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Integral Geometry and Radon Transforms

Sigurdur Helgason
Integral Geometry and Radon Transforms
1 C

Sigurdur Helgason Department of Mathematics Massachusetts Institute of Technology Cambridge, MA 02139 USA helgason@mit.edu

ISBN 978-1-4419-6054-2

e-ISBN 978-1-4419-6055-9

DOI 10.1007/978-1-4419-6055-9

Springer New York Dordrecht Heidelberg London

Library of Congress Control Number: 2010938299

Mathematics Subject Classification (2010): 53C65, 44A12

© Springer Science+Business Media, LLC 2011 All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.

Printed on acid-free paper

Springer is part of Springer Science+Business Media (www.springer.com)

TO ARTIE

Preface

This book deals with a special subject in the wide field of Geometric Analysis. The subject has its origins in results by Funk [1913] and Radon [1917] determining, respectively, a symmetric function on the two-sphere S2 from its great circle integrals and an integrable function on R2 from its straight line integrals. (See References.) The first of these is related to a geometric theorem of Minkowski [1911] (see Ch. III, §1).
While the above work of Funk and Radon lay dormant for a while, Fritz John revived the subject in important papers during the thirties and found significant applications to differential equations. More recent applications to X-ray technology and tomography have widened interest in the subject.
This book originated with lectures given at MIT in the Fall of 1966, based mostly on my papers during 1959–1965 on the Radon transform and its generalizations. The viewpoint of these generalizations is the following.
The set of points on S2 and the set of great circles on S2 are both acted on transitively by the group O(3). Similarly, the set of points in R2 and the set P2 of lines in R2 are both homogeneous spaces of the group M(2) of rigid motions of R2. This motivates our general Radon transform definition from [1965a] and [1966a], which forms the framework of Chapter II:
Given two homogeneous spaces X = G/K and Ξ = G/H of the same group G, two elements x = gK and ξ = γH are said to be incident (denoted x#ξ) if gK ∩ γH = ∅ (as subsets of G). We then define the abstract Radon
ˇ transform f → f from Cc(X) to C(Ξ) and the dual transform ϕ → ϕ from
Cc(Ξ) to C(X) by

f (ξ) = f (x) dm(x) ,
x#ξ

ˇϕ(x) = ϕ(ξ) dμ(ξ)
ξ#x

with canonical measures dm and dμ. These geometrically dual operators
ˇ f → f and ϕ → ϕ are also adjoint operators relative to the G-invariant
measures dgK , dgH on G/K and G/H. In the example R2, one takes G = M(2) and K the subgroup O(2) fixing
the origin xo and H the subgroup mapping a line ξo into itself. Thus we have
X = G/K = R2 , Ξ = G/H = P2

and here it turns out x ∈ X is incident to ξ ∈ Ξ if and only if their distance equals the distance p between xo and ξo. It is important not just to consider the case p = 0. Also the abstract definition does not require the members
of Ξ to be subsets of X. Some natural questions arise for the operators
ˇ f → f, ϕ → ϕ, namely:

viii
(i) Injectivity
(ii) Inversion formulas
(iii) Ranges and kernels for specific function spaces on X and on Ξ
(iv) Support problems (does f of compact support imply f of compact support?)
We investigate these problems for a variety of examples, mainly in Chapter II. Interesting analogies and differences appear. One such instance is when the classical Poisson integral for the unit disk turns out to be a certain Radon transform and offers wide ranging analogies with the X-ray transform in R3. See Table II.1 in Chapter II, §4.
In the abstract framework indicated above, a specific result for a single example automatically raises a host of conjectures.
The problems above are to a large extent solved for the X-ray transform and for the horocycle transform on Riemannian symmetric spaces. When G/K is a Euclidean space (respectively, a Riemannian symmetric space) and G/H the space of hyperplanes (respectively, the space of horocycles) the transform f → f has applications to certain differential equations. If L is a natural differential operator on G/K, the map f → f transfers it into a more manageable operator L on G/H by the relation
(Lf ) = Lf .
Then the support theorem
f compact support ⇒ f compact support
implies the existence theorem L C∞(G/K) = C∞(G/K) for G-invariant differential operators L on G/K.
On the other hand, the applications of the original Radon transform on R2 to X-ray technology and tomography are based on the fact that for an unknown density f , X-ray attenuation measurements give f directly and thus yield f itself via Radon’s inversion formula. More precisely, let B be a planar convex body, f (x) its density at the point x, and suppose a thin beam of X-rays is directed at B along a line ξ. Then, as observed by Cormack, the line integral f (ξ) of f along ξ equals log(I0/I) where I0 and I, respectively, are the intensities of the beam before hitting B and after leaving B. Thus while f is at first unknown, the function f (and thus f ) is determined by the X-ray data. See Ch. I, §7,B. This work, initiated by Cormack and Hounsfield and earning them a Nobel Prize, has greatly increased interest in Radon transform theory. The support theorem brings in a certain refinement that the density f (x) outside a convex set C can be determined by only using X-rays that do not enter C. See Ch. I, §7, B.

ix

This book includes and recasts some material from my earlier book, “The Radon Transform”, Birkh¨auser (1999). It has a large number of new examples of Radon transforms, has an extended treatment of the Radon transform on constant curvature spaces, and contains full proofs for the antipodal Radon transform on compact two-point homogeneous spaces. The X-ray transform on symmetric spaces is treated in detail with explicit inversion formulas.
In order to make the book self-contained we have added three chapters at the end of the book. Chapter VII treats Fourier transforms and distributions, relying heavily on the concise treatment in Ho¨rmander’s books. We call particular attention to his profound Theorem 4.9, which in spite of its importance does not seem to have generally entered distribution theory books. We have found this result essential in our study [1994b] of the Radon transform on a symmetric space. Chapter VIII contains a short treatment of basic Lie group theory assuming only minimal familiarity with the concept of a manifold. Chapter IX is a short exposition of the basics of the theory of Cartan’s symmetric spaces. Most chapters end with some Exercises and Further Results with explicit references.
Although the Bibliography is fairly extensive no completeness is attempted. In view of the rapid development of the subject the Bibliographical Notes can not be up to date. In these notes and in the text my books [1978] and [1984] and [1994b] are abbreviated to DS and GGA and GSS.
I am indebted to T.H. Danielsen, S. Jensen and J. Orloff for critical reading of parts of the manuscript, to R. Melrose and R. Seeley for suggestions, to F. Gonzalez, J. Hilgert, A. Kurusa, F. Rouvi`ere and H. Schlichtkrull for concrete contributions mentioned at specific places in the text, and for various textual suggestions. Once more my sincere thanks to Jan Wetzel for skillful preparation of the manuscript and to Kaitlin Leach at Springer for her patient cooperation.

MIT May 2009

Sigurdur Helgason

Contents

Preface

vii

CHAPTER I The Radon Transform on Rn

§1 Introduction

1

§2 The Radon Transform of the Spaces D(Rn) and S(Rn). The

Support Theorem

2

§3 The Inversion Formula. Injectivity Questions

16

§4 The Plancherel Formula

25

§5 Radon Transform of Distributions

27

§6 Integration over d-planes. X-ray Transforms. The Range of the

d-plane Transform

32

§7 Applications

46

A. Partial Differential Equations. The Wave Equation

46

B. X-ray Reconstruction

52

Exercises and Further Results

56

Bibliographical Notes

60

CHAPTER II

A Duality in Integral Geometry

§1 Homogeneous Spaces in Duality

63

§2 The Radon Transform for the Double Fibration

67

(i) Principal Problems

68

(ii) Ranges and Kernels. General Features

71

(iii) The Inversion Problem. General Remarks

72

§3 Orbital Integrals

75

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality 77

A. The Funk Transform

77

B. The X-ray Transform in H2

80

C. The Horocycles in H2

82

D. The Poisson Integral as a Radon Transform

86

E. The d-plane Transform

88

F. Grassmann Manifolds

90

G. Half-lines in a Half-plane

91

H. Theta Series and Cusp Forms

94

I. The Plane-to-Line Transform in R3. The Range

95

J. Noncompact Symmetric Space and Its Family of Horocycles 103

Exercises and Further Results

104

Bibliographical Notes

108

xii

CHAPTER III

The Radon Transform on Two-Point Homogeneous Spaces

§1 Spaces of Constant Curvature. Inversion and Support Theorems 111

A. The Euclidean Case Rn

114

B. The Hyperbolic Space

118

C. The Spheres and the Elliptic Spaces

133

D. The Spherical Slice Transform

145

§2 Compact Two-Point Homogeneous Spaces. Applications

147

§3 Noncompact Two-Point Homogeneous Spaces

157

§4 Support Theorems Relative to Horocycles

159

Exercises and Further Results

167

Bibliographical Notes

168

CHAPTER IV

The X-Ray Transform on a Symmetric Space

§1 Compact Symmetric Spaces. Injectivity and Local Inversion.

Support Theorem

171

§2 Noncompact Symmetric Spaces. Global Inversion and General

Support Theorem

178

§3 Maximal Tori and Minimal Spheres in Compact Symmetric Spaces 180

Exercises and Further Results

182

Bibliographical Notes

183

CHAPTER V

Orbital Integrals and the Wave Operator for Isotropic Lorentz Spaces

§1 Isotropic Spaces

185

A. The Riemannian Case

186

B. The General Pseudo-Riemannian Case

186

C. The Lorentzian Case

190

§2 Orbital Integrals

190

§3 Generalized Riesz Potentials

199

§4 Determination of a Function from Its Integral over Lorentzian

Spheres

202

§5 Orbital Integrals and Huygens’ Principle

206

Bibliographical Notes

208

CHAPTER VI

The Mean-Value Operator

§1 An Injectivity Result

209

§2 A´ sgeirsson’s Mean-Value Theorem Generalized

211

§3 John’s Identities

215

Exercises and Further Results

217

Bibliographical Notes

219

xiii

CHAPTER VII

Fourier Transforms and Distributions. A Rapid Course

§1 The Topology of Spaces D(Rn), E(Rn), and S(Rn)

221

§2 Distributions

223

§3 Convolutions

224

§4 The Fourier Transform

226

§5 Differential Operators with Constant Coefficients

234

§6 Riesz Potentials

236

Exercises and Further Results

248

Bibliographical Notes

250

CHAPTER VIII

Lie Transformation Groups and Differential Operators

§1 Manifolds and Lie Groups

253

§2 Lie Transformation Groups and Radon Transforms

261

CHAPTER IX Symmetric Spaces

§1 Definition and Examples

265

§2 Symmetric Spaces of the Noncompact Type

267

§3 Symmetric Spaces of the Compact Type

273

Bibliography

275

Notational Conventions

295

Frequently Used Symbols

297

Index

299

CHAPTER I

THE RADON TRANSFORM ON Rn

§1 Introduction

It was proved by J. Radon in 1917 that a differentiable function on R3 can be determined explicitly by means of its integrals over the planes in R3. Let J(ω, p) denote the integral of f over the hyperplane x, ω = p, ω denoting a unit vector and , the inner product. Then

1 f (x) = − 8π2 Lx

J(ω, ω, x ) dω ,

S2

where L is the Laplacian on R3 and dω the area element on the sphere S2 (cf. Theorem 3.1).
We now observe that the formula above has built in a remarkable du-
ality: first one integrates over the set of points in a hyperplane, then one integrates over the set of hyperplanes passing through a given point. This
ˇ suggests considering the transforms f → f , ϕ → ϕ defined below. The formula has another interesting feature. For a fixed ω the integrand x → J(ω, ω, x ) is a plane wave, that is a function constant on each plane perpendicular to ω. Ignoring the Laplacian the formula gives a continuous decomposition of f into plane waves. Since a plane wave amounts
to a function of just one variable (along the normal to the planes) this decomposition can sometimes reduce a problem for R3 to a similar problem for R. This principle has been particularly useful in the theory of partial
differential equations.
The analog of the formula above for the line integrals is of importance in radiography where the objective is the description of a density function
by means of certain line integrals. In this chapter we discuss relationships between a function on Rn and
its integrals over k-dimensional planes in Rn. The case k = n − 1 will be the one of primary interest. We shall occasionally use some facts about Fourier transforms and distributions. This material will be developed in
sufficient detail in Chapter VII so the treatment should be self-contained.
Later chapters involve some Lie groups and symmetric spaces so the needed background is developed in Chapter VIII and Chapter IX.
Following Schwartz [1966] we denote by E(Rn) and D(Rn), respectively, the space of complex-valued C∞ functions (respectively C∞ functions of compact support) on Rn. The space S(Rn) of rapidly decreasing functions on Rn is defined in connection with (6) below. Cm(Rn) denotes the space of m times continuously differentiable functions. We write C(Rn) for C0(Rn), the space of continuous functions on Rn.

S. Helgason, Integral Geometry and Radon Transforms, DOI 10.1007/978-1-4419-6055-9_1, © Springer Science+Business Media, LLC 2011

2 Chapter I. The Radon Transform on Rn
For a manifold M , Cm(M ) (and C(M )) is defined similarly and we write D(M ) for Cc∞(M ) and E(M ) for C∞(M ).
§2 The Radon Transform of the Spaces D(Rn) and S(Rn). The Support Theorem
Let f be a function on Rn, integrable on each hyperplane in Rn. Let Pn denote the space of all hyperplanes in Rn, Pn being furnished with the obvious topology. The Radon transform of f is defined as the function f on Pn given by
f (ξ) = f (x)dm(x) ,
ξ
where dm is the Euclidean measure on the hyperplane ξ. Along with the
ˇ transformation f → f we consider also the dual transform ϕ → ϕ which ˇ to a continuous function ϕ on Pn associates the function ϕ on Rn given
by
ˇϕ(x) = ϕ(ξ) dμ(ξ)
x∈ξ
where dμ is the measure on the compact set {ξ ∈ Pn : x ∈ ξ} which is invariant under the group of rotations around x and for which the measure of the whole set is 1 (see Fig. I.1). We shall relate certain function spaces
ˇ on Rn and on Pn by means of the transforms f → f, ϕ → ϕ; later we
obtain explicit inversion formulas.

x FIGURE I.1.

ξ

x

}〈x, ω〉

ω

0

FIGURE I.2.

§2 The Radon Transform . . . The Support Theorem 3
Each hyperplane ξ ∈ Pn can be written ξ = {x ∈ Rn : x, ω = p}, where , is the usual inner product, ω = (ω1, . . . , ωn) a unit vector and p ∈ R (Fig. I.2). Note that the pairs (ω, p) and (−ω, −p) give the same ξ; the mapping (ω, p) → ξ is a double covering of Sn−1 × R onto Pn. Thus Pn has a canonical manifold structure with respect to which this covering map is differentiable and regular. We thus identify continuous (differentiable) function on Pn with continuous (differentiable) functions ϕ on Sn−1 × R satisfying the symmetry condition ϕ(ω, p) = ϕ(−ω, −p). Writing f (ω, p) instead of f (ξ) and ft (with t ∈ Rn) for the translated function x → f (t + x) we have

ft(ω, p) =

f (x + t) dm(x) =

f (y) dm(y)

x,ω =p

y,ω =p+ t,ω

so

(1)

ft(ω, p) = f (ω, p + t, ω ) .

Taking limits we see that if ∂i = ∂/∂xi

(2)

(∂if

)b(ω,

p)

=

ωi

∂f ∂p

(ω,

p)

.

Let L denote the Laplacian Σi∂i2 on Rn and let denote the operator

∂2

ϕ(ω, p)

→

ϕ(ω, p) , ∂p2

which is a well-defined operator on E(Pn) = C∞(Pn). It can be proved that if M(n) is the group of isometries of Rn, then L (respectively ) generates the algebra of M(n)-invariant differential operators on Rn (respectively Pn).

ˇ Lemma 2.1. The transforms f → f, ϕ → ϕ intertwine L and , i.e., ˇ (Lf )b = (f ) , ( ϕ)∨ = Lϕ .

Proof. The first relation follows from (2) by iteration. For the second we just note that for a certain constant c,

(3)

ˇϕ(x) = c ϕ(ω, x, ω ) dω ,

Sn−1

where dω is the usual measure on Sn−1.

4 Chapter I. The Radon Transform on Rn

The Radon transform is closely connected with the Fourier transform

f (u) = f (x)e−i x,ω dx u ∈ Rn.

Rn

In fact, if s ∈ R, ω a unit vector,

∞
f (sω) = dr

f (x)e−is x,ω dm(x)

so (4)

−∞

x,ω =r

∞
f (sω) = f (ω, r)e−isr dr .

−∞

This means that the n-dimensional Fourier transform is the 1-dimensional Fourier transform of the Radon transform. From (4) it follows that the Radon transform of the convolution

is the convolution

f (x) = f1(x − y)f2(y) dy
Rn

(5)

f (ω, p) = f1(ω, p − q)f2(ω, q) dq .

R

Formula (5) can also be proved directly:

f (ω, p) =

f1(x − y) dm(x) f2(y) dy

Rn x,ω =p

=

f1(z) dm(z) f2(y) dy

Rn z,ω =p− y,ω

= f1(ω, p − y, ω )f2(y) dy = f1(ω, p − q)f2(ω, q) dq .

Rn

R

We consider now the space S(Rn) of complex-valued rapidly decreasing functions on Rn. We recall that f ∈ S(Rn) if and only if for each
polynomial P and each integer m ≥ 0,

(6)

sup | |x|mP (∂1, . . . , ∂n)f (x)| < ∞ ,

x

|x| denoting the norm of x. We now formulate this in a more invariant fashion.

§2 The Radon Transform . . . The Support Theorem 5
Lemma 2.2. A function f ∈ E(Rn) belongs to S(Rn) if and only if for each pair k, ∈ Z+
sup |(1 + |x|)k(L f )(x)| < ∞ .
x∈Rn

This is easily proved just by using the Fourier transforms.
In analogy with S(Rn) we define S(Sn−1 × R) as the space of C∞ functions ϕ on Sn−1 × R which for any integers k, ≥ 0 and any differential operator D on Sn−1 satisfy

(7)

sup

(1 + |r|k) d (Dϕ)(ω, r) < ∞ .

ω∈Sn−1,r∈R

dr

The space S(Pn) is then defined as the set of ϕ ∈ S(Sn−1 × R) satisfying ϕ(ω, p) = ϕ(−ω, −p).

Lemma 2.3. For each f ∈ S(Rn) the Radon transform f (ω, p) satisfies the following condition: For k ∈ Z+ the integral

f (ω, p)pk dp
R
can be written as a kth degree homogeneous polynomial in ω1, . . . , ωn. Proof. This is immediate from the relation

(8)

f (ω, p)pk dp = pk dp

f (x) dm(x) = f (x) x, ω k dx .

R

R

x,ω =p

Rn

In accordance with this lemma we define the space

SH (Pn) =

F ∈ S(Pn) :

For each k ∈ Z+, R F (ω, p)pk dp is a homogeneous polynomial

.

in ω1, . . . , ωn of degree k

With the notation D(Pn) = Cc∞(Pn) we write

DH (Pn) = SH (Pn) ∩ D(Pn) .

According to Schwartz [1966], p. 249, the Fourier transform f → f maps the space S(Rn) onto itself. See Ch. VII, Theorem 4.1. We shall now settle the analogous question for the Radon transform.
Theorem 2.4. (The Schwartz theorem) The Radon transform f → f is a linear one-to-one mapping of S(Rn) onto SH (Pn).

6 Chapter I. The Radon Transform on Rn

Proof. Since

d

n

f (sω) = ds

ωi(∂if )

i=1

it is clear from (4) that for each fixed ω the function r → f (ω, r) lies in S(R). For each ω0 ∈ Sn−1 a subset of (ω1, . . . , ωn) will serve as local coordinates on a neighborhood of ω0 in Sn−1. To see that f ∈ S(Pn), it therefore suffices to verify (7) for ϕ = f on an open subset N ⊂ Sn−1 where
ωn is bounded away from 0 and ω1, . . . , ωn−1 serve as coordinates, in terms of which D is expressed. Since

(9) u1 = sω1, . . . , un−1 = sωn−1 , un = s(1 − ω12 − · · · − ωn2−1)1/2 ,

we have

∂ (f (sω))
∂ωi

=

∂f s
∂ui

−

sωi(1

−

ω12

−

···

−

ωn2 −1 )−1/2

∂f ∂un

.

It follows that if D is any differential operator on Sn−1 and if k, ∈ Z+ then

(10)

sup (1 + s2k) d (Df )(ω, s) < ∞ .

ω∈N,s∈R

ds

We can therefore apply D under the integral sign in the inversion formula

to (4),

1 f (ω, r) =

f (sω)eisr ds

2π

R

and obtain

(1+r2k) d dr

1

Dω(f (ω, r))

= 2π

1+(−1)k

d2k ds2k

(is) Dω(f (sω)) eisrds .

Now (10) shows that f ∈ S(Pn) so by Lemma 2.3, f ∈ SH (Pn). Because of (4) and the fact that the Fourier transform is one-to-one it
only remains to prove the surjectivity in Theorem 2.4. Let ϕ ∈ SH (Pn). In order to prove ϕ = f for some f ∈ S(Rn) we put
∞
ψ(s, ω) = ϕ(ω, r)e−irs dr .
−∞
Then ψ(s, ω) = ψ(−s, −ω) and ψ(0, ω) is a homogeneous polynomial of degree 0 in ω1, . . . , ωn, hence constant. Thus there exists a function F on Rn such that
F (sω) = ϕ(ω, r)e−irs dr .
R

§2 The Radon Transform . . . The Support Theorem 7

While F is clearly smooth away from the origin we shall now prove it to be
smooth at the origin too; this is where the homogeneity condition in the definition of SH (Pn) enters decisively. Consider the coordinate neighborhood N ⊂ Sn−1 above and if h ∈ C∞(Rn − 0) let h∗(ω1, . . . , ωn−1, s) be the function obtained from h by means of the substitution (9). Then

∂h

= n−1 ∂h∗ ∂ωj

∂h∗ +

·

∂s

(1 ≤ i ≤ n)

∂ui j=1 ∂ωj ∂ui ∂s ∂ui

and

∂ωj ∂ui

=

1 s (δij

−

uiuj s2

)

(1 ≤ i ≤ n ,

1 ≤ j ≤ n − 1) ,

∂s ∂ui

=

ωi

(1

≤

i

≤

n − 1),

∂s ∂un

=

(1 − ω12 − · · · − ωn2−1)1/2 .

