes of test functions so that dilational and rotational invariances of the image R_S,µf of such a test functionf determines dilational and rotational invariances of rotational Radon transform R_S,µ. Then we de- termine the defining flowerS and weightµ of a conformal Radon transform R_S,µ in terms of the image R_S,µf of an unknown function that is a sum of an L²-function and finitely many Dirac distributions if the flower S is not selftangent.

1. Introduction

LetSω be a set of hypersurfacesS_ω,t in Rⁿ so thatω∈Sⁿ⁻¹ andt∈[0,∞).

TheRadon transformR_S,µof functionsf:Rⁿ →Rintegrable on eachSω,tis defined by

(1.1) R_S,µf(ω, t) =

Z

Sω,t

f(x)µ_ω,t(x) dx,

where dx is the natural surface measure on Sω,t and µω,t is a strictly positive continuous function onSω,tthat depends continuously onωandt. In this definition, the hypersurfacesSω,tare called thepetals, the setS=S

ω∈Sⁿ⁻¹Sωof them is called flowerandµω,t is called theweighton the petalSω,t.

In terms of this definition the “classic” Radon transform R_H,µ is defined by the flower H={Hω,t:ω ∈Sⁿ⁻¹, t∈[0,∞)} of the petalsHω,t={x:hx, ωi=t}

with weight µ_ω,t≡1.

AMS Subject Classification(2000): 44A12.

Key words and phrases: invariant Radon transform, identification of Radon transform, rotational Radon transform, conformal Radon transform.

∗Supported by the European Union and co-funded by the European Social Fund under the project “Telemedicine-focused research activities on the field of Matematics, Informatics and Medical sciences” of project number “T ´AMOP-4.2.2.A-11/1/KONV-2012-0073”.

The problem considered in this article is to identify an a priori unknown Radon transform R_S,µ or the properties of its flower and/or weight by means of the known images R_S,µf of (partially) (un)known test functionsf.

In the special case where the petals are total geodesics of a certain Riemannian manifold Mukhometov proves in [9] and [10] that R_S,11 determinesS.

Considering theattenuated Radon transformR_H,µ which was arisen in single photon emission computed tomography Natterer shows in [11] and [12] that the weight µ can be computed up to a multiplicative function in the support of the test functionf iff is known to be a finite sum of Dirac measures. Hertle proves in [5] that forconstantly attenuated Radon transformsR_H,µthe map (f, µ)7→R_H,µfis injective on the set of the compactly supported, not radial distributionsf. Solmon shows in [16] that the weight µ of an exponential Radon transform (which is a synonym for constantly attenuated Radon transforms) R_H,µcan be computed from R_H,µf if and only if the compactly supported distributionf is not radial.

For generalized Radon transforms J. Boman justifies in [1] that if R_H,µf = R_H,νgfor the compactly supported measuresf, g, andf is a sum of an L²-function and a finite sumf₀ of Dirac measures, theng is also a sum of an L²-function and a finite sum g0 of Dirac measures andf0µ≡g0ν.

In this article, first we describe the class of Radon transforms we are studying then present some preliminary results on appropriate flowers and on the dual Radon transforms.

In Section 3 we show classes of test functions so that dilational and rotational invariances of the image R_S,µf of such a test functionf determine dilational and rotational properties of the defining flowerS and weight µ.

Finally in Section 4 we determine the defining flowerS and weightµ of any conformal Radon transform R_S,µ in terms of the image R_S,µf of an unknown function that is a sum of an L²-function and finitely many Dirac distributions if the flowerS has no two tangent petals.

2. Preliminaries

We call a flowerS in Rⁿ (2≤n∈N)

• symmetricifx∈Sω,t is equivalent to 2hx, ωiω−x∈Sω,t, i.e., the petalsSω,t

are invariant against the reflection onto the straight line through the origin with directionω,

• rotationalif S_Φω,t = Φ(S_ω,t) for any rotation Φ∈ SO(n), i.e., any rotation around the origin rotates a petal into a petal with the same second parameter and rotated first parameter, and

• dilationalif ¯tS_ω,t =tS_ω,¯tfor all t,¯t >0, i.e., any homogeneous dilation with the origin as homothetic centre takes a petal into a petal with the same first parameter and dilated second parameter.

Observe, that in dimensions n ≥ 3 the petals Sω,t of a rotational flower S are rotationally symmetric around the straight line through the origin with directionω and therefore they are also symmetric. If a flower is rotational and dilational, then we call it conformal.

A weightµon a flowerS is called

• symmetricifS is symmetric andµ_ω,t(x) =µ_ω,t(2hx, ωiω−x) for allx∈S_ω,t,

• rotational if S is rotational and µ_Φω,t(Φx) = µ_ω,t(x) for all Φ ∈ SO(n), ω∈Sⁿ⁻¹,t∈Randx∈Sω,t, and

• dilationalifS is dilational andµ_ω,pt(px) =µ_ω,t(x) for all p >0.

If a weight is rotational and dilational, then it is called conformal. Observe, that a rotational weightµhas necessarily the form

(2.1) µω,r(x) = (

¯ µ_r

|x|_|hx,ω^hx,ω⊥⊥ⁱi|

, ifn= 2,

¯

µr(|x|), ifn >2 or it is symmetric inn= 2, for a suitable function ¯µ:R+×R→R.

We call a Radon transform of form (1.1)symmetric,rotational, dilationalor conformalif both the flower and the weight are symmetric, rotational, dilational or conformal, respectively. To mention some examples, the classic Radon transform is symmetric conformal, while the exponential Radon transform is symmetric and rotational but not dilational.

