Reconstruction from integral data 978-1-4987-1011-4

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MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Reconstruction from Integral Data

Victor Palamodov Tel-Aviv University Israel

MONOGRAPHS AND RESEARCH NOTES IN MATHEMATICS

Series Editors John A. Burns Thomas J. Tucker Miklos Bona Michael Ruzhansky

Published Titles Application of Fuzzy Logic to Social Choice Theory, John N. Mordeson, Davender S. Malik and Terry D. Clark Blow-up Patterns for Higher-Order: Nonlinear Parabolic, Hyperbolic Dispersion and Schrödinger Equations, Victor A. Galaktionov, Enzo L. Mitidieri, and Stanislav Pohozaev Complex Analysis: Conformal Inequalities and the Bieberbach Conjecture, Prem K. Kythe Computational Aspects of Polynomial Identities: Volume l, Kemer’s Theorems, 2nd Edition Alexei Kanel-Belov, Yakov Karasik, and Louis Halle Rowen Cremona Groups and Icosahedron, Ivan Cheltsov and Constantin Shramov Diagram Genus, Generators, and Applications, Alexander Stoimenow Difference Equations: Theory, Applications and Advanced Topics, Third Edition, Ronald E. Mickens Dictionary of Inequalities, Second Edition, Peter Bullen Iterative Optimization in Inverse Problems, Charles L. Byrne Line Integral Methods for Conservative Problems, Luigi Brugnano and Felice Iavernaro Lineability: The Search for Linearity in Mathematics, Richard M. Aron, Luis Bernal González, Daniel M. Pellegrino, and Juan B. Seoane Sepúlveda Modeling and Inverse Problems in the Presence of Uncertainty, H. T. Banks, Shuhua Hu, and W. Clayton Thompson Monomial Algebras, Second Edition, Rafael H. Villarreal Nonlinear Functional Analysis in Banach Spaces and Banach Algebras: Fixed Point Theory Under Weak Topology for Nonlinear Operators and Block Operator Matrices with Applications, Aref Jeribi and Bilel Krichen Partial Differential Equations with Variable Exponents: Variational Methods and Qualitative Analysis, Vicenţiu D. Rădulescu and Dušan D. Repovš A Practical Guide to Geometric Regulation for Distributed Parameter Systems, Eugenio Aulisa and David Gilliam Reconstruction from Integral Data, Victor Palamodov Signal Processing: A Mathematical Approach, Second Edition, Charles L. Byrne Sinusoids: Theory and Technological Applications, Prem K. Kythe Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume l, Victor H. Moll Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume ll, Victor H. Moll

Forthcoming Titles Actions and Invariants of Algebraic Groups, Second Edition, Walter Ferrer Santos and Alvaro Rittatore Analytical Methods for Kolmogorov Equations, Second Edition, Luca Lorenzi Geometric Modeling and Mesh Generation from Scanned Images, Yongjie Zhang Groups, Designs, and Linear Algebra, Donald L. Kreher Handbook of the Tutte Polynomial, Joanna Anthony Ellis-Monaghan and Iain Moffat Microlocal Analysis on Rˆn and on NonCompact Manifolds, Sandro Coriasco Practical Guide to Geometric Regulation for Distributed Parameter Systems, Eugenio Aulisa and David S. Gilliam Stochastic Cauchy Problems in Infinite Dimensions: Generalized and Regularized Solutions, Irina V. Melnikova and Alexei Filinkov Symmetry and Quantum Mechanics, Scott Corry

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Contents

Preface

ix

Notations

xi

1 Radon Transform 1.1 Radon Transform and Inversion . . . . . . . . . . . . 1.1.1 Classical Results . . . . . . . . . . . . . . . . . 1.1.2 Reconstruction from Fan Beam Data . . . . . . 1.2 Range Conditions and Frequency Analysis . . . . . . 1.2.1 Moment Conditions . . . . . . . . . . . . . . . 1.2.2 Frequency Analysis, Case n = 2 . . . . . . . . . 1.2.3 Case n = 3 . . . . . . . . . . . . . . . . . . . . 1.3 Support Theorem . . . . . . . . . . . . . . . . . . . . 1.4 Reconstruction of Functions from Attenuated Integrals 1.5 Reconstruction of Differential Forms . . . . . . . . . . 1.6 Bibliographic Notes . . . . . . . . . . . . . . . . . . . 2 Ray and Line Integral Transforms 2.1 Introduction . . . . . . . . . . . . . . . . 2.2 Reconstruction from Line Integrals . . . . 2.2.1 Relations to the Radon Transform 2.2.2 Completeness Condition . . . . . . 2.2.3 Nonlocality . . . . . . . . . . . . . 2.3 Range Conditions . . . . . . . . . . . . . 2.4 Shift-Invariant FBP Reconstruction . . . 2.5 Backprojection Filtration Method . . . . 2.6 Tuy’s Regularized Method . . . . . . . . 2.6.1 Allocation Function . . . . . . . . 2.7 Ray Integrals of Differential Forms . . . . 2.8 Symmetric Tensors and Differentials . . . 2.9 Reconstruction from Ray Integrals . . . . 2.10 Bibliographic Notes . . . . . . . . . . . .

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Contents

3 Factorization Method 3.1 Factorable Maps . . . . . . . . . . . . . . . 3.2 Spaces of Constant Curvature . . . . . . . 3.3 Funk Transform on the Orthogonal Group 3.4 Reconstruction from Non-Redundant Data 3.5 Range Conditions . . . . . . . . . . . . . . 3.6 Bibliographic Notes . . . . . . . . . . . . .

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4 General Method of Reconstruction 4.1 Geometric Integral Transforms . . 4.2 Reconstruction . . . . . . . . . . . 4.2.1 Continuation of the Proof . 4.2.2 Odd n . . . . . . . . . . . . 4.2.3 Calculation the Coefficient 4.3 Integral Transforms with Weights 4.4 Resolved Generating Functions . . 4.5 Analysis of Convergence . . . . . . 4.6 Wave Front of Integral Transform 4.6.1 Transformation of the Wave 4.7 Bibliographic Notes . . . . . . . .

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5 Applications to Classical Geometries 5.1 Minkowski–Funk Transform . . . . . . . . . . . . 5.1.1 Funk’s Method . . . . . . . . . . . . . . . 5.2 Nongeodesic Hyperplane Sections of a Sphere . . 5.3 Totally Geodesic Transform in Hyperbolic Spaces 5.3.1 Alternative Formulas . . . . . . . . . . . . 5.4 Horospherical Transform . . . . . . . . . . . . . 5.4.1 Parallel Spheres in a Hyperbolic Space . . 5.4.2 Alternative Inversion . . . . . . . . . . . . 5.5 Hyperboloids . . . . . . . . . . . . . . . . . . . . 5.6 Cormack’s Curves . . . . . . . . . . . . . . . . . 5.6.1 Generalization . . . . . . . . . . . . . . . 5.7 Confocal Paraboloids . . . . . . . . . . . . . . . 5.8 Cassini Ovals and Ovaloids . . . . . . . . . . . . 5.8.1 Polynomial Ovals . . . . . . . . . . . . . . 5.8.2 Cassini Ovaloids . . . . . . . . . . . . . . 5.9 Bibliography Notes . . . . . . . . . . . . . . . .

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73 73 74 76 78 80 81 82 84 84 86 87 88 91 93 94 95

6 Applications to the Spherical 6.1 Oscillatory Sets . . . . . . 6.2 Reconstruction . . . . . . . 6.3 Examples . . . . . . . . . . 6.3.1 Andersson’s Formula 6.4 Time Reversal Structure .

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97 97 101 106 108 109

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Mean Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents 6.5 6.6 6.7 6.8 6.9

vii

Boundary Isometry for Waves in a Cavity . Range Conditions . . . . . . . . . . . . . . Spheres Tangent to a Hyperplane . . . . . 6.7.1 Regularization of Improper Integrals Summary of Spherical Mean Transform . . Bibliography Notes . . . . . . . . . . . . .

Appendix A.1 Distributions and Generalized Functions A.1.1 Operations . . . . . . . . . . . . A.1.2 Tempered Distributions . . . . . A.2 Fourier Transform . . . . . . . . . . . . A.2.1 Basic Properties . . . . . . . . . A.2.2 Convolution . . . . . . . . . . . . A.3 Manifolds and Differential Forms . . . . A.3.1 Manifolds . . . . . . . . . . . . . A.3.2 Tangent Vectors and Covectors . A.3.3 Differential Forms . . . . . . . . A.3.4 Contraction . . . . . . . . . . . . A.3.5 Multiplication and Division . . . A.3.6 Odd Forms . . . . . . . . . . . . A.3.7 Integral . . . . . . . . . . . . . . A.3.8 Distributions on Manifolds . . . A.3.9 Geometric Operations . . . . . . A.3.10 Riemannian Metric . . . . . . . . A.4 Homogeneous Distributions . . . . . . . A.4.1 Singular Degrees . . . . . . . . . A.5 Principal Value of Integrals . . . . . . . A.5.1 Singular Functions . . . . . . . . A.5.2 Delta-Functions and Derivatives A.6 Integrals of Rational Functions . . . . . A.7 Interpolation of Bandlimited Functions A.8 Cauchy Integral Equation on Interval . A.9 Bibliographic Notes . . . . . . . . . . .

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121 121 124 124 126 127 128 130 130 132 132 133 134 136 136 137 137 138 139 142 143 143 144 146 148 150 152

Bibliography

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Index

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Preface

Reconstruction of a function from data of integrals is the mathematical model for a number of problems arising in diagnostic techniques. These are x-ray, positron radiography, ultrasound, scattering, sonar, seismic, impedance, and wave tomography, crystallographic, thermoacoustic, photoacoustic, photoelastic, strain tomography, and other methods. The objective of this small book is to represent, in a uniform way, some old and recent mathematical results. We focus on exact explicit analytic formulas for reconstruction of a function or a vector field from data of integrals over lines, rays, circles, arcs, parabolas, hyperbolas, planes, hyperplanes, spheres, paraboloids, etc., with the motivation in both applications and pure mathematics. Range characterizations are also of interest. Chapter 1 contains mostly known facts on the classical and attenuated Radon transform. Chapters 2 and 3 are devoted to reconstructions from data of the ray (circle) integral. These mathematical studies are motivated by the problem of three-dimensional reconstruction of a body (object) from data obtained by various physical modalities. In Chapters 5 and 6, reconstructions in classical and new geometries are collected that are based on the method developed in Chapter 4. Much of the material in Chapters 1–6 has not yet appeared in any books. The appendix gathers necessary definitions and elementary facts from geometry and analysis that are not always included in textbooks.

ix

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Notations

Z Set of integers R Set of real numbers R+ Set of positive numbers C Set of complex numbers √ j + 2πı, ı + −1 Γ (z) Gamma function V Real vector space of dimension n E = En Euclidean space of dimension n n−1 S The unit sphere in E n−1 S = 2π n/2 /Γ (n/2) Area of the unit sphere dx = dx1 ∧ ... ∧ dxn Volume form on E n e = Σxi ∂/∂xi Euler tangent field in E Ω = e x dx Euclidean volume form in S (E) R −jhξ,xi ˆ F (f dx) = f = e f (x) dx Fourier transform of function f C k (X) Space of k-smooth functions on X C0k (X) , D (X) = C0∞ (X) Space of k-smooth functions on X with compact support K (X) Space of ∞-smooth densities on X with compact support MΦ geometric integral transform

xi

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Chapter 1 Radon Transform

1.1 1.1.1

Radon Transform and Inversion Classical Results

Let E be an Euclidean space of dimension n > 1 with the scalar product h·, ·i . We denote by H(p, ω) the hyperplane of E given by equation hω, xi = p, where ω is a unit vector in E and p ∈ R. This parameterization is two-sheeted, since H (−p, −ω) = H (p, ω). The Radon transform of an integrable function f on E is the family of its integrals over the hyperplanes H = H (p, ω), Z Z . Rf (H) = f dH, Rf (p, ω) = f dH. H