Hence

⎛

⎞

∂h ∂ui

=

1 s

∂h∗ ∂ωi

+

ωi

⎝ ∂h∗ ∂s

−

1 s

n−1
ωj
j=1

∂h∗ ⎠ ∂ωj

(1 ≤ i ≤ n − 1)

⎛

⎞

∂h ∂un

=

(1

− ω12

−

···

−

ωn2 −1

)1/2

⎝

∂h∗ ∂s

−

1 s

n−1 j=1

ωj

∂ ∂

h∗ ωj

⎠

.

In order to use this for h = F we write

∞

∞

F (sω) = ϕ(ω, r) dr + ϕ(ω, r)(e−irs − 1) dr .

−∞

−∞

By assumption the first integral is independent of ω. Thus using (7) we have for constant K > 0

1 ∂ (F (sω)) ≤ K (1 + r4)−1s−1|e−isr − 1| dr ≤ K s ∂ωi

1

|r| + r4

dr

and a similar estimate is obvious for ∂F (sω)/∂s. The formulas above there-

fore imply that all the derivatives ∂F/∂ui are bounded in a punctured ball 0 < |u| < so F is certainly continuous at u = 0.

More generally, we prove by induction that

(11)

∂qh

∂i+j h∗

∂ui1 . . . ∂uiq = 1≤i+j≤q,1≤k1,··· ,ki≤n−1 Aj,k1...ki (ω, s) ∂ωk1 . . . ∂ωki ∂sj ,

where the coefficients A have the form

(12)

Aj,k1...ki (ω, s) = aj,k1...ki (ω)sj−q .

8 Chapter I. The Radon Transform on Rn

For q = 1 this is in fact proved above. Assuming (11) for q we calculate

∂ q+1 h ∂ui1 . . . ∂uiq+1

using the above formulas for ∂/∂ui. If Aj,k1...ki (ω, s) is differentiated with respect to uiq+1 we get a formula like (12) with q replaced by q + 1. If on the other hand the (i + j)th derivative of h∗ in (11) is differentiated with
respect to uiq+1 we get a combination of terms

s−1

∂ i+j +1 h∗

,

∂ i+j +1 h∗ ,

∂ωk1 . . . ∂ωki+1 ∂sj ∂ωk1 . . . ∂ωki ∂sj+1

and in both cases we get coefficients satisfying (12) with q replaced by q +1, noting sj−q = sj+1−(q+1). This proves (11)–(12) in general. Now

(13)

F (sω) =

∞

q−1 (−isr)k

ϕ(ω, r)

dr +

k!

∞
ϕ(ω, r)eq(−irs) dr ,

−∞

0

−∞

where

tq

tq+1

eq(t) = q! + (q + 1)! + · · ·

Our assumption on ϕ implies that the first integral in (13) is a polynomial
in u1, . . . , un of degree ≤ q−1 and is therefore annihilated by the differential operator (11). If 0 ≤ j ≤ q, we have

(14)

|sj−q

∂j ∂sj

(eq (−irs))|

=

|(−ir)q

(−irs)j−q eq−j

(−irs)|

≤

kj rq

,

where kj is a constant because the function t → (it)−pep(it) is obviously bounded on R (p ≥ 0). Since ϕ ∈ S(Pn) it follows from (11)–(14) that each qth order derivative of F with respect to u1, . . . , un is bounded in a punctured ball 0 < |u| < . Thus we have proved F ∈ E(Rn). That F is
rapidly decreasing is now clear from (7), Lemma 2.2 and (11). Finally, if f
is the function in S(Rn) whose Fourier transform is F then

∞
f (sω) = F (sω) = ϕ(ω, r)e−irs dr ;

−∞

hence by (4), f = ϕ and the theorem is proved.
To make further progress we introduce some useful notation. Let Sr(x) denote the sphere {y : |y − x| = r} in Rn and A(r) its area. Let Br(x) denote the open ball {y : |y − x| < r}. For a continuous function f on Sr(x) let (M rf )(x) denote the mean value

(M rf )(x) = 1 A(r)

f (ω) dω ,

Sr (x)

§2 The Radon Transform . . . The Support Theorem 9
where dω is the Euclidean measure. Let K denote the orthogonal group O(n), dk its Haar measure, normalized by dk = 1. If y ∈ Rn, r = |y| then

(15) (M rf )(x) = f (x + k · y) dk .
K
(Fig. I.3) In fact, for x, y fixed both sides represent rotation-invariant functionals on C(Sr(x)), having the same value for the function f ≡ 1. The rotations being transitive on Sr(x), (15) follows from the uniqueness of such invariant functionals. Formula (3) can similarly be written

k⋅y

0

y

x+k⋅y x

FIGURE I.3.

ˇ (16) ϕ(x) = ϕ(x + k · ξ0) dk
K

if ξ0 is some fixed hyperplane through the origin. We see then that if f (x) =

0(|x|−n), Ωk

the

area

of

the

unit

sphere

in

Rk,

i.e.,

Ωk

=

2

πk/2 Γ(k/2)

,

(f )∨(x) = f (x + k · ξ0) dk =

f (x + k · y) dm(y) dk

K

K ξ0

∞

= (M |y|f )(x) dm(y) = Ωn−1

rn−2

1 Ωn

f (x + rω) dω dr,

ξ0

0

S n−1

so

(17)

(f )∨(x) = Ωn−1 |x − y|−1f (y) dy .

Ωn

Rn

ˇ We consider now the analog of Theorem 2.4 for the transform ϕ → ϕ. ˇ But ϕ ∈ SH (Pn) does not imply ϕ ∈ S(Rn). (If this were so and we
by Theorem 2.4 write ϕ = f , f ∈ S(Rn) then the inversion formula in Theorem 3.1 for n = 3 would imply f (x) dx = 0.) On a smaller space we
shall obtain a more satisfactory result. Let S∗(Rn) denote the space of all functions f ∈ S(Rn) which are or-
thogonal to all polynomials, i.e.,

f (x)P (x) dx = 0 for all polynomials P .
Rn

10 Chapter I. The Radon Transform on Rn

0A

ξ d(0, ξ) > A

FIGURE I.4.
Similarly, let S∗(Pn) ⊂ S(Pn) be the space of ϕ satisfying
ϕ(ω, r)p(r) dr = 0 for all polynomials p .
R
Note that under the Fourier transform the space S∗(Rn) corresponds to the subspace S0(Rn) ⊂ S(Rn) of functions all of whose derivatives vanish at 0.
ˇ Corollary 2.5. The transforms f → f , ϕ → ϕ are bijections of S∗(Rn)
onto S∗(Pn) and of S∗(Pn) onto S∗(Rn), respectively.
The first statement is clear from (8) if we take into account the elementary fact that the polynomials x → x, ω k span the space of homogeneous
ˇ polynomials of degree k. To see that ϕ → ϕ is a bijection of S∗(Pn) onto
S∗(Rn) we use (17), knowing that ϕ = f for some f ∈ S∗(Rn). The right hand side of (17) is the convolution of f with the tempered distribution |x|−1 whose Fourier transform is by Chapter VII, §6 a constant multiple of |u|1−n. (Here we leave out the trivial case n = 1.) By Chapter VII, §4 this convolution is a tempered distribution whose Fourier transform is a constant multiple of |u|1−nf (u). But, by Lemma 6.6, Chapter VII this lies in
ˇ the space S0(Rn) since f does. Now (17) implies that ϕ = (f )∨ ∈ S∗(Rn) ˇ ˇ and that ϕ ≡ 0 if ϕ ≡ 0. Finally we see that the mapping ϕ → ϕ is
surjective because the function
((f )∨)e(u) = c|u|1−nf (u)
(where c is a constant) runs through S0(Rn) as f runs through S∗(Rn). We now turn to the space D(Rn) and its image under the Radon trans-
form. We begin with a preliminary result. (See Fig. I.4.)
Theorem 2.6. (The support theorem.) Let f ∈ C(Rn) satisfy the following conditions:

§2 The Radon Transform . . . The Support Theorem 11
(i) For each integer k > 0, |x|kf (x) is bounded. (ii) There exists a constant A > 0 such that

f (ξ) = 0 for d(0, ξ) > A ,

d denoting distance. Then

f (x) = 0 for |x| > A .

Proof. Replacing f by the convolution ϕ∗f , where ϕ is a radial C∞ function with support in a small ball B (0), we see that it suffices to prove the theorem for f ∈ E(Rn). In fact, ϕ ∗ f is smooth, it satisfies (i) and by (5) it satisfies (ii) with A replaced by A+ . Assuming the theorem for the smooth case we deduce that support (ϕ ∗ f ) ⊂ BA+ (0) so letting → 0 we obtain support (f ) ⊂ Closure BA(0).
To begin with we assume f is a radial function. Then f (x) = F (|x|)
where F ∈ E(R) and even. Then f has the form f (ξ) = F (d(0, ξ)) where
F is given by

F (p) =

F ((p2 + |y|2)1/2) dm(y) , (p ≥ 0)

Rn−1

because of the definition of the Radon transform. Using polar coordinates in Rn−1 we obtain

(18)

∞
F (p) = Ωn−1 F ((p2 + t2)1/2)tn−2 dt .
0

Here we substitute s = (p2 + t2)−1/2 and then put u = p−1. Then (18) becomes
u
un−3F (u−1) = Ωn−1 (F (s−1)s−n)(u2 − s2)(n−3)/2 ds .
0

We write this equation for simplicity

(19)

u
h(u) = g(s)(u2 − s2)(n−3)/2 ds .
0

This integral equation is very close to Abel’s integral equation (WhittakerWatson [1927], Ch. XI) and can be inverted as follows. Multiplying both

12 Chapter I. The Radon Transform on Rn

sides by u(t2 − u2)(n−3)/2 and integrating over 0 ≤ u ≤ t, we obtain

t
h(u)(t2 − u2)(n−3)/2u du

0
=

⎡

⎤

tu

⎣ g(s)[(u2 − s2)(t2 − u2)](n−3)/2 ds⎦ u du

0 0⎡

⎤

t

t

=

g(s) ⎣ u[(t2 − u2)(u2 − s2)](n−3)/2 du⎦ ds .

0

u=s

The substitution (t2 − s2)V = (t2 + s2) − 2u2 gives an explicit evaluation of the inner integral and we obtain

t

t

h(u)(t2 − u2)(n−3)/2u du = C g(s)(t2 − s2)n−2 ds ,

0

0

where

C

=

21−nπ

1 2

Γ((n−

1)/2)/Γ(n/2).

Here

we

apply

the

operator

d d(t2)

=

1 2t
of

dtdt−1(gn(−t).1H) etnimceeswewhoebrteabiny

the

right

hand

side

gives

a

constant

multiple

(20)

F (t−1)t−n = ct

d d(t2)

n−1

t
(t2 − u2)(n−3)/2un−2F (u−1) du

0

where c−1 = (n − 2)!Ωn/2n. By assumption (ii) we have F (u−1) = 0 if u−1 ≥ A, that is if u ≤ A−1. But then (20) implies F (t−1) = 0 if t ≤ A−1, that is if t−1 ≥ A. This proves the theorem for the case when f is radial.
We consider next the case of a general f . Fix x ∈ Rn and consider the function
gx(y) = f (x + k · y) dk
K
as in (15). Then gx satisfies (i) and

(21)

gx(ξ) = f (x + k · ξ) dk ,

K

x + k · ξ denoting the translate of the hyperplane k · ξ by x. The triangle inequality shows that

d(0, x + k · ξ) ≥ d(0, ξ) − |x| , x ∈ Rn, k ∈ K .

§2 The Radon Transform . . . The Support Theorem 13

Hence we conclude from assumption (ii) and (21) that

(22)

gx(ξ) = 0 if d(0, ξ) > A + |x| .

But gx is a radial function so (22) implies by the first part of the proof that

(23)

f (x + k · y) dk = 0 if |y| > A + |x| .

K

Geometrically, this formula reads: The surface integral of f over S|y|(x) is 0 if the ball B|y|(x) contains the ball BA(0). The theorem is therefore a consequence of the following lemma.
Lemma 2.7. Let f ∈ C(Rn) be such that for each integer k > 0,

sup |x|k|f (x)| < ∞ .
x∈Rn

Suppose f has surface integral 0 over every sphere S which encloses the unit ball. Then f (x) ≡ 0 for |x| > 1.

Proof. The idea is to perturb S in the relation

(24)

f (s) dω(s) = 0

S

slightly, and differentiate with respect to the parameter of the perturbations, thereby obtaining additional relations. (See Fig. I.5.) Replacing, as above, f with a suitable convolution ϕ ∗ f we see that it suffices to prove the lemma for f in E(Rn). Writing S = SR(x) and viewing the exterior of the ball BR(x) as a union of spheres with center x we have by the assumptions,

R

1

x

0

S = SR(x)

f (y) dy = f (y) dy ,

BR (x)

Rn

FIGURE I.5.

which is a constant. Differentiating with respect to xi we obtain

(25)

(∂if )(x + y) dy = 0 .

BR (0)

14 Chapter I. The Radon Transform on Rn
We use now the divergence theorem

(26)

(divF )(y) dy =

F, n (s) dω(s)

BR (0)

SR(0)

for a vector field F on Rn, n denoting the outgoing unit normal and dω

the

surface

element

on

SR(0).

For

the

vector

field

F

(y)

=

f (x

+

y

)

∂ ∂yi

we

obtain from (25) and (26), since n = R−1(s1, . . . , sn),

(27)

f (x + s)si dω(s) = 0 .

SR (0)

But by (24)

so by adding

f (x + s)xi dω(s) = 0
SR (0)

f (s)si dω(s) = 0 .
S

This means that the hypotheses of the lemma hold for f (x) replaced by the function xif (x). By iteration

f (s)P (s) dω(s) = 0
S
for any polynomial P , so f ≡ 0 on S. This proves the lemma as well as Theorem 2.6.

Corollary 2.8. Let f ∈ C(Rn) satisfy (i) in Theorem 2.6 and assume

f (ξ) = 0

for all hyperplanes ξ disjoint from a certain compact convex set C. Then

(28)

f (x) = 0 for x ∈/ C .

In fact, if B is a closed ball containing C we have by Theorem 2.6, f (x) = 0 for x ∈/ B. But C is the intersection of such balls. In fact, if x ∈/ C, x and C are separated by a hyperplane ξ. Let z ∈ ξ. The half space bounded by ξ, containing C is the union of an increasing sequence of balls B tangential to ξ at z. Finitely many cover C and the largest of those contains C. Thus, x ∈/ B⊃C B, so (28) follows.

§2 The Radon Transform . . . The Support Theorem 15

Remark 2.9. While condition (i) of rapid decrease entered in the proof of Lemma 2.7 (we used |x|kf (x) ∈ L1(Rn) for each k > 0) one may wonder whether it could not be weakened in Theorem 2.6 and perhaps even dropped in Lemma 2.7.
As an example, showing that the condition of rapid decrease can not be dropped in either result consider for n = 2 the function

f (x, y) = (x + iy)−5

made smooth in R2 by changing it in a small disk around 0. Using Cauchy’s theorem for a large semicircle we have f (x) dm(x) = 0 for every line outside the unit circle. Thus (ii) is satisfied in Theorem 2.6. Hence (i) cannot be dropped or weakened substantially.

This same example works for Lemma 2.7. In fact, let S be a circle

|z

− z0|

=

r

enclosing

the

unit

disk.

Then

dω(s)

=

−ir

dz z−z0

,

so,

by

ex-

panding the contour or by residue calculus,

z−5(z − z0)−1 dz = 0 ,
S
(the residue at z = 0 and z = z0 cancel) so we have in fact

f (s) dω(s) = 0 .
S
We recall now that DH (Pn) is the space of symmetric C∞ functions ϕ(ξ) = ϕ(ω, p) on Pn of compact support such that for each k ∈ Z+, R ϕ(ω, p)pk dp is a homogeneous kth degree polynomial in ω1, . . . , ωn. Combining Theorems 2.4, 2.6 we obtain the following characterization of the Radon transform of the space D(Rn). This can be regarded as the analog for the Radon transform of the Paley-Wiener theorem for the Fourier transform (see Chapter VII).
Theorem 2.10. (The Paley-Wiener theorem.) The Radon transform is a bijection of D(Rn) onto DH (Pn).
We conclude this section with a variation and a consequence of Theorem 2.6.
Lemma 2.11. Let f ∈ Cc(Rn), A > 0, ω0 a fixed unit vector and N ⊂ S a neighborhood of ω0 in the unit sphere S ⊂ Rn. Assume
f (ω, p) = 0 for ω ∈ N, p > A .

Then

(29)

f (x) = 0 in the half-space x, ω0 > A .

16 Chapter I. The Radon Transform on Rn

Proof. Let B be a closed ball around the origin containing the support of f . Let > 0 and let H be the union of the half spaces x, ω > A + as ω runs through N . Then by our assumption,

(30)

f (ξ) = 0 if ξ ⊂ H .

Now choose a ball B with a center on the ray from 0 through −ω0, with the point (A + 2 )ω0 on the boundary, and with radius so large that any hyperplane ξ intersecting B but not B must be in H . Then by (30),

f (ξ) = 0 whenever ξ ∈ Pn, ξ ∩ B = ∅ .

Hence by Theorem 2.6, f (x) = 0 for x ∈/ B . In particular, f (x) = 0 for x, ω0 > A + 2 ; since > 0 is arbitrary, the lemma follows.
Corollary 2.12. Let N be any open subset of the unit sphere Sn−1. If f ∈ Cc(Rn) and
f (ω, p) = 0 for p ∈ R, ω ∈ N,
then f ≡ 0.
Since f(−ω, −p) = f (ω, p) this is obvious from Lemma 2.11.

§3 The Inversion Formula. Injectivity Questions

We shall now establish explicit inversion formulas for the Radon transform
ˇ f → f and its dual ϕ → ϕ.
Theorem 3.1. The function f can be recovered from the Radon transform by means of the following inversion formula

(31)

cf = (−L)(n−1)/2((f )∨), f ∈ E(Rn) ,

provided f (x) = 0(|x|−N ) for some N > n − 1. Here c is the constant

c = (4π)(n−1)/2Γ(n/2)/Γ(1/2) ,

and the power of the Laplacian L is given in Chapter VII, §6.

Proof. We first give a geometric proof of (31) for n odd. We start with some general useful facts about the mean value operator M r. It is a familiar fact that if f ∈ C2(Rn) is a radial function, i.e., f (x) = F (r), r = |x|, then

d2F n − 1 dF

(32)

(Lf )(x) = +

.

dr2

r dr

This is immediate from the relations

∂2f ∂2f ∂r 2 ∂f ∂2r ∂x2i = ∂r2 ∂xi + ∂r ∂x2i .

§3 The Inversion Formula. Injectivity Questions 17

Lemma 3.2. (i) LM r = M rL for each r > 0.

(ii) For f ∈ C2(Rn) the mean value (M rf )(x) satisfies the “Darboux equation”

Lx (M rf )(x) =

∂2 n − 1 ∂ +

(M rf (x)) ,

∂r2

r ∂r

that is, the function F (x, y) = (M |y|f )(x) satisfies

Lx(F (x, y)) = Ly(F (x, y)) .
Proof. We prove this group theoretically, using expression (15) for the mean value. For z ∈ Rn, k ∈ K, let Tz denote the translation x → x + z and Rk the rotation x → k · x. Since L is invariant under these transformations, we have if r = |y|,

(LM rf )(x) = Lx(f (x + k · y)) dk = (Lf )(x + k · y) dk = (M rLf )(x)

K

K

= [(Lf ) ◦ Tx ◦ Rk](y) dk = [L(f ◦ Tx ◦ Rk)](y) dk

K
⎛

K
⎞

= Ly ⎝ f (x + k · y)⎠ dk ,
K

which proves the lemma.
Now suppose f ∈ S(Rn). Fix a hyperplane ξ0 through 0, and an isometry g ∈ M(n). As k runs through O(n), gk · ξ0 runs through the set of hyperplanes through g · 0, and we have

ˇϕ(g · 0) = ϕ(gk · ξ0) dk
K
and therefore

(f )∨(g · 0) =

f (gk · y) dm(y) dk

K ξ0
= dm(y) f (gk · y) dk = (M |y|f )(g · 0) dm(y) .

ξ0

K

ξ0

Hence

∞

(33)

((f ))∨(x) = Ωn−1 (M rf )(x)rn−2 dr ,

0

18 Chapter I. The Radon Transform on Rn

where Ωn−1 is the area of the unit sphere in Rn−1. Applying L to (33), using (32) and Lemma 3.2, we obtain

∞

(34)

L((f )∨) = Ωn−1

d2F dr2

+

n − 1 dF r dr

rn−2 dr,

0

where F (r) = (M rf )(x). Integrating by parts and using

we get

F (0) = f (x) , lim rkF (r) = 0 ,
r→∞

L((f )∨) =

−Ωn−1f (x)

if n = 3 ,

−Ωn−1(n − 3)

∞ 0

F

(r)rn−4

dr

(n > 3) .

More generally,

∞

Lx

(M rf )(x)rk dr

=

−(n−2)f (x) if k = 1 ,

−(n−1 − k)(k−1)

∞ 0

F

(r)rk−2 dr,

(k

>

1)

.

0

If n is odd, the formula in Theorem 3.1 follows by iteration. Although we assumed f ∈ S(Rn) the proof is valid under much weaker assumptions.
Another proof (for all n) uses the Riesz potential in Ch. VII. In fact, (17) implies

(35)

(f )∨

=

2n−1

π

n 2

−1

Γ(n/2)I

n−1f

,

so the inversion formula follows from Theorem 6.11 in Ch. VII. However, since the fractional power (−L)(n−1)/2 is only defined by means
of holomorphic continuation of the Riesz potential, the geometric proof above for n odd and the alternative proof of Theorem 1.4 in Ch. III have
to be considered much more explicit and direct inversion formulas.

Remark 3.3. It is interesting to observe that while the inversion formula requires f (x) = 0(|x|−N ) for one N > n − 1 the support theorem requires f (x) = 0(|x|−N ) for all N as mentioned in Remark 2.9.
In connection with the inversion formula we shall now discuss the weaker question of injectivity.
Proposition 3.4. Let f ∈ L1(Rn). Then f (ω, p) exists for almost all (ω, p) ∈ Sn−1 × R . Also the map f → f is injective on L1(Rn).
Proof. For a fixed ω ∈ Sn−1, x = x + pω where x , ω = 0, p ∈ R. This gives a product representation Rn = R×H with H Rn−1. By the Fubini theorem f (ω, p) exist for almost all p, and

f (x) dx = f (ω, p) dp .

Rn

R

§3 The Inversion Formula. Injectivity Questions 19

Then

|f (ω, p)| dω dp ≤

|f |b(ω, p) dω dp

Sn−1×R

Sn−1×R

proving the first statement.