A piecewise differentiable image cr of an interval in the plane is called an inner(outer)nicecurve if in a coordinate system

(1) cr has a unique point Pr farthest from (closest to) the origin, so that Pr = (r,0),

(2) each point c⁺_r(%) of the curve c_r in the closed upper halfplane is uniquely determined by its distance%=|c⁺_r(%)| ≥0 to the origin,

(3) each pointc⁻_r(%) of the curvec_r in the closed downward halfplane is uniquely determined by its distance%=|c⁻_r(%)| ≥0 to the origin, and

(4) the signed angle ϕ^±_r(%) of c^±_r(%) to the x-axis as a function is such that ϕ^±_r(%)/p

|r−%| is piecewise C² on suppϕ^±_r = (inf_∃c^±

r(%)%, r] (suppϕ^±_r = [r,sup_∃c±

r(%)%)).

We call a nice curvesymmetric^(2:1)if the curvesc^±_r are the reflection of each other with respect to the x-axis. If this happens, we defineϕr(%) :=±ϕ^±_r(%).

(2:1) This corresponds to the symmetry term defined at the beginning of this section.

A surface inRⁿ is calledinner(outer)nice, if it is obtained by spinning such a symmetric inner (outer) nice plane curvecraround its axis for which

(5) the angle ϕ_r(%) of c_r(%) to the axis of c_r as a function is such that ϕr(%)/p

|r−%|is piecewise C^[(n+2)/2]on suppϕr= (inf_∃cr(%)%, r] (suppϕr= [r,sup_∃c

r(%)%)).

Note, that a nice surface inR²is a symmetric nice curve. Obviously,His an outer nice flower, and the set of spheres passing through the origin is an inner nice flower.

According to [7], if a conformal Radon transformR_S,1 is invertible, then the petals of its flower are nice (see [2], [3] for special nice flowers). Based on this result, without further notice in this article we restrict our considerations to rotational Radon transforms with nice flowers, and generally work out the calculations only for inner nice flowers.

On a nice curvec_r the arc length measure at c_r(%) is q

1 + (%ϕ˙^±r(%))²d%[7], therefore a rotational Radon transform in the plane takes the form

(2.2) R_S,µf(ωα, r) = Z r

0

X

◦∈±

f(%ω_α+ϕ^◦

r(%))µα,r(%ω_α+ϕ^◦

r(%))p

1 +%²( ˙ϕ^◦_r(%))²d%,

where we have used the practical notationωαfor the unit vector (cosα,sinα).

On a nice surface S_r in Rⁿ (n ≥3) the surface measure at the point %ω is (%sinϕr(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%) dωd%[7], hence a rotational Radon transform inRⁿ (n≥3) takes the form

(2.3)

R_S,µf(¯ω, r)

= Z r

0

Z

Sⁿ⁻²ω¯

f(%(ωsinϕr(%) + ¯ωcosϕr(%)))×

×µω,r¯ (%(ωsinϕr(%) + ¯ωcosϕr(%))) dω×

×(%sinϕ_r(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%) d%, where ¯ω is a unit vector andSⁿ⁻²ω¯ denotes the (n−2)-dimensional unit sphere in the subspace Rⁿ⁻¹ perpendicular to ¯ω.

Throughout this article for any function g:Rⁿ → R we use the notations g^Φ(x) := g(Φx) and g_p(x) := g(px) for any Φ∈ GL(n) and for any 0 6=p ∈R, respectively. Also the function spaces

L²(Rⁿ, r^k) ={f:Rⁿ→R:f(x)|x|^k/2∈L²(Rⁿ)}, L²_m(Rⁿ, r^k) ={f ∈L²(Rⁿ, r^k) :|x| ≤m⇒f(x) = 0},

L²_∗(Rⁿ, r^k) =S

m>0L²_m(Rⁿ, r^k),

where k ∈Z and m > 0, appear regularly. For the Dirac delta distribution δon Rⁿ we use the notationδx(y) :=δ(x−y). In the planeω^⊥ is used to denote the unit vector obtained by rotating the unit vector ω anticlockwise with angle π/2, i.e., ω_ϕ^⊥=ωϕ+π/2 with the frequently used notationωα= (cosα,sinα).

We close this section with general observations on the flowers, weights and the Radon transforms they define.

Lemma 2.1. A nice rotational flower inRⁿ (n≥2) is dilational if and only if (2.4) sinⁿ⁻²ϕ^±_r(yr)

q

1 +y²r²( ˙ϕ^±r)²(yr) = sinⁿ⁻²ϕ^±₁(y) q

1 +y²( ˙ϕ^±₁)²(y) for all y∈[0,1]andr∈[0,∞).

Proof. If the flower is conformal, then clearlyϕ^±_r(yr) =ϕ^±₁(y). Differentiation of this byy gives ˙ϕ^±₁(y) =rϕ˙^±_r(yr). These two equations prove (2.4).

Now we assume (2.4), and letψ(y) = sinⁿ⁻²ϕr(yr)p

1 +y²r²ϕ˙²_r(yr).

Our assumption in dimension 2 implies the independence ofφ(y) :=rϕ˙_r(yr) from rthat leads toϕr(yr) =R

φ(y)dy, whence the statement follows.

In higher dimensions, differentiating the square ofψ(y) with respect tor, we obtain

(2.5) sin²ⁿ⁻⁵ϕr(yr)(νr(y)gr(y) +λr(y) ˙gr(y)) =∂rψ(y)≡0, where

gr(y) =∂r(ϕr(yr)) =yϕ˙r(yr) +∂rϕr(yr), νr(y) = (2n−4) cosϕr(yr)(1 +y²r²ϕ˙²_r(yr)), λr(y) = sinϕr(yr)2ry²ϕ˙r(yr).