H(p,ω)

Here, dH is the Euclidean volume form on H. The integral converges for almost any pair (p, ω) and Rf (−p, −ω) = Rf (p, ω) . The Fourier transform gives the straightforward method of inversion of the Radon transform. Proposition 1.1 [Slice theorem] For an arbitrary integrable function f on E, the following identity holds for all points (σ, ω) , σ ∈ R: Z Z exp (−j hσω, xi) f (x) dx = exp (−jσp) Rf (p, ω)dp. A proof is based on Fubini’s theorem.  Corollary 1.2 The inversion of the Radon transform can be implemented by inverting the Fourier transform: Z Z f (x) = exp (j hx, ξi) exp (−jσp) Rf (p, ω) dpdξ. σω=ξ

Here and later we use the notation j = 2πi. Theorem 1.3 For arbitrary even n and any function f ∈ C n (E) with com-

1

2

Reconstruction from integral data

pact support, we have Z Z ∞ 1 Rf (p, ω)dp Ω (1.1) f (x) = n j S −∞ (p − hω, xi)n Z Z ∞  n−1 1 ∂ dp = n Rf (p, ω) Ω j (n − 1)! S(E) −∞ ∂p hω, xi − p  n Z Z ∞ ∂ 1 log |hω, xi − p| Rf (p, ω)dp Ω, = n j (n − 1)! S(E) −∞ ∂p and for odd n and f ∈ C0n−1 (E), f (x) =

1 2jn−1



Z S(E)

∂ ∂p

n−1 Rf (hω, xi , ω)Ω,

(1.2)

where S (E) is the unit sphere in E and Ω is the Euclidean volume form on the sphere. There are many proofs of this classical result, one can find one more in Chapter 4, where it is shown that (1.1) and (1.2) hold for f ∈ L2 (X) . The Radon transform and the reconstruction have sense for an arbitrary generalized function f with compact support; see §4.7. Remark. The formula (1.2) is local: For reconstruction of the value of f at a point x, we only need to know the values of ∂pn−1 Rf for hyperplanes H(ω, p) containing the point x, whereas (1.1) is nonlocal, since knowledge of ∂pn−1 Rf for all hyperplanes is necessary. Both formulae can be written  n−1 H ∂ ∗ f =R − Rf, (1.3) 2π ∂p where H is the Hilbert operator in variable p (§7.2) and R∗ is the backprojection operator Z ∗ R g (x) = g (hω, xi , ω) Ω. (1.4) S/2

We denote by S/2 an arbitrary unit hemisphere (recall that Rf is an even function). This formula reduces to (1.1) and (1.2) if we take in account that the operators H and ∂/∂p commute. This is the filtered backprojection (FBP) type formula, since filtration n−1 operator (H∂/∂p) is applied first and backprojection operator R∗ is the second. Next is the backprojection filtration (BPF) type formula, where filtration is the second step: Corollary 1.4 Equation (1.3) can written in the form  f=

−∆ 4π 2

(n−1)/2

R∗ Rf,

(1.5)

Radon Transform

3

where the fractional power of −∆ is defined as follows: 

−∆ 4π 2

λ

  2λ g = F ∗ |ξ| F g (ξ) .

Proof. For an arbitrary function h on R with compact support, we have 

−∆ 4π 2

λ

 h (hω, xi) =

H ∂ 2π ∂p

2λ h (p)|p=hω,xi .

Applying this equation to (1.4) yields 

−∆ 4π 2

λ

R∗ g = R∗



H ∂ 2π ∂p

2λ g.

For λ = (n − 1) /2, we obtain (1.5).  For n = 2, (1.3) gives Z πZ ∞ 1 ∂p g (p, ϕ) dp f (x1 , x2 ) = dϕ. 2 2π 0 −∞ cos ϕ x1 + sin ϕ x2 − p

(1.6)

In the case n = 3, formula Z Z 1 1 ∂p2 g(hω, xi , ω)Ω = − 2 ∆x g(hω, xi , ω)Ω f (x) = − 2 4π S2 /2 4π S2 /2 is attributed to H. Lorentz.

1.1.2

Reconstruction from Fan Beam Data

The Radon transform g of a function f on E 2 can be parametrized by pairs of points (y, z) , y = 6 z, namely g = h (y, z) , where h (y, z) is the line integral of f over the line L (y, z) that contain these points. If y and z run over a closed curve Γ in E 2 , a function f can be recovered from h at any point x surrounded by Γ. In the case of convex curve Γ, a reconstruction looks quite simple: Proposition 1.5 Let X be a bounded convex domain with piecewise smooth boundary Γ = ∂X in an oriented plane E. Then for any function f ∈ C 1 (E) with support in X and any point x ∈ X, we have Z Z 1 z−y dy h (y, z) ∧ dz − dz h (y, z) ∧ dy f (x) = − 2 × , (1.7) 4π y∈Γ z∈Γ (x − y) × (x − z) |z − y| where a × b + a1 b2 − a2 b1 . Proof. Take number q and angle ϕ as parameters of a line L (y, z) , where (sin ϕ, − cos ϕ) =

z−y (x − y) × (x − z) , q= : |z − y| |z − y|

4

Reconstruction from integral data

We have dϕ =

d (z − y) z−y × , |z − y| |z − y|

which implies dy h (y, z) ∧ dz − dz h (y, z) ∧ dy z−y 1 × = dh ∧ dϕ. (x − y) × (x − z) |z − y| q Therefore, integral (1.7) is equal to Z πZ ∞ Z Z dp g (p, ϕ) dh ∧ dϕ =2 dϕ. q p − cos ϕ x1 − sin ϕ x2 −∞ 0 Γ Γ The coefficient 2 appears, since each line L counts twice in the left-hand side. We deduce (1.7) from (1.6). 

1.2

Range Conditions and Frequency Analysis

1.2.1

Moment Conditions −q

Proposition 1.6 Let f be a function on E = E n satisfying |f (x)| ≤ Cq |x| for any q ≥ 0 and some Cq . Then for arbitrary integer k ≥ 0, the function Z . mk (ω) = pk Rf (p, ω)dp

is equal to the restriction to the unit sphere S (E) of a homogeneous polynomial of degree k. Proof. By Fubini’s theorem, Z

k

Z

p Rf (p, ω)dp =

k

!

Z

p

f dS

Z dp =

k

(hω, xi f (x)dx.

H(p,ω) k

The function ξ 7→ hξ, xi is a homogeneous polynomial of degree k, hence k the integral of the function hξ, xi f (x) is also a polynomial of degree k. The restriction of this polynomial to the unit sphere coincides with mk .  This is the complete set of range conditions: Theorem 1.7 Let g = g (p, ω) be a smooth  function on R × S (E), such that −q all derivatives of g are equal to O |p| for any q ≥ 0 as |p| → ∞. If the moment function Z . mk (ω) = pk g(p, ω)dp, ω ∈ S (E)

Radon Transform

5

is equal to the restriction of a homogeneous polynomial on E of degree k for k = 0, 1, 2, .., then there exists a function f belonging to the Schwartz space L (E), such that Rf = g. Proof. The integral . φ(ξ) =

∞

  ξ exp(−j|ξ|p)g p, dp |ξ| −∞

Z

(1.8)

converges, since g decreases fast at infinity. Let us show that φ ∈ L (E) . For an arbitrary natural k, we can write k

exp(j|ξ|p) = 1 − j|ξ|p +

(−j) j2 2 k (|ξ|p) + ... + (|ξ|p) + rk (|ξ|p) . 2 k!

By the Lagrange theorem, the remainder rk is a smooth function that vanishes at the origin with its derivatives up to the order k. Substituting this equation in (1.8), we get φ(ξ) =

k i X (−j) 0

i!

|ξ|i

Z

    Z ξ ξ pi g p, dp + rk (|ξ|p)g p, dp. |ξ| |ξ|

(1.9)

The general term of the sum equals ci |ξ|i mi (|ξ|−1 ξ). By the assumption, it is a homogeneous polynomial of degree i of the variables ξ. Therefore, the first term of (1.9) is a polynomial of order ≤ k. All derivatives of the second term up to the order k are continuous. Therefore, the sum has continuous derivatives up to order, at least, k. Function φ is infinitely differentiable, since the number k is arbitrary. Integrating by parts in (1.8), we find that this function decays fast as |ξ| → ∞. The same is true for derivatives of φ, since we can integrate by parts in (1.9). This implies that φ ∈ L (E) . By Proposition 1.1 and (1.8), we have for f + F ∗ φ Fp→σ Rf (p, ω) = Fx→ξ f (σω) = φ (σω) = Fp7→σ g. Applying the inverse Fourier transform Fσ7∗→p to both sides, we obtain Rf = g. 

1.2.2

Frequency Analysis, Case n = 2

The range conditions for the Radon transform are related to its harmonic decomposition. The Radon transform of a function f with compact support is an even function on the commutative group G + R × S1 . Its harmonic decomposition is the combination of the Fourier decomposition on R and on S1 . The set of coefficients is a function on the dual group G∗ ∼ = R × Z. We show that for any integrable function f on E 2 with compact support, the Fourier transform F (g) of the Radon transform g = Rf decreases fast as |σ| → ∞ and |m| > (1 + ε) |σ| for any ε > 0. This implies that values of g

6

Reconstruction from integral data

on cone Ξ + {(σ, m) ∈ G∗ : |m| ≤ |σ|} are informative enough for reasonable approximation of f . First, we find an upper bound for the Fourier coefficients of g : Z ∞ Z 2π 1 exp(−i(σp + mϕ))g(p, ϕ)dp. dϕ gˆ(σ, m) = 2π 0 0 Theorem 1.8 For any integrable function f on a plane with support in the closed unit disc B and any positive number ε, the Fourier coefficients satisfy the inequality Z |ˆ g (σ, m)| ≤ exp(|σ| sinh ε − ε|m|) |f | dxdy. (1.10) Proof. We have Z gˆ(σ, m) =

2π

0

1

Z

Z

0

f (x, y) dqe−i(σp+mϕ) dpdϕ,

cos ϕ x+sin ϕ y=p

where p = cos ϕ x + sin ϕ y, q = − sin ϕ x + cos ϕ y. Changing variables yield Z 2π  Z gˆ(σ, m) = f (x, y) exp(−iF (x, y; ϕ)) dϕ dxdy, (1.11) B

0

where F (x, y, ϕ) = σ (cos ϕ x + sin ϕ y) + mϕ. Function F has a holomorphic continuation for ϕ ∈ C, whereas exp (−iF ) is 2π-periodic and has holomorphic . continuation into the cylinder C/2πZ. Let ζ = ϕ+ıψ (mod 2π) be the complex coordinate in the cylinder. Choose ε > 0 and move the integration circle {−π ≤ ϕ ≤ π} of the inner integral in (1.11) to the circle {ζ : ψ = − sign(m)ε} . By Cauchy’s theorem, we have Z π Z π exp(−iF (x, y; ϕ)) dϕ = exp(−iF (x, y; ϕ + ıψ)) dϕ, −π

−π

where |exp(−iF (x, y; ϕ + ıψ))| = exp (σq sinh ψ + mψ) and |σq sinh ψ + mψ| ≤ − |m| ε + |σ| sinh ε for (x, y) ∈ B. Therefore, Z π exp(−iF (x, y; ϕ + ıψ)) dϕ ≤ 2π exp (|σ| sinh ε − |m| ε) . −π

This and (1.11) imply (1.10). 