= dω |f (x)| dx < ∞,

Sn−1

Rn

Using this on the function f (x)e−i x,ω , we derive (4) for f ∈ L1(Rn). Thus f (ω, p) = 0 for almost all ω, p implies f = 0 a.e. so f = 0 by the injectivity of the Fourier transform.

An example of non-injectivity We shall now give an example of a smooth function f ≡ 0 on R2 which is integrable on each line ξ ⊂ R2, yet f (ξ) = 0 for all ξ. Such an example was first constructed by Zalcman [1982], using a delicate approximation theorem by Arakelyan [1964]. A more elementary construction, which we follow below, was given by Armitage [1994].
Lemma 3.5. Let z1, z2 ∈ C and |z1 − z2| < 1. If f1 is holomorphic on C − {z1} and > 0 there exists a function f2 holomorphic on C − {z2} such that

(36)

|f1(z) − f2(z)| < (1 + |z|)2 for |z − z2| > 1 .

Proof. We have a Laurent expansion of f1 centered at z2:
∞
f1(z) = f0(z) + aj(z − z2)−j , |z − z2| > |z1 − z2|
j=1

f0 being an entire function. For m ∈ Z+ define the function
m
f2(z) = f0(z) + aj(z − z2)−j z = z2
j=1
so

∞

(37)

f1(z) − f2(z) = aj(z − z2)−j .

m+1

We shall choose m such that (36) holds. The coefficients aj are given by

1 aj = 2πi

f1(ζ) dζ (ζ − z2)j+1

|ζ−z2|=r

20 Chapter I. The Radon Transform on Rn

with any r > |z1 − z2|. In order to estimate (37) it is convenient to take r such that r|z1 − z2| = 2. Let M be the maximum of f1(ζ) on the circle |ζ − z2| = r. Then
M |aj| ≤ rj .
We can then estimate the right hand side of (37) by a geometric series, noting that r|z − z2| − 1 > r|z1 − z2| − 1. Then

M

(38)

|f1(z) − f2(z)| ≤ (r|z − z2|)m .

Given K > 1 we can choose m = m1 such that the right hand side is ≤ (1 + |z|)−2 for |z − z2| > K. We can also find m = m2 such that it is bounded by (1 + |z|)−2 for 1 < |z − z2| ≤ K. With m = m1 + m2, (36) holds.

Theorem 3.6. There exists a holomorphic function f ≡ 0 in C such that each derivative f (n) is integrable on every line ξ and

(39)

f (n)(z) dm(z) = 0 , n = 0, 1, 2, . . . .

ξ

Proof. Choose a sequence (ζk) on the parabolic arc P = {t + it2|t ≥ 0} such that
ζ0 = 0 , |ζk − ζk−1| < 1 k ≥ 1 , ζk → ∞ .

Let g0(z) = 1/z2. By iteration of Lemma 3.5 we obtain gk holomorphic on C − {ζk} and

(40)

|gk (z )

−

gk−1(z)|

≤

2k(1

1 +

|z|)2

,

k ≥ 1 , |z − ζk| > 1 .

For each z0 there is an integer N and a neighborhood |z − z0| < δ on which (40) holds for k > N . Thus

|gk+p(z) − gk(z)|

≤

1 (1 + |z|)2

1 2k

,

|z − z0| < δ , p > 0 .

Thus gk(z) converges uniformly to an entire function g(z). Let

Pa = {z : infw∈P |z − w| > a} .

If z ∈ P1 then by (40)

∞

1

|g(z) − g0(z)| ≤ |gk(z) − gk−1(z)| ≤ 1 + |z|2 < |g0(z))| ,

k=1

§3 The Inversion Formula. Injectivity Questions 21

so g(z) ≡ 0 and

1 |g(ζ)| ≤ 2 |ζ|2

(ζ ∈ P1) .

For z ∈ P2 consider a circle S1(z) and express g(n)(z) by Cauchy’s formula

g(n)(z) = n! 2πi

g(ζ) (ζ − z)n+1 dζ .

S1(z)

Since S1(z) ⊂ P1 we have

1

1

|g(ζ)|

≤

2 |ζ |2

≤

2 (|z| − 1)2

,

so

|g(n)(z)|

≤

2n! (|z| − 1)2

,

z ∈ P2 .

Since ξ P2 is bounded for each line ξ we have

|g(n+1)(z)| dm(z) < ∞
ξ
and g(n)(z) → 0 as z → ∞ on ξ. Then the function f (z) = g (z) satisfies (39).

ˇ We shall now prove an inversion formula for the dual transform ϕ → ϕ
on the subspace S∗(Pn) similar to Theorem 3.1.

Theorem 3.7. We have

ˇ cϕ = (− )(n−1)/2(ϕ)b, ϕ ∈ S∗(Pn) ,

where c is the constant (4π)(n−1)/2Γ(n/2)/Γ(1/2).

Here

denotes

as

before

the

operator

d2 dp2

and

its

fractional

powers

are

again defined in terms of the Riesz’ potentials on the 1-dimensional p-space.

If n is odd our inversion formula follows from the odd-dimensional case

ˇ in Theorem 3.1 if we put f = ϕ and take Lemma 2.1 and Corollary 2.5 into
account. Suppose now n is even. We claim that

(41)

((−L)

n−1 2

f

)b =

(−

n−1
)2 f

f ∈ S∗(Rn) .

By Lemma 6.6 in Chapter VII, (−L)(n−1)/2f belongs to S∗(Rn). Taking the 1-dimensional Fourier transform of ((−L)(n−1)/2f )b, we obtain

(−L)(n−1)/2f e(sω) = |s|n−1f (sω) .

22 Chapter I. The Radon Transform on Rn

On the other hand, for a fixed ω, p → f (ω, p) is in S∗(R). By the lemma quoted, the function p → ((− )(n−1)/2f )(ω, p) also belongs to S∗(R) and its Fourier transform equals |s|n−1f (sω). This proves (41). Now Theorem 3.7 follows from (41) if we put in (41)

ϕ = g , f = (g)∨ , g ∈ S∗(Rn) ,

because, by Corollary 2.5, g belongs to S∗(Pn) . Because of its theoretical importance we now prove the inversion theo-
rem (3.1) in a different form. The proof is less geometric and involves just the one variable Fourier transform.
Let H denote the Hilbert transform

∞

(HF )(t) = i π

F (p) dp,
t−p

−∞

F ∈ S(R),

the integral being considered as the Cauchy principal value (see Lemma 3.9 below). For ϕ ∈ S(Pn) let Λϕ be defined by

(42)

(Λϕ)(ω, p) =

dn−1 dpn−1

ϕ(ω

,

p)

Hp

dn−1 dpn−1

ϕ(ω,

p)

n odd, n even.

Note that in both cases (Λϕ)(−ω, −p) = (Λϕ)(ω, p) so Λϕ is a function on Pn.

Theorem 3.8. Let Λ be as defined by (42). Then

cf = (Λf )∨ , f ∈ S(Rn) ,

where as before

c = (−4π)(n−1)/2Γ(n/2)/Γ(1/2) .

Proof. By the inversion formula for the Fourier transform and by (4),

f (x) = (2π)−n

∞∞

dω

e−ispf (ω, p) dp eis x,ω sn−1 ds ,

Sn−1

0 −∞

which we write as

f (x) = (2π)−n

F (ω, x) dω = (2π)−n

1 2

(F

(ω,

x)+

F

(−ω,

x))

dω

.

Sn−1

Sn−1

Using f (−ω, p) = f (ω, −p) this gives the formula

(43)

∞

∞

f (x)

=

1 2

(2π)−n

dω |s|n−1eis x,ω ds e−ispf (ω, p) dp .

Sn−1

−∞

−∞

§3 The Inversion Formula. Injectivity Questions 23

If n is odd the absolute value on s can be dropped. The factor sn−1 can

be

removed

by

replacing

f (ω, p)

by

(−i)n−1

dn−1 dpn−1

f

(ω,

p).

The

inversion

formula for the Fourier transform on R then gives

f (x)

=

1 2

(2π)−n

(2π)+1(−i)n−1

Sn−1

dn−1 dpn−1 f (ω, p) p= x,ω dω

as desired.

In order to deal with the case n even we recall some general facts.

Lemma 3.9. Let S denote the Cauchy principal value

S : ψ → lim
→0

ψ(x) dx .
x

|x|≥

Then S is a tempered distribution and S is the function

S(s) = −πi sgn(s) =

−πi πi

s≥0 s<0

.

Proof. If supp(ψ) ⊂ (−K, K) we can write

ψ(x) dx =
x

ψ(x) − ψ(0) dx + x

ψ(0) dx
x

|x|≥

K≥|x|≥

K≥|x|≥

and the last term is 0. Thus S is indeed a distribution. To see that S is tempered let ϕ ∈ E(R) be such that ϕ(x) = 0 for |x| ≤ 1 and ϕ(x) = 1 for |x| ≥ 2. For ψ ∈ S(R) we put

ψ = ϕψ + (1 − ϕ)ψ = ξ + η .

Then ξ ∈ S(R) but ξ = 0 near 0 and η has compact support. If ψi → 0 in S(R), then ξi → 0 in S and S(ξi) → 0. Also S(ηi) → 0 is obvious. Thus S is tempered. Also xS = 1 so
2πδ = 1 = (xS)e = i(S) .

But sgn = 2δ, so S = −πi sgn +C. But S and sgn are odd, so C = 0. By Ch. VII, Proposition 4.4, HF is a tempered distribution and by The-
orem 4.6,

(44)

(HF )e(s) = sgn(s)F (s) .

This

in

turn

implies

(HF

)(s)

=

d ds

HF

(s)

since

both

sides

have

the

same

Fourier transform. For n even we write in (43), |s|n−1 = sgn(s)sn−1 and

then (43) implies

(45)

f (x) = c0

dω

sgn(s)eis x,ω ds

dn−1 dpn−1

f

(ω,

p)e−isp

dp

,

Sn−1

R

R

24 Chapter I. The Radon Transform on Rn

where

c0

=

1 2

(−i)n−1(2π)−n.

Now

we

have

for

each

F

∈

S(R)

the

identity

sgn(s)eist F (p)e−ips dp ds = 2π(HF )(t) .

R

R

In fact, if we apply both sides to ψ with ψ ∈ S(R), the left hand side is by (44)

sgn(s)eistF (s) ds ψ(t) dt
RR
= sgn(s)F (s)2πψ(s) ds = 2π(HF )e(ψ) = 2π(HF )(ψ) .
R

Putting

F (p)

=

dn−1 dpn−1

f

(ω,

p)

and

using

(45),

Theorem

3.8

follows

also

for

n even.

For later use we add here a few remarks concerning H. Let F ∈ D have support contained in (−R, R). Then

−iπ(HF )(t) = lim
→0

F (p)

F (p)

dp = lim

dp ,

t−p

→0 t − p

<|t−p|

I

where I = {p : |p| < R, < |t − p|}. We decompose this last integral

F (p) t−p

dp

=

F

(p) − F t−p

(t)

dp

+

F

(t)

t

dp −p

.

I

I

I

The last term vanishes for |t| > R and all > 0. The first term on the right is majorized by

F (t) − F (p) t−p

dp ≤ 2R sup |F | .

|p|<R

Thus by the dominated convergence theorem

lim (HF )(t) = 0 .
|t|→∞

Also if J ⊂ (−R, R) is a compact subset the mapping F → HF is continuous from DJ into E(R) (with the topologies in Chapter VII, §1).
Later we prove one more version of the inversion formula from the point
of view of double fibrations. See Theorem 1.4 in Chapter III.

§4 The Plancherel Formula 25
§4 The Plancherel Formula
We recall that the functions on Pn have been identified with the functions ϕ on Sn−1 × R which are even: ϕ(−ω, −p) = ϕ(ω, p). The functional

(46)

ϕ→

ϕ(ω, p) dω dp ϕ ∈ Cc(Pn) ,

Sn−1 R

is therefore a well defined measure on Pn, denoted dω dp. The group M(n) of rigid motions of Rn acts transitively on Pn: it also leaves the measure dω dp invariant. It suffices to verify this latter statement for the translations T in M(n) because M(n) is generated by them together with the rotations around 0, and these rotations clearly leave dω dp invariant. But

(ϕ ◦ T )(ω, p) = ϕ(ω, p + q(ω, T ))

where q(ω, T ) ∈ R is independent of p so

(ϕ ◦ T )(ω, p) dω dp = ϕ(ω, p + q(ω, T )) dω dp = ϕ(ω, p) dp dω ,

proving the invariance.
In accordance with the definition of (−L)p in Ch. VII the fractional power k is defined on S(Pn) by

(47)

(−

k)ϕ(ω, p) = 1 H1(−2k)

ϕ(ω, q)|p − q|−2k−1 dq

R

and then the 1-dimensional Fourier transform satisfies

(48)

((− )kϕ)e(ω, s) = |s|2kϕ(ω, s) .

Now, if f ∈ S(Rn) we have by (4)

f (ω, p) = (2π)−1 f (sω)eisp ds

and

(49)

(−

n−1
)4

f (ω,

p)

=

(2π)−1

|s|

n−1 2

f

(sω)eisp

ds

.

R

Theorem 4.1. The mapping f →

f n−1 4

extends

to

an

isometry

of

L2(Rn) onto the space L2e(Sn−1 × R) of even functions in L2(Sn−1 × R), the measure on Sn−1 × R being

1 2

(2π)1−n

dω

dp

.

26 Chapter I. The Radon Transform on Rn

Proof. By (49) we have from the Plancherel formula on R

(2π)

|(−

n−1
)4

f (ω,

p)|2

dp

=

|s|n−1|f (sω)|2 ds ,

R

R

so by integration over Sn−1 and using the Plancherel formula for f (x) →

f (sω) we obtain

|f (x)|2

dx

=

1 2

(2π)1−n

|

n−1 4

f

(ω

,

p)|2

dω

dp

.

Rn

Sn−1 ×R

It remains to prove that the mapping is surjective. For this it would suffice to prove that if ϕ ∈ L2(Sn−1 × R) is even and satisfies

ϕ(ω, p)(−

n−1
) 4 f (ω, p) dω dp = 0

Sn−1 R
for all f ∈ S(Rn), then ϕ = 0. Taking Fourier transforms we must prove that if ψ ∈ L2(Sn−1 × R) is even and satisfies

(50)

ψ(ω,

s)|s|

n−1 2

f (sω)

ds

dω

=

0

Sn−1 R
for all f ∈ S(Rn), then ψ = 0. Using the condition ψ(−ω, −s) = ψ(ω, s), we see that

0

ψ(ω,

s)|s|

1 2

(n−1)

f

(sω)

ds

dω

Sn−1 −∞

∞

=

ψ(ω,

t)|t|

1 2

(n−1)f

(tω)

dt

dω

Sn−1 0

so (50) holds with R replaced with the positive axis R+. But then the

function

Ψ(u) = ψ

u |u|

,

|u|

|u|−

1 2

(n−1)

,

u ∈ Rn − {0}

satisfies

Ψ(u)f (u) du = 0 , f ∈ S(Rn)

Rn
so Ψ = 0 almost everywhere, whence ψ = 0. If we combine the inversion formula in Theorem 3.8 with (51) below we
obtain the following version of the Plancherel formula

c f (x)g(x) dx = (Λf )(ξ)g(ξ) dξ .

Rn

Pn

§5 Radon Transform of Distributions 27
§5 Radon Transform of Distributions
It will be proved in a general context in Chapter II (Proposition 2.2) that

(51)

ˇ f(ξ)ϕ(ξ) dξ = f (x)ϕ(x) dx

Pn

Rn

for f ∈ Cc(Rn), ϕ ∈ C(Pn) if dξ is a suitable fixed M(n)-invariant measure on Pn. Thus dξ = γ dω dp where γ is a constant, independent of f and
ϕ. With applications to distributions in mind we shall prove (51) in a
somewhat stronger form.

ˇ Lemma 5.1. Formula (51) holds (with f and ϕ existing almost anywhere)
in the following two situations:

(a) f ∈ L1(Rn) vanishing outside a compact set; ϕ ∈ C(Pn) .

(b) f ∈ Cc(Rn), ϕ locally integrable.

Also dξ = Ω−n 1 dω dp.

Proof. We shall use the Fubini theorem repeatedly both on the product Rn × Sn−1 and on the product Rn = R × Rn−1. Since f ∈ L1(Rn) we have (as noted before) that for each ω ∈ Sn−1, f (ω, p) exists for almost all p and

f (x) dx = f (ω, p) dp .

Rn

R

We also proved that f (ω, p) exists for almost all (ω, p) ∈ Sn−1 × R. Next we consider the measurable function

(x, ω) → f (x)ϕ(ω, ω, x ) on Rn × Sn−1 .

We have

|f (x)ϕ(ω, ω, x )| dω dx
Sn−1 ×Rn

=

|f (x)ϕ(ω, ω, x )| dx dω

Sn−1 Rn

=

|f |b(ω, p)|ϕ(ω, p)| dp dω ,

Sn−1 R

which in both cases is finite. Thus f (x) · ϕ(ω, ω, x ) is integrable on Rn × Sn−1 and its integral can be calculated by removing the absolute

28 Chapter I. The Radon Transform on Rn
values above. This gives the left hand side of (51). Reversing the integra-
ˇ tions we conclude that ϕ(x) exists for almost all x and that the double
integral reduces to the right hand side of (51). The formula (51) dictates how to define the Radon transform and its
dual for distributions (see Chapter VII). In order to make the definitions
ˇ formally consistent with those for functions we would require S(ϕ) = S(ϕ), ˇΣ(f ) = Σ(f) if S and Σ are distributions on Rn and Pn, respectively. But
while f ∈ D(Rn) implies f ∈ D(Pn) a similar implication does not hold
ˇ for ϕ; we do not even have ϕ ∈ S(Rn) for ϕ ∈ D(Pn) so S cannot be
defined as above even if S is assumed to be tempered. Using the notation E (resp. D) for the space of C∞ functions (resp. of compact support) and D (resp. E ) for the space of distributions (resp. of compact support) we make the following definition.
Definition. For S ∈ E (Rn) we define the functional S by
ˇ S(ϕ) = S(ϕ) for ϕ ∈ E(Pn) ; ˇ for Σ ∈ D (Pn) we define the functional Σ by
ˇΣ(f ) = Σ(f ) for f ∈ D(Rn) . ˇ Lemma 5.2. (i) For each Σ ∈ D (Pn) we have Σ ∈ D (Rn).
(ii) For each S ∈ E (Rn) we have S ∈ E (Pn).
Proof. For A > 0 let DA(Rn) denote the set of functions f ∈ D(Rn) with support in the closure of BA(0). Similarly let DA(Pn) denote the set of functions ϕ ∈ D(Pn) with support in the closure of the “ball”
βA(0) = {ξ ∈ Pn : d(0, ξ) < A} .
ˇ The mapping of f → f from DA(Rn) to DA(Pn) being continuous (with
the topologies defined in Chapter VII, §1) the restriction of Σ to each DA(Rn) is continuous so (i) follows. That S is a distribution is clear from (3). Concerning its support select R > 0 such that S has support inside
ˇ BR(0). Then if ϕ(ω, p) = 0 for |p| ≤ R we have ϕ(x) = 0 for |x| ≤ R ˇ whence S(ϕ) = S(ϕ) = 0.
Lemma 5.3. For S ∈ E (Rn), Σ ∈ D (Pn) we have
ˇ (LS)b = S , ( Σ)∨ = LΣ .
Proof. In fact by Lemma 2.1,
ˇ ˇ (LS)b(ϕ) = (LS)(ϕ) = S(Lϕ) = S(( ϕ)∨) = S( ϕ) = ( S)(ϕ) .
The other relation is proved in the same manner.

§5 Radon Transform of Distributions 29

We shall now prove an analog of the support theorem (Theorem 2.6) for distributions. For A > 0 let βA(0) be defined as above and let supp denote support.
Theorem 5.4 (Support theorem for distributions). Let T ∈ E (Rn) satisfy the condition

supp T ⊂ C (βA(0)) , (C = closure) .

Then

supp(T ) ⊂ C (BA(0)) .

Proof. For f ∈ D(Rn), ϕ ∈ D(Pn) we can consider the “convolution”

(f × ϕ)(ξ) = f (y)ϕ(ξ − y) dy ,
Rn
where for ξ ∈ Pn, ξ − y denotes the translate of the hyperplane ξ by −y. Then
ˇ (f × ϕ)∨ = f ∗ ϕ .
In fact, if ξ0 is any hyperplane through 0,

(f × ϕ)∨(x) = dk f (y)ϕ(x + k · ξ0 − y) dy

K

Rn

=

ˇ dk f (x − y)ϕ(y + k · ξ0) dy = (f ∗ ϕ)(x) .

K

Rn

By the definition of T , the support assumption on T is equivalent to

ˇ T (ϕ) = 0
for all ϕ ∈ D(Pn) with support in Pn−C (βA(0)). Let > 0, let f ∈ D(Rn) be a symmetric function with support in C (B (0)) and let ϕ = D(Pn) have support contained in Pn − C (βA+ (0)). Since d(0, ξ − y) ≤ d(0, ξ) + |y| it follows that f × ϕ has support in Pn − C (βA(0)); thus by the formulas above, and the symmetry of f ,

ˇ ˇ (f ∗ T )(ϕ) = T (f ∗ ϕ) = T ((f × ϕ)∨) = 0 .

But then

ˇ (f ∗ T )b(ϕ) = (f ∗ T )(ϕ) = 0 ,

which means that (f ∗T )bhas support in C (βA+ (0)). But now Theorem 2.6 implies that f ∗ T has support in C (BA+ )(0). Letting → 0 we obtain by Prop. 3.4, Ch. VII the desired conclusion, supp(T ) ⊂ C (BA(0)).