By (5) we have a function h(r, %) ∈C^[(n+2)/2]({(r, %)∈ R² : 0≤%≤ r}) so that ϕr(%) =h(r, %)√

r−%. This implies gr

% r

= %

rϕ˙r(%) +∂rϕr(%)

= % r

∂2h(r, %)√

r−%− h(r, %) 2√

r−%

+

∂1h(r, %)√

r−%+ h(r, %) 2√

r−%

= (2%∂₂h(r, %) + 2r∂₁h(r, %) +h(r, %))

√r−% 2r ,

hence g_r is differentiable, and g_r(1) = 0. Since the factors in equation (2.5) are continuous andϕrdoes not vanish on any interval, dividing (2.5) by sin²ⁿ⁻⁵ϕr(yr) yields the equivalent equation

(2.6) ν_r(y)g_r(y) +λ_r(y) ˙g_r(y)≡0.

Ifλ_r(y)6= 0, thenλ_r does not vanish on an open intervalI 3y, and on that interval one can multiply (2.6) by

κ_r(y) := expZ νr(y)−λ˙r(y) λr(y) dy to arrive at the equivalent differential equation

∂_y(λ_r(y)κ_r(y)g_r(y)) =∂_y(λ_r(y)κ_r(y))g_r(y) +λ_r(y)κ_r(y) ˙g_r(y)

=ν_r(y)κ_r(y)g_r(y) +λ_r(y)κ_r(y) ˙g_r(y)≡0.

This means thatλr(y)κr(y)gr(y) does not depend ony in the intervalI, therefore ifλ_r vanishes on any endpoint ofI, thenλ_r(y)κ_r(y)g_r(y)≡0 on the wholeI.

Thus, ifλr has a root in [0,1], thenλrκrgr vanishes on the whole [0,1]. If on the contraryλrhas no root in [0,1], thengr(1) = 0 implies the vanishing ofλrκrgr

on the whole [0,1].

Asλrκrgr vanishes on the whole interval [0,1], if λr(y)6= 0 then κr(y)>0 implies g_r(y) = 0. If on the contrary λ_r(y) = 0, then equation (2.6) implies gr(y) = 0, becauseνr(y)>0.

We concludeg_r≡0 that means∂_r(ϕ_r(yr))≡0. The proof is complete.

In the following theorem both inner and outer nice flowers appear. It shows that the petals of the flower of the dual of a rotational Radon transform R_S,µare the inversions of the petals of S in the unit sphere, which is a generalization of Cormack’s observation in [3].

In the theorem and its proof we identify the functionsf, g:Rⁿ →Rby their counterpartsf◦M andg◦M onSⁿ⁻¹×R+, respectively, by the mapM:Sⁿ⁻¹× R+→Rⁿ with (ω, r)7→rω.

Theorem 2.2. Let S be a conformal flower defined by the curves c^◦_r(%) =%ω_ϕ^◦

r(%), where (◦ ∈ ±), and define the flower S^∗ by the curves c^◦∗_% (r) = rω_ψ◦

%(r), where ψ_%^◦(r) =−ϕ^◦_r(%). Letµbe a rotational weight on the flowerS and define the weight µ^∗ on S^∗ by%µ^∗_ω,%(r¯ω) =rµ_ω,r¯ (%ω). Then the Radon transformR_S∗,µ^∗ is the dual R^∗_S,µof the Radon transformR_S,µ.

Proof. Since the flowerS is dilational, we observe that (2.7) rψ˙%(r) =r∂

∂r −ϕ1

% r

= ˙ϕ1

% r

% r =% ∂

∂%

ϕ1

% r

=%ϕ˙r(%).

We are proving that hR_S,µf, gi=hf,R_S∗,µ^∗gi.

For dimensionn= 2, by equation (2.2) we have

hR_S,µf, gi= Z ∞

0

Z

S¹

R_S,µf(¯ω, r)g(rω)r¯ d¯ωdr

= Z ∞

0

Z π

−π

Z r 0

X

◦∈±

f(%ωα+ϕ^◦_r(%))µωα,r(%ωα+ϕ^◦_r(%))×

×p

1 +%²( ˙ϕ^◦_r(%))²d% g(rω_α)rdαdr.

Using (2.7) this implies

hR_S,µf, gi= Z ∞

0

Z ∞

%

Z π

−π

X

◦∈±

f(%ω_α+ϕ◦

r(%))g(rω_α)µ^∗_ω

α+ϕ◦

r(%),%(rω_α) dα×

×p

1 +%²( ˙ϕ^◦_r(%))² dr %d%

= Z ∞

0

Z π

−π

f(%ω_β) Z ∞

%

X

◦∈±

g(rωβ+ψ^◦_%(r))µ^∗_ωβ,%(rωβ+ψ^◦_%(r))×

× q

1 +r²( ˙ψ_%^◦(r))² drdβ %d%

= Z ∞

0

Z π

−π

f(%ω_β)R_S∗,µ^∗g(ωβ, %) dβ %d%=hf,R_S∗,µ^∗gi,

which means R^∗_S,µ= R_S∗,µ^∗ as was stated.

For higher dimensionsn≥3, we use equation (2.3) to get

hR_S,µf, gi= Z ∞

0

Z

Sⁿ⁻¹

R_S,µf(¯ω, r)g(r¯ω)rⁿ⁻¹d¯ωdr

= Z ∞

0

Z

Sⁿ⁻¹

Z r 0

Z

Sⁿ⁻²ω¯

f(%(ωsinϕr(%) + ¯ωcosϕr(%)))×

×µω,r¯ (%ˆωsinϕr(%) + ¯ωcosϕr(%))) dω×

×%ⁿ⁻²sinⁿ⁻²ϕ_r(%)p

1 +%²ϕ˙²_r(%) d% g(r¯ω)rⁿ⁻¹d¯ωdr

= Z ∞

0

Z ∞

%

Z

Sⁿ⁻¹

Z

Sⁿ⁻¹

f(%ˆω)g(rω)δ¯ _cosϕr(%)(hˆω,ωi)µ¯ _ω,r¯ (%ˆω) dˆωd¯ω×

×%ⁿ⁻²rⁿ⁻¹sinⁿ⁻²ϕr(%)p

1 +%²ϕ˙²_r(%) drd%.