1.2.3

Case n = 3

Let f be an integrable function on E 3 with support in the closed unit ball and g = Rf . Write the harmonic decomposition of g XZ 1 g(s, ω) = gˆ(σ, l, m) exp(−ıσs) dσ Ylm (ω), l,m

0

Radon Transform

7

where Ylm are spherical harmonics. We show that the function gˆ(σ, l, m) is . essentially supported in the cone Σ = {|σ| ≥ l}, where l is the number of the corresponding irreducible representation of the rotation group O(3). Theorem 1.9 For any integrable function f on E 3 with support in the unit ball, the inequality  |ˆ g (σ, l, m)| ≤ Cl,k

|σ| l(l + 1) − σ 2

−k Z |f | dx

holds for the Fourier coefficients gˆ of Rf for any l, m, k, and σ, such that 1 ≤ σ 2 ≤ l(l + 1) and some constants Cl,k . For a proof, see [104], Theorem 2.13.

1.3

Support Theorem

Theorem 1.10 Let f be a continuous function on E n \B, where n ≥ 2 and B is convex compact set, and f satisfies the following conditions δ (i) For each δ ≥ 0, |x| f (x) is bounded. (ii) Rf (H) = 0 for any hyperplane H ⊂ E n \B. Then f (x) = 0. Here is a version of this statement. Theorem 1.11 Let f be a continuous function on E 2 \B, where B be convex compact set, and f satisfies conditions (i) for some δ > 1 and (ii) for any line H ⊂ E 2 \B. Then for any circle S that encloses B, we have Z f (x) p (x) ds = 0 (1.12) S

for any polynomial p of degree < δ. This statement implies theorem 1.10 for the case n = 2, since by (i) f is orthogonal to any polynomial on S. Therefore, f vanishes on any circle S, hence f ≡ 0 on E 2 \B. The case n > 2 follows by the descent method. Remark. Equation (1.12) does not hold for δ = k, since the function f (z) = Re z −k , k ≥ 2 fulfils (i) for δ = k and (ii), but f p > 0 for p = Re z k . Proof. We assume that B is the unit ball and expand f in harmonics: f (x) =

∞ X −∞

exp (ikθ) fk (r) , x = r (cos θ, sin θ) , r ≥ 1.

8

Reconstruction from integral data

The coefficients 1 fk (r) = 2π

Z

2π

f (exp (iθ) r) exp (−ikθ) dθ 0

are continuous functions on [1, ∞) and satisfy |fk (r)| ≤ Cr−δ , k ∈ Z

(1.13)

for a constant C. Condition (ii) is fulfilled for each term of this series, which means
Z ∞ cos k cos−1 (a/r) fk (r)  1/2 dr = 0 2 a 1 − (a/r) for any a ≥ 1. Following Cormack’s method [20], we multiply this equation by
cosh k cosh−1 (a/s) da  1/2 a 2 (a/s) − 1 and integrate over the ray 1 ≤ s ≤ a

Z ∞

Z ∞ cosh k cosh−1 as cos k cos−1 ar da 
1/2  1/2 fk (r) dr a = 0.
2 s a a 2 −1 1 − ar s

(1.14)

The integral of the absolute value is bounded by
Z ∞Z r a |k|−1 s

Ck s

s

Z ≤ Ck




a 2 s

1/2  −1 1−


1/2 a 2 r

da |fk (r)| dr

∞

|fk (r)| r|k|−1 log rdr,

1 −1


(·)  is the Chebyshev  polynomial of degree |k| and the in|k| ner integral equals O (r/s) log (r/s) . Now (1.13) implies that the integral (1.14) is absolutely convergent on [1, ∞) , since |k| < δ. Changing the order of integration we get



Z ∞ Z r cosh k cosh−1 ab cos k cos−1 ar da = 0. fk (r) dr 
1/2 
1/2 a a b a 2 a 2 − 1 1 − b r since cosh k cosh

By [20], the inner integral is equal to π/2, which implies Z ∞ fk (r) dr = 0 a

for any a ≥ 1. Differentiation yields fk = 0 for any k, |k| < δ, and (1.12) follows for any polynomial of degree < δ. 

Radon Transform

1.4

9

Reconstruction of Functions from Attenuated Integrals

Let E be an oriented Euclidean plane with the basis (e1 , e2 ) and the scalar product hu, vi = u1 v1 + u2 v2 . For arbitrary complex-valued integrable functions a and f on E, the integral  Z ∞  Z ∗ Ra f (p, θ) + f (x) exp − a (x + tθ ) dt ds (1.15) hθ,xi=p

0

is called the attenuated Radon (line) transform of f with the attenuation coefficient a. We denote here θ∗ = (−θ2 , θ1 ) and Z ∞ Ra (s, θ) = a (sθ + uθ∗ ) du, −∞ Z Z 1 ∞ ∞ a (tθ + uθ∗ ) dt HRa (s, θ) = du, π −∞ −∞ s−t Z 1 ∞ Sa (s, t, θ) = a (sθ + uθ∗ ) sgn (t − u) du, 2 −∞ i C± a (s, t, θ) = Sa (s, t, θ) ± HRa (s, θ) , 2 Z ∞
± Ra f (s, θ) = f (sθ + tθ∗ ) exp C± a (s, t, θ) dt. (1.16) −∞

By the assumptions, these functions are well-defined for almost all values of the arguments. Comparing with (1.15), we have   1 i HRa (s, θ) + Ra (s, θ) Ra f (s, θ) . (1.17) R± f (s, θ) = exp ± a 2 2 Theorem 1.12 For arbitrary functions a and f as above, Z 2π X 1 f (x) = hθ, ∇x i exp (−Cε a (s, t, θ)) HRεa f (s, θ) dϕ, 8π 0 ε=±

(1.18)

where s = hθ, xi , t = hθ∗ , xi , θ∗ = (−θ2 , θ1 ) . Remark. By (1.17), Equation (1.18) gives a reconstruction of f in terms of the attenuated line transform (1.15). This equation can be written in the form Z 2π 1 ∂ f (x) = (exp (−Sa (s, t, θ)) Ha Ra f (s, θ)) dϕ, 4π 0 ∂s where θ = (cos ϕ, sin ϕ) and Z 1 cos (1/2 (HRa (s, θ) − HRa (σ, θ))) Ha g (s, θ) = g (σ, θ) dσ. π R s−σ

10

Reconstruction from integral data

Lemma 1.13 For any x ∈ E, U± (x, θ) + C± a (s, t, θ) are odd functions of z = cos θ + i sin θ that have holomorphic, respectively, antiholomorphic continuation to the unit disc |z| < 1. Proof of Lemma. By the construction, C± a (−s, −t, −θ) = −C± a (s, t, θ), hence is U± is odd. We have Z Z a (y) dy 1 i a (sθ + uθ∗ ) sgn (t − u) du + U+ (x, θ) = 2 R 2π E hθ, x − yi Z Z  Z 1 a (y) dy i = , (1.19) − a (sθ + uθ∗ ) du + 2 2π hθ, x − yi ut E where s = hθ, xi , t = hθ∗ , xi . By A.4, we can write Z

Z

∗

j

0

a (x + vθ∗ ) dv

a (sθ + uθ ) du = −∞

u 0}, Z Z Z a (y) dy a (y) dy ∗ j a (sθ + uθ ) du = − , hθ, y − xi − i0 hθ, y − xi + i0 u>t E− E− and

Z 2 E

a (y) dy = hθ, x − yi

Z E

a (y) dy + hθ, x − yi + i0

Z E

a (y) dy . hθ, x − yi − i0

Substituting in (1.19), we get −1

+

Z

U (x, θ) = −j

E+

The function

a (y) dy + hθ, x − yi + i0

Z E−

a (y) dy hθ, x − yi − i0

! .

(1.20)

1 2 = hθ, x − yi wz ¯ + w/z

has meromorphic continuation to the unit disc D with poles z = ±iw on the boundary, where x1 − y1 + i (x2 − y2 ) w+ . |x − y| The boundary value of this function from the unit disc is equal to lim

λ→1−

2 e+ (θ) e− (θ) = + , wλz ¯ + w/λz hθ, x − yi + i0 hθ, x − yi − i0

Radon Transform

11

where e± is the indicator function of the set     ∂ w θ : ± Im w ¯ (1 − ε) z + >0 . ∂ε (1 − ε) z ε=0
This derivative is equal to −wz ¯ + wz −1 /2 = hθ∗ , x − yi, hence e± = 1 in E± . Therefore, we can write (1.20) in the form Z 2a (y) dy + a (x, θ) = lim −1 , ε→+0 w ¯ (1 − ε) z + w ((1 − ε) z) E2 which implies that a+ is value on the boundary of a holomorphic function in D. This proves the lemma for U+ .The similar arguments work for U− .  Lemma 1.14 Let h = h (z) be an arbitrary continuous function on the closed disc D ⊂ C holomorphic in the interior. For an arbitrary unit vector ξ = (ξ1 , ξ2 ) , we have Z 2π 1 h (θ) i dϕ = (h (−iw) − h (iw)) , (1.21) 2π 0 hξ, θi 2 Z
i
1 h (θ) ↑ θ dϕ = h (0) t ξ + it ξ ∗ − (h (iw) + h (−iw)) t ξ ∗ , (1.22) 2π |z|=1 hξ, θi 2 where the integrals take its principal value, θ = (cos ϕ, sin ϕ), and w = ξ1 +iξ2 . Notation t θ means 2-column transposed to the row θ. For an antiholomorpic function g, Equation (1.21) holds with the opposite sign and (1.22) holds with i replaced by −i.
Proof. We have 2 hξ, θi = w ¯ wz ¯ + wz −1 , 2 cos ϕ = z + z −1 , 2 sin ϕ =
−i z − z −1 , and
Z 2π Z z 2 + 1 h (z) dz h (θ) cos ϕdϕ = −i . hξ, θi wz ¯ 2+w z 0 |z|=1 The integrand β is a meromorphic form on the unit disc with poles in z = 0, ±iw. Its integral over the boundary equals 2πires0 β + πiresiw β + πires−iw β   w ¯−w = 2π h (0) w ¯− (h (iw) + h (−iw)) . 4 Similarly, Z 2π 0

h (θ) sin ϕdϕ = hξ, θi

Z |z|=1


1 − z 2 h (z) dz wz ¯ 2+w z

  w ¯+w = 2πi h (0) w ¯− (h (iw) + h (−iw)) , 4

12

Reconstruction from integral data

which implies (1.22), since (w, ¯ iw) ¯ = ξ + iξ ∗ . By the same arguments, we obtain Z 2π Z h (θ) 2h (z) dϕ = −i dz = −πi (h (iw) − h (−iw)) hξ, θi ¯ 2+w 0 |z|=1 wz and (1.21) follows. For an antiholomorphic function h, we apply (1.21) and ¯  (1.22) to g = h. Proof of Theorem. We have Z Z ds 1 ± f (sθ + tθ∗ ) exp C± a (s, t, θ) dt Hga (hθ, xi , θ) = π hθ, xi − s Z f (y) 1 exp U± (y, θ) dy, = π R2 hθ, x − yi where y = sθ + tθ∗ , s = hθ, yi , t = hθ∗ , yi. It follows that Z
1 f (y) exp −U± (x, θ) Hga± (s, θ) = exp U± (y, x, θ) dy, π hθ, x − yi where U± (y, x, θ) = U± (y, θ) − U± (x, θ) . Set ξ = (x − y) / |x − y| and have U± (y, x, ξ ∗ ) = U± (y, ξ ∗ ) − U± (x, ξ ∗ ) =

1 2

Z

(1.23)

a (sξ ∗ − tξ) (sgn (hy, ξi − t) − sgn (hx, ξi − t)) dt,

R

where s = hy, ξ ∗ i = hx, ξ ∗ i . By Lemma 1.13 and Lemma 1.14, Z 2π t 1 θ dϕ exp U± (y, x, θ) 2π 0 hθ, x − yi t t ∗ (ξ + iξ ∗ ) ξ = −i cosh U± (y, x, ξ ∗ ) , |x − y| |x − y| where the sinh term vanishes, since U± is odd. By (1.23), Z t t t ∗ 1 θ (ξ + iξ ∗ ) ξ exp U+ (y, x, θ) dϕ = −i 2π hθ, x − yi |x − y| |x − y| Z 1 × cosh a (sz ∗ − tz) (sgn (hy, ξi − t) − sgn (hx, ξi − t)) dt. 2

(1.24)

Similarly, t t ∗ θ (ξ − iξ ∗ ) ξ exp U− (y, x, θ) dy = +i hθ, x − yi |x − y| |x − y| Z 1 × cosh a (sξ ∗ + tξ) (sgn (hy, ξi − t) − sgn (hx, ξi − t)) . 2

1 2π

Z

(1.25)

Radon Transform

13

Combining (1.24), (1.25), we get Z t
1 θ exp U+ (y, x, θ) + exp U− (y, x, θ) dϕ 2π hθ, x − yi t ξ dy, =2 |x − y| hence, hθ, ∇x i

X

exp (−Cε a (θ, hθ, xi , hθ∗ , xi)) Hgaε

ε=±

Z X 1 dϕ hθ, ∇x i f (y) dy exp Uε (y, x, θ) = 2π hθ, x − yi ε=± Z t ξ f (y) dy. = 2 divx |x − y| By Example 7.5, the right-hand side equals 4πf (x) , which proves (1.18). 