30 Chapter I. The Radon Transform on Rn
We can now extend the inversion formulas for the Radon transform to distributions. First we observe that the Hilbert transform H can be extended to distributions T on R of compact support. It suffices to put
H(T )(F ) = T (−HF ) , F ∈ D(R) .
In fact, as remarked at the end of §3, the mapping F −→ HF is a continuous mapping of D(R) into E(R). In particular H(T ) ∈ D (R).
Theorem 5.5. The Radon transform S −→ S (S ∈ E (Rn)) is inverted by the following formula
cS = (ΛS)∨ , S ∈ E (Rn) ,
where the constant c = (−4π)(n−1)/2Γ(n/2)/Γ(1/2). In the case when n is odd we have also
c S = L(n−1)/2((S)∨) .
Remark 5.6. Since S has compact support and since Λ is defined by means of the Hilbert transform, the remarks above show that ΛS ∈ D (Pn), so the right hand side is well defined.
Proof. Using Theorem 3.8 we have
(ΛS)∨(f ) = (ΛS)(f ) = S(Λf ) = S((Λf )∨) = cS(f ) .
The other inversion formula then follows, using the lemma. In analogy with βA we define the “sphere” σA in Pn as σA = {ξ ∈ Pn : d(0, ξ) = A} .
From Theorem 5.5 we can then deduce the following complement to Theorem 5.4. Corollary 5.7. Suppose n is odd. Then if S ∈ E (Rn) ,
supp(S) ⊂ σR ⇒ supp(S) ⊂ SR(0) .
To see this let > 0 and let f ∈ D(Rn) have supp(f ) ⊂ BR− (0). Then supp f ∈ βR− and since Λ is a differential operator, supp(Λf ) ⊂ βR− . Hence
cS(f ) = S((Λf )∨) = S(Λf ) = 0 , so supp(S) ∩ BR− (0) = ∅. Since > 0 is arbitrary,
supp(S) ∩ BR(0) = ∅ .
On the other hand, by Theorem 5.4, supp(S) ⊂ C (BR(0)). This proves the corollary.

§5 Radon Transform of Distributions 31

Let M be a manifold and dμ a measure such that on each local coordinate patch with coordinates (t1, . . . , tn) the Lebesque measure dt1, . . . , dtn and dμ are absolutely continuous with respect to each other. If h is a function on M locally integrable with respect to dμ the distribution ϕ → ϕh dμ will be denoted Th.
Proposition 5.8. (a) Let f ∈ L1(Rn) vanish outside a compact set. Then the distribution Tf has Radon transform given by

(52)

Tf = Tfb .

(b) Let ϕ be a locally integrable function on Pn. Then

(53)

ˇ (Tϕ)∨ = Tϕ .

ˇ Proof. The existence and local integrability of f and ϕ was established
during the proof of Lemma 5.1. The two formulas now follow directly from Lemma 5.1.
As a result of this proposition the smoothness assumption can be dropped in the inversion formula. In particular, we can state the following result.

Corollary 5.9. (n odd.) The inversion formula

cf = L(n−1)/2((f )∨) ,

c = (−4π)(n−1)/2Γ(n/2)/Γ(1/2), holds for all f ∈ L1(Rn) vanishing outside a compact set, the derivative interpreted in the sense of distributions.
Examples. If μ is a measure (or a distribution) on a closed submanifold S of a manifold M , the distribution on M given by ϕ → μ(ϕ|S) will also be denoted by μ.

(a) Let δ0 be the delta distribution f → f (0) on Rn. Then

ˇ δ0(ϕ) = δ0(ϕ) = Ω−n 1

ϕ(ω, 0) dω ,

Sn−1

so

(54)

δ0 = Ω−n 1mSn−1 ,

the normalized measure on Sn−1 considered as a distribution on Sn−1 × R.

(b) Let ξ0 denote the hyperplane xn = 0 in Rn, and δξ0 the delta distribution ϕ → ϕ(ξ0) on Pn. Then

(δξ0 )∨ (f ) = f (x) dm(x) ,
ξ0

32 Chapter I. The Radon Transform on Rn

so

(55)

(δξ0 )∨ = mξ0 ,

the Euclidean measure of ξ0.
(c) Let χB be the characteristic function of the unit ball B ⊂ Rn. Then by (52),

χB(ω, p) =

Ωn−1 n−1

(1

−

p2)(n−1)/2

, |p| ≤ 1 .

0

, |p| > 1

(d) Let Ω be a bounded convex region in Rn whose boundary is a smooth
surface. We shall obtain a formula for the characteristic funtion of Ω in
terms of the areas of its hyperplane sections. For simplicity we assume n
odd. The characteristic function χΩ is a distribution of compact support and (χΩ)b is thus well defined. Approximating χΩ in the L2-norm by a sequence (ψn) ⊂ D(Ω) we see from Theorem 4.1 that ∂p(n−1)/2ψn(ω, p) converges in the L2-norm on Pn. Since

ˇ ψ(ξ)ϕ(ξ) dξ = ψ(x)ϕ(x) dx

it follows from Schwarz’ inequality that ψn −→ (χΩ)b in the sense of distributions and accordingly ∂(n−1)/2ψn converges as a distribution to ∂(n−1)/2((χΩ)b ). Since the L2 limit is also a limit in the sense of distributions this last function equals the L2 limit of the sequence ∂(n−1)/2ψn.
From Theorem 4.1 we can thus conclude the following result:

Theorem 5.10. Let Ω ⊂ Rn (n odd) be a convex region as above. Let A(ω, p) denote the (n − 1)-dimensional area of the intersection of Ω with the hyperplane x, ω = p. Then

(56)

c0χΩ(x) = L(x12 )(n−1)

A(ω, (x, ω)) dω ,

Sn−1

in the sense of distributions.

c0 = 2(2πi)n−1

§6 Integration over d -planes. X-ray Transforms. The Range of the d -plane Transform

Let d be a fixed integer in the range 0 < d < n. We define the d dimensional Radon transform f → f by

(57)

f(ξ) = f (x) dm(x) , where ξ is a d-plane .

ξ

§6 Integration over d-planes. X-ray Transforms 33
Because of the applications to radiology indicated in § 7,B) the 1-dimensional Radon transform is often called the X-ray transform. Since a hyperplane can be viewed as a disjoint union of parallel d-planes parameterized by Rn−1−d it is obvious from (4) that the transform f → f is injective. Similarly we deduce the following consequence of Theorem 2.6.
Corollary 6.1. Let f, g ∈ C(Rn) satisfy the rapid decrease condition: For each m > 0, |x|mf (x) and |x|mg(x) are bounded on Rn. Assume for the d-dimensional Radon transforms
f (ξ) = g(ξ)
whenever the d-plane ξ lies outside the unit ball. Then
f (x) = g(x) for |x| > 1 .
We shall now generalize the inversion formula in Theorem 3.1. If ϕ is
ˇ a continuous function on the space of d-planes in Rn we denote by ϕ the
point function
ˇϕ(x) = ϕ(ξ) dμ(ξ) ,
x∈ξ
where μ is the unique measure on the (compact) space of d-planes passing through x, invariant under all rotations around x and with total measure 1. If σ is a fixed d-plane through the origin we have in analogy with (16),

(58)

ˇϕ(x) = ϕ(x + k · σ) dk ,

K

where again K = O(n).
Theorem 6.2. The d-dimensional Radon transform in Rn is inverted by the formula

(59)

cf = (−L)d/2((f )∨) ,

where c = (4π)d/2Γ(n/2)/Γ((n − d)/2). Here it is assumed that f (x) = 0(|x|−N ) for some N > d.

Proof. We have, in analogy with (34),

(f )∨(x) =

f (x + k · y) dm(y) dk

Kσ

= dm(y) f (x + k · y) dk = (M |y|f )(x) dm(y) .

σ

K

σ

34 Chapter I. The Radon Transform on Rn

Hence

∞
(f )∨(x) = Ωd (M rf )(x)rd−1 dr .

0
Using polar coordinates around x, in the integral below, we obtain

(60)

(f )∨(x) = Ωd |x − y|d−nf (y) dy .

Ωn

Rn

The theorem now follows from Theorem 6.11 in Chapter VII.

As a consequence of Theorem 2.10 we now obtain a generalization, char-

acterizing the image of the space D(Rn) under the d-dimensional Radon

transform.

The set G(d, n) of d-planes in Rn is a manifold, in fact a homogeneous

space of the group M(n) of all isometries of Rn. Let Gd,n denote the manifold of all d-dimensional subspaces (d-planes through 0) of Rn. The

parallel translation of a d-plane to one through 0 gives a mapping π of

G(d, n) onto Gd,n. The inverse image π−1(σ) of a member σ ∈ Gd,n is naturally identified with the orthogonal complement σ⊥. Let us write

ξ = (σ, x ) = x + σ if σ = π(ξ) and x = σ⊥ ∩ ξ .

(See Fig. I.6.) Then (57) can be written (61)
f (x + σ) = f (x + x ) dx .
σ
For k ∈ Z+ we consider the polynomial (62)
Pk(u) = f (x) x, u k dx .
Rn

x′′

ξ = x′′ + σ

0
σ σ
FIGURE I.6.

If u = u ∈ σ⊥ this can be written

f (x) x, u k dx =

f (x + x ) x , u k dx dx ,

Rn

σ⊥ σ

so the polynomial

Pσ,k(u ) = f (x + σ) x , u k dx
σ⊥

§6 Integration over d-planes. X-ray Transforms 35
is the restriction to σ⊥ of the polynomial Pk. In analogy with the space DH (Pn) in §2 we define the space
DH (G(d, n)) as the set of C∞ functions
ϕ(ξ) = ϕσ(x ) = ϕ(x + σ) (if ξ = (σ, x ))
on G(d, n) of compact support satisfying the following condition. (H) : For each k ∈ Z+ there exists a homogeneous kth degree polynomial
Pk on Rn such that for each σ ∈ Gd,n the polynomial
Pσ,k(u ) = ϕ(x + σ) x , u k dx , u ∈ σ⊥ ,
σ⊥
coincides with the restriction Pk|σ⊥.
Theorem 6.3 (The Range Theorem). The d-dimensional Radon transform is a bijection of
D(Rn) onto DH (G(d, n)) .
Proof. For d = n − 1 this is Theorem 2.10. We shall now reduce the case of general d ≤ n − 2 to the case d = n − 1. It remains just to prove the surjectivity in Theorem 6.3.
Let ϕ ∈ D(G(d, n)). Let ω ∈ Rn be a unit vector. Choose a d-dimensional subspace σ perpendicular to ω and consider the (n − d − 1)-dimensional integral

(63)

Ψσ(ω, p) =

ϕ(x + σ)dn−d−1(x ) ,

ω,x =p,x ∈σ⊥

p ∈R.

We claim that this is independent of the choice of σ. In fact

Ψσ(ω, p)pk dp = pk ϕ(x + σ)dn−d−1(x ) dp

R

R

= ϕ(x + σ) x , ω k dx = Pk(ω) .

σ⊥

If we had chosen another σ, say σ1, perpendicular to ω, then, by the above, Ψσ(ω, p) − Ψσ1(ω, p) would have been orthogonal to all polynomials in p; having compact support it would have been identically 0. Thus we have a
well-defined function Ψ(ω, p) = Ψσ(ω, p) to which Theorem 2.10 applies. In fact, the smoothness of Ψ(ω, p) is clear from (63) since for ω in a neighbor-
hood of a fixed ω0 we can let σ depend smoothly on ω. From this theorem we get a function f ∈ D(Rn) such that

(64)

Ψ(ω, p) =

f (x) dm(x) .

x,ω =p

36 Chapter I. The Radon Transform on Rn

It remains to prove that

(65)

ϕ(x + σ) = f (x + x ) dx .

σ

But as x runs through an arbitrary hyperplane in σ⊥ it follows from (63) and (64) that both sides of (65) have the same integral. By the injectivity of the (n − d − 1)-dimensional Radon transform on σ⊥, equation (65) follows. This proves Theorem 6.3.
Modifying the argument we shall now prove a stronger statement.

Theorem 6.4. Let ϕ ∈ D(G(d, n)) have the property: For each pair σ, τ ∈ Gd,n and each k ∈ Z+ the polynomials

Pσ,k(u) = ϕ(x + σ) x , u k dx
σ⊥

u ∈ Rn

Pτ,k(u) = ϕ(y + τ ) y , u k dy
τ⊥

u ∈ Rn

agree for u ∈ σ⊥ ∩ τ ⊥. Then ϕ = f for some f ∈ D(Rn).

Proof. Let ϕ = D(G(d, n)) have the property above. Let ω ∈ Rn be a unit vector. Let σ, τ ∈ Gd,n be perpendicular to ω. Consider again the (n − d − 1)-dimensional integral

(66)

Ψσ(ω, p) =

ϕ(x + σ)dn−d−1(x ) , p ∈ R .

ω,x =p, x ∈σ⊥

We claim that

Ψσ(ω, p) = Ψτ (ω, p) .

To see this consider the moment

Ψσ(ω, p)pk dp
R

= pk
R

ϕ(x + σ)dn−d−1(x ) dp = ϕ(x + σ) x , ω k dx
σ⊥

= ϕ(y + τ ) y , ω k dy = Ψτ (ω, p)pk dp .

τ⊥

R

Thus Ψσ(ω, p) − Ψτ (ω, p) is perpendicular to all polynomials in p; having compact support it would be identically 0. We therefore put Ψ(ω, p) = Ψσ(ω, p). Observe that Ψ is smooth; in fact for ω in a neighborhood of a fixed ω0 we can let σ depend smoothly on ω, so by (66), Ψσ(ω, p) is smooth.

§6 Integration over d-planes. X-ray Transforms 37

Writing

x ,ω k =

pα(x )ωα ,

|α|=k

ωα = ω1α1 . . . ωnαn

we have where

Ψ(ω, p)pk dp =

Aαωα ,

R

|α|=k

Aα = ϕ(x + σ)pα(x ) dx .

σ⊥
Here Aα is independent of σ if ω ∈ σ⊥; in other words, viewed as a function of ω, Aα has for each σ a constant value as ω varies in σ⊥ ∩ S1(0). To see that this value is the same as the value on τ ⊥ ∩ S1(0) we observe that there exists a ρ ∈ Gd,n such that ρ⊥ ∩ σ⊥ = 0 and ρ⊥ ∩ τ ⊥ = 0. (Extend the 2-plane spanned by a vector in σ⊥ and a vector in τ ⊥ to an (n − d)plane.) This shows that Aα is constant on S1(0) so Ψ ∈ DH (Pn). Thus by Theorem 2.10,

(67)

Ψ(ω, p) =

f (x) dm(x)

x,ω =p

for some f ∈ D(Rn). It remains to prove that

(68)

ϕ(x + σ) = f (x + x ) dx .

σ
But as x runs through an arbitrary hyperplane in σ⊥ it follows from (63) and (64) that both sides of (68) have the same integral. By the injectivity of the (n−d−1)-dimensional Radon transform on σ⊥ equation (68) follows. This proves Theorem 6.4.
Theorem 6.4 raises the following elementary question: If a function f on Rn is a polynomial on each k-dimensional subspace, is f itself a polynomial? The answer is no for k = 1 but yes if k > 1. See Proposition 6.13 below, kindly communicated by Schlichtkrull.
We shall now prove another characterization of the range of D(Rn) under the d-plane transform (for d ≤ n − 2). The proof will be based on Theorem 6.4.
Given any d + 1 points (x0, . . . , xd) in general position let ξ(x0, . . . , xd) denote the d-plane passing through them. If ϕ ∈ E(G(d, n)) we shall write ϕ(x0, . . ., xd) for the value ϕ(ξ(x0, . . ., xd)). We also write V ({xi − x0}i=1,d) for the volume of the parallelepiped spanned by vectors (xi − x0), (1 ≤ i ≤ d). The mapping

(λ1, . . . , λd) → x0 + Σdi=1λi(xi − x0)

38 Chapter I. The Radon Transform on Rn

is a bijection of Rd onto ξ(x0, . . . , xd) and

(69) f (x0, . . . , xd) = V ({xi − x0}i=1,d) f (x0 + Σiλi(xi − x0)) dλ .
Rd
The range D(Rn) can now be described by the following alternative to Theorem 6.4. Let xki denote the kth coordinate of xi.
Theorem 6.5. If f ∈ D(Rn) then ϕ = f satisfies the system

(70) (∂i,k∂j, − ∂j,k∂i, ) ϕ(x0, . . . , xd)/V ({xi − x0}i=1,d) = 0 ,

where

0 ≤ i, j ≤ d , 1 ≤ k, ≤ n , ∂i,k = ∂/∂xki .

Conversely, if ϕ ∈ D(G(d, n)) satisfies (70) then ϕ = f for some f ∈ D(Rn).

The validity of (70) for ϕ = f is obvious from (69) just by differentiation under the integral sign. For the converse we first prove a simple lemma.

Lemma 6.6. Let ϕ ∈ E(G(d, n)) and A ∈ O(n). Let ψ = ϕ ◦ A. Then if ϕ(x0, . . . , xd) satisfies (70) so does the function

ψ(x0, . . . , xd) = ϕ(Ax0, . . . , Axd) .

Proof. Let yi = Axi so yi = Σpa pxpi . Then, if Di,k = ∂/∂yik,

(71)

(∂i,k ∂j, − ∂j,k∂i, ) = Σnp,q=1apkaq (Di,pDj,q − Di,qDj,p) .

Since A preserves volumes, the lemma follows.

Suppose now ϕ satisfies (70). We write σ = (σ1, . . . , σd) if (σj ) is an orthonormal basis of σ. If x ∈ σ⊥, the (d + 1)-tuple

(x , x + σ1, . . . , x + σd) represents the d-plane x + σ and the polynomial

(72) Pσ,k(u ) = ϕ(x +σ) x , u k dx
σ⊥

= ϕ(x , x +σ1,. . ., x +σd) x , u k dx , u ∈ σ⊥ ,
σ⊥

depends only on σ. In particular, it is invariant under orthogonal transfor-
mations of (σ1, . . . , σd). In order to use Theorem 6.4 we must show that for any σ, τ ∈ Gd,n and any k ∈ Z+,

(73)

Pσ,k(u) = Pτ,k(u) for u ∈ σ⊥ ∩ τ ⊥ , |u| = 1 .

The following lemma is a basic step towards (73).

§6 Integration over d-planes. X-ray Transforms 39

Lemma 6.7. Assume ϕ ∈ G(d, n) satisfies (70). Let

σ = (σ1, . . . , σd), τ = (τ1, . . . , τd)

be two members of Gd,n. Assume

σj = τj for 2 ≤ j ≤ d .

Then

Pσ,k(u) = Pτ,k(u) for u ∈ σ⊥ ∩ τ ⊥ , |u| = 1 .

Proof. Let ei (1 ≤ i ≤ n) be the natural basis of Rn and = (e1, . . . , ed). Select A ∈ O(n) such that

σ = A , u = Aen .

Let η = A−1τ = (A−1τ1, . . . , A−1τd) = (A−1τ1, e2, . . . , ed) .
The vector E = A−1τ1 is perpendicular to ej (2 ≤ j ≤ d) and to en (since u ∈ τ ⊥). Thus

n−1
E = a1e1 + aiei
d+1

(a21 + a2i = 1) .
i

In (72) we write Pσϕ,k for Pσ,k. Putting x = Ay and ψ = ϕ ◦ A we have

Pσϕ,k(u) =

ϕ(Ay, A(y

+ e1), . . . , A(y

+ ed))

y, en

k

dy

=

P

ψ ,k

(en)

⊥

and similarly

Pτϕ,k(u) = Pηψ,k(en) .

Thus, taking Lemma 6.6 into account, we have to prove the statement:

(74)

P ,k(en) = Pη,k(en) ,

where = (e1, . . . , ed), η = (E, e2, . . . , ed), E being any unit vector perpendicular to ej (2 ≤ j ≤ d) and to en. First we take

E = Et = sin t e1 + cos t ei (d < i < n)

and put t = (Et, e2, . . . , ed). We shall prove

(75)

P t,k(en) = P ,k(en) .

With no loss of generality we can take i = d + 1. The space

⊥ t

consists

of

the vectors

(76)

n

xt = (− cos t e1 + sin t ed+1)λd+1 +

λiei ,

i=d+2

λi ∈ R .

40 Chapter I. The Radon Transform on Rn

Putting P (t) = P t,k(en) we have

(77)

P (t) =

ϕ(xt, xt + Et, xt + e2, . . . , xt + ed)λkn dλn . . . dλd+1 .

Rn−d

In order to use (70) we replace ϕ by the function

ψ(x0, . . . , xd) = ϕ(x0, . . . , xd)/V ({xi − x0}i=1,d) .

Since the vectors in (77) span volume 1, replacing ϕ by ψ in (77) does not change P (t). Applying ∂/∂t we get (with dλ = dλn . . . dλd+1),

(78)

P (t)

=
Rn−d

d
λd+1(sin t ∂j,1ψ + cos t ∂j,d+1ψ)
j=0

+ cos t ∂1,1ψ − sin t ∂1,d+1ψ λkn dλ .

Now ϕ is a function on G(d, n). Thus for each i = j it is invariant under the substitution

yk = xk (k = i), yi = sxi + (1 − s)xj = xj + s(xi − xj ) , s > 0 ,

whereas the volume changes by the factor s. Thus ψ(y0, . . . , yd) = s−1ψ(x0, . . . , xd) .
Taking ∂/∂s at s = 1 we obtain

n

(79)

ψ(x0, . . . , xd) + (xki − xkj )(∂i,kψ)(x0, . . . , xd) = 0 .

k=1

Note that in (78) the derivatives are evaluated at

(80)

(x0, . . . , xd) = (xt, xt + Et, xt + e2, . . . , xt + ed) .

Using (79) for (i, j) = (1, 0) and (i, j) = (0, 1) and adding we obtain

(81)

sin t (∂0,1ψ + ∂1,1ψ) + cos t (∂0,d+1ψ + ∂1,d+1ψ) = 0 .

For i ≥ 2 we have

xi − x0 = ei, xi − x1 = − sin t e1 − cos t ed+1 + ei , and this gives the relations (for j = 0 and j = 1)

(82)

ψ(x0, . . . , xd) + (∂i,iψ)(x0, . . . , xd) = 0 ,

(83)

ψ − sin t (∂i,1ψ) − cos t (∂i,d+1ψ) + ∂i,iψ = 0 .

§6 Integration over d-planes. X-ray Transforms 41

Thus by (81)–(83) formula (78) simplifies to

P (t) =

[cos t (∂1,1ψ) − sin t (∂1,d+1ψ)] λkn dλ .

Rn−d

In order to bring in 2nd derivatives of ψ we integrate by parts in λn,

(84) (k + 1)P (t) =

∂ −
∂λn

[cos t

(∂1,1ψ)

−

sin

t

(∂1,d+1ψ)]

λkn+1

dλ

.

Rn−d

Since the derivatives ∂j,kψ are evaluated at the point (80) we have in (84)

(85)

∂

d

∂λn (∂j,kψ) = i=0 ∂i,n(∂j,kψ)

and also, by (76) and (80),

(86)

∂ ∂ λd+1

(∂j,k ψ )

=

−

cos

t

d 0

∂i,1(∂j,kψ) + sin t

d
∂i,d+1(∂j,kψ) .
0

We now plug (85) into (84) and then invoke equations (70) for ψ which give

(87)

d

d

∂i,n∂1,1ψ = ∂1,n ∂i,1ψ ,

0

0

d

d

∂i,n∂1,d+1ψ = ∂1,n ∂i,d+1ψ .