Using (2.7) this results in hR_S,µf, gi=

Z ∞ 0

Z ∞

%

Z

Sⁿ⁻¹

Z

Sⁿ⁻¹

f(%ˆω)g(rω)δ¯ _cosϕr(%)(hˆω,ωi)µ¯ ^∗_ω,%ˆ (r¯ω) d¯ωdˆω×

×%ⁿ⁻¹rⁿ⁻²sinⁿ⁻²ψ%(r) q

1 +r²ψ˙_%²(r) drd%

= Z

Sⁿ⁻¹

Z ∞ 0

f(%ˆω) Z ∞

%

Z

Sⁿ⁻¹

g(r¯ω)δ_cosψ%(r)(hˆω,ωi)µ¯ ^∗_ω,%ˆ (r¯ω) d¯ω×

×rⁿ⁻²sinⁿ⁻²ψ_%(r) q

1 +r²ψ˙²_%(r) dr %ⁿ⁻¹d%dˆω

= Z

Sⁿ⁻¹

Z ∞ 0

f(%ˆω)R_S∗,µ^∗g(ˆω, %)%ⁿ⁻¹d%dˆω=hf,R_S∗,µ^∗gi that shows R^∗_S,µ= R_S∗,µ^∗.

3. Identifying properties of rotational Radon transforms

A Radon transform R_S,µ is called rotationally invariant if R_S,µf(Φ¯ω, r) = R_S,µf^Φ(¯ω, r) for all Φ∈SO(n) and for allf ∈L²(Rⁿ), and it is calleddilationally invariant if R_S,µf(¯ω, pr) =pⁿ⁻¹R_S,µf_p(¯ω, r) for allp >0 and for allf ∈L²(Rⁿ).

We call a rotationally and dilationally invariant Radon transform R_S,µconformally invariant.

Quinto proved in [13] that rotationally invariant generalized Radon trans- forms R_H,µ have rotational weight µ. Here we generalize and sharpen Quinto’s result using a different method and theattenuated hyper Pompeiu propertywhich is defined in the Appendix of this article.

Theorem 3.1. Let S be a rotational flower, and the function f ∈L²(Rⁿ)be inte- grable on every petal of S. Define the functions f_%(ω) :=f(%ω) on Sⁿ⁻¹ for all

% >0 and the functions g`(x) :=|x|<sup>`</sup>f(x)for all `∈N. Assume that (1) ifn≥3, then each f% has the attenuated hyper Pompeiu property, and (2) ifn= 2, then R

S¹f%(ω) dω6= 0 and the equationf%(ωα+β) =f%(ωα)has only finitely many solutions forαat any β∈(−π, π].

If we have

(3) R_S,µg`(Φ¯ω, r) = R<sub>S,µ</sub>g<sup>Φ</sup><sub>`</sub>(¯ω, r)for allΦ∈SO(n)and`∈N, then R_S,µ is rotational.

Proof. We have only to prove that the weightµis rotational.

If the dimensionnis at least 3, then equation (2.3) and the conditions give for any g_`that

(3.1)

R_S,µg`(Φ¯ω, r)

= Z r

0

Z

Sⁿ⁻²ω¯

g_`(%Φ(ωsinϕ_r(%) + ¯ωcosϕ_r(%)))×

×µ_ω,r¯ (%(ωsinϕ_r(%) + ¯ωcosϕ_r(%)))×

×(%sinϕr(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%) dωd%

= Z r

0

Z

Sⁿ⁻¹

f%(Φω)δcosϕ_r(%)(hω,ωi)µ¯ ω,r¯ (%ω) dω×

×(%sinϕr(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%)%^`d%, where δ_cosϕ

r(%) is the Dirac delta distribution supported at cosϕr(%) and h., .i denotes the usual inner product onRⁿ.

By the Stone–Weierstrass Theorem the set of functions%<sup>`</sup> (`∈N) is a com- plete base in L²[0, r], whence equations (3.1) (`∈N) determine the function

ψ(Φ¯ω, r, %) :=

Z

Sⁿ⁻¹

f%(Φω)δ_cosϕ

r(%)(hω,ωi)µ¯ ω,r¯ (%ω) dω (0< %≤r).

Substituting ¯ω= Φ⁻¹ωˆ andω= Φ⁻¹ω˜ results in (3.2) ψ(ˆω, r, y) =

Z

Sⁿ⁻²_ωˆ

f_%(ω)µ_Φ−1ω,rˆ (%(Φ⁻¹ωsinϕ_r(%) + Φ⁻¹ωˆcosϕ_r(%))) dω.

Considering Φ inK_ωⁿ_¯ :={Φ∈SO(n) : Φˆω= ˆω} ∼=SO(n−1), we arrive at (3.3) ψ(ˆω, r, y) =

Z

Sⁿ⁻²_ωˆ

f%(ω)µω,rˆ (%(Φωsinϕr(%) + ˆωcosϕr(%))) dω

for all Φ∈ Kⁿ_ω¯. Sincef%has the attenuated hyper Pompeiu property, its restriction on any great sphere in an everywhere dense set of great spheresSⁿ⁻²ωˆ has the Pom- peiu property, which means thatµˆω,r(%(Φωsinϕr(%) + ˆωcosϕr(%))) is independent from Φ∈ Kⁿ_ωˆ on an everywhere dense set of ˆω ∈Sⁿ⁻¹. Sinceµ is continuous, this implies that the function µω,rˆ (%(ωsinϕr(%) + ˆωcosϕr(%))) is independent fromω for every ˆω∈Sⁿ⁻¹. By this and equation (3.2) we conclude

ψ(ˆω, r, y) =µ_Φ⁻¹_ω,rˆ (%Φ⁻¹(ˆω^⊥sinϕr(%) + ˆωcosϕr(%))) Z

Sⁿ⁻²_ωˆ

f%(ω) dω

for any unit vector ˆω^⊥ that is perpendicular to ˆω. This shows thatµ_{Φ ˆω,r}(%Φω) is independent from Φ that is what was to be proved for higher dimensionsn≥3.