1.5

Reconstruction of Differential Forms

If the attenuation coefficient a does not vanish, it is possible to reconstruct a vector field v = (v1 , v2 ) in a plane from only data of attenuated line integrals Ra hθ, vi of the normal component hθ, vi of the field v. Alternatively, one can use the axial integrals Ra hθ∗ , vi . If a = 0, then data of all integrals is necessary. We start with the Helmholtz decomposition of the form v = ∇p + ∇∗ q. Functions p and q are called scalar and pseudoscalar potential of v, respectively, and the terms ∇p, ∇∗ q are the irrotational and the solenoidal parts of v. The decomposition is unique, if the potentials are o (|x|) at infinity, which follows from the uniqueness theorem for harmonic functions. Any field v allowing such a decomposition with integrable terms has zero integrals, since Z Z Z v1 dx = (∂1 p − ∂2 q) dx = 0, v2 dx = ... = 0. E

E

The following version of the Helmholtz theorem guarantees the existence of “almost” integrable potentials. Theorem 1.15 Any bounded vector field v on E satisfying condition Z ε (|x| + 1) |v| dx < ∞ (1.26) E

for some ε > 0 can be written in the form v = ∇p + ∇∗ q for some bounded functions p and q on E, such that

14

Reconstruction from integral data Z (|x| + 1)

−ε

Z |∇p| dx +

(|x| + 1)

−ε

|∇∗ q| dx < ∞.

(1.27)

We postpone the proof to the end of the section. Theorem 1.16 Let a be a bounded integrable function on E and v be a bounded field on E satisfying (1.26). The potentials p and q of v can be reconstructed from the attenuated line transform of the normal component hθ, vi by Z 2π X 1 p (x) = exp (−Cε a (s, t, θ)) HRεa hθ, vi (s, θ) dϕ, (1.28) 8π 0 ε=± Z 1 ∂ X exp (−Cε a (s, t, θ)) HRεa hθ, vi (s, θ) dϕ, (1.29) a (x) q (x) = 8π ∂s ε=± ∗ where R± a is as in (1.16) and s = hθ, xi , t = hθ , xi .

Corollary 1.17 Let a and v be as above. The potentials p, q of v can be reconstructed from the attenuated line transform of the axial component hθ∗ , vi of v by Z 2π 1 ∂ X ap = − (1.30) exp (−Cε a (s, t, θ)) HRεa hθ∗ , vi (s, θ) dϕ, 8π 0 ∂s ε=± Z 2π X 1 q= exp (−Cε a (s, t, θ)) HRεa hθ∗ , vi (s, θ) dϕ, (1.31) 8π 0 ε=± where R± a , s, and t are as above. Proof of Corollary. By Theorem 1.16 applied to the field v ∗ , we get v ∗ = ∇˜ p + ∇∗ q˜, where Z 2π X 1 exp (−Cε a (s, t, θ)) HRεa hθ, v ∗ i (s, θ) dϕ, (1.32) p˜ = 8π 0 ε=± Z 1 ∂ X a˜ q= exp (−Cε a (s, t, θ)) HRεa hθ, v ∗ i (s, θ) dϕ. (1.33) 8π ∂s ε=± ∗

We have hθ∗ , vi = − hθ, v ∗ i and v = − (v ∗ ) = −∇∗ p˜ + ∇˜ q , that is, q˜ = p, p˜ = −q. Therefore, (1.32) and (1.33) imply (1.30) and (1.31).  Remark. For the zero attenuation, the scalar potential cannot be found from axial integral data but can be reconstructed from data of normal integrals, and vice versa for the pseudoscalar potential.

Radon Transform

15

Proof of Theorem 1.16. By Theorem 1.12, we have Z X v = hθ, ∇x i exp (−Cε a (s, t, θ)) HRεa v (s, θ) dϕ ε=±

Z Z =

hθ, ∇x i

X ε=±

exp U± (y, x, θ)

hθ, vi dydϕ. hθ, x − yi

(1.34)

Write exp U± = cosh U± + sinh U± in the right-hand side. The factor cosh U± makes the integrand in (1.34) an odd form with respect to θ, hence the integral vanishes. Calculate the second term by means of Lemma 1.14 applied to h (z) = z −1 sinh U± : Z t 1 θ 1 2π dϕ = (1.35) hθ, ∇x i sinh U± (y, x, θ) π 0 hθ, x − yi 2π |x − y| Z t


sinh U± (y, x, θ) θ × z 2 + 1 ∂1 + i − iz 2 ∂2 dϕ z hθ, x − yi t (ξ + iξ ∗ ) = (∂1 + i∂2 ) Z± (y, x) |x − y| t ∗ ξ , ∓ i (−ξ2 ∂1 + ξ1 ∂2 ) sinh U± (±ξ ∗ , y, x) |x − y| where ∂ ∂ ± sinh U± (y, x, θ)z=0 = U (y, x, θ)z=0 ∂z ∂z ∂ ± ∂ ± = U (y, θ)z=0 − U (x, θ)z=0 . ∂z ∂z

Z± (y, x) =

We have ∂ ± U (x, θ)z=0 = ∂z

 Z  i Sa (s, t, θ) ± HRa (s, θ) e−iϕ dϕ, 2

where Z

Sa (s, t, θ) e−iϕ dϕ Z Z 1 2π ∞ a (sx + uθ∗ ) sgn (t − u) due−iϕ dϕ = 2 0 −∞ Z Z ∞ 1 e−iϕ = (a (x − vθ∗ ) − a (x + vθ∗ )) vdv dϕ 2 v Z 0 1 iη1 + η2 = (a (x − η) − a (x + η)) dη 2 2 E |η| Z iη1 + η2 =− a (x + η) dη, 2 |η| E

16

Reconstruction from integral data

and s = hθ, xi , t = hθ∗ , xi . Here, we changed the variables v = ± (t − u), η = vθ∗ , and have iη1 + η2 = e−iϕ v. The result can be written in the form Z Sa (s, t, θ) e−iϕ dϕ = −iA1 (x) − A2 (x) , where

Z Ai (x) +

a (x + η) ηi E

dη

2,i

|η|

= 1, 2.

Further, Z

1 = π

Z Z Z

1 =− π

HRa (s, θ) e−iϕ dϕ

a (σθ + τ θ∗ ) dτ

Z Z a (x + η) E

dσ e−iϕ dϕ hθ, xi − σ

dη −iϕ e dϕ hθ, ηi

Z Z 2π 1 θ1 − iθ2 =− a (x + η) dϕdη π E hθ, ηi 0 Z η1 − iη2 = −2 a (x + η) dη = 2 (A1 (x) − iA2 (x)) , 2 |η| where η = (σ − s) θ + (τ − t) θ∗ = σθ + τ θ∗ − x; dη = dσdτ. It follows that  Z Z  i HRa (s, θ) + Sa (s, t, θ) e−iϕ dϕ U+ (x, θ) e−iϕ dϕ = 2 = (−i (A1 (x) − iA2 (x)) − (iA1 (x) − A2 (x))) = −2 (iA1 (x) + A2 (x)) . and Z

U+ (y, x, θ) e−iϕ dϕ = 2 (i (A1 (x) − A1 (y)) + A2 (x) − A2 (y)) .

Combining (1.35) and (1.36), we obtain Z Z t θ hθ, ∇x i exp U± (y, x, θ) dydϕ hθ, x − yi S1 E ∂ (A2 (x) − A2 (y)) (t ξ) − (A1 (x) − A1 (y)) (t ξ ∗ ) = ∂x1 |x − y| ∂ (A1 (x) − A1 (y)) (t ξ) − (A2 (x) − A2 (y)) (t ξ ∗ ) + + B± ∂x2 |x − y| t ∗ ξ = 2πa (x) dy + B± , |x − y|

(1.36)

Radon Transform

17

where B± = ∓iπ (−ξ2 ∂1 + ξ1 ∂2 ) sinh U± (±ξ ∗ , y, x)

t ∗




ξ . |x − y|

This yields Z t
θ 1 hθ, ∇x i exp U+ (y, x, θ) + exp U− (y, x, θ) dϕ 2π S1 hθ, x − yi t ∗ ξ . = −4a (x) |x − y| Applying both parts of the equation to v = ∇p + ∇∗ q and integrating by parts, we get Z Z X 1 hθ, v (y)i hθ, ∇x i dydϕ exp Uε (y, x, θ) 2π S1 E hθ, x − yi ε=± Z t ∗ ξ = −4a (∇p (y) + ∇∗ q (y)) dy = 8πaq, |x − y| according to Example 7.5. This proves (1.29). By Theorem 1.12 and by loc. cit., we have Z Z
hθ, vi 1 exp U+ (y, x, θ) + exp U− (y, x, θ) dydϕ 2π S1 E hθ, x − yi Z t ξ (∇p (y) + ∇∗ q (y)) dy = 4πp (x) , = −4π |x − y| E which implies (1.28).  Proof of Theorem 1.15. Function F = − log |x| /2π is a fundamental solution for the Laplace operator on E, that is, ∆F = δ0 . Because of (1.26), the convolution p + ∇F ∗v = ∂1 F ∗v1 +∂2 F ∗v2 is well defined on E and is bounded. We can write ∇p = G ∗ v + H ∗ v, where G = e∇∇F, H = (1 − e) ∇∇F , and e is the indicator function of the unit disc D. Convolution H ∗ v is well defined, since v is integrable and (1 − e) ∇∇F is bounded. Function w (x) = |x| + 1 fulfils w (x − y) ≤ w (x) w (y) , which implies Z w−ε (x) |(H ∗ v) (x)| dx E Z Z ≤ w−ε (x − y) wε (y) |H (x − y) v (y)| dydx E E Z Z −ε ≤ C w (ξ) |H (ξ)| dξ wε (y) |v (y)| dy. (1.37) The right-hand side is finite for some ε > 0, since of (1.26) and of estimate |H| ≤ Cw−2 (x) . The function G has R compact support and is a homogeneous of degree −2 in D and satisfies ∂D Gds = 0. Therefore, by Calderon– Ziegmund’s theorem [17], Z Z Z r r |G ∗ v| dx ≤ Cr |v| dx ≤ Cr |v| dx < ∞

18

Reconstruction from integral data

for arbitrary r > 1. On the other hand, by the H¨older inequality, 1/r Z 1/s Z Z r −ε −sε < ∞, w |G ∗ v| dx ≤ w dx |G ∗ v| dx where s > 2/ε is arbitrary and r = s/ (s − 1) > 1. This and (1.37) imply (1.27) for ∇p. Vector ∇∗ q = ∇∗ F ∗ v is estimated in the same way. Finally, we have ∇p + ∇∗ q = (∇∇ + ∇∗ ∇∗ ) F ∗ v = ∆F ∗ v = δ0 ∗ v = v. 