0

0

Using (85) and (87) we see that (84) becomes

−(k + 1)P (t) =

[∂1,n(cos t Σi∂i,1ψ −sin t Σi∂i,d+1ψ)] (xt, xt +Et, . . . , xt +ed)λkn+1 dλ ,
Rn−d

so by (86)

(k + 1)P (t) =

∂ ∂ λd+1

(∂1,nψ)λkn+1

dλ

.

Rn−d

Since d + 1 < n, the integration in λd+1 shows that P (t) = 0, proving (75). This shows that without changing P ,k(en) we can pass from =
(e1, . . . , ed) to t = (sin t e1 + cos t ed+1, e2, . . . , ed) .

By iteration we can replace e1 by

sin tn−d−1 . . . sin t1e1 + sin tn−d−1 . . . sin t2 cos t1ed+1+· · · +cos tn−d−1en−1,

but keeping e2, . . . , ed unchanged. This will reach an arbitrary E, so (74) is proved.

42 Chapter I. The Radon Transform on Rn

We shall now prove (73) in general. We write σ and τ in orthonormal
bases, σ = (σ1, . . . , σd), τ = (τ1, . . . , τd). Using Lemma 6.7 we shall pass from σ to τ without changing Pσ,k(u), u being fixed.
Consider τ1. If two members of σ, say σj and σk, are both not orthogonal to τ1 that is ( σj , τ1 = 0, σk, τ1 = 0) we rotate them in the (σj , σk)-plane so that one of them becomes orthogonal to τ1. As remarked after (72) this has no effect on Pσ,k(u). We iterate this process (with the same τ1) and end up with an orthogonal frame (σ1∗, . . . , σd∗) of σ in which at most one entry σi∗ is not orthogonal to τ1. In this frame we replace this σi∗ by τ1. By Lemma 6.7 this change of σ does not alter Pσ,k(u).
We now repeat this process with τ2, τ3 . . ., etc. Each step leaves Pσ,k(u) unchanged (and u remains fixed) so this proves (73) and the theorem.
We consider now the case d = 1, n = 3 in more detail. Here f → f is the X-ray transform in R3. We also change the notation and write ξ for x0, η for x1 so V ({x1 − x0}) equals |ξ − η|. Then Theorem 6.5 reads as follows.
Theorem 6.8. The X-ray transform f → f in R3 is a bijection of D(R3) onto the space of ϕ ∈ D(G(1, 3)) satisfying

∂ ∂ ∂ ∂ ϕ(ξ, η)

(88)

− ∂ξk ∂η ∂ξ ∂ηk

|ξ − η| = 0 , 1 ≤ k, ≤ 3 .

Now let G (1, 3) ⊂ G(1, 3) denote the open subset consisting of the non-

horizontal lines. We shall now show that for ϕ ∈ D(G(1, n)) (and even

for ϕ ∈ E(G (1, n))) the validity of (88) for (k, ) = (1, 2) implies (87)

for general (k, ). Note that (79) (which is also valid for ϕ ∈ E(G (1, n)))

implies

ϕ(ξ, η) 3

∂ ϕ(ξ, η)

+ |ξ − η|

1 (ξi − ηi) ∂ξi

|ξ − η|

= 0.

Here we apply ∂/∂ηk and obtain

3

∂2

∂∂

(ξi
i=1

− ηi) ∂ξi∂ηk

−

∂ξk

+

∂ηk

ϕ(ξ, η) = 0.
|ξ − η|

Exchanging ξ and η and adding we derive

3

∂2

∂2

ϕ(ξ, η)

(89)

(ξi − ηi)
i=1

− ∂ξi∂ηk ∂ξk∂ηi

=0 |ξ − η|

for k = 1, 2, 3. Now assume (88) for (k, ) = (1, 2). Taking k = 1 in (89) we derive (88) for (k, ) = (1, 3). Then taking k = 3 in (89) we deduce (88) for (k, ) = (3, 2). This verifies the claim above.
We can now put this in a simpler form. Let (ξ, η) denote the line through the points ξ = η. Then the mapping

(ξ1, ξ2, η1, η2) → ((ξ1, ξ2, 0), (η1, η2, −1))

§6 Integration over d-planes. X-ray Transforms 43

is a bijection of R4 onto G (1, 3). The operator

(90)

Λ = ∂2 − ∂2

∂ξ1∂η2 ∂ξ2∂η1

is a well defined differential operator on the dense open set G (1, 3). If ϕ ∈ E(G(1, 3)) we denote by ψ the restriction of the function (ξ, η) → ϕ(ξ, η)/|ξ − η| to G (1, 3). Then we have proved the following result.

Theorem 6.9. The X-ray transform f → f is a bijection of D(R3) onto the space

(91)

{ϕ ∈ D(G(1, 3)) : Λψ = 0} .

We shall now rewrite the differential equation (91) in Plu¨cker coordinates.
The line joining ξ and η has Plu¨cker coordinates (p1, p2, p3, q1, q2, q3) given by

i
ξ1 η1

j
ξ2 η2

k
ξ3 η3

= p1i + p2j + p3k ,

qi =

ξi ηi

1 1

which satisfy

(92)

p1q1 + p2q2 + p3q3 = 0 .

Conversely, each ratio (p1 : p2 : p3 : q1 : q2 : q3) determines uniquely a line
provided (92) is satisfied. The set G (1, 3) is determined by q3 = 0. Since
the common factor can be chosen freely we fix q3 as 1. Then we have a bijection τ : G (1, 3) → R4 given by

(93)

x1 = p2 + q2, x2 = −p1 − q1, x3 = p2 − q2, x4 = −p1 + q1

with inverse

(p1, p2, p3, q1, q2) =

1 2

(−x2

−

x4

),

1 2

(x1

+

x3),

1 4

(−x21

−

x22

+

x23

+

x24

),

1 2

(−x2

+

x4

),

1 2

(x1

−

x3

)

.

Theorem 6.10. If ϕ ∈ D(G(1, 3)) satisfies (91) then the restriction ϕ|G (1, 3) (with q3 = 1) has the form

(94)

ϕ(ξ, η) = |ξ − η| u(p2 + q2, −p1 − q1, p2 − q2, −p1 + q1)

where u satisfies

∂2u ∂2u ∂2u ∂2u

(95)

∂x21 + ∂x22 − ∂x23 − ∂x24 = 0 .

On the other hand, if u satisfies (95) then (94) defines a function ϕ on G (1, 3) which satisfies (91).

44 Chapter I. The Radon Transform on Rn

Proof. First assume ϕ ∈ D(G(1, 3)) satisfies (91) and define u ∈ E(R4) by

(96)

u(τ (

))

=

ϕ(

)(1 + q12

+

q22)−

1 2

,

where ∈ G (1, 3) has Plu¨cker coordinates (p1, p2, p3, q1, q2, 1). On the line consider the points ξ, η for which ξ3 = 0, η3 = −1 (so q3 = 1). Then since

p1 = −ξ2, p2 = ξ1, q1 = ξ1 − η1, q2 = ξ2 − η2

we have ϕ(ξ, η)
(97) |ξ − η| = u(ξ1 + ξ2 − η2, −ξ1 + ξ2 + η1, ξ1 − ξ2 + η2, ξ1 + ξ2 − η1) .

Now (91) implies (95) by use of the chain rule. On the other hand, suppose u ∈ E(R4) satisfies (95). Define ϕ by (96).
Then ϕ ∈ E(G (1, 3)) and by (97),

ϕ(ξ, η)

Λ

= 0.

|ξ − η|

As shown before the proof of Theorem 6.9 this implies that the whole system (88) is verified.
We shall now see what implications A´ sgeirsson’s mean-value theorem Chapter VI, §2 has for the range of the X-ray transform. We have from (95),

2π

2π

(98)

u(r cos ϕ, r sin ϕ, 0, 0) dϕ = u(0, 0, r cos ϕ, r sin ϕ) dϕ .

0

0

The first points (r cos ϕ, r sin ϕ, 0, 0) correspond via (93) to the lines with

(p1, p2, p3, q1, q2, q3)

=

(−

r 2

sin

ϕ,

r 2

cos

ϕ,

−

r2 4

,

−

r 2

sin

ϕ,

r 2

cos

ϕ,

1)

containing the points

(ξ1, ξ2, ξ3)

=

(

r 2

cos

ϕ,

r 2

sin

ϕ,

0)

(η1, η2, η3)

=

(

r 2

(sin

ϕ

+

cos

ϕ),

+

r 2

(sin

ϕ

−

cos

ϕ),

−1)

with

|ξ

−

η|2

=

1

+

r2 4

.

The

points

(0,

0,

r

cos

ϕ,

r

sin

ϕ)

correspond

via

(93)

to the lines with

(p1,

p2,

p3,

q1,

q2,

q3)

=

(−

r 2

sin

ϕ,

r 2

cos

ϕ,

r2 4

,

r 2

sin

ϕ,

−

r 2

cos

ϕ,

1)

containing the points

(ξ1, ξ2, ξ3)

=

(

r 2

cos

ϕ,

r 2

sin

ϕ,

0),

(η1, η2, η3)

=

(

r 2

(cos

ϕ

−

sin

ϕ),

r 2

(cos

ϕ

+

sin

ϕ),

−1),

§6 Integration over d-planes. X-ray Transforms 45

with

|ξ

−

η|2

=

1

+

r2 4

.

Thus

(98)

takes

the

form

2π

(99)

ϕ(

r 2

cos

θ,

r 2

sin

θ,

0,

r 2

(sin

θ

+

cos

θ),

r 2

(sin

θ

−

cos

θ),

−1)

dθ

0

2π

=

ϕ(

r 2

cos

θ,

r 2

sin

θ,

0,

r 2

(cos

θ

−

sin

θ),

r 2

(cos

θ

+

sin

θ),

−1)

dθ

.

0

The lines forming the arguments of ϕ in these integrals are the two families of generating lines for the hyperboloid (see Fig. I.7)

x2

+ y2

=

r2 4

(z

2

+ 1) .

Definition. A function ϕ ∈ E(G (1, 3)) is said to be a harmonic line function if

ϕ(ξ, η)

Λ

= 0.

|ξ − η|

Theorem 6.11. A function ϕ ∈ E(G (1, 3)) is a harmonic line function if and only if for each hyperboloid of revolution H of one sheet and vertical axis the mean values of ϕ over the two families of generating lines of H are equal. (The variable of integration is the polar angle in the equatorial plane of H.).

FIGURE I.7.

The proof of (99) shows that ϕ harmonic implies the mean value property for ϕ. The converse is also true since A´ sgeirsson’s theorem has a converse; for Rn this is obvious from the relation between L and M r in Ch. VII, §6.
Corollary 6.12. Let ϕ ∈ D(G(1, 3)). Then ϕ is in the range of the Xray transform if and only if ϕ has the mean value property for arbitrary hyperboloid of revolution of one sheet (and arbitrary axis).
We conclude this section with the following result due to Schlichtkrull mentioned in connection with Theorem 6.4.
Proposition 6.13. Let f be a function on Rn and k ∈ Z+, 1 < k < n. Assume that for each k-dimensional subspace Ek ⊂ Rn the restriction f |Ek is a polynomial on Ek. Then f is a polynomial on Rn.
For k = 1 the result is false as the example f (x, y) = xy2/(x2 + y2), f (0, 0) = 0 shows. We recall now the Lagrange interpolation formula. Let

46 Chapter I. The Radon Transform on Rn

a0, . . . , am be distinct numbers in C. Then each polynomial P (x) (x ∈ R) of degree ≤ m can be written

P (x) = P (a0)Q0(x) + · · · + P (am)Qm(x) ,

where

m

Qi(x) = (x − aj)/(x − ai) (ai − aj) .

j=0

j=i

In fact, the two sides agree at m + 1 distinct points. This implies the following result.

Lemma 6.14. Let f (x1, . . . , xn) be a function on Rn such that for each i with xj(j = i) fixed the function xi → f (x1, . . . , xn) is a polynomial. Then f is a polynomial.

For this we use Lagrange’s formula on the polynomial x1 −→ f (x1, x2, . . . , xn) and get

m
f (x1, . . . , xn) = f (aj, x2, . . . , xm)Qj(x1) .
j=0

The lemma follows by iteration.
For the proposition we observe that the assumption implies that f re-
stricted to each 2-plane E2 is a polynomial on E2. For a fixed (x2, . . . , xn) the point (x1, . . . , xn) is in the span of (1, 0, . . . , 0) and (0, x2, . . . , xn) so f (x1, . . . , xn) is a polynomial in x1. Now the lemma implies the result.

§7 Applications

A. Partial Differential Equations. The Wave Equation
The inversion formula in Theorem 3.1 is very well suited for applications to partial differential equations. To explain the underlying principle we write the inversion formula in the form

(100)

n−1
f (x) = γ Lx 2

f (ω, x, ω ) dω ,

Sn−1

where

the

constant

γ

equals

1 2

(2πi)1−n.

Note

that

the

function

fω (x)

=

f (ω, x, ω ) is a plane wave with normal ω, that is, it is constant on

each hyperplane perpendicular to ω.

Consider now a differential operator

D=

ak1...kn ∂1k1 . . . ∂nkn

(k)

§7 Applications 47

with constant coefficients ak1,...,kn , and suppose we want to solve the differential equation

(101)

Du = f ,

where f is a given function in S(Rn). To simplify the use of (100) we assume n to be odd. We begin by considering the differential equation

(102)

Dv = fω ,

where fω is the plane wave defined above and we look for a solution v which is also a plane wave with normal ω. But a plane wave with normal ω is just a function of one variable; also if v is a plane wave with normal ω so is the function Dv. The differential equation (102) (with v a plane wave) is therefore an ordinary differential equation with constant coefficients. Suppose v = uω is a solution and assume that this choice can be made smoothly in ω. Then the function

(103)

n−1
u=γL 2

uω dω

Sn−1

is

a

solution

to

the

differential

equation

(101).

In

fact,

since

D

and

L n−1 2

commute we have

n−1
Du = γL 2

n−1
Duω dω = γL 2

fω dω = f .

Sn−1

Sn−1

This method only assumes that the plane wave solution uω to the ordinary differential equation Dv = fω exists and can be chosen so as to depend smoothly on ω. This cannot always be done because D might annihilate all plane waves with normal ω. (For example, take D = ∂2/∂x1∂x2 and ω = (1, 0).) However, if this restriction to plane waves is never 0 it
follows from a theorem of Tr`eves [1963] that the solution uω can be chosen depending smoothly on ω. Thus we can state the following result.

Theorem 7.1. If the restriction Dω of D to the space of plane waves with normal ω is = 0 for each ω, then formula (103) gives a solution to the differential equation Du = f (f ∈ S(Rn)).

The method of plane waves can also be used to solve the Cauchy problem
for hyperbolic differential equations with constant coefficients. We illustrate this method by means of the wave equation Rn,

(104)

∂2u Lu = ∂t2 , u(x, 0) = f0(x), ut(x, 0) = f1(x) ,

f0, f1 being given functions in D(Rn).

48 Chapter I. The Radon Transform on Rn
Lemma 7.2. Let h ∈ C2(R) and ω ∈ Sn−1. Then the function
v(x, t) = h( x, ω + t) satisfies Lv = (∂2/∂t2)v.
The proof is obvious. It is now easy, on the basis of Theorem 3.8, to write down the unique solution of the Cauchy problem (104). Theorem 7.3. The solution to (104) is given by

(105) where

u(x, t) = (Sf )(ω, x, ω + t) dω ,
Sn−1

Sf =

c(∂n−1f0 + ∂n−2f1) , n odd c H(∂n−1f0 + ∂n−2f1) , n even .

Here ∂ = ∂/∂p and the constant c equals

c

=

1 2

(2πi)1−n

.

Lemma 7.2 shows that (105) is annihilated by the operator L − ∂2/∂t2, so we just have to check the initial conditions in (104).

(a) If n > 1 is odd then ω → (∂n−1f0)(ω, x, ω ) is an even function on Sn−1 but the other term in Sf , that is the function ω → (∂n−2f1)(ω, x, ω ), is odd. Thus by Theorem 3.8, u(x, 0) = f0(x). Applying ∂/∂t to (105) and putting t = 0 gives ut(x, 0) = f1(x), this time because the function ω → (∂nf0)(ω, x, ω ) is odd and the function ω → (∂n−1f1)(ω, x, ω ) is even.

(b) If n is even the same proof works if we take into account the fact that H interchanges odd and even functions on R.

Definition. For the pair f = {f0, f1} we refer to the function Sf in (105) as the source.
In the terminology of Lax-Philips [1967] the wave u(x, t) is said to be

(a) outgoing if u(x, t) = 0 in the forward cone |x| < t; (b) incoming if u(x, t) = 0 in the backward cone |x| < −t.

The notation is suggestive because “outgoing” means that the function x → u(x, t) vanishes in larger balls around the origin as t increases.
Corollary 7.4. The solution u(x, t) to (104) is

§7 Applications 49
(i) outgoing if and only if (Sf )(ω, s) = 0 for s > 0, all ω.
(ii) incoming if and only if (Sf )(ω, s) = 0 for s < 0, all ω.
Proof. For (i) suppose (Sf )(ω, s) = 0 for s > 0. For |x| < t we have x, ω +t ≥ −|x|+t > 0 so by (105) u(x, t) = 0 so u is outgoing. Conversely, suppose u(x, t) = 0 for |x| < t. Let t0 > 0 be arbitrary and let ϕ(t) be a smooth function with compact support contained in (t0, ∞).
Then if |x| < t0 we have

0 = u(x, t)ϕ(t) dt = dω (Sf )(ω, x, ω + t)ϕ(t) dt

R

Sn−1

R

=

dω (Sf )(ω, p)ϕ(p − x, ω ) dp .

Sn−1

R

Taking arbitrary derivative ∂k/∂xi1 . . . ∂xik at x = 0 we deduce

(Sf )(ω, p)ωi1 . . . ωik dω (∂kϕ)(p) dp = 0
R Sn−1
for each k and each ϕ ∈ D(t0, ∞). Integrating by parts in the p variable we conclude that the function

(106)

p → (Sf )(ω, p)ωi1 . . . ωik dω , p ∈ R
Sn−1

has its kth derivative ≡ 0 for p > t0. Thus it equals a polynomial for p > t0. However, if n is odd the function (106) has compact support so it must vanish identically for p > t0.
On the other hand, if n is even and F ∈ D(R) then as remarked at the end of §3, lim|t|→∞(HF )(t) = 0. Thus we conclude again that expression (106) vanishes identically for p > t0.
Thus in both cases, if p > t0, the function ω → (Sf )(ω, p) is orthogonal to all polynomials on Sn−1, hence must vanish identically.

One can also solve (104) by means of the Fourier transform (see Ch. VII)
f (ζ) = f (x)e−i x,ζ dx .
Rn
Assuming the function x → u(x, t) in S(Rn), for a given t we obtain

utt(ζ, t) + ζ, ζ u(ζ, t) = 0 .

50 Chapter I. The Radon Transform on Rn

Solving this ordinary differential equation with initial data given in (104), we get

(107)

u(ζ,

t)

=

f0 (ζ )

cos(|ζ|t)

+

sin(|ζ|t) f1(ζ) |ζ|

.

The function ζ → sin(|ζ|t)/|ζ| is entire of exponential type |t| on Cn in the sense of (18), Ch. VII. In fact, if ϕ(λ) is even, holomorphic on C and satisfies the exponential type estimate (18) in Theorem 4.7, Ch. VII, then the same holds for the function Φ on Cn given by Φ(ζ) = Φ(ζ1, . . . , ζn) = ϕ(λ) where λ2 = ζ12 + · · · + ζn2. To see this put
λ = μ + iν , ζ = ξ + iη μ, ν ∈ R , ξ , η ∈ Rn .

Then so

μ2 − ν2 = |ξ|2 − |η|2 , μ2ν2 = (ξ · η)2 , |λ|4 = (|ξ|2 − |η|2)2 + 4(ξ · η)2

and 2|Im λ|2 = |η|2 − |ξ|2 + (|ξ|2 − |η|2)2 + 4(ξ · η)2 1/2 .

Since |(ξ · η)| ≤ |ξ| |η| this implies |Im λ| ≤ |η| so the estimate (18) follows

for Φ. Thus by Chapter VII, §4 there exists a Tt ∈ E (Rn) with support in

B|t|(0) such that

sin(|ζ|t) |ζ|

=

e−i ζ,x dTt(x) .

Rn

Theorem 7.5. Given f0, f1 ∈ E(Rn) the function

(108)

u(x, t) = (f0 ∗ Tt )(x) + (f1 ∗ Tt)(x)

satisfies (104). Here Tt stands for ∂t(Tt).
Note that (104) implies (108) if f0 and f1 have compact support. The converse holds without this support condition.

Corollary 7.6. If f0 and f1 have support in BR(0) then u has support in the region
|x| ≤ |t| + R .
In fact, by (108) and support property of convolutions (Ch. VII, §3), the function x → u(x, t) has support in BR+|t|(0)−. While Corollary 7.6 implies that for f0, f1 ∈ D(Rn) u has support in a suitable solid cone, we shall now see that Theorem 7.3 implies that if n is odd u has support in a conical shell (see Fig. I.8).

Corollary 7.7. Let n be odd. Assume f0 and f1 have support in the ball BR(0).

§7 Applications 51

R

n

0

FIGURE I.8.

(i) Huygens’ Principle. The solution u to (104) has support in the conical shell

(109)

|t| − R ≤ |x| ≤ |t| + R ,

which is the union for |y| ≤ R of the light cones,

Cy = {(x, t) : |x − y| = |t|} .

(ii) The solution to (104) is outgoing if and only if

∞

(110)

f0(ω, p) = f1(ω, s) ds , p > 0 , all ω ,
p

and incoming if and only if

p

f0(ω, p) = − f1(ω, s) ds , p < 0 , all ω .
−∞

Note that Part (ii) can also be stated: The solution is outgoing (incoming) if and only if

f0 = f1

f0 = − f1

π

Hπ

π

Hπ

for an arbitrary hyperplane π (0 ∈/ π) Hπ being the halfspace with boundary π which does not contain 0.

To verify (i) note that since n is odd, Theorem 7.3 implies

(111)

u(0, t) = 0 for |t| ≥ R .

If z ∈ Rn, F ∈ E(Rn) we denote by F z the translated function y → F (y +z). Then uz satisfies (104) with initial data f0z, f1z which have support contained in BR+|z|(0). Hence by (111),

(112)

u(z, t) = 0 for |t| > R + |z| .