Now, we proceed to the case of dimensionn= 2.

Equation (2.2) and the conditions imply for anyg` that

(3.4)

R_S,µg_`(ω_α+β, r)

= R_S,µg_`^Φ(ω_α, r)

= Z r

0

X

◦∈±

g`(%ω_α+β+ϕ◦

r(%))µω_α,r(%ω_α+ϕ◦ r(%))p

1 +%²( ˙ϕ^◦_r(%))²d%

= Z r

0

X

◦∈±

f%(ω_α+β+ϕ◦

r(%))µω_α,r(%ω_α+ϕ◦ r(%))p

1 +%²( ˙ϕ^◦_r(%))²%^`d%,

where Φ ∈ SO(2) is the anticlockwise rotation by angle β. By the Stone–

Weierstrass Theorem the set of functions %<sup>`</sup>(`∈N) is a complete base in L²[0, r], whence equations (3.4) (`∈N) determine the function

(3.5) ψ(α+β, r, %) :=X

◦∈±

f%(ωα+β+ϕ^◦_r(%))µω_α,r(%ωα+ϕ^◦_r(%))p

1 +%²( ˙ϕ^◦_r(%))²

for%∈(0, r]. Integrating this by β on (−π, π] we get (3.6) ψ(r, %) :=¯

R

S¹ψ(α+β, r, %) dβ R

S¹f_%(ω_α+β+ϕ◦

r(%)) dβ =X

◦∈±

µω_α,r(%ω_α+ϕ^◦

r(%))p

1 +%²( ˙ϕ^◦_r(%))².

Using this, equation (3.5) gives ψ(α+β, r, %)

= ¯ψ(r, %)f%(ω_α+β+ϕ− r(%))+

+ (f_%(ω_α+β+ϕ+

r(%))−f_%(ω_α+β+ϕ−

r(%)))µ_ωα,r(%ω_α+ϕ+ r(%))

q

1 +%²( ˙ϕ⁺r(%))², which by (2) and the continuity ofµimplies

ψ(α+β, r, %)−ψ(r, %)f¯ _%(ω_α+β+ϕ− r(%)) f%(ω_α+β+ϕ+

r(%))−f%(ω_α+β+ϕ⁻

r(%)) =µωα,r(%ω_α+ϕ+ r(%))

q

1 +%²( ˙ϕ⁺r(%))².

The right-hand side shows that the left-hand side does not depend on β, whence the value β = −α implies that the right-hand side does not depend onα. This implies via (3.6) that µω_α,r(%ω_α+ϕ−

r(%)) is also independent fromα.

The theorem is proved.

Our result implies immediately that each rotationally invariant Radon trans- form with rotational flower is rotational, which is a generalization and sharpening of Quinto’s result in [13, Proposition 2.2].

The following result clarifies why Quinto’s [13, Proposition 2.2] differs from our Theorem 3.1 in the plane. The difference occurs because Quinto uses O(2) while we used SO(2) in the definition of rotational invariance, and because His not only rotational but also symmetric.

Theorem 3.2. Let R_S,µ be a rotational Radon transform in the plane, and let the function f ∈ L²(R²) be integrable on every petal of S. Define the functions f%(ω) := f(%ω) on S¹ for all % > 0 and the functions g`(x) := |x|<sup>`</sup>f(x) for all

`∈N. Assume that

(1) the equationf%(ωα+β) =f%(ωα)has only finitely many solutions forαat any β∈(−π, π].

If S is symmetric and

(2) R_S,µg_{`</sub>(Γ¯ω, r) = R<sub>S,µ</sub>g<sup>Γ</sup><sub>`}(¯ω, r) for all ` ∈ N, where Γ is the reflection to the x-axis,

then R_S,µ is symmetric.

Proof. We only need to prove thatµis symmetric.

As the flower is symmetric, there is a functionϕr(%) so thatϕ^±_r(%) =±ϕr(%).

Now equation (3.4) at β= 0 reads

(3.7)

R_S,µg_`(ω_α, r)

= Z r

0

X

i∈{−1,1}

f%(ω_α+iϕ

r(%))µω₀,r(%ω_iϕ

r(%))p

1 +%²ϕ˙²_r(%)%^`d%, because µis rotational. Using condition (2), we get in the same way that

(3.8)

R_S,µg_`(ω_α, r)

= R_S,µg^Γ_`(ω_−α, r)

= Z r

0

X

i∈{−1,1}

f%(ω_α−iϕ

r(%))µω₀,r(%ω_iϕ

r(%))p

1 +%²ϕ˙²_r(%)%^`d%.

By the Stone–Weierstrass Theorem the set of functions %<sup>`</sup> (` ∈N) is a complete base in L²[0, r], whence the difference of equations (3.7) and (3.8) implies

(3.9) 0 = X

i∈{−1,1}

(f%(ωα+iϕ_r(%))−f%(ω_α−iϕr(%)))µω0,r(%ωiϕ_r(%))

for all%∈(0, r]. By condition (1) the coefficientsf_%(ω_α+iϕ

r(%))−f_%(ω_α−iϕ

r(%)) in (3.23) may only vanish for finitely many α, therefore except for finitely many α we haveµ_ωα,r(%ω_α+ϕ

r(%)) =µ_ωα,r(%ω_α−ϕ

r(%)). Asµ is continuous, this proves the theorem.