1.6

Bibliographic Notes

H. Minkowski has shown that any even function f on the Euclidean 2-sphere is uniquely determined by its integrals Mf over big circles. P. Funk 1913–1915 [36] found a reconstruction formula. He noted that the average A (Mf ) of Mf over big circles tangent to a circle in the sphere is related to the average A (f ) of f by the Abel transform. His method was adapted by J. Radon [122] for Euclidean planes. The following historical remarks are due to A. Cormack [21] and A. Gr¨ unbaum “In 1906 H. B. A. Bockwinkel used a reconstruction formula of H. A. Lorentz in a paper on propagation of light in biaxial crystals. Lorentz’s result was generalized by G. Uhlenbeck in 1925. In 1936, H. Cram´er and H. Wold proved their theorem on marginal distribution (which is widely used in the probability theory). Also, in 1936, Victor Ambartsumyan found reconstruction methods (for the plane and space cases) and used these to calculate the distribution of velocities of stars from their radial velocities in various directions. This is the first numerical inversion of the Radon transform and it gives the lie to the often made statement that computed tomography would be impossible without computers. In 1947, J. Szarski and T. Wa˙zewski presented an abstract addressing the problem proposed by physician M. S. Majerek from Krakow for the reconstruction of the human head from X-ray pictures taken from various directions. R. N. Bracewell ran into Radon’s problem investigating the sun in about 1956.” §1.1 Inversion and properties of the Radon transform were studied by many authors, in particular, John [63], Gel’fand–Graev–Vilenkin [39], Helgason [56], and Ludwig [79]. An elementary introduction to the mathematical theory can be found in Gel’fand–Gindikin–Graev [43], Chapter 1 and in Epstein [30] in the context of applications to medical imaging. The advanced theory of the Radon transform and detailed historical surveys are contained in Helgason [58] and Natterer [83]. See Helgason [58] for generalizations of the Radon transform for spaces of constant curvature and for more general homogeneous spaces.

Radon Transform

19

§1.2 Theorem 1.7 containing characterization of the range of the Radon transform is due to Helgason [56]. Theorem 1.8 is a form of the LindgrenRattey property of the Fourier series of the Radon transform in parallel and in fan-beam geometries. This property is exploited for the efficient reconstruction algorithms by Natterer [84]. §1.3 The support theorem is due to Helgason (1964) [58]. See further results and a survey in Quinto [121]. §1.4 An inversion formula for the exponential ray transform was found by Tretiak and Metz [137] for constant attenuation coefficient. Other methods were applied by Kuchment and Shneiberg [73]. The range description was obtained in [3]. An approximate inversion formula for variable attenuation was proposed by Natterer [84]. Reconstruction formulas for arbitrary real attenuation were found by R. Novikov [94],[95] and by Kazantsev and Bukhgeim [68],[69] by different methods. Previous papers are surveyed in [69]. Another version of Novikov’s method is due to Natterer [84],[85]. You [142] extended arguments of Natterer [85] for reconstruction with arbitrary complex-valued attenuation coefficient. In this Section we follow [85] and [142]. A previous result is due to Boman and Str¨omberg [14]. Range characterizations of the attenuated ray transform were given in Natterer [83] and Novikov [96]. Noo and Wagner [91] proposed a reconstruction algorithm that uses only half of data of the attenuated ray integrals. Bal [7] studied inversion of the attenuated ray transforms for geodesics and horocycles in the hyperbolic plane geometry. An approximate reconstruction method from half of data of the attenuated line transform was proposed by Rullg˚ ard [126]. Puro and Garin [119] applied Cormack’s harmonic series method for reconstruction of the attenuated transform with axially symmetric attenuation. Boman [13] gave an example of a weighted Radon transform in plane that is not injective. §1.5 Attenuated ray transform of vector fields (differential forms) was studied by Kazantsev and Bukhgeim [69]. They have shown that a vector field on a plane can be reconstructed from attenuated normal line integrals. This is not possible for the zero attenuation. Natterer [86] gave another derivation of the reconstruction formulas. We combine methods of You [142] and Natterer [86] to obtain a reconstruction of a field from the attenuated normal or the axial line transform with arbitrary complex-valued non vanishing attenuation.

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Chapter 2 Ray and Line Integral Transforms

2.1

Introduction

The ray transform of a function f with compact support in a vector space V is defined by Z ∞ Xf (y, v) = f (y + tv) dt, v ∈ V \0, 0

where y is called the vertex or source point of the integration ray. The ray integral Xf (y, v) is a homogeneous function of v of degree -1. The line transform of a function f is defined in the similar way: Z ∞ Lf (y, v) = f (y + tv) dt = Xf (y, v) + Xf (y, −v) . (2.1) −∞

Reconstruction problem is to find a reconstruction formula Lf |Σ 7→ f for functions f supported in a compact set K ⊂ V , where Σ is a family of lines in V. We call the family Σ non redundant if dim Σ = n. In particular the manifold L (V ) of all lines in V has dimension 2n − 2. It is redundant for n > 2 and non redundant for n = 2.

2.2 2.2.1

Reconstruction from Line Integrals Relations to the Radon Transform

Radon’s formula for E 2 provides a reconstruction a function f from its line integrals g = Lf. For a function f defined on E n , n > 2, there are many ways to get a reconstruction by choosing different planes F ⊂ E n . For any vector v ∈ Sn−1 , any vector ω orthogonal to v, any point y ∈ E n and natural k, we

21

22

Reconstruction from integral data

denote k ∂;ω Xf (y, v) +

∂k Xf (y, v + tω)|t=0 k Z∂t ∞

(2.2)

k

hω, ∇x i f (y + tv) tk dt.

= 0

Proposition 2.1 Let h be a homogeneous function of degree 1 − n on R. For an arbitrary n − 1-smooth function f on E n with compact support and any ω ∈ S, Z Z Xf (y, v) h (hω, vi) Ω = Rf (p, ω) h (p − hω, yi) dp. (2.3) v∈S(E)

This equation holds also for h (t) = δ (n−2) (t), i.e. Z n−2 ∂;ω Xf (y, v) Ωn−2 (v) = ∂pn−2 Rf (hω, yi , ω) ,

(2.4)

v∈S(E), hω,vi=0

where Ωn−2 is the volume form on Sn−2 . Proof. For a continuous homogeneous function h, the left hand side of (2.3) equals Z Z ∞ Z Z ∞ Ω f (x) dt h (hω, vi) = f (x) h (hω, tvi) tn−1 dt Ω 0 0 Z = f (x) h (hω, x − yi) dx ZE = Rf (p, ω) h (p − hω, yi) dp, where we changed variables x = h y + tv and p = hω, xi i. By §7.5, we have 1−n 1−n (n−2) h (t) = δ (t) = (n − 2)!/j (t − i0) − (t + i0) . Therefore, (2.4) reduces to (2.3).  Note that for even n, the quantities Xf (a, v) and Xf (a, −v) enter in (2.4) symmetrically. Therefore, we can replace Xf by Lf /2. Equation (2.4) shows the way to reconstruction from the ray transform by reduction to the Radon transform. Proposition 2.2 For n = 3, the equation holds

Fv7→ω (Lf (a, v) dv) =

1 2π 2

Z

∂ dp Rf (p, ω) . ∂p p − hω, ai

(2.5)

Proof. For any point a ∈ E, distribution ua (v) = Lf (a, v) dv is even and homogeneous of degree 2. By Corollary 7.15, equations (2.1) and (2.3) with

Ray and Line Integral Transforms

23

h (t) = t−2 , we have F (ua ) (ω) = j

−2

−2

=j

Z hω, ∂v i Lf (a, v) ZS

Ω hω, vi

1 Lf (a, v) 2 = − 2π 2 hω, vi S Ω

Z

Rf (p, ω)

2 dp

(p − hω, ai)

for any ω ∈ S where j = 2πi. Now (2.5) follows by partial integration. 

2.2.2

Completeness Condition

Let V be a vector space, K ⊂ V be a compact set. A set of lines Σ ⊂ L (V ) is called complete in K if for any point x ∈ K and any hyperplane H through x, there exists a line σ ∈ Σ such that x ∈ σ ⊂ H. This condition is necessary for existence of an operator I : g|Σ 7→ f which is continuous for some reasonable norm or a topology in the set of integral data {g = Lf (σ) , σ ∈ Σ, suppf ⊂ K} . Let Γ ⊂ V and Σ (Γ) be the set of rays in V with vertices y ∈ Γ. The family Σ (Γ) is complete in K ⊂ V \Γ if any hyperplane H that meets K has a common point with Γ. Let now V = E n be an Euclidean space and Sn−1 be the unit sphere. We call allocation function for Γ an arbitrary real bounded function ε on Γ × Sn−1 that fulfils the equation X ε (y, ω) = 1 (2.6) y;hy,ωi=hx,ωi

for any ω ∈ Sn−1 and x ∈ K. It is easy to see that an allocation function exists, if the set Σ (Γ) is complete in K. Theorem 2.3 Suppose that the set Σ (Γ) of rays with vertices in Γ is complete in K and let ε be a allocation function for Γ. For an arbitrary function f with support in K, the Fourier transform fˆ can be reconstructed from data of ray integrals Xf by Z 1 ˆ f (σω) = ε (y, ω) exp (−jσ hy, ωi) (2.7) n−2 (jσ) y∈Γ Z n−2 × ∂q Xf (y, v) Ωn−2 hω, dyi |v|=1,hω,vi=q q=0

for any ω ∈ Sn−1 and σ ∈ R. Proof. The euclidean volume form on the hypersurface H (p, ω) = {x : hx, ωi = p} is equal to dH = tn−2 dt Ωn−2 , t ≥ 0. Equation (2.2) yields for arbitrary y ∈ Γ ∩ H (p, ω) Z Z n−2 n−2 n−2 ∂q Xf (y, v) Ω = ∂;ω Xf (y, v) Ωn−2 hω,vi=q hω,vi=p q=p

24 Z =

Reconstruction from integral data Z n−2 n−2 hω, ∇x i f dH = ∂p f dH.

H(p,ω)

By (2.8) and (2.6), for any ω ∈ Sn−1 , Z Z X n−2 n−2 n−2 ε (y, ω) ∂;ω Xf (y, v) Ω = ∂p y∈Γ∩H(p,ω)

(2.8)

H(p,ω)

v∈Sω

f dH,

H(p,ω)

where the right-hand side vanishes if K ∩ H (p, ω) = ∅. This yields Z Z n−2 ε (y, ω) ∂;ω Xf (y, v) Ωn−2 exp (jσ hy, ωi) hω, dyi y∈Γ hω,vi=0 Z Z f dHdp. exp (−jσp) ∂pn−2 = H(p,ω)

R

Integrating by parts in the right-hand side we get Z Z n−2 (jσ) f dH exp (−jσp) dp R H(p,ω) Z n−2 n−2 ˆ = (jσ) f exp (−jσ hω, xi) dx = (jσ) f (σω) , E

and (2.7) follows.  Remark. This method allows reconstruction of fˆ on a conic frequency domain generated by a set Ω ⊂ Sn−1 . If f has a compact support data of fˆ can be extrapolated see §7.7.