52 Chapter I. The Radon Transform on Rn

The other inequality in (109) follows from Corollary 7.6.

For the final statement in (i) we note that if |y| ≤ R and (x, t) ∈ Cy then |x − y| = t so |x| ≤ |x − y| + |y| ≤ |t| + R and |t| = |x − y| ≤ |x| + R

proving (109). Conversely, if (x, t) satisfies (109) then (x, t) ∈ Cy with

y

=

x

−

|t|

x |x|

=

x |x|

(|x|

−

t)

which

has

norm

≤

R.

For (ii) we just observe that since fi(ω, p) has compact support in p, (110) is equivalent to (i) in Corollary 7.4.

Thus (110) implies that for t > 0, u(x, t) has support in the thinner shell

|t| ≤ |x| ≤ |t| + R.

B. X-ray Reconstruction

The classical interpretation of an X-ray picture is an attempt at reconstructing properties of a 3-dimensional body by means of the X-ray projection on a plane.

In modern X-ray technology

the picture is given a more re-

fined mathematical interpreta-

B

tion. Let B ⊂ R3 be a body (for example a part of a human

I

body) and let f (x) denote its density at a point x. Let ξ be a line in R3 and suppose a thin

ξ x

beam of X-rays is directed at

B along ξ. Let I0 and I re-

I0

spectively, denote the intensity

of the beam before entering B and after leaving B (see

FIGURE I.9.

Fig. I.9). As the X-ray traverses the distance Δx along ξ it will undergo the relative intensity loss ΔI/I = f (x) Δx. Thus dI/I = −f (x) dx whence

(113)

log(I0/I) = f (x) dx ,
ξ

the integral f (ξ) of f along ξ. Since the left hand side is determined by the X-ray picture, the X-ray reconstruction problem amounts to the determination of the function f by means of its line integrals f(ξ). The inversion formula in Theorem 3.1 gives an explicit solution of this problem.
If B0 ⊂ B is a convex subset (for example the heart) it may be of interest to determine the density of f outside B0 using only X-rays which do not intersect B0. The support theorem (Theorem 2.6, Cor. 2.8 and Cor. 6.1) implies that f is determined outside B0 on the basis of the integrals f (ξ) for which ξ does not intersect B0. Thus the

§7 Applications 53

density outside the heart can be determined by means of X-rays
which bypass the heart. In practice one can of course only determine the integrals f(ξ) in (113)
for finitely many directions. A compensation for this is the fact that only an approximation to the density f is required. One then encounters the mathematical problem of selecting the directions so as to optimize the approximation.
As before we represent the line ξ as the pair ξ = (ω, z) where ω ∈ Rn is a unit vector in the direction of ξ and z = ξ ∩ ω⊥ (⊥ denoting orthogonal complement). We then write

(114)

f (ξ) = f (ω, z) = (Pωf )(z) .

The function Pωf is the X-ray picture or the radiograph in the direction ω. Here f is a function on Rn vanishing outside a ball B around the origin,
and for the sake of Hilbert space methods to be used it is convenient to
assume in addition that f ∈ L2(B). Then f ∈ L1(Rn) so by the Fubini theorem we have: for each ω ∈ Sn−1, Pωf (z) is defined for almost all z ∈ ω⊥. Moreover, we have in analogy with (4),

(115)

f (ζ) = (Pωf )(z)e−i z,ζ dz (ζ ∈ ω⊥) .
ω⊥

Proposition 7.8. An object is determined by any infinite set of radiographs.
In other words, a compactly supported function f is determined by the functions Pωf for any infinite set of ω.

Proof. Since f has compact support f is an analytic function on Rn. But if f (ζ) = 0 for ζ ∈ ω⊥ we have f (η) = ω, η g(η) (η ∈ Rn) where g is
also analytic. If Pω1 f, . . . , Pωk f . . . all vanish identically for an infinite set ω1, . . . , ωk . . . , we see that for each k

k
f (η) = ωi, η gk(η) ,
i=1

where gk is analytic. But this would contradict the power series expansion of f which shows that for a suitable ω ∈ Sn−1 and integer r ≥ 0, limt→0 f (tω)t−r = 0.
If only finitely many radiographs are used we get the opposite result.

Proposition 7.9. Let ω1, . . . , ωk ∈ Sn−1 be an arbitrary finite set. Then there exists a function f ∈ D(Rn), f ≡ 0, such that

Pωi f ≡ 0 for all 1 ≤ i ≤ k .

54 Chapter I. The Radon Transform on Rn

Proof. We have to find f ∈ D(Rn), f ≡ 0, such that f (ζ) = 0 for ζ ∈ ωi⊥(1 ≤ i ≤ k). For this let D be the constant coefficient differential operator such that
k
(Du)e(η) = ωi, η u(η) η ∈ Rn .
1
If u ≡ 0 is any function in D(Rn) then f = Du has the desired property.
We next consider the problem of approximate reconstruction of the
function f from a finite set of radiographs Pω1 f, . . . , Pωk f . Let Nj denote the null space of Pωj and let Pj denote the orthogonal
projection of L2(B) on the plane f + Nj; in other words,

(116)

Pjg = Qj(g − f ) + f ,

where Qj is the (linear) projection onto the subspace Nj ⊂ L2(B). Put P = Pk . . . P1. Let g ∈ L2(B) be arbitrary (the initial guess for f ) and form the sequence P mg, m = 1, 2, . . .. Let N0 = ∩k1Nj and let P0 (resp. Q0) denote the orthogonal projection of L2(B) on the plane f + N0 (subspace N0). We shall prove that the sequence P mg converges to the projection P0g. This is natural since by P0g − f ∈ N0, P0g and f have the same radiographs in the directions ω1, . . . , ωk.
Theorem 7.10. With the notations above,

P mg −→ P0g as m −→ ∞

for each g ∈ L2(B).

Proof. We have, by iteration of (116)

(Pk . . . P1)g − f = (Qk . . . Q1)(g − f ) and, putting Q = Qk . . . Q1 we obtain
P mg − f = Qm(g − f ) .

We shall now prove that Qmg −→ Q0g for each g; since
P0g = Q0(g − f ) + f
this would prove the result. But the statement about Qm comes from the following general result about abstract Hilbert space.
Theorem 7.11. Let H be a Hilbert space and Qi the projection of H onto a subspace Ni ⊂ H(1 ≤ i ≤ k). Let N0 = ∩k1 Ni and Q0 : H −→ N0 the projection. Then if Q = Qk . . . Q1,
Qmg −→ Q0g for each g ∈ H .

§7 Applications 55

Since Q is a contraction ( Q ≤ 1) we begin by proving a simple lemma about such operators.
Lemma 7.12. Let T : H −→ H be a linear operator of norm ≤ 1. Then

H = C ((I − T )H) ⊕ Null space (I − T )

is an orthogonal decomposition, C denoting closure, and I the identity. Proof. If T g = g then since T ∗ = T ≤ 1 we have
g 2 = (g, g) = (T g, g) = (g, T ∗g) ≤ g T ∗g ≤ g 2 ,

so all terms in the inequalities are equal. Hence
g − T ∗g 2 = g 2 − (g, T ∗g) − (T ∗g, g) + T ∗g 2 = 0 ,
so T ∗g = g. Thus I −T and I −T ∗ have the same null space. But (I −T ∗)g = 0 is equivalent to (g, (I − T )H) = 0, so the lemma follows.

Definition. An operator T on a Hilbert space H is said to have property S if

(117)

fn ≤ 1, T fn −→ 1 implies (I − T )fn −→ 0 .

Lemma 7.13. A projection, and more generally a finite product of projections, has property (S).

Proof. If T is a projection then

(I − T )fn 2 = fn 2 − T fn 2 ≤ 1 − T fn 2 −→ 0

whenever

fn ≤ 1 and T fn −→ 1 .

Let T2 be a projection and suppose T1 has property (S) and T1 ≤ 1. Suppose fn ∈ H and fn ≤ 1, T2T1fn −→ 1. The inequality implies
T1fn ≤ 1 and since

T1fn 2 = T2T1fn 2 + (I − T2)(T1fn) 2

we also deduce T1fn −→ 1. Writing

(I − T2T1)fn = (I − T1)fn + (I − T2)T1fn

we conclude that T2T1 has property (S). The lemma now follows by induction.

56 Chapter I. The Radon Transform on Rn
Lemma 7.14. Suppose T has property (S) and T ≤ 1. Then for each f ∈H
T nf −→ πf as n −→ ∞ ,
where π is the projection onto the fixed point space of T .
Proof. Let f ∈ H. Since T ≤ 1, T nf decreases monotonically to a limit α ≥ 0. If α = 0 we have T nf −→ 0. By Lemma 7.12 πT = T π so πf = T nπf = πT nf so πf = 0 in this case. If α > 0 we put gn = T nf −1(T nf ). Then gn = 1 and T gn → 1. Since T has property (S) we deduce
T n(I − T )f = T nf (I − T )gn −→ 0 .
Thus T nh −→ 0 for all h in the range of I − T . If g is in the closure of this range then given > 0 there exist h ∈ (I − T )H such that g − h < . Then
T ng ≤ T n(g − h) + T nh < + T nh , whence T ng −→ 0. On the other hand, if h is in the null space of I − T then T h = h so T nh −→ h. Now the lemma follows from Lemma 7.12.
In order to deduce Theorem 7.11 from Lemmas 7.13 and 7.14 we just have to verify that N0 is the fixed point space of Q. But if Qg = g, then
g = Qk . . . Q1g ≤ Qk−1 . . . Q1g ≤ . . . ≤ Q1g ≤ g
so equality signs hold everywhere. But the Qi are projections so the norm identities imply
g = Q1g = Q2Q1g = . . . = Qk . . . Q1g
which shows g ∈ N0. This proves Theorem 7.11.

Exercises and Further Results

1. Radon Transform on Measures. (Hertle [1979], Boman-Lindskog
[2009])
Let C0(Rn) denote the space of continuous complex-valued functions on Rn vanishing at ∞, taken with the uniform norm. The dual space is the space M(Rn) of bounded measures on Rn. The Radon transform can be defined on M(Rn) in analogy with the distribution definition of §5. The spaces C0(Pn) and M(Pn) are defined in the obvious fashion.

ˇ (i) If ϕ ∈ C0(Pn) then ϕ ∈ C0(Rn).
(ii) For μ ∈ M (Rn) define μ by

Then μ ∈ M(Pn).

ˇ μ(ϕ) = μ(ϕ) .

Exercises and Further Results 57

(iii) Given μ ∈ M(Rn) its Fourier transform μ is defined by

μ(ξ) = e−i x,ξ dμ(x) ,
Rn
Then if σ ∈ R, |ω| = 1,

ξ ∈ Rn .

μ(σω) = μ(ω, p)e−iσp dp .
R
(iv) The map μ → μ is injective. (v) The measure μ ∈ M(Rn) with norm μ is said to be rapidly decreasing at ∞ if χrμ = O(r−m) as r → ∞ for each m > 0. Here χr is a continuous approximation to the characteristic function of Rn − Br(0) (i.e., χr(x) = 0 for |x| ≤ r − 1, χr = 1 for |x| ≥ r and 0 ≤ χr(x) ≤ 1 for all x).
(vi) Let K ⊂ Rn be a compact and convex subset. Assume μ rapidly decreasing and vanishing on the open set of hyperplanes not intersecting K. Then μ = 0 on Rn\K. (Use the convolution method of Theorem 5.4.)

2. The Hilbert Transform.

(i) H extends to an isometry of L2(R) into L2(R). (ii) (Titchmarsh [1948]). For f ∈ L2(R) put

Then

g(x) = i Hf (x) .

f (x) = −i Hg(x) . The functions f and g are then called conjugate. Thus H2 = I on L2(R). Moreover, the following conditions are equivalent:
(a) A function Φ0 ∈ L2(R) is the limit of a holomorphic function Φ(z) in Im z > 0 satisfying
|Φ(x + iy)|2 dx < K .
R
(b) Φ0(x) = f (x) − ig(x) where f, g ∈ L2 are conjugate. (c) The Fourier transform of Φ0,
Φ0(x)e−ixξ dx ,
vanishes for ξ > 0.

58 Chapter I. The Radon Transform on Rn

(iii) Let ϕa be the characteristic function of (0, a). Then

1 a+y

ϕa(x)

and

log π

a−y

are conjugate; so are the functions

1 1+x2
sin(ax) x

and and

y 1+y2

,

cos(ay)−1 y

.

(iv) The Hilbert transform has the following properties:
(a) It commutes with translations. (b) It commutes with dilations x → tx, t > 0. (c) Anticommutes with f (x) → f (−x). Conversely, a bounded operator on L2(R) satisfying (a), (b) and (c) is a constant multiple of H, (see e.g. Stein [1970], p. 55).

3. The Inversion Formula Interpreted in Terms of Distributions.
Given ω ∈ Sn−1, p ∈ R let Tp,ω denote the distribution on Rn given by

Tp,ω(f ) = f (ω, p) .

Then Tp,ω has support in the hyperplane

ω, x

=

p

and

d dp

(Tp,ω

)

is

the

normal

derivative

of

this

distribution.

For

p

=

0

we

write

this

as

∂ ∂ν

Tω

.

For

n odd Theorem 3.8 can be written

δ=c

∂ n−1 ∂νn−1 Tω dω

Sn−1

which is a decomposition of the delta function into plane supported distributions.

4. Convolutions.
With the method of proof of (5) show that with × denoting the convolution in (5),
ˇϕ ∗ f = (ϕ × f )∨
for ϕ ∈ D(Pn), f ∈ D(Rn). (Natterer [1986].)

5. Exterior Problem.
The transform f → f is a one-to-one mapping of L2(Rn, |x|n−1 dx) into L2(Sn−1 × R). If B is the exterior of a ball with center 0 in Rn (n ≥ 3), then the null space of f → f on L2(B, |x|n−1 dx) is the closure of the span of functions of the form f (x) = |x|−n−kY (x/|x|) where 0 ≤ k ≤ , − k is even, and Y is a homogeneous spherical harmonic of degree (Quinto
[1982]).

Exercises and Further Results 59

6. Support Property. This is a refinement of Lemma 2.7 (Quinto [2008]).
(i) Let f ∈ C∞(Rn) be rapidly decreasing. Fix x0 ∈ Rn and a neighborhood U of x0. Let r0 > 0 and assume
(M rf )(x) = 0 for all x ∈ U , and all r > r0 .
Then f ≡ 0 outside Br0(x0). (ii) The statement holds for an analytic Riemannian manifold M with infinite injectivity radius, provided f ∈ Cc∞(M ).
7. Geometric Inversion. For d even prove the inversion formula in Theorem 6.2 by the geometric
method used for the hyperplane case (Theorem 3.1).
8. Density in P2. When parametrizing the set of lines in R2 by using ux + vy = 1, the
M(2) invariant measure is given by
du dv .
(u2 + v2)3/2
(Cartan [1896].)
9. Generalized Radon Transform. Let μ ∈ C∞(Rn × Sn−1 × R) be a strictly positive function such that
μ(z, ω, p) = μ(z, −ω, −p), and Rμ the corresponding Radon transform

(Rμf )(ω, p) =

f (x)μ(x, ω, p) dx .

ω,x =p

Call Rμ invariant if for fa(x) = f (x + a) (Rμfa)(ω, p) = ν(a, ω, p)(Rμf )(ω, p + a, ω )

a ∈ Rn ,

where ν(a, ω, p) = μ(−a, ω, p)/μ(0, ω, p + a, ω ). Then Rμ is invariant if and only if
μ(x, ω, p) = μ1(ω, p)e μ2(ω),x ,
with μ1 ∈ E(Sn−1 × R), μ2 : Sn−1 → R, C∞ maps. In this case, Rμ is injective on D(Rn) and if (Rμf )(ω, p) ≡ 0 for p > r then supp(f ) ⊂ Br(0) (Kurusa [1991c]).

60 Chapter I. The Radon Transform on Rn

10. Topological Properties of the Radon Transform.

(i) The range D(Rn)b is closed in DH (Pn) and the range E (Rn)b is closed in E (Pn) (Helgason [GGA], Chapter I, Ex. B4, B5).
(ii) The transform f → f is a homeomorphism of S(Rn) onto SH (Rn). (The closedness follows as in (i) and since both sides are Fr´echet spaces, the result follows.)
(iii) E(Pn)∨ = E(Rn) (Hertle [1984a]). (This follows from (i) and the Fr´echet space result that if α : E → F is a continuous mapping and E and F Fr´echet spaces then α is surjective iff tα(F ) is weak∗ closed in E and
ˇ α : F → E injective; take α as and tα as .) ˇ (iv) The mapping S → S of D (Sn−1 × R) into D (Rn) is not surjective.
In fact, Hertle shows [1984a] that the distribution

∞
T (x) =
j=1

∂ ∂x2

jnδ(x − xj )

(xj = j, 0, . . . 0)

is not in the range. Parts (iv) and (v) have not been verified by the author.
(v) The mapping f → f is not a homeomorphism of D(Rn) onto its image. (Hertle [1984].) (By the results under (iii) for the LF-space D, the
ˇ injectivity of f → f implies that the dual map S → S of D (Pn) → D (Rn) ˇ has a dense range. Thus by (iv) the range of S → S is not closed. Thus by
Schaefer [1971, Ch. IV, 7.5], f → f is not a homeomorphism.)

Bibliographical Notes

§§1-2. The inversion formulas

(i)

f (x)

=

1 2

(2πi)1−n

Lx(n−1)/2

Sn−1 J (ω,

ω, x ) , dω

(ii)

f (x)

=

1 2

(2πi)−n

Lx(n−2)/2

Sn−1 dω

∞ dJ(ω,p) −∞ p−(ω,x)

(n odd) (n even)

for a function f ∈ D(X) in terms of its plane integrals J(ω, p) go back to Radon [1917] and John [1934], [1955]. According to Bockwinkel [1906] the case n = 3 had been proved before 1906 by H.A. Lorentz, but fortunately, both for Lorentz and Radon, the transformation f (x) → J(ω, p) was not baptized “Lorentz transformation”. In John [1955] the proofs are based on the Poisson equation Lu = f . Other proofs, using distributions, are given in Gelfand-Shilov [1960]. See also Nievergelt [1986]. The dual transforms,
ˇ f → f , ϕ → ϕ, the unified inversion formula and its dual,
ˇ cf = L(n−1)/2((f )∨) , c ϕ = (n−1)/2((ϕ)b) ,

Bibliographical Notes 61
were given by the author in [1964]. The first proof of Theorem 3.1 in the text is from the author’s paper [1959]. It is valid for constant curvature spaces as well. The version in Theorem 3.8 is also proved in Ludwig [1966].
The support theorem, the Paley-Wiener theorem and the Schwartz theorem (Theorems 2.4,2.6, 2.10) are from Helgason [1964], [1965a]. The example in Remark 2.9 was also found by D.J. Newman, cf. Weiss’ paper [1967], which gives another proof of the support theorem. See also Droste [1983]. The local result in Corollary 2.12 goes back to John [1935]; our derivation is suggested by the proof of a similar lemma in Flensted-Jensen [1977], p. 83. Another proof is in Ludwig [1966].See Palamodov and Denisjuk [1988] for a related inversion formula.
The simple geometric Lemma 2.7 is from the author’s paper [1965a] and is extended to hyperbolic spaces in [1980b]. In the Proceedings containing [1966a] the author raised the problem (p. 174) to extend Lemma 2.7 to each complete simply connected Riemannian manifold M of negative curvature. If in addition M is analytic this was proved by Quinto [1993b] and Grinberg and Quinto [2000]. This is an example of injectivity and support results obtained by use of the techniques of microlocal analysis and wave front sets. As further samples involving very general Radon transforms we mention Boman [1990], [1992], [1993], Boman and Quinto [1987], [1993], Quinto [1983], [1992], [1993b], [1994a], [1994b], Agranovsky and Quinto [1996], Gelfand, Gindikin and Shapiro [1979].
Corollary 2.8 is derived by Ludwig [1966] in a different way. There he proposes alternative proofs of the Schwartz- and Paley-Wiener theorems by expanding f (ω, p) in spherical harmonics in ω. However, the proof fails because the principal point—the smoothness of the function F in the proof of Theorem 2.4 (the Schwartz theorem)—is overlooked. Theorem 2.4 is from the author’s papers [1964] [1965a].
This proof of the Schwartz theorem in [1965a] is adopted in Palamodov [2004] and Gelfand–Gindikin–Graev [2003]. These versions do not seem to me to take sufficiently into account the needed relationship in (11) and (12), §2. See also Carton–Lebrun [1984] for a generalization.
Since the inversion formula is rather easy to prove for odd n it is natural to try to prove Theorem 2.4 for this case by showing directly that
ˇ if ϕ ∈ SH (Pn) then the function f = cL(n−1)/2(ϕ) for n odd belongs ˇ to S(Rn) (in general ϕ ∈ S(Rn)). This approach is taken in Gelfand-
Graev-Vilenkin [1966], pp. 16-17. However, this method seems to offer some unresolved technical difficulties. For some generalizations see Kuchment and Lvin [1990], Aguilar, Ehrenpreis and Kuchment [1996] and Katsevich [1997]. Cor. 2.5 is stated in Semyanisty [1960].
§5. The approach to Radon transforms of distributions adopted in the text is from the author’s paper [1966a]. Other methods are proposed in Gelfand-Graev-Vilenkin [1966] and in Ludwig [1966]. See also Ramm [1995].

62 Chapter I. The Radon Transform on Rn

§6. Formula (60) for the d-plane transform was proved by Fuglede

[1958]. The inversion in Theorem 6.2 is from Helgason [1959], p. 284. The

range characterization for the d-plane transform in Theorem 6.3 is from our

book [1980c] and was used skillfully by Kurusa [1991] to prove Theorem 6.5,

which generalizes John’s range theorem for the X-ray transform in R3

[1938]. The geometric range characterization (Corollary 6.12) is also due

to John [1938]. Papers devoted to the d-plane range question for S(Rn)

are Gelfand-Gindikin and Graev [1982], Grinberg [1987], Richter [1986]

and Gonzalez [1991]. This last paper gives the range as the kernel of a

single 4th order differential operator on the space of d-planes. As shown by

Gonzalez, the analog of Theorem 6.3 fails to hold for S(Rn). An L2-version

of Theorem 6.3 was given by Solmon [1976], p. 77. Proposition 6.13 was

communicated to me by Schlichtkrull.