Theorem 3.3. Let R_S,µ be a symmetric rotational Radon transform, and let f ∈ L²_∗(Rⁿ)be a radial function of compact support without vanishing moments so that R_S,µf(¯ω, pr) =pⁿ⁻¹R_S,µf_p(¯ω, r)for all p >0. If either S or µis dilational, then R_S,µis conformal.

Proof. Sinceµ is a rotational weight, we have a real function ¯µ_r so that (2.1) is fulfilled. Since f is radial, there is a real function ˆf that satisfiesf(x) = ˆf(|x|) for allx∈Rⁿ.

AsS is rotational and symmetric, equations (2.3) and (2.2) give

(3.10)

R_S,µf(¯ω, pr)

=pⁿ⁻¹R_S,µfp(¯ω, r)

=|Sⁿ⁻²|pⁿ⁻¹ Z r

0

fˆ(p%)¯µr(%)(%sinϕr(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%) d%

for allp, r∈[0,∞), where we set|S⁰|= 2.

Integrating (3.10) over [0,∞) with respect to q^−kdq (2 ≤ k ∈ N), where q=rp, then substituting t=p%and finally substituting%=yrwe get

1

|Sⁿ⁻²| Z ∞

0

R_S,µf(¯ω, q)q^−kdq

= Z ∞

0

fˆ(t)t^n−1−kdt Z 1

0

¯

µ_r(yr)y^k−2sinⁿ⁻²ϕ_r(yr)p

1 + (yr)²ϕ˙²_r(yr) dy . Dividing the equation by R∞

0 f(t)t^n−1−kdt results in c_k =

Z 1 0

y^k−2µ¯_r(yr) sinⁿ⁻²ϕ_r(yr)p

1 +y²r²ϕ˙²_r(yr) dy, 2≤k∈N, where ck is a real number for eachk.

By the Stone–Weierstrass Theorem the set of functionsy^k (k∈N) is a com- plete base in L²[0,1], whence the constants c_k (k ∈ N) determine the function ψ(y) := ¯µr(yr) sinⁿ⁻²ϕr(yr)p

1 +y²r²ϕ˙²_r(yr) (0< y≤1) which is therefore inde- pendent fromr.

Now, ifSis dilational, then sinⁿ⁻²ϕ_r(yr)p

1 +y²r²ϕ˙²_r(yr) depends only ony, therefore ¯µr(yr) is independent from r, that proves the conformality ofµ. If µis dilational, then ¯µ_r(yr) depends only ony, whence sinⁿ⁻²ϕ_r(yr)p

1 +y²r²ϕ˙²_r(yr) is independent from r, which by Lemma 2.1 implies the conformality ofS.

The theorem is proved.

The previous two theorems imply that a conformally invariant Radon trans- form with conformal flower is conformal, but as radial functions are not suitable for Theorem 3.1, to check conformality we need different test functions for The- orems 3.1 and 3.3. In the following theorem we use only one function to get the conformality of a Radon transform.

Theorem 3.4. Let S be a conformal flower and let f ∈ L²_∗(Rⁿ) be of compact support such that

(1) the coefficientsf_l,min the spherical harmonic expansion off are independent inL²(R),

(2) for each`∈N the functiong(x) :=|x|<sup>`</sup>f(x)satisfies (2a) R_S,µg(Φ¯ω, r) = R_S,µg^Φ(¯ω, r)for allΦ∈SO(n), and (2b) R_S,µg(¯ω, pr) =pⁿ⁻¹R_S,µgp(¯ω, r)for allp >0.

ThenR_S,µis conformal.

Proof. Before entrenching ourself into the proof let us consider some corollaries of condition (1). First of all, there is an 0 < R ∈ R so that f ∈ L²_∗(RBⁿ), where Bⁿ is the unit ball of Rⁿ, therefore the coefficients f_l,m in the spherical harmonic expansion f(qω) = P∞

l,mfl,m(q)Yl,m(ω) are all in L²[0, R]. These co- efficients are independent in L²[0, R], therefore the infinite-dimensional analogue of the Gram–Schmidt process gives an orthogonal system of compactly supported functions f_l,m^∗ ∈L²[0, R] so that

Z ∞ 0

f_l^∗0,m⁰(q)f_l,m(q) dq=

1, if (l⁰, m⁰) = (l, m), 0, if (l⁰, m⁰)6= (l, m).

Then, by the Stone–Weierstrass Theorem there is a sequence of polynomials ˜p^j_l0,m⁰

(j∈N) tending uniformly to f_l^∗0,m⁰ on [0, R], and therefore

(3.11) F_h˜pj

l0,m0i(ω) =

∞

X

l,m

Z ∞ 0

˜

p^j_l0,m⁰(q)fl,m(q) dq Yl,m(ω)

→Yl⁰,m⁰(ω).

where F_hˆpi(ω) :=R∞

0 p(q)fˆ (qω) dqfor all polynomials ˆp.

Having this result we start proving our theorem in dimensions higher than 2.