2.2.3

Nonlocality

All above methods are not local, that is, for reconstruction of f (x) at a point x ∈ E n , line integrals Lf are used for lines that are far from x. We show that there is no local reconstruction method. Theorem 2.4 The equation Z f (0) = Sn−1

a (v, D) Lf (q, v)|q=0 Ω

(2.9)

can not valid for all test function f on E n , where a is a differential operator on the manifold of pairs (q, v) , q ∈ E n , v ∈ Sn−1 such that hq, vi = 1. Proof. Suppose that (2.9) is possible. We can write X a (v, D) = ai (v) Dqi , where ai (v) are differential operators on Sn−1 and the sum is finite. Rescale the metric on E n by replacing ds by λds, λ > 0. This makes λLf from Lf and (2.9) implies that ai = 0 except for the case |i| = 1. The involution (q, v) 7→ (−q, v) preserves (2.9) but changes the sigh of Dqi , |i| = 1. Taking the sum of the two equations we get 2f (0) = 0 which is a contradiction. 

Ray and Line Integral Transforms

2.3

25

Range Conditions

Let V be a vector space of dimension n ≥ 2. Denote by L (q, v) the line in V parametrized by x = q + tv, t ∈ R, where q ∈ V, v ∈ V \0. Any transform (q, v) 7→ (q + av, bv) , a ∈ R, b ∈ R\0 does not change the line. Given an integrable function f on V , the line transform Z ∞ g (q, v) = Lf (q, v) = f (q + tp) dt −∞

is defined for almost all values of the parameters q, v. The function g of 2n variables satisfies a system of differential equations which express redundancy of the parametrization: −1

g (q + av, bv) = |b|

g (q, v) .

(2.10)

If g has continuous first derivatives, then (2.10) can be also written as a system of differential equations     ∂g ∂g = 0, v, = −g. (2.11) v, ∂q ∂v Proposition 2.5 Let V =  E be an Euclidean space. If the second order  −2−ε derivatives of f are equal O (|x| + 1) for some ε > 0, then g = Lf satisfies John’s system ∂2g ∂2g = , i, j = 1, ..., n, ∂qi ∂vj ∂qj ∂vi

(2.12)

where q = (q1 , ..., qn ) , v = (v1 , ..., vn ). Proof. It is easy to check that Z ∞ ∂2g ∂2f = (q + tv)tdt ∂qi ∂vj −∞ ∂yi ∂yj for any i and j. The same integral represents the derivative ∂ 2 g/∂qj ∂vi .  F. John proved sufficiency of the above conditions for Euclidean space E 3 under additional smoothness assumptions. The vector of Pl¨ ucker coordinates −1

p (q, v) = |v|

v×q

depends only on the line L (q, v) and we can write g (q, v) = g (p) . The number |p (q, v)| is equal to the distance of L (q, v) to the origin.

26

Reconstruction from integral data

Theorem 2.6 Let g = g (p) be a function defined for all lines L ⊂ E that has continuous first and second order derivatives in E satisfying (2.12-2.11) and     −1−ε −2−ε ∇p,v g = O |p| , ∇2p.v g = O |p| as |p| → ∞ for some  ε > 0. Then g = Lf for a function f (x) on E that is −1−ε equal to O |x| as |x| → ∞. This fact holds also for Euclidean spaces of arbitrary dimension n ≥ 3. Note that there is a gap between necessary (Proposition 2.5) and sufficient (Theorem 2.6) conditions. An invertible characterization of the image of the operator L can be given for more narrow functional classes. If f belongs to the Schwartz space L (E) , then its line transform   g = Lf fulfils (2.12,2.11) −b i,j and satisfies the estimate ∇p,v g (q, v) = O |p| for any i, j and b ≥ 0. the inverse is also true Theorem 2.7 If n > 2, then any smooth function g on E×E\0 satisfying these conditions is equal to the line transform of any function f ∈ L (E) .

2.4

Shift-Invariant FBP Reconstruction

Theorem 2.8 Let Γ be a 1-smooth curve in E = E 3 and x ∈ E\Γ. Suppose that a bounded function w (x, y) and a smooth unit vector field u (x, y) orthog−1 onal to z (x, y) = |x − y| (x − y) are chosen for y ∈ Γ, w (x, y) 6= 0. For an arbitrary function f with compact support in E 3 , we have Z Z −1 w (x, y (s)) 2π ∂ dθ f (x) = g (y (q) , v (x, y, θ))|q=s ds, (2.13) 2 2π |x − y (s)| 0 ∂q sin θ where y = y (s) is a parameterization of Γ and v (x, y, θ) = cos θz (x, y) + sin θu (x, y) , provided X w (x, y) sgn hξ, u (x, y)i sgn hξ, y 0 i = 1 (2.14) y∈Γ,hx−y,ξi=0

for almost any ξ ∈ S2 . It is assumed that the number of points y ∈ Γ such that hx − y, ξi = 0 is finite for almost all ξ and y ∈ Γ. Proof. Let H (y) be the plane spanned by z and u. The vector h (t, θ) = au + bz runs through H (y) where a = t sin θ, b = t cos θ. We have dh = tdtdθ

Ray and Line Integral Transforms

27

and Z

2π

g (y, v (x, y, θ)) 0

dθ = sin θ

Z

2π

Z

∞

f (y + h (t, θ)) dt Z0

=

0

f (y + h) H(y)

dθ sin θ

da db. a

This yields Z 2π ∂ 1 dθ g (y (q) , v (x, y, θ))|q=s |x − y| 0 ∂q sin θ Z da ∂ 1 f (y (q) + h)|q=s db = |x − y| Hy ∂q a Z Z ∂ 1 da = fˆ (ξ) exp (j hξ, (y (q) + h)i) dbdξ |x − y| E a Hy ∂q 2 Z −2π = fˆ (ξ) hξ, y 0 i exp (j hξ, yi) sgn hξ, ui δ (hξ, zi) dξ |x − y| E Z = −2π 2 fˆ (ξ) hξ, y 0 i sgn hξ, ui δ (hξ, x − yi) dξ ZE 2 = −2π exp (j hξ, yi) fˆ (ξ) hξ, y 0 i sgn hξ, ui δ (hξ, x − yi) dξ,

(2.15)

E −1

since |x − y| δ (hξ, zi) = δ (hξ, x − yi) . The right-hand side of (2.13) can be written in the form Z 2 −2π exp (j hξ, yi) fˆ (ξ) K (x, ξ) dξ, (2.16) E3

where

Z K (x, ξ) =

w (x, y) hξ, y 0 i sgn hξ, ui δ (hξ, x − yi) ds.

Γ

Change variable s to η = hξ, x − y (s)i in (2.15) and get the integral Z w (x, y) sgn hξ, ui sgn hξ, y 0 i δ (η) dη, Γ

since dη = |η 0 | ds = |hξ, y 0 i| ds. It reduces to the sum X w (x, y) sgn hξ, ui sgn hξ, y 0 i . y∈Γ,hx−y,ξi=0

The sum equals 1 by (2.14), which implies K (x, ξ) = 1 for almost all ξ ∈ E. It follows that (2.16) equals −2π 2 f (x). 

28

Reconstruction from integral data

2.5

Backprojection Filtration Method

Let Γ be a connected 1-smooth curve in an Euclidean space E n of dimension n ≥ 2, f be a function on E n with compact support vanishing on Γ, and g (y, v) = Xf (y, v) be its ray transform. Let y = y (s) , a ≤ s ≤ b be a 1smooth parametrization of Γ. For a function f on E n and a point x ∈ E 3 \Γ, we take the integral along the curve arc between these some points y (s1 ) and y (s2 ) : Z s2 ∂ ds. g (y (s) , x − y (s) + εy 0 (s)) G (x, y1 , y2 ) + s1 ∂ε ε=0 Let L be the line that contains the chord [y1 , y2 ] . Proposition 2.9 Suppose that a function f vanishes on L\ [y1 , y2 ] . If x belongs to L, that is, x = ry1 + (1 − r) y2 for some r, 0 ≤ r ≤ 1, then Z 1 dt 2 f (ty1 + (1 − t) y2 ) = −g (y1 , x − y1 ) + g (y2 , x − y2 ) t − τ 0 + G (x, y1 , y2 ) . (2.17) Proof. We have Z ∞ ∂ ∂ 0 = g (y, x − y + εy ) f (y + t (x − y + εy 0 )) dt ∂ε ∂ε 0 ε=0 ∞

Z

h∇f (tx + (1 − t) y) , y 0 i tdt =

= 0

=−

∂ ∂s

Z

1

f (x + q (y − x)) −∞

∂ ∂s Z

dq ∂ − q ∂s

Z

f (tx + (1 − t) y) tdt 1−t

∞

f (tx + (1 − t) y) dt, 0

where q = 1 − t and the last term equals −∂g (y, x − y) /∂s. Integrating both sides for s ∈ [s1 , s2 ] yields Z s2 Z 1 ∂ dq G (x, y1 , y2 ) = − f (x + q (y − x)) ds q s1 ∂s −∞ Z s2 ∂ − g (y, x − y) ds = −g (y2 , x − y2 ) + g (y1 , x − y1 ) (2.18) s1 ∂s Z 1 Z 1 dq dq − f (x + q (y2 − x)) + f (x + q (y1 − x)) . q q −∞ −∞ The last two terms are integrals over rays in L with vertex at x. Substitute −1 q = (1 − r) (t − r) in the last two integrals and obtain Z 1 dt , (2.19) 2 f (ty1 + (1 − t) y2 ) t − τ 0

Ray and Line Integral Transforms

29

since f (x) vanishes for x ∈ L\ [y1 , y2 ] . Rearranging (2.18) we come up to (2.17).  Corollary 2.10 A function f (x) can be reconstructed on any chord [y1 , y2 ] from data of ray integrals Xf (y, v) , y ∈ Γ. Proof. The left hand side of ((2.17) is equal to the integral (2.19), which equals πHf (x) , where H is the Hilbert transform. The right-hand side and the function Hf are known for any t, 0 ≤ t ≤ 1, that is, for any point x in the interval [y1 , y2 ] and f is supported by this interval. This function can be recovered by the method of A.8. 

2.6

Tuy’s Regularized Method

Theorem 2.11 Let Γ be a 1-smooth curve in an Euclidean space E 3 with a 6 0, s ∈ [0, 1] that satisfies the completeness parameterization y = y (s), ys0 = condition for a point x ∈ E 3 \Γ : almost any plane through x meets Γ at a point y transversely. Then arbitrary real 2-smooth function f on E 3 with compact support can be reconstructed at x from its ray transform g by Z 1 Gs (y, ξ) Ω, (2.20) f (x) = j ξ∈S2 hys0 , ξi where Ω is the volume form on the unit sphere S2 and, for any ξ ∈ S2 , y is an arbitrary point in Γ such that hy, ξi = hx, ξi and hys0 , ξi 6= 0. The numerator is equal to Gs (y, ξ) =

∂G (y, ξ) , G (y, ξ) = Fv→ξ (g (y, v) dv) , ∂s

where dv is the volume form on E.

30

Reconstruction from integral data

Figure 2.1: Source curve Γ. Proof. Write Z

g (y, v) e−jhξ,vi dv

G (y, ξ) = E

and rearrange the right-hand side Z Z Z ∞ −jhξ,vi g (y, v) e dv = f (y + tv) e−jhξ,vi dtdv E E 0 Z ∞Z = f (y + u) e−jrhξ,ui rdudr 0

Z

∞

Z

=

−jrhξ,xi

f (x) e 0

E

jrhξ,yi

dxe

E

Z rdr =

∞

rfˆ (rξ) ejrhξ,yi dr,

0

by changing the variables v = ru, t = r−1 , x = y + u. This yields Z ∞ G (y, ξ) = rfˆ (rξ) ejrhξ,yi dr. 0

It follows that G depends on y through the product hξ, yi = hξ, xi . Differentiating we obtain Z ∞ Gs (y, ξ) = r2 fˆ (rξ) ejrhξ,xi dr. j hy 0 , ξi 0 Integrating over the unit sphere yields Z Z Z ∞ 1 Gs (y, ξ) dS = dS r2 fˆ (rξ) ejrhξ,xi dr = f (x) j ξ∈S2 hy 0 , ξi S2 0

Ray and Line Integral Transforms

31

and (2.20) follows.  Divergence of (2.20) can be avoided taking into account that the distribution g (·, v) dv on E is homogeneous of degree α = 2. Therefore, Theorem A.13 implies Z g (y, v) Ω 1 , ξ 6= 0. G (y, ξ) = − 2 4π v∈S2 (hξ, vi − i0)2 Substituting in (2.20) we obtain an integral with a complex valued kernel Z Z 1 Ω (ξ) gs (y, v) Ω2 f (x) = − 3 , 8π i S2 hy 0 , ξi S2 (hξ, vi − i0)2 where gs = ∂g/∂s and Ω is the area form on S2 . We assume that f is real and take the real part of both sides Z Z 1 Ω f (x) = − 2 gs (y, v) δ 0 (hξ, vi) Ω (v) , 8π S2 hy 0 , ξi S2 by applying the formula −2

(t − i0)

−2

− (t + i0)

= −jδ 0 (t) .