Some difficulties with the d-plane transform on L2(Rn) are pointed out

by Smith and Solmon [1975] and Solmon [1976], p. 68. In fact, the function

f

(x)

=

|x|−

1 2

n

(log

|x|)−1

(|x| ≥ 2), 0 otherwise, is square integrable on Rn

but

is

not

integrable

over

any

plane

of

dimension

≥

n 2

.

Nevertheless,

see

for

example Rubin [1998a], Strichartz [1981], Markoe [2006] for Lp-extensions

of the d-plane transform.

§7. The applications to partial differential equations go in part back to

Herglotz [1931]; see John [1955]. Other applications of the Radon transform

to partial differential equations with constant coefficients can be found in

Courant-Lax [1955], Gelfand-Shapiro [1955], John [1955], Borovikov [1959],

G˚arding [1961] and Ludwig [1966]. Our discussion of the wave equation

(Theorem 7.3 and Corollary 7.4) is closely related to the treatment in Lax-

Phillips [1967], Ch. IV, where however, the dimension is assumed to be

odd. Applications to general elliptic equations were given by John [1955].

While the Radon transform on Rn can be used to “reduce” partial differ-

ential equations to ordinary differential equations one can use a Radon type

transform on a symmetric space X to “reduce” an invariant differential op-

erator D on X to a partial differential operator with constant coefficients.

For an account of these applications see the author’s monograph [1994b],

Chapter V.

While the applications to differential equations are perhaps the most

interesting to mathematicians, the tomographic applications of the X-ray

transform (see §7, B) have revolutionized medicine. These applications orig-

inated with Cormack [1963], [1964] and Hounsfield [1973]. See §7, B for

the medical relevance of the support theorem. For the approximate recon-

struction problem, including Propositions 7.8 and 7.9 and refinements of

Theorems 7.10, 7.11 see Smith, Solmon and Wagner [1977], Solmon [1976]

and Hamaker and Solmon [1978]. Theorem 7.11 is due to Halperin [1962],

the proof in the text to Amemiya and Ando [1965]. For an account of some

of those applications see e.g. Deans [1983], Natterer [1986], Markoe [2006]

and Ramm and Katsevich [1996]. Applications in radio astronomy appear

in Bracewell and Riddle [1967].

CHAPTER II

A DUALITY IN INTEGRAL GEOMETRY

§1 Homogeneous Spaces in Duality

The inversion formulas in Theorems 3.1, 3.7, 3.8 and 6.2, Ch. I suggest the general problem of determining a function on a manifold by means of its integrals over certain submanifolds. This is essentially the title of Radon’s paper. In order to provide a natural framework for such problems
ˇ we consider the Radon transform f → f on Rn and its dual ϕ → ϕ from
a group-theoretic point of view, motivated by the fact that the isometry group M(n) acts transitively both on Rn and on the hyperplane space Pn. Thus

(1)

Rn = M(n)/O(n) , Pn = M(n)/Z2 × M(n − 1) ,

where O(n) is the orthogonal group fixing the origin 0 ∈ Rn and Z2 × M(n − 1) is the subgroup of M(n) leaving a certain hyperplane ξ0 through 0 stable. (Z2 consists of the identity and the reflection in this hyperplane.)
We observe now that a point g1O(n) in the first coset space above lies on a plane g2(Z2 × M(n − 1)) in the second if and only if these cosets, considered as subsets of M(n), have a point in common. In fact

g1 · 0 ⊂ g2 · ξ0 ⇔ g1 · 0 = g2h · 0 for some h ∈ Z2 × M(n − 1) ⇔ g1k = g2h for some k ∈ O(n) .
This leads to the following general setup. Let G be a locally compact group, X and Ξ two left coset spaces of G,

(2)

X = G/K , Ξ = G/H ,

where K and H are closed subgroups of G. Let L = K ∩ H. We assume that the subset KH ⊂ G is closed. This is automatic if one of the groups K or H is compact.
Two elements x ∈ X, ξ ∈ Ξ are said to be incident if as cosets in G they intersect. We put (see Fig. II.1)

ˇx = {ξ ∈ Ξ : x and ξ incident}
ξ = {x ∈ X : x and ξ incident} .

Let x0 = {K} and ξ0 = {H} denote the origins in X and Ξ, respectively.
ˇ If Π : G → G/H denotes the natural mapping then since x0 = K · ξ0 we
have
ˇ Π−1(Ξ − x0) = {g ∈ G : gH ∈/ KH} = G − KH .

S. Helgason, Integral Geometry and Radon Transforms, DOI 10.1007/978-1-4419-6055-9_2, © Springer Science+Business Media, LLC 2011

64 Chapter II. A Duality in Integral Geometry

X
ξ^ x

Ξ
^
x
ξ

x

∈

^
ξ

⇔

^
ξ ∈x

FIGURE II.1.

ˇ ˇ In particular Π(G − KH) = Ξ − x0 so since Π is an open mapping, x0 is a
closed subset of Ξ. This proves the following:
ˇ Lemma 1.1. Each x and each ξ is closed.
Using the notation Ag = gAg−1 (g ∈ G, A ⊂ G) we have the following lemma.
Lemma 1.2. Let g, γ ∈ G, x = gK, ξ = γH. Then

ˇx is an orbit of Kg, ξ is an orbit of Hγ ,
and
ˇx = Kg/Lg , ξ = Hγ/Lγ .
Proof. By definition

(3)

ˇx = {δH : δH ∩ gK = ∅} = {gkH : k ∈ K} ,

which is the orbit of the point gH under gKg−1. The subgroup fixing gH is gKg−1 ∩ gHg−1 = Lg. Also (3) implies

ˇ ˇ x = g · x0 , ξ = γ · ξ0 ,
where the dot · denotes the action of G on X and Ξ. We often write τ (g) for the maps x → g · x, ξ → g · ξ and

f τ (g)(x) = f (g−1 · x) , Sτ (g)(f ) = S(f τ (g−1))

for f a function, S a distribution.

Lemma 1.3. Consider the subgroups
KH = {k ∈ K : kH ∪ k−1H ⊂ HK} , HK = {h ∈ H : hK ∪ h−1K ⊂ KH} .
The following properties are equivalent:

§1 Homogeneous Spaces in Duality 65

(a) K ∩ H = KH = HK .
ˇ (b) The maps x → x (x ∈ X) and ξ → ξ (ξ ∈ Ξ) are injective.
We think of property (a) as a kind of transversality of K and H.

ˇ ˇ ˇ Proof. Suppose x1 = g1K, x2 = g2K and x1 = x2. Then by (3) g1 · x0 = ˇ ˇ ˇ ˇ g1 · x0 so g · x0 = x0 if g = g1−1g2. In particular g ·ξ0 ⊂ x0 so g ·ξ0 = k ·ξ0 for ˇ ˇ some k ∈ K. Hence k−1g = h ∈ H so h· x0 = x0, that is hK ·ξ0 = K ·ξ0. As
a relation in G, this means hKH = KH. In particular hK ⊂ KH. Since
ˇ ˇ ˇ ˇ h · x0 = x0 implies h−1 · x0 = x0 we have also h−1K ⊂ KH so by (a) h ∈ K
which gives x1 = x2.
ˇ On the other hand, suppose the map x → x is injective and suppose
h ∈ H satisfies h−1K ∪ hK ⊂ KH. Then
hK · ξ0 ⊂ K · ξ0 and h−1K · ξ0 ⊂ K · ξ0 .
ˇ ˇ ˇ ˇ ˇ ˇ By Lemma 1.2, h · x0 ⊂ x0 and h−1 · x0 ⊂ x0. Thus h · x0 = x0 whence by
the assumption, h · x0 = x0 so h ∈ K. Thus we see that under the transversality assumption a) the elements ξ
ˇ can be viewed as the subsets ξ of X and the elements x as the subsets x
of Ξ. We say X and Ξ are homogeneous spaces in duality. The maps are also conveniently described by means of the following dou-
ble fibration

(4)

|xxxpxxxxxG/LFFFFFπFFF"

G/K

G/H

where p(gL) = gK, π(γL) = γH. In fact, by (3) we have

ˇx = π(p−1(x)) ξ = p(π−1(ξ)) .
We now prove some group-theoretic properties of the incidence, supplementing Lemma 1.3.
Theorem 1.4. (i) We have the identification

G/L = {(x, ξ) ∈ X × Ξ : x and ξ incident}

via the bijection τ : gL → (gK, gH).

(ii) The property

KHK = G

is equivalent to the property:

Any two x1, x2 ∈ X are incident to some ξ ∈ Ξ. A similar statement holds for HKH = G.

66 Chapter II. A Duality in Integral Geometry

(iii) The property

HK ∩ KH = K ∪ H

is equivalent to the property:
For any two x1 = x2 in X there is at most one ξ ∈ Ξ incident to both. By symmetry, this is equivalent to the property:
For any ξ1 = ξ2 in Ξ there is at most one x ∈ X incident to both.

Proof. (i) The map is well-defined and injective. The surjectivity is clear because if gK ∩ γH = ∅ then gk = γh and τ (gkL) = (gK, γH).

(ii) We can take x2 = x0. Writing x1 = gK, ξ = γH we have
x0, ξ incident ⇔ γh = k (some h ∈ H, k ∈ K) x1, ξ incident ⇔ γh1 = g1k1 (some h1 ∈ H, k1 ∈ K) .
Thus if x0, x1 are incident to ξ we have g1 = kh−1h1k1−1. Conversely if g1 = k h k we put γ = k h and then x0, x1 are incident to ξ = γH.
(iii) Suppose first KH ∩ HK = K ∪ H. Let x1 = x2 in X. Suppose ξ1 = ξ2 in Ξ are both incident to x1 and x2. Let xi = giK, ξj = γjH. Since xi is incident to ξj there exist kij ∈ K, hij ∈ H such that

(5)

gikij = γjhij i = 1, 2 ; j = 1, 2 .

Eliminating gi and γj we obtain

(6)

k2−21k2 1h−2 11h1 1 = h−2 21h1 2k1−21k1 1 .

This being in KH ∩ HK it lies in K ∪ H. If the left hand side is in K then h−2 11h1 1 ∈ K, so

g2K = γ1h2 1K = γ1h1 1K = g1K ,

contradicting x2 = x1. Similarly if expression (6) is in H, then k1−21k1 1 ∈ H, so by (5) we get the contradiction

γ2H = g1k1 2H = g1k1 1H = γ1H .
Conversely, suppose KH ∩ HK = K ∪ H. Then there exist h1, h2, k1, k2 such that h1k1 = k2h2 and h1k1 ∈/ K ∪ H. Put x1 = h1K, ξ2 = k2H. Then x1 = x0, ξ0 = ξ2, yet both ξ0 and ξ2 are incident to both x0 and x1.

§2 The Radon Transform for the Double Fibration 67
Examples
(i) Points outside hyperplanes. We saw before that if in the coset space representation (1) O(n) is viewed as the isotropy group of 0 and Z2M(n − 1) is viewed as the isotropy group of a hyperplane through 0 then the abstract incidence notion is equivalent to the naive one: x ∈ Rn is incident to ξ ∈ Pn if and only if x ∈ ξ.
On the other hand we can also view Z2M(n − 1) as the isotropy group of a hyperplane ξδ at a distance δ > 0 from 0. (This amounts to a different embedding of the group Z2M(n−1) into M(n).) Then we have the following generalization.
Proposition 1.5. The point x ∈ Rn and the hyperplane ξ ∈ Pn are incident if and only if distance (x, ξ) = δ.
Proof. Let x = gK , ξ = γH where K = O(n), H = Z2M(n − 1). Then if gK ∩ γH = ∅, we have gk = γh for some k ∈ K, h ∈ H. Now the orbit H · 0 consists of the two planes ξδ and ξδ parallel to ξδ at a distance δ from ξδ. The relation
g · 0 = γh · 0 ∈ γ · (ξδ ∪ ξδ )
together with the fact that g and γ are isometries shows that x has distance δ from γ · ξδ = ξ.
On the other hand if distance (x, ξ) = δ, we have g·0 ∈ γ·(ξδ∪ξδ ) = γH·0, which means gK ∩ γH = ∅.

(ii) Unit spheres. Let σ0 be a sphere in Rn of radius one passing through the origin. Denoting by Σ the set of all unit spheres in Rn, we have the
dual homogeneous spaces

(7)

Rn = M(n)/O(n) ; Σ = M(n)/O∗(n)

where O∗(n) is the set of rotations around the center of σ0. Here a point x = gO(n) is incident to σ0 = γO∗(n) if and only if x ∈ σ.

§2 The Radon Transform for the Double Fibration
With K, H and L as in §1 we assume now that K/L and H/L have positive measures dμ0 = dkL and dm0 = dhL invariant under K and H, respectively. This is for example guaranteed if L is compact.
Lemma 2.1. Assume the transversality condition (a). Then there exists a
ˇ ˇ measure on each x coinciding with dμ0 on K/L = x0 such that whenever ˇ ˇ ˇ ˇ g · x1 = x2 the measures on x1 and x2 correspond under g. A similar
statement holds for dm on ξ.

68 Chapter II. A Duality in Integral Geometry
ˇ ˇ ˇ Proof. If x = g · x0 we transfer the measure dμ0 = dkL over on x by the ˇ ˇ map ξ → g · ξ. If g · x0 = g1 · x0 then (g · x0)∨ = (g1 · x0)∨ so by Lemma 1.3,
g · x0 = g1 · x0 so g = g1k with k ∈ K. Since dμ0 is K-invariant the lemma follows.
ˇ The measures defined on each x and ξ under condition (a) are denoted
by dμ and dm, respectively.
ˇ Definition. The Radon transform f → f and its dual ϕ → ϕ are defined
by

(8)

ˇ f (ξ) = f (x) dm(x) , ϕ(x) = ϕ(ξ) dμ(ξ) ,

ξb

ˇx

whenever the integrals converge. Because of Lemma 1.1, this will always happen for f ∈ Cc(X), ϕ ∈ Cc(Ξ).

In the setup of Proposition 1.5, f (ξ) is the integral of f over the two
ˇ hyperplanes at distance δ from ξ and ϕ(x) is the average of ϕ over the set
of hyperplanes at distance δ from x. For δ = 0 we recover the transforms of Ch. I, §1.
Formula (8) can also be written in the group-theoretic terms,

(9)

ˇ f (γH) = f (γhK) dhL , ϕ(gK) = ϕ(gkH) dkL .

H/L

K/L

Note that (9) serves as a definition even if condition (a) in Lemma 1.3 is not satisfied. In this abstract setup the spaces X and Ξ have equal status. The theory in Ch. I, in particular Lemma 2.1, Theorems 2.4, 2.10, 3.1 thus raises the following problems:

Principal Problems:

A. Relate function spaces on X and on Ξ by means of the transforms
ˇ f → f, ϕ → ϕ. In particular, determine their ranges and kernels. ˇ B. Invert the transforms f → f, ϕ → ϕ on suitable function spaces.
C. In the case when G is a Lie group so X and Ξ are manifolds let D(X) and D(Ξ) denote the algebras of G-invariant differential operators on X and Ξ, respectively. Is there a map D → D of D(X) into D(Ξ) and a map
ˇ E → E of D(Ξ) into D(X) such that
ˇˇ (Df )b= Df , (Eϕ)∨ = Eϕ ?

D. Support Property: Does f of compact support imply that f has compact support?

§2 The Radon Transform for the Double Fibration 69

Although weaker assumptions would be sufficient, we assume now that the groups G, K, H and L all have bi-invariant Haar measures dg, dk, dh and d . These will then generate invariant measures dgK , dgH, dgL, dkL, dhL on G/K, G/H, G/L, K/L, H/L, respectively. This means that

(10)

F (g) dg =

F (gk) dk dgK

G

G/K K

and similarly dg and dh determine dgH , etc. Then

(11)

Q(gL) dgL = c dgK Q(gkL) dkL

G/L

G/K

K/L

for Q ∈ Cc(G/L) where c is a constant. In fact, the integrals on both sides of (11) constitute invariant measures on G/L and thus must be proportional. However,

(12)

F (g) dg =

F (g ) d dgL

G

G/L L

and

(13)

F (k) dk =

F (k ) d dkL .

K

K/L L

We use (13) on (10) and combine with (11) taking Q(gL) = F (g ) d . Then we see that from (12) the constant c equals 1.

ˇ We shall now prove that f → f and ϕ → ϕ are adjoint operators. We
write dx for dgK and dξ for dgH .

ˇ Proposition 2.2. Let f ∈ Cc(X), ϕ ∈ Cc(Ξ). Then f and ϕ are continu-
ous and

ˇ f (x)ϕ(x) dx = f(ξ)ϕ(ξ) dξ .

X

Ξ

Proof. The continuity statement is immediate from (9). We consider the function
P = (f ◦ p)(ϕ ◦ π)
on G/L. We integrate it over G/L in two ways using the double fibration (4). This amounts to using (11) and its analog with G/K replaced by G/H with Q = P . Since P (gk L) = f (gK)ϕ(gkH), the right hand side of (11) becomes
ˇ f (gK)ϕ(gK) dgK .
G/K
If we treat G/H similarly, the lemma follows.

70 Chapter II. A Duality in Integral Geometry

The result shows how to define the Radon transform and its dual for measures and, in case G is a Lie group, for distributions.

Definition. Let s be a measure on X of compact support. Its Radon transform is the functional s on Cc(Ξ) defined by

(14)

ˇ s(ϕ) = s(ϕ) .

ˇ Similarly σ is defined by

(15)

ˇσ(f ) = σ(f ) , f ∈ Cc(X) ,

if σ is a compactly supported measure on Ξ.

Lemma 2.3. (i) If s is a compactly supported measure on X, s is a measure on Ξ.
ˇ (ii) If s is a bounded measure on X and if x0 has finite measure then s
as defined by (14) is a bounded measure.

Proof. (i) The measure s can be written as a difference s = s+ − s− of two positive measures, each of compact support. Then s = s+ − s− is a difference of two positive functionals on Cc(Ξ).
Since a positive functional is necessarily a measure, s is a measure.

(ii) We have

ˇ ˇ sup |ϕ(x)| ≤ sup |ϕ(ξ)| μ0(x0) ,

x

ξ

so for a constant K,

ˇ ˇ ˇ |s(ϕ)| = |s(ϕ)| ≤ K sup |ϕ| ≤ Kμ0(x0) sup |ϕ| ,

and the boundedness of s follows.

If G is a Lie group then (14), (15) with f ∈ D(X) , ϕ ∈ D(Ξ) serve
ˇ to define the Radon transform s → s and the dual σ → σ for distri-
butions s and σ of compact support. We consider the spaces D(X) and E(X) (= C∞(X)) with their customary topologies (Chapter VII, §1). The duals D (X) and E (X) then consist of the distributions on X and the distributions on X of compact support, respectively.
Proposition 2.4. The mappings
f ∈ D(X) → f ∈ E(Ξ)
ˇ ϕ ∈ D(Ξ) → ϕ ∈ E(X)
are continuous. In particular,
s ∈ E (X) ⇒ s ∈ D (Ξ)
ˇ σ ∈ E (Ξ) ⇒ σ ∈ D (X) .

Proof. We have

§2 The Radon Transform for the Double Fibration 71

(16)

f (g · ξ0) = f (g · x) dm0(x) .

ξb0

Let g run through a local cross section through e in G over a neighborhood of ξ0 in Ξ. If (t1, . . . , tn) are coordinates of g and (x1, . . . , xm) the coordinates of x ∈ ξ0 then (16) can be written in the form

F (t1, . . . , tn) = F (t1, . . . , tn ; x1, . . . , xm) dx1 . . . dxm .

Now it is clear that f ∈ E(Ξ) and that f → f is continuous, proving the proposition.

The result has the following refinement.
Proposition 2.5. Assume K compact. Then
(i) f → f is a continuous mapping of D(X) into D(Ξ).
ˇ (ii) ϕ → ϕ is a continuous mapping of E(Ξ) into E(X).
A self-contained proof is given in the author’s book [1994b], Ch. I, § 3. The result has the following consequence. Corollary 2.6. Assume K compact. Then E (X)b⊂ E (Ξ), D (Ξ)∨ ⊂ D (X).

Ranges and Kernels. General Features
It is clear from Proposition 2.2 that the range R of f → f is orthogonal to
ˇ the kernel N of ϕ → ϕ. When R is closed one can often conclude R = N ⊥,
also when is extended to distributions (Helgason [1994b], Chapter IV, §2, Chapter I, §2). Under fairly general conditions one can also deduce
ˇ that the range of ϕ → ϕ equals the annihilator of the kernel of T → T for
distributions (loc. cit., Ch. I, §3).
In Chapter I we have given solutions to Problems A, B, C, D in some cases. Further examples will be given in § 4 of this chapter and Chapter III will include their solution for the antipodal manifolds for compact two-point homogeneous spaces.
The variety of the results for these examples make it doubtful that the individual results could be captured by a general theory. Our abstract setup in terms of homogeneous spaces in duality is therefore to be regarded as a framework for examples rather than as axioms for a general theory.

72 Chapter II. A Duality in Integral Geometry

Nevertheless, certain general features emerge from the study of these examples. If dim X = dim Ξ and f → f is injective the range consists of functions which are either arbitrary or at least subjected to rather weak conditions. As the difference dim Ξ − dim X increases more conditions are imposed on the functions in the range. (See the example of the d-plane transform in Rn.)
In case G is a Lie group there is a group-theoretic explanation for this. Let X be a manifold and Ξ a manifold whose points ξ are submanifolds of X. We assume each ξ ∈ Ξ to have a measure dm and that the set {ξ ∈ Ξ : ξ x} has a measure dμ. We can then consider the transforms

(17)

ˇ f(ξ) = f (x) dm(x) , ϕ(x) = ϕ(ξ) dμ(ξ) .

ξ

ξx

If G is a Lie transformation group of X permuting the members of Ξ
ˇ including the measures dm and dμ, the transforms f → f, ϕ → ϕ commute
with the G-actions on X and Ξ

(18)

(f )τ (g) = (f τ (g))b

ˇ (ϕτ(g))∨ = (ϕ)τ(g) .

Let λ and Λ be the homomorphisms

λ : D(G) → E(X) Λ : D(G) → E(Ξ)

in Ch. VIII, §2. Using (13) in Ch. VIII we derive

(19)

ˇ (λ(D)f ) = Λ(D)f , (Λ(D)ϕ)∨ = λ(D)ϕ .