Then, by (2.3) and the conditions we have for any g that R_S,µg(Φ¯ω, pr)

=pⁿ⁻¹R_S,µg_p^Φ(¯ω, r)

=pⁿ⁻¹ Z r

0

Z

Sⁿ⁻²ω¯

g(p%Φ(ωsinϕ_r(%) + ¯ωcosϕ_r(%)))×

×µ_ω,r¯ (%(ωsinϕ_r(%) + ¯ωcosϕ_r(%)))×

×(%sinϕr(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%) dωd%, that is

(3.12)

R_S,µg(Φ¯ω, pr)

=pⁿ⁻¹ Z r

0

Z

Sⁿ⁻¹

g(p%Φω)δ_cosϕ

r(%)(hω,ωi)µ¯ _ω,r¯ (%ω) dω×

×(%sinϕ_r(%))ⁿ⁻²p

1 +%²ϕ˙²_r(%) d%,

= (pr)ⁿ⁻¹ Z 1

0

Z

Sⁿ⁻¹

g(pyrΦω)δ_cosϕ

1(y)(hω,ωi)µ¯ ω,r¯ (yrω) dω×

×yⁿ⁻²sinⁿ⁻²ϕ₁(y) q

1 +y²ϕ˙²₁(y) dy, where δ_cosϕ

r(%) is the Dirac delta distribution offset to cosϕ_r(%) andh., .idenotes the usual inner product onRⁿ.

Integrating R_S,µg(Φ¯ω, t) by t^−kdt (2 ≤ k ∈ N) over [0,∞), where t = pr, then using (3.12) in which substitutingq=pyr leads to

hk(Φ¯ω) :=

Z ∞ 0

R_S,µg(Φ¯ω, t)t^−kdt=r^1−k Z ∞

0

R_S,µg(Φ¯ω, pr)p^−kdp

= Z 1

0

Z

Sⁿ⁻¹

Z ∞ 0

q^n−1−kg(qΦω) dq δ_cosϕ

1(y)(hω,ωi)µ¯ ω,r¯ (yrω) dω×

×y^k−2sinⁿ⁻²ϕ₁(y) q

1 +y²ϕ˙²₁(y) dy.

Choosing g(x) :=|x|<sup>`+k−n+1</sup>f(x) (`∈N) we get

(3.13)

hk,`(Φ¯ω) :=

Z 1 0

Z

Sⁿ⁻¹

Z ∞ 0

q^`f(qΦω) dq δ_cosϕ

1(y)(hω,ωi)µ¯ ω,r¯ (yrω) dω×

×y^k−2sinⁿ⁻²ϕ₁(y) q

1 +y²ϕ˙²₁(y) dy.

Since this is valid for all integersk≥2, again the Stone–Weierstrass Theorem gives that

Z

Sⁿ⁻¹

Z ∞ 0

q^`f(qΦω) dq δ_cosϕ

1(y)(hω,ωi)µ¯ ω,r¯ (yrω) dω

depends only on y ∈ [0,1] and on Φ¯ω, hence it is a function h^∗_`(Φ¯ω, y) (this can be determined from the functions hk,k(Φ¯ω) by (3.13)). We thus have for any polynomial ˆponRthe function

(3.14) h^∗_hˆpi(Φ¯ω, y) = Z

Sⁿ⁻¹

Z ∞ 0

ˆ

p(q)f(qΦω) dq δ_cosϕ

1(y)(hω,ωi)µ¯ ω,r¯ (yrω) dω, where (¯ω, y)∈Sⁿ⁻¹×[0,1]. Substituting Φ = Ψ⁻¹, ¯ω= Ψˆω,ω= Ψ˜ω, then ˆω= ¯ω and finally ˜ω=ω, equation (3.14) can be written as

(3.15) h^∗_hˆpi(¯ω, y) = Z

Sⁿ⁻¹

F_hˆpi(ω)δ_cosϕ

1(y)(hω,ωi)µ¯ _{Ψ ¯ω,r}(ryΨω) dω, where F_hˆpi(ω) = R∞

0 p(q)fˆ (qω) dq. Taking into account the left-hand side’s inde- pendence from Ψ we get

0 = Z

Sⁿ⁻¹

F_hˆpi(ω)δ_cosϕ

1(y)(hω,ωi)(µ¯ _ω,r¯ (ryΨω)−µ_ω,r¯ (ryω)) dω, Ψ∈ Kⁿ_ω¯, where Kⁿ_ω¯ :={Ψ∈SO(n) : Ψ¯ω= ¯ω} ∼=SO(n−1). Substituting the sequence of polynomials ˜p^j_l0,m⁰ (j∈N) leads to

0 = Z

Sⁿ⁻¹

Y_l⁰_,m⁰(ω)δ_cosϕ

1(y)(hω,ωi)(µ¯ _ω,r¯ (ryΦω)−µ_ω,r¯ (ryω)) dω by equation (3.11).

As the spherical harmonicsYl⁰,m⁰ constitute a base in L²(Sⁿ⁻¹), this implies µω,r¯ (ryΦω) =µω,r¯ (ryω) for all ω ∈Sⁿ⁻¹ satisfyinghω,ωi¯ = cosϕ₁(y) and for all Φ∈ Kⁿ_ω¯. Taking this into account in considering equation (3.15) we get

h^∗_hˆpi(¯ω, y) =µΨ ¯ω,r(ryΨ¯ω^⊥) Z

Sⁿ⁻¹

F_hˆpi(ω)δcosϕ₁(y)(hω,ωi) dω,¯

where ¯ω^⊥ is any unit vector perpendicular to ¯ω. This implies thatµis conformal, that proves the statement in dimensions higher than 2.

Now we proceed to the case of the plane, i.e., dimensionn= 2. By (2.2) and the conditions we have for any g and β ∈ (−π, π], where Φ is the anticlockwise rotation by β, that

(3.16)

R_S,µg(ωα+β, pr)

=pR_S,µg^Φ_p(ωα, r)

=p Z r

0

X

◦∈±

g(p%ω_α+β+ϕ◦

r(%))µ_ωα,r(%ω_α+ϕ◦ r(%))p

1 +%²( ˙ϕ^◦_r(%))²d%

=pr Z 1

0

X

◦∈±

g(pryω_α+β+ϕ^◦

1(y))µω_α,r(ryω_α+ϕ^◦

1(y)) q

1 +y²( ˙ϕ^◦₁(y))²dy.