The interior integral equals Z Z gs (y, v) δ 0 (hξ, vi) dS = −∂q S2

hξ,vi=q

gs (y, v) dθ|q=0 ,

where dθ (v) is the angular measure on the circle {v : hv, ξi = 0} . This results in
Corollary 2.12 Let x ∈ E\Γ be as above. Any function f ∈ C02 E 3 can be reconstructed at x by means of integral Z Z Ω 1 f (x) = − 2 ∂ gs (y, v) dθ|q=0 . (2.21) q 8π S2 hy 0 , ξi hξ,vi=q

2.6.1

Allocation Function

Now we state a reconstruction formula in terms of convergent integrals. Take a point x ∈ E\Γ that fulfils the completeness condition with respect to Γ. The set  O + y ∈ Γ, ξ ∈ S2 : hy − x, ξi = 0 is a smooth oriented manifold of dimension 2. Consider the projection π : O → S2 , (y, ξ) 7→ ξ. By Sard’s theorem, the set Cπ ⊂ S2 of critical values of . π has zero measure. For any ξ ∈ S2 \Cπ , the set O (ξ) = π −1 (ξ) is finite and hy 0 , ξi 6= 0 for any choice y = y (s) ∈ O (ξ) . Choose an allocation function ε defined on Γ × S2 \Cπ such that for any ξ ∈ S2 \Cπ , X ε (y, ξ) = 1. (2.22) y,hy−x,ξi=0

32

Reconstruction from integral data


Theorem 2.13 Let ε be an allocation function. For any function f ∈ C02 E 3 and an arbitrary point x ∈ E 3 \Γ such that any plane P through x meets Γ at a point transversely, the equation holds Z 1 ds f (x) = − 2 (2.23) 8π y∈Γ |y − x| ! Z Z ×

∂q hy−x,ξi=0

hξ,vi=q

gs (y, v) dθ|q=0

ε˜dϕ,

where dϕ is the angular measure on the circle  O (y) = ξ ∈ S2 : hy − x, ξi = 0 , y ∈ Γ and ε˜ (y, ξ) = sgn hy 0 , ξi ε (y, ξ) . Remark. Note that if ε is even in ξ, then ε˜ is odd and the product ...˜ ε (y, ξ) is an even function. The circle hy − x, ξi = 0 in (2.23) can be replaced by any half-circle with a factor 2. Proof. Let U by an open neighborhood of Cπ ; we integration in (2.21) over S2 \U and insert the left hand side of (2.22) in the integral. This yields the function Z 1 (2.24) fU (x) = − 2 8π S2 \U Z X Ω × ε (y, ξ) 0 ∂q gs (y, v) dθ|q=0 , hy , ξi hξ,vi=q R

y∈O(ξ)

which converges to f (x) as mes U → 0, since hy 0 , ξi 6= 0 on S2 \U. Equation d hy − x, ξi = hy 0 , ξi ds + hy − x, dξi = 0 holds on O which yields hy 0 , ξi ds = − hy − x, dξi . Therefore, the quotient in (2.21) restricted to O can be represented in the form Ω ds ∧ Ω ds ∧ Ω = 0 =− hy 0 , ξi hy , ξi ds hy − x, dξi dϕ Ω = ds ∧ , = ds ∧ hy − x, dξi |y − x|

(2.25)

where the form dϕ + |y − x| Ω/ hy − x, dξi is equal to the even form on the circle O (y) = ξ ∈ S2 , hy − x, ξi = 0 . Orientation in O (y) is defined by the normal covector hy − x, dξi . By (2.25), the orientation S2 defined by Ω coincides with the orientation defined by sgn hy 0 , ξi ds ∧ dϕ. This and (2.24) yield Z 1 ds fU (x) = − 2 4π y∈Γ |y − x| ! Z Z ×

∂q ξ∈O(y)\U

hξ,vi=q

gs (y, v) dθ|q=0

ε˜dϕ,

Ray and Line Integral Transforms

33

where ε˜ = ε˜ (y, ξ) = sgn hy 0 , ξi ε (y, ξ) . The right-hand side tends to the righthand side of (2.23), since the kernel has no singularity in Cπ . This implies 2.23), since fU (x) → f (x). 

2.7

Ray Integrals of Differential Forms

Let E be an Euclidean space of dimension n P ≥ 2 and x1 , ..., xn be a system of linear coordinates. A differential form f = fi (x) dxi an be written as a P function f (x; v) = fi (x) vi of x ∈ E and of a vector v = (v1 , ..., vn ) ∈ E. The line transform of the form f is the function Z ∞ Lf (λ) + Lf (y, v) = f (y + tv; v) dt −∞

defined for oriented lines λ = λ (y, v) ⊂ E. We have L (da) = 0 for any function a ∈ C 1 (E) vanishing at infinity. The reconstruction problem can be formulated as follows: to recover 2-form df from non redundant data of line integrals Lf (λ) . We consider here the cases n = 2, 3. In the case n = 2, by Stokes’ theorem Z Lf (λ) = df, H(λ)

where H (λ) is the half plane such that ∂H (λ) = L (λ) . We write df = F dx and have Z Z ∗ hv , ∇v i df = F ds, H(λ)

λ

where v = (v1 , v2 ) is the unit direction vector of λ and v ∗ = (−v2 , v1 ) . The function F can be recovered by Radon’s formula. In the case n = 3, we suppose that a sampling of ray integrals Z ∞ g (y; v) = Xf (y, v) = f (y + tv; v) dt 0

is known, where y ∈ E and v ∈ E\0. For any vector ξ ∈ E, we denote ∂ξ g (x; v) = hξ, ∇x i g (x; v) , ∂;ξ g (x; v) = hξ, ∇v i g (x; v) . The differential df is a 2-differential form in E. An arbitrary 2-form can be written as a function h (x; u, v) of variables x, u, v ∈ E. It is a bilinear function of u and v and h (x; v, u) = −h (x; u, v) . For p ∈ R, ω ∈ S2 , H (p, ω) denotes the plane {hω, xi = p} in E.

34

Reconstruction from integral data

Proposition 2.14 Let f be a 1-form on E with coefficients in C01 (E) . For any hyperplane H = H (p, ω) , any point x ∈ H and any vector s parallel to H, we have Z Z 2 ∂p df (y; s, ω) dH (y) = ∂s ∂;ω g (x; v) ΩH (v) , (2.26) H(p,ω)

SH

where SH is the unit circle in H (0, ω) and dH, ΩH are volume forms on H and SH , respectively. The form df is integrated for constant arguments s and ω, the left hand side is a linear function of s depending on ω ∈ S2 and p ∈ R. Proof. The second derivative of g in the direction ω is as follows Z ∞ Z ∞ ∂;ω g (x; v) = ∂ω f (x + tv; v) tdt + f (x + tv; ω) dt, 0 Z0 ∞ Z ∞ 2 ∂;ω g (x; v) = ∂ω2 f (x + tv; v) t2 dt + 2 ∂ω f (x + tv; ω) tdt. 0

0

Next, we take the s-derivative of both parts:

2 ∂s ∂;ω g (x; v) =

Z

∂s ∂ω2 f (x + tv; v) t2 dt + 2

Z ∂s ∂ω f (x + tv; ω) tdt.

Integrating against ΩH (x) over the circle SH in the last equation, we get Z XZ 2 ∂s ∂;ω g (x; v) ΩH = ∂s ∂ω2 f (z; y) dH (z) SH

i Z

+2

H

∂s ∂ω f (z; ω) dH (z) , H

where y = tv, z = x + y, tf (z; v) = f (z; y) and tdt ΩH (x) = dH (z) . Integrating by parts in the right hand side yields Z ∂s ∂ω f (z; ω) dH = 0, H Z Z X 2 ∂s ∂ω f (z; y) dH = − ∂;s ∂ω2 f (z; y) dH H

i

Z =−

H

∂ω2 f (z; s) dH = −∂p2

Z f (z; s) dH.

H

H(p,ω)

where ∂p = d/∂p. Finally, Z Z 2 ∂s ∂;ω g (x; v) ΩH = −∂p2 S

H(p,ω)

f (z; s) dH.

(2.27)

Ray and Line Integral Transforms

35

For arbitrary y, s, v ∈ E, we have df (y; s, v) = ∂s f (y; v) − ∂v f (y; s) . If v = ω and hs, ωi = 0, this yields Z Z df (y; s, ω) dH = − H(p,ω)

(2.28)

∂ω f (z; s) dH

H(p,ω)

Z = −∂p

f (z; s) dH, H(p,ω)

since the integral of ∂s f (z; ω) dH vanishes. This together with (2.27) imply (2.26).  Theorem 2.15 Let f be a 1-form on E and Γ ⊂ E be a set such that any hyperplane H that meets the support of f has, at least, one common point with Γ. The form df can be reconstructed from data of first x-derivatives of ray integrals g (x; v) = Xf (x; v) for x ∈ Γ. Proof. For arbitrary vectors s, v ∈ E and any plane H = H (p, ω) , we denote Z RH (s, v) = ∂p df (y; s, v) dH. H(p,ω)

This is a bilinear and skew-symmetric functional of vectors s and v. If both vectors are parallel to H, the function RH (s, v) vanishes in view of (2.28) and partial integration. If v = ω we have RH (s, ω) = RH (s0 , ω) , where s0 = s − hs, ωi ω is parallel to H. By (2.26), Z 2 RH (s, ω) = hs0 , ∇x i ∂;ω g (x; v) ΩH v∈S

for arbitrary x ∈ H, hence this integral is known. For arbitrary vectors (s, v) , we have RH (s, v) = RH (hv, ωi s, ω) − RH (hs, ωi v, ω) . (2.29) Indeed, if v = ω the second term vanishes and the first term equals RH (s, ω) which fulfils this equation. The same true if s = ω. If s and v are parallel to H, then both sides vanish and (2.29) follows. By (2.26) and (2.29), Z ∂p df (y; s, v) dH = (2.30) H

Z

2 [hv, ωi hs, ∇y i − hs, ωi hv, ∇y i] ∂;ω g (x; v) ΩH ,

SH

hence we know the Radon data ∂p Rf (·, s, v) (H) for arbitrary vectors s, v and any hyperplane H = H (p, ω) in E. These data are sufficient for reconstruction of df by using Radon’s inversion formula. 

36

Reconstruction from integral data

Remark. These arguments are generalized to arbitrary n > 3. It n is even, n−1 n−1 n−1 ∂;ω g (x; v) + ∂;ω g (x; −v) = ∂;ω g (x; v) , hence g can be replaced by Lf in the right hand side of (2.30) replacing the sphere SH by an arbitrary hemisphere. An analog of John’s system (2.12) is a forth order system of equations for line integrals g of a 1-form f on E. Proposition 2.16 For arbitrary n > 1, the function g (q, v) = Lf (q, v) defined on E×E\0 satisfies the system of equation    ∂2 ∂2 ∂2 ∂2 − − g = 0, i, j, k, l = 1, ..., n ∂qi ∂vj ∂qj ∂vi ∂qk ∂vl ∂ql ∂vk and

−1

g (q + av, bv) = |b|

g (q, v) .