Therefore Λ(D) annihilates the range of f → f if λ(D) = 0. In some cases, including the case of the d-plane transform in Rn, the range is characterized as the null space of these operators Λ(D) (with λ(D) = 0). This is illustrated by Theorems 6.5 and 6.8 in Ch. I and even more by theorems of Richter, Gonzalez which characterized the range as the null space of certain explicit invariant operators ([GSS, I, §3]). Much further work in this direction has been done by Gonzalez and Kakehi (see Part I in Ch. II, §4). Examples of (17)–(18) would occur with G a group of isometries of a Riemannian manifold, Ξ a suitable family of geodesics. The framework (8) above fits here too but goes further in that Ξ does not have to consist of subsets of X. We shall see already in the next Theorem 4.1 that this feature is significant.

The Inversion Problem. General Remarks

In Theorem 3.1 and 6.2 in Chapter I as well as in several later results the Radon transform f → f is inverted by a formula

(20)

f = D((f )∨) ,

§2 The Radon Transform for the Double Fibration 73

where D is a specific operator on X, often a differential operator. Rouvi`ere has in [2001] outlined an effective strategy for producing such a D.
Consider the setup X = G/K, Ξ = G/H from §1 and assume G, K and H are unimodular Lie groups and K compact. On G we have a convolution (in the style of Ch. VII),

(u ∗ v)(h) = u(hg−1)v(g) dg = u(g)v(g−1h) dg ,

G

G

provided one of the functions u, v has compact support. Here dg is Haar measure. More generally, if s, t are two distributions on G at least one of compact support the tensor product s ⊗ t is a distribution on G × G given by
(s ⊗ t)(u(x, y)) = u(x, y) ds(x) dt(y) u ∈ D(G × G) .

G×G

Note that s⊗t = t⊗s because they agree on the space spanned by functions of the type ϕ(x)ψ(y) which is dense in D(G × G). The convolution s ∗ t is defined by

(s ∗ t)(v) =

v(xy) ds(x) dt(y) .

GG

Lifting a function f on X to G by f = f ◦ π where π : G → G/K is the natural map we lift a distribution S on X to a S ∈ D (G) by S(u) = S(u) where
u˙ (gK) = u(gk) dk .

K

Thus S(f ) = S(f ) for f ∈ D(X). If S, T ∈ D (X), one of compact support the convolution × on X is defined by

(21)

(S × T )(f ) = (S ∗ T )(f ) .

If one of these is a function f , we have

(22)

(f × S)(g · x0) = f (gh−1 · x0) dS(h) ,

G

(23)

(S × f )(g · x0) = f (h−1g · x0) dS(h) .

G

The first formula can also be written

(24)

f × S = f (g · x0)Sτ(g) dg

G

74 Chapter II. A Duality in Integral Geometry

as distributions. In fact, let ϕ ∈ D(X). Then

(f × S)(ϕ) = (f × S)(g · x0)ϕ(g · x0) dg
G

=

f (gh−1 · x0) dS(h) ϕ(g · x0) dg

GG

=

f (g · x0)ϕ(gh) dg dS(h)

GG

=

f (g · x0)(ϕτ(g−1))∼(h) dg dS(h)

GG

= f (g · x0)S(ϕτ(g−1)) dg = f (g · x0)Sτ(g)(ϕ) dg .

G

G

Now let D be a G-invariant differential operator on X and D∗ its adjoint. It is also G-invariant. If ϕ = D(X) then the invariance of D∗ and (24)
imply

(D(f × S))(ϕ) = (f × S)(D∗ϕ) = f (g · x0)Sτ(g)(D∗ϕ) dg

G

= f (g · x0)S(D∗(ϕ ◦ τ (g))) dg = f (g · x0)(DS)τ(g)(ϕ) dg,

G

G

so

(25)

D(f × S) = f × DS .

Let D denote the distribution f → (D∗f )(x0). Then Df = f × D ,

because by (24)

(f × D)(ϕ) =

f (g

·

x0)

τ (g) D

(ϕ)

G

= f (g · x0)D∗(ϕτ(g−1))(x0) dg = f (g · x0)(D∗ϕ)(g · x0) dg

G

G

= f (x)(D∗ϕ)(x) dx = (Df )(x)ϕ(x) dx .

X

X

We consider now the situation where the elements ξ of Ξ are subsets of X (cf. Lemma 1.3).

§3 Orbital Integrals 75

Theorem 2.7 (Rouvi`ere). Under the assumptions above (K compact) there exists a distribution S on X such that

(26)

(f )∨ = f × S , f ∈ D(X) .

Proof. Define a functional S on Cc(X) by

S(f ) = (f )∨(x0) =

f (kh · x0) dh dk .

KH

Then S is a measure because if f has compact support C the set of h ∈ H for which kh·x0 ∈ C for some k is compact. The restriction of S to D(X) is a distribution which is clearly K-invariant. By (24) we have for ϕ ∈ D(X)

(f × S)(ϕ) = f (g · x0)S(ϕτ(g−1)) dg ,
G
which, since the operations and ∨ commute with the G action, becomes

f (g · x0)(ϕ)∨(g · x0) dg = (f )∨(x)ϕ(x) dx ,

G

X

because of Proposition 2.2. This proves the theorem.

Corollary 2.8. If D is a G-invariant differential operator on X such that DS = δ (delta function at x0) then we have the inversion formula

(27)

f = D((f )∨) , f ∈ D(X) .

This follows from (26) and f × δ = f .

§3 Orbital Integrals
As before let X = G/K be a homogeneous space with origin o = (K). Given x0 ∈ X let Gx0 denote the subgroup of G leaving x0 fixed, i.e., the isotropy subgroup of G at x0.
Definition. A generalized sphere is an orbit Gx0 · x in X of some point x ∈ X under the isotropy subgroup at some point x0 ∈ X.
Examples. (i) If X = Rn, G = M(n) then the generalized spheres are just the spheres.

76 Chapter II. A Duality in Integral Geometry

(ii) Let X be a locally compact subgroup L and G the product group L×L

acting on L on the right and left, the element ( 1, 2) ∈ L × L inducing

action

→1

−1 2

on

L.

Let

ΔL

denote

the

diagonal

in

L

×

L.

If

0∈L

then the isotropy subgroup of 0 is given by

(28)

(L × L) 0 = ( 0, e)ΔL( −0 1, e)

and the orbit of under it by

(L × L) 0 ·

=

0(

−1 0

)L ,

which is the left translate by

0 of the conjugacy class of the element

−1 0

.

Thus the generalized spheres in the group L are the left (or right) translates

of its conjugacy classes.

Coming back to the general case X = G/K = G/G0 we assume that G0, and therefore each Gx0 , is unimodular. But Gx0 ·x = Gx0 /(Gx0 )x so (Gx0 )x unimodular implies the orbit Gx0 · x has an invariant measure determined up to a constant factor. We can now consider the following general problem
(following Problems A, B, C, D above).

E. Determine a function f on X in terms of its integrals over generalized spheres.

Remark 3.1. In this problem it is of course significant how the invariant measures on the various orbits are normalized.
(a) If G0 is compact the problem above is rather trivial because each orbit Gx0 · x has finite invariant measure so f (x0) is given as the limit as x → x0 of the average of f over Gx0 · x.
(b) Suppose that for each x0 ∈ X there is a Gx0-invariant open set Cx0 ⊂ X containing x0 in its closure such that for each x ∈ Cx0 the isotropy group (Gx0 )x is compact. The invariant measure on the orbit Gx0 · x (x0 ∈ X, x ∈ Cx0 ) can then be consistently normalized as follows: Fix a Haar measure dg0 on G0. If x0 = g · o we have Gx0 = gG0g−1 and can carry dg0 over to a measure dgx0 on Gx0 by means of the conjugation z → gzg−1 (z ∈ G0). Since dg0 is bi-invariant, dgx0 is independent of the choice of g satisfying x0 = g · o, and is bi-invariant. Since (Gx0 )x is compact it has a unique Haar measure dgx0,x with total measure 1 and now dgx0 and dgx0,x determine canonically an invariant measure μ on the orbit Gx0 · x = Gx0 /(Gx0)x. We can therefore state Problem E in a more specific form.
E . Express f (x0) in terms of integrals

(29)

f (p) dμ(p) ,
Gx0 ·x

x ∈ Cx0 .

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality 77
For the case when X is an isotropic Lorentz manifold the assumptions above are satisfied (with Cx0 consisting of the “timelike” rays from x0) and we shall obtain in Ch. V an explicit solution to Problem E (Theorem 4.1, Ch. V).
(c) If in Example (ii) above L is a semisimple Lie group Problem E is a basic step (Gelfand–Graev [1955], Harish-Chandra [1954], [1957]) in proving the Plancherel formula for the Fourier transform on L.

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality
In this section we discuss some examples of the abstract formalism and problems set forth in the preceding sections §1–§2.

A. The Funk Transform
This case goes back to Funk [1913], [1916] (preceding Radon’s paper [1917]) where he proved, inspired by Minkowski [1911], that a symmetric function on S2 is determined by its great circle integrals. This is carried out in more detail and in greater generality in Chapter III, §1. Here we state the solution of Problem B for X = S2, Ξ the set of all great circles, both as homogeneous spaces of O(3). Given p ≥ 0 let ξp ∈ Ξ have distance p from the North Pole o, Hp ⊂ O(3) the subgroup leaving ξp invariant and K ⊂ O(3) the subgroup fixing o. Then in the double fibration

O(3)/(K vmmmmmmmmmmmmm

∩

HQQpQ)QQQQQQQQQQ(

X = O(3)/K

Ξ = O(3)/Hp

x ∈ X and ξ ∈ Ξ are incident if and only if d(x, ξ) = p. The proof is
ˇ the same as that of Proposition 1.5. We denote by fp and ϕp the Radon
transforms (9) for the double fibration. Then fp(ξ) the integral of f over
ˇ ˇ two circles at distance p from ξ and ϕp is the average of ϕ(x) over the great
circles ξ that have distance p from x. (See Fig. II.2.) We need fp only for p = 0 and put f = f0. Note that (f )∨p (x) is the average of the integrals of f over the great circles ξ at distance p from x (see Figure II.2). As a special
case of Theorem 1.22, Chapter III, we have the following inversion.

Theorem 4.1. The Funk transform f → f is (for f even) inverted by

(30)

1 f (x) =
2π

u

d du

(f )∨cos−1(v)(x)v(u2

−

v2)−

1 2

dv

0

.
u=1

78 Chapter II. A Duality in Integral Geometry o

x

C

C′

FIGURE II.2.

FIGURE II.3.

We shall see later that this formula can also be written

π

2

(31)

f (x) =

1 f (w) dw − 2π

d dp

(f )∨p (x)

dp ,
sin p

Ex

0

where dw is the normalized measure on the equator Ex corresponding to x. In this form the formula holds in all dimensions.
Also Theorem 1.26, Ch. III shows that if f is even and if all its derivatives vanish on the equator then f vanishes outside the “arctic zones” C and C if and only if f (ξ) = 0 for all great circles ξ disjoint from C and C (Fig. II.3).

The Hyperbolic Plane H2

We now introduce the hyperbolic plane. This formulation fits well into Klein’s Erlanger Program under which geometric properties of a space should be understood in terms of a suitable transformation group of the space.

Theorem 4.2. On the unit disk D : |z| < 1 there exists a Riemannian metric g which is invariant under all conformal transformations of D. Also g is unique up to a constant factor.

For

this

consider

a

point

a

∈

D.

The

mapping

ϕ

:

z

→

a−z 1−a¯z

is

a

conformal

transformation of D and ϕ(a) = 0. The invariance of g requires

ga(u, u) = g0(dϕ(u), dϕ(u))
for each u ∈ Da (the tangent space to D at a) dϕ denoting the differential of ϕ. Since g0 is invariant under rotations around 0, g0(z, z) = c|z|2, where c is a constant. Here z ∈ D0 (= C). Let t → z(t) be a curve with z(0) = a,

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality 79

z (0) = u ∈ C. Then dϕ(u) is the tangent vector

d

dϕ dz

|a|2 − 1

ϕ(z(t)) =

=

u,

dt

t=0

dz a dt 0

(1 − a¯z)2 z=a

so

ga(u,

u)

=

c

(1

−

1 |a|2)2

|u|2

,

and the proof shows that g is indeed invariant. Thus we take the hyperbolic plane H2 as the disk D with the Riemannian
structure

(32)

ds2

=

(1

|dz|2 − |z|2)2

.

This remarkable object enters into several fields in mathematics. In particular, it offers at least two interesting cases of Radon transforms. The Laplace-Beltrami operator for (32) is given by

L = (1 − x2 − y2)2

∂2 ∂2 ∂x2 + ∂y2

.

The group G = SU(1, 1) of matrices

a b : |a|2 − |b|2 = 1 ba

acts transitively on the unit disk by

(33)

ab ba

· z = az + b bz + a

and leaves the metric (32) invariant. The length of a curve γ(t) (α ≤ t ≤ β) is defined by

β

(34)

L(γ) = ( γ (t), γ (t) γ(t))1/2 dt .

α

In particular take γ(t) = (x(t), y(t)) such that γ(α) = 0, γ(β) = x (0 < x <

1), and let γ0(τ ) = τ x, 0 ≤ τ ≤ 1, so γ and γ0 have the same endpoints.

Then

β

β

L(γ) ≥

1

|x −

(t)| x(t)2

dt

≥

x (t) 1 − x(t)2 dt ,

α

α

80 Chapter II. A Duality in Integral Geometry

which by τ = x(t)/x, dτ /dt = x (t)/x becomes

1
x 1 − τ 2x2 dτ = L(γ0) .
0
Thus L(γ) ≥ L(γ0) so γ0 is a geodesic and the distance d satisfies

1

(35)

d(o, x) =

1

|x| − t2x2

dt

=

1 2

log

1 1

+ −

|x| |x|

.

0

Since G acts conformally on D the geodesics in H2 are the circular arcs in |z| < 1 perpendicular to the boundary |z| = 1.
We consider now the following subgroups of G where sh t = sinh t etc.:

K

= {kθ =

eiθ 0 0 e−iθ

: 0 ≤ θ < 2π}

M = {k0, kπ} ,

M

=

{k0,

kπ

,

k−

π 2

,

k

π 2

}

A = {at =

ch t sh t sh t ch t

: t ∈ R},

N

= {nx =

1 + ix, −ix ix, 1 − ix

Γ = CSL(2, Z)C−1 ,

: x ∈ R}

where C is the transformation w → (w − i)/(w + i) mapping the upper half-plane onto the unit disk.
The orbits of K are the circles around 0. To identify the orbit A · z we use this simple argument by Reid Barton:

cht z + sh t z + th t at · z = sh t z + cht = th t z + 1 .

Under

the

map

w

→

z+w zw+1

(w

∈

C)

lines

go

into

circles

and

lines.

Taking

w = th t we see that A · z is the circular arc through −1, z and 1. Barton’s

argument also gives the orbit nx · t (x ∈ R) as the image of iR under the

map

w

→

w(t w(t

− 1) + t − 1) + 1

.

They are circles tangential to |z| = 1 at z = 1. Clearly N A · 0 is the whole disk D so G = N AK (and also G = KAN ).

B. The X-ray Transform in H2
The (unoriented) geodesics for the metric (32) were mentioned above. Clearly the group G permutes these geodesics transitively (Fig. II.4). Let

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality 81

Ξ be the set of all these geodesics. Let o denote the origin in H2 and ξo the horizontal geodesic through o. Then

(36) X = G/K , Ξ = G/M A .
We can also fix a geodesic ξp at distance p from o and write (37) X = G/K , Ξ = G/Hp ,

geodesics in D

where Hp is the subgroup of G leaving ξp stable. Then for the homogeneous

spaces (37), x and ξ are

incident if and only if

d(x, ξ) = p. The transform

FIGURE II.4.

ˇ f → f is inverted by means of the dual transform ϕ → ϕp for (37). The
inversion below is a special case of Theorem 1.11, Chapter III, and is the

analog of (30). Observe also that the metric ds is renormalized by the factor

2 (so curvature is −1).

Theorem 4.3. The X-ray transform in H2 with the metric

ds2 = 4|dz|2 (1 − |z|2)2

is inverted by

⎧

⎫

(38)

f

(z)

=

−

⎨ ⎩

d dr

∞

⎬

(t2

−

r2

)−

1 2

t(f )∨s(t)(z)

dt⎭

,

r

r=1

where s(t) = cosh−1(t).

Another version of this formula is

∞

(39)

1 f (z) = − π

d dp

(f )∨p (z)

dp sinh p

0

and in this form it is valid in all dimensions (Theorem 1.12, Ch. III). One more inversion formula is

(40)

f = − 1 LS((f )∨ ) ,

4π

where S is the operator of convolution on H 2 with the function x → coth(d(x, o)) − 1, (Theorem 1.16, Chapter III).

82 Chapter II. A Duality in Integral Geometry

C. The Horocycles in H2
Consider a family of geodesics with the same limit point on the boundary B. The horocycles in H2 are by definition the orthogonal trajectories of such families of geodesics. Thus the horocycles are the circles tangential to |z| = 1 from the inside (Fig. II.5).

One such horocycle is ξ0 = N · o, the orbit of the origin o under the action of N . Now we take H2 with the metric (32). Since at · ξ is the horocycle with diameter (tanh t, 1) G acts transitively on the set Ξ of horocycles. Since G = KAN it is easy to see that M N is the subgroup leaving ξo invariant. Thus we have here (41) X = G/K , Ξ = G/M N .

geodesics and horocycles in D
FIGURE II.5.

Furthermore each horocycle has the form ξ = kat · ξ0 where kM ∈ K/M and t ∈ R are unique. Thus Ξ ∼ K/M × A, which is also evident from the figure.
We observe now that the maps
ψ : t → at · o , ϕ : x → nx · o
of R onto γ0 and ξ0, respectively, are isometries. The first statement follows from (35) because
d(o, at · o) = d(o, tanh t) = t .
For the second we note that
ϕ(x) = x(x + i)−1 , ϕ (x) = i(x + i)−2
so ϕ (x), ϕ (x) ϕ(x) = (x2 + 1)−2(1 − |x(x + i)−1|2)−2 = 1 .
Thus we give A and N the Haar measures d(at) = dt and d(nx) = dx. Geometrically, the Radon transform on X relative to the horocycles is
defined by

(42)

f (ξ) = f (x) dm(x) ,

ξ

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality 83

where dm is the measure on ξ induced by (32). Because of our remarks about ϕ, (42) becomes

(43)

f (g · ξ0) = f (gn · o) dn ,

N

so the geometric definition (42) coincides with the group-theoretic one in (9). The dual transform is given by

(44)

ˇϕ(g · o) = ϕ(gk · ξo) dk , (dk = dθ/2π) .

K

In order to invert the transform f → f we introduce the non-Euclidean analog of the operator Λ in Chapter I, §3. Let T be the distribution on R given by

(45)

Tϕ

=

1 2

(sh t)−1ϕ(t) dt ,

ϕ ∈ D(R) ,

R

considered as the Cauchy principal value, and put T = dT /dt. Let Λ be the operator on D(Ξ) given by

(46)

(Λϕ)(kat · ξ0) = ϕ(kat−s · ξ0)e−s dT (s) .

R

Theorem 4.4. The Radon transform f → f for horocycles in H2 is inverted by

(47)

f = 1 (Λf )∨ ,

π

We begin with a simple lemma.

f ∈ D(H2) .

Lemma 4.5. Let τ be a distribution on R. Then the operator τ on D(Ξ) given by the convolution

(τ ϕ)(kat · ξ0) = ϕ(kat−s · ξ0) dτ (s)
R
is invariant under the action of G.
Proof. To understand the action of g ∈ G on Ξ ∼ (K/M ) × A we write gk = k at n . Since each a ∈ A normalizes N we have
gkat · ξ0 = gkatN · o = k at n atN · o = k at+t · ξ0 .
Thus the action of g on Ξ (K/M ) × A induces this fixed translation at → at+t on A. This translation commutes with the convolution by τ , so the lemma follows.

84 Chapter II. A Duality in Integral Geometry
Since the operators Λ, , ∨ in (47) are all G-invariant, it suffices to prove the formula at the origin o. We first consider the case when f is K-invariant, i.e., f (k · z) ≡ f (z). Then by (43),

(48)

f (at · ξ0) = f (atnx · o) dx .

R

Because of (35) we have

(49)

|z| = tanh d(o, z) , cosh2 d(o, z) = (1 − |z|2)−1 .

Since

atnx · o = (sh t − ix et)/(ch t − ix et)

(49) shows that the distance s = d(o, atnx · o) satisfies

(50)

ch2s = ch2t + x2e2t .

Thus defining F on [1, ∞) by

(51)

F (ch2s) = f (tanh s) ,

we have

F (ch2s) = f (tanh s)(2sh s ch3s)−1

so, since f (0) = 0, limu→1 F (u) exists. The transform (48) now becomes (with xet = y)

(52)

etf (at · ξ0) = F (ch2t + y2) dy .

R

We put ϕ(u) = F (u + y2) dy
R
and invert this as follows:

ϕ (u + z2) dz = F (u + y2 + z2) dy dz

R

R2

∞

∞

= 2π F (u + r2)r dr = π F (u + ρ) dρ ,

0

0

so

−πF (u) = ϕ (u + z2) dz .

R

§4 Examples of Radon Transforms for Homogeneous Spaces in Duality 85

In particular,

f (o) = − 1 ϕ (1 + z2) dz = − 1 ϕ (ch2τ )chτ dτ ,

π

π

R

R

1 = −π

F (ch2t + y2) dy ch t dt ,

RR

so

1 f (o) = − 2π

d dt

(etf (at

·

dt ξ0)) sh t

.

R

Since (etf )(at · ξ0) is even (cf. (52)), its derivative vanishes at t = 0, so the integral is well defined. With T as in (45), the last formula can be written

(53)

f (o)

=

1 π

Tt

(et

f

(at

·

ξ0)) ,

the prime indicating derivative. If f is not necessarily K-invariant we use (53) on the average

2π

1

f (z) =

f (k · z) dk = 2π

f (kθ · z) dθ .

K

0

Since f (o) = f (o), (53) implies

(54)

1 f (o) =
π

[et(f )b(at · ξ0)] dT (t) .

R

This can be written as the convolution at t = 0 of (f )b(at · ξ0) with the image of the distribution etTt under t → −t. Since T is even the right hand
side of (54) is the convolution at t = 0 of f with e−tTt . Thus by (46),

1

f (o)

=

(Λf π

)(ξ0) .

Since Λ and commute with the K action this implies

1 f (o) =
π

(Λf )(k

·

ξ0)

=

1 (Λf )∨(o) π

K

and this proves the theorem. Theorem 4.4 is of course the exact analog to Theorem 3.6 in Chapter I,
although we have not specified the decay conditions for f needed in generalizing Theorem 4.4.