Integrating R_S,µg(ωα+β, t) byt^−kdt (2 ≤k ∈N) over [0,∞), wheret =pr, then using (3.16) in which substitutingq=pyr leads to

hk(ωα+β) :=

Z ∞ 0

R_S,µg(ωα+β, t)t^−kdt=r^1−k Z ∞

0

R_S,µg(ωα+β, pr)p^−kdp

= Z 1

0

X

◦∈±

Z ∞ 0

g(qω_α+β+ϕ^◦

1(y))q^1−kdq×

×µω_α,r(ryω_α+ϕ^◦

1(y))y^k−2 q

1 +y²( ˙ϕ^◦₁(y))² dy.

Choosing g(x) :=|x|<sup>`+k−1</sup>f(x) (`∈N) we get

(3.17)

hk,`(ωα+β) :=

Z 1 0

X

◦∈±

Z ∞ 0

q^`f(qω_α+β+ϕ^◦

1(y)) dq×

×µω_α,r(ryω_α+ϕ^◦

1(y)) q

1 +y²( ˙ϕ^◦₁(y))²y^k−2dy.

Since this is valid for all integersk≥2, again the Stone–Weierstrass Theorem gives that

X

◦∈±

Z ∞ 0

q^`f(qω_α+β+ϕ^◦

1(y)) dq µω_α,r(ryω_α+ϕ^◦

1(y)) q

1 +y²( ˙ϕ^◦₁(y))²

depends only on y ∈[0,1] and on ωα+β, hence it is a function h^∗_`(ωα+β, y) (this can be determined from the functionsh_k,k(ω_α+β) by (3.17)). We thus have for any polynomial ˆponRthe function

(3.18) h^∗_hˆpi(ω_α+β, y) =X

◦∈±

F_hˆpi(ω_α+β+ϕ^◦

1(y))µ_ωα,r(ryω_α+ϕ^◦

1(y)) q

1 +y²( ˙ϕ^◦₁(y))²,

where y ∈[0,1]. Substituting β =−γ, α= ˆα+γ, then ˆα=αand finallyγ=β, equation (3.18) can be written as

(3.19) h^∗_hˆpi(ωα, y) =X

◦∈±

F_hˆpi(ω_α+ϕ^◦

1(y))µω_α+β,r(ryω_α+β+ϕ^◦

1(y)) q

1 +y²( ˙ϕ^◦₁(y))².

Fixing αand y, this is a linear equation of the unknownsµω_α+β,r(ryωα+β+ϕ^◦₁(y)) for each polynomial ˆp. The system of these linear equations can be degenerated for a pair (α, y) if and only if the vectors (F_hˆpi(ω_α+γ), F_hˆpi(ω_α)) are real multiples of each other for all polynomials ˆp, where γ=ϕ⁺₁(y)−ϕ⁻₁(y)6= 0.

As an indirect step, assume now that system (3.19) degenerates at someαand γ6= 0, which means that (F_hˆpi(ωα+γ), F_hˆpi(ωα)) are real multiples of each other for all polynomials ˆp. Choosing the sequences of polynomials ˆpj := ˜p^j_l0,m⁰ (j∈N) and

¯

pk := ˜p^k_l,m (k∈N), and takingk→ ∞, we obtain

(F_hˆpji(ωα+γ), F_hˆpji(ωα)) =λjk(α, γ)(F_h¯pki(ωα+γ), F_h¯pki(ωα))

→λj;l,m(α, γ)(Yl,m(ωα+γ), Yl,m(ωα))

for some functionsλ_jkandλ_j;l,mthat may become zero or infinity at finitely many arguments (α, γ) only. Takingj→ ∞leads to

(F_hˆpji(ω_α+γ), F_hˆpji(ω_α)) =λ_j;l,m(α, γ)(Y_l,m(ω_α+γ), Y_l,m(ω_α))

↓ ↓

(Yl⁰,m⁰(ωα+γ), Yl⁰,m⁰(ωα)) =λl⁰,m⁰;l,m(α, γ)(Yl,m(ωα+γ), Yl,m(ωα)) (3.20)

for a function λ_l⁰_,m⁰_;l,m that may become zero or infinity at finitely many argu- ments (α, γ) only.

Ifλl⁰,m⁰;l,m(α, γ) is zero or infinity, then either Yl,m(ωα+γ) = 0 = Yl,m(ωα) or Yl⁰,m⁰(ωα+γ) = 0 = Yl⁰,m⁰(ωα), whence the argument (α, γ) must be in the set {(iπ/`, iπ/`) :i, j ∈ Z, ` ∈ {2l,2l⁰}}. Therefore, the coefficientλ_i0,j⁰;i,j(α, γ) can not be zero or infinity, if both i and i⁰ are relative prime to both num- bers 2l and 2l⁰. Thus, for every pair (α, γ) there is an infinite set Lα,γ so that λl⁰,m⁰;l,m(α, γ)∈R\ {0}for allm, m⁰ ∈ {−1,1}. Ifl⁰, l∈ Lα,γ, then (3.20) implies Y_l⁰_,m⁰(ω_α+γ)Y_l,m(ω_α) =Y_l,m(ω_α+γ)Y_l⁰_,m⁰(ω_α) for allm, m⁰∈ {−1,1}, which means











cos(l⁰(α+γ)) sin(lα) = sin(l(α+γ)) cos(l⁰α), if−m=m⁰= 1, sin(l⁰(α+γ)) cos(lα) = cos(l(α+γ)) sin(l⁰α), ifm=−m⁰= 1, cos(l⁰(α+γ)) cos(lα) = cos(l(α+γ)) cos(l⁰α), ifm=m⁰ = 1, sin(l⁰(α+γ)) sin(lα) = sin(l(α+γ)) sin(l⁰α), ifm=m⁰ =−1