A proof is by direct calculation.

2.8

Symmetric Tensors and Differentials

Let E be an Euclidean space of dimension 3. Choose an euclidean coordinate system (x1 , x2 , x3 ) . Tensor fields of rank 1 and 2 are expressions of the form X X f= fi dxi , g = gij dxi · dxj , respectively, where the symbol · means a symmetric product that is dxj ·dxi = dxi · dxj . The components fi and gij = gji are functions in E which transform under coordinate changes as vectors and bivectors, respectively. We can write any 2-tensor field g defined in E as a function in E × E × E which is bilinear and symmetric in vector variables X g (x; u, v) = gij (x) ui vj , u = (u1 , u2 , u3 ) , v = (v1 , v2 , v3 ) . The spaces of smooth or singular (generalized) tensor fields in E of rank 1 and 2 we denote by Σ1 and Σ2 respectively. The symmetric differential D : Σ1 → Σ2 reads Df = g, gii = ∂i fi , gij =

1 (∂i fj + ∂j fi ) , ∂i = ∂/∂xi , 2

2 where no summation on repeating indices P is assumed. Let Λ 4 be the space of skew symmetric 2-differential forms a = aij dxi ∧dxj and B be the space of P 2 × 2-tensor fields b = bij,kl dxi ∧ dxj · dxk ∧ dxl whose components bij,kl are

Ray and Line Integral Transforms

37

functions in U that are skew symmetric in (i, j) and in (k, l) and symmetric with respect to permutation (i, j) ↔ (k, l) . The Saint-Venant operator V : Σ2 → B 4 is defined for a tensor field g ∈ Σ2 by (Vg)ij,kl + ∂i ∂k gjl − ∂i ∂l gjk − ∂j ∂k gil + ∂j ∂l gik . The fields ∂i , ∂j , ∂k , ∂l can be replaced here by arbitrary tangent vectors α, β, γ, δ in E: (Vg)αβ,γδ = ∂α ∂γ g (β, δ) − ∂α ∂δ g (β, γ) − ∂β ∂γ g (α, δ) + ∂β ∂δ g (α, γ) . (2.31) The tensor Vg vanishes for any potential field g = Df, since VD = 0.

2.9

Reconstruction from Ray Integrals

For a 2-tensor field g in E with components in C01 (E), the ray integral Z ∞ Xg (y; v) = g (y + tv; v, v) dt 0

is well defined for any point y ∈ E (source point of the ray) and any direction vector v ∈ E\0. For an arbitrary vector ξ ∈ E, we denote ∂ξ Xg (y; v) = hξ, ∇y i Xg (y; v) , ∂ Xg (y; v + tξ)|t=0 . ∂;ξ Xg (y; v) = ∂t Let S2 denote the unit sphere in E and Ω be area form in S2 . Theorem 2.17 Let Γ ⊂ E be a piecewise smooth curve and g be a smooth 2-tensor with compact support in E\Γ. For an arbitrary point x ∈ E\Γ such that any plane H through x meets Γ transversely at a point y, tensor Vg can be reconstructed from data of ray integrals with source points in Γ by Z 1 (Vg)αβ;γδ (x) = − 2 ∂p RH(p,ω) |p=hx,ωi (α, β; γ, δ) Ω, (2.32) 8π ω∈S2 where α, β, γ, δ ∈ E are arbitrary, and for any plane H containing a point y ∈ Γ, RH (α, β; γ, δ) = RH (hβ, ωi α, ω; hδ, ωi γ, ω) − RH (hα, ωi β, ω; hδ, ωi γ, ω) − RH (hβ, ωi α, ω; hγ, ωi δ, ω) + RH (hα, ωi β, ω; hγ, ωi δ, ω) , and RH (α, ω; β, ω) =

1 2

Z 0

2π 3 ∂α ∂β ∂;ω Xg (y; v) dθ.

(2.33)

38

Reconstruction from integral data

Remark. This acquisition geometry is essentially the same as in the previous section, where a differential 1-form is reconstructed from first derivatives of its ray integrals for sources on Γ. For reconstruction of a 2-tensor, we need to know second derivatives ∂ 2 Xg (y; v) /∂y i ∂y j of the ray integrals for y ∈ Γ which is a reasonable volume of the integral data. Lemma 2.18 For an arbitrary 2-tensor g, any plane H, any point y ∈ H and arbitrary vectors α, β parallel to H, we have Z Z 1 3 ∂α ∂β ∂;ω Xg (y; v) dθ, (2.34) ∂p (Vg)ωα,ωβ dH = 2 v∈S(ω) H(p,ω) where H = H (p, ω) , S (ω) is the unit circle in H (0, ω) , dθ is the angular measure in S (ω) and dH is the area density in the plane H (p, ω). Proof of Lemma. By (2.31), Z Z (Vg)ωα,ωβ dH = ∂p2 H(p,ω)

g (y; α, β) dH.

(2.35)

H(p,ω)

For arbitrary y ∈ H and any unit vector v parallel to H, Z ∞ Z ∞ ∂;ω Xg (y; v) = ∂ω g (y + tv; v, v) tdt + 2 g (y + tv; ω, v) dt, 0

0

... 3 ∂;ω Xg (y; v) =

Z

∞

∂ω3 g (y + tv; v, v) t3 dt

(2.36)

0

Z ∞ Z ∞ +6 ∂ω2 g (y + tv; ω, v) t2 dt + 6 ∂ω g (y + tv; ω, ω) tdt 0 0 Z ∞ Z ∞ = ∂ω3 g (x; z, z) tdt + 6 ∂ω2 g (x; ω, z) tdt 0 0 Z ∞ +6 ∂ω g (x; ω, ω) tdt 0

where z = tv, x = y + z. For arbitrary vectors α, β ∈ E parallel to H, we apply a derivative ∂α and ∂β to both sides of (2.36) and integrate by parts two times: Z 2π Z 3 ∂β ∂α ∂;ω Xg (y; v) dθ = ∂β ∂α ∂ω3 g (x; z, z) dH 0

Z

H

∂β ∂α ∂ω2 g (x; ω, z) dH

+6 ZH =2 H

Z +6

∂β ∂α ∂ω g (x; ω, ω) dH 0

∂ω3 g (x; α, β) dH + 0 + 0

∞

Ray and Line Integral Transforms

39

The integral of ∂β ∂α ∂ω g (x; ω, ω) vanishes, since g (x; ω, ω) depends only on the first argument. The integral of ∂β ∂α ∂ω2 g (x; ω, z) turns into the integral of −∂β ∂;α ∂ω2 g (x; ω, z) = −∂β ∂ω2 g (x; ω, α) after the first partial integration and vanishes after the second one. This and (2.35) implies (2.34).  Proof of Theorem. For arbitrary α, β, γ, δ ∈ E and plane H ⊂ E, we set for arbitrary vectors α, β, γ, δ, Z VH (α, β; γ, δ) = ∂p (Vg)αβ,γδ dH, (2.37) H

and note that by (2.33), VH (α, ω; β, ω) = RH (α, ω; β, ω) for arbitrary vector α, β parallel to H. On the other hand this tensor fulfils the identity for arbitrary vectors α, β, γ, δ: VH (α, β; γ, δ) = VH (hβ, ωi α, ω; hδ, ωi γ, ω) − VH (hα, ωi β, ω; hδ, ωi γ, ω) − VH (hβ, ωi α, ω; hγ, ωi δ, ω) + VH (hα, ωi β, ω; hγ, ωi δ, ω) ,

(2.38)

where ω is a unit normal vector to H. Both sides are skew symmetric in (α, β) and symmetric in pairs (α, β) and (γ, δ) . If α and β are parallel to H, then the right-hand side of (2.38) vanish. The same is true for the left hand side since of (2.31). The same arguments work if γ and δ are parallel to H. In the general case, we can write α = α0 + aω, β = β 0 + bω, where α0 , β 0 are parallel to H and a, b ∈ R. Then we have VH (α, β; ·, ·) = VH (α0 , β 0 ; ·, ·) + aVH (ω, β 0 ; ·, ·) + bVH (α0 , ω; ·, ·) , where the first term vanishes. A similar equation holds for the right hand side of (2.38). Finally, it is sufficient to consider the case α = γ = ω and β and δ are parallel to H. Then all terms of the right-hand side of (2.38) vanish except the last one, which coincides with the left hand side. The tensor RH (α, β, γ, δ) fulfils the same identity by the definition. Therefore, VH = RH holds for all α, β, γ, δ and by (2.35) is known from Xg. Finally, (2.32) follows from H. Lorentz’s formula Z Z 1 2 Ω ∂p (Vg)αβ,γδ dH (Vg)αβ;γδ (x) = − 2 8π ω∈S2 H(p,ω) p=hω,xi

and substitute RH = VH . 

2.10

Bibliographic Notes

§§2.1-2 The problem of reconstruction of a function from a sampling of line or ray integrals has a long story. Hamaker et al [52] studied uniqueness of

40

Reconstruction from integral data

reconstruction of a function on Rn . For the family of rays with sources on a curve Γ in R3 , the completeness condition was formulated by Orlov [98] and by Tuy [139]. The reconstruction formula (2.20) was proposed in [139]. The special case of this geometry with a curve at infinity was used earlier by Orlov [99]. Grangeat’s method [47] provides reconstruction of a derivative of the Radon transform of the function from a sampling of ray integrals. Smith [133] and Gel’fand-Goncharov [41] applied methods equivalent to (2.5) for reconstruction of a function on R3 . A reconstruction from integrals of rays tangent to a surface was published in [103]; an earlier version is due to Denisjuk. Denisjuk [24] also found an analog of this theorem for lines in R5 . Finch [32] studied reconstruction from a sampling of ray integrals which do not satisfy Tuy’s condition. For analysis of reconstructions from incomplete data, see [83] and references therein. Paper of Gel’fand and Graev [42] is devoted to reconstructions in Rn from data of line integrals. A method of reconstruction from non redundant data of line integrals was proposed in [110]. Gel’fand-Graev [38] and Kirillov [70] obtained local reconstructions from data of integrals over complex lines in Cn . For data of real line integrals, a local reconstruction is not possible. §2.3 For Theorem 2.6, see John [62] Theorem 6. A characterization of the image of the line transform was given by Solmon [134]. Theorem 2.7 is a particular case of the more general result of Richter [123] concerning k-plane integral transforms on Rn . This problem was studied for general homogeneous spaces in papers E. Grinberg [48] and Gonzalez [46]. See therein more complete list of references. §2.4 The result of this section is essentially due to Katsevich [64],[66] who proposed a choice of the field u for the case of helical curve. Any point x in the cylinder inside the helical curve belongs to a chord and w (x, y) is the indicator function of the corresponding ”PI segment” of the helical curve. A more general reconstruction method is given in [65]. Pack and Noo [101] proposed reconstructions based on the formula (2.13) and on a different choice of the field u. Further results were obtained in Zhuang and Chen [144], Katsevich [65] and Katsevich and Kapralov [67]. See more references in [67]. §2.5 This reconstruction method was proposed by Pack, Noo and Clackdoyle [102]. §2.6 Theorem 2.11 is due to Tuy [139] and Corollary 2.12 to Zang-ClackGullberg [142]. Theorem 2.13 is new. §2.7 Integrals of a vector field model tomographic measurements of moving medium, tumor detection, optics and plasma physics etc. see Braun and Hauck [15], Norton [92], Prince [118], Osman and Prince [100], Desbat and Mennesier [28]. In this context, data of the longitudinal (axial) line integrals of a vector field is called the Doppler transform according to the Doppler shift law. A longitudinal (axial) measurement of a field can be interpreted as the line integral of the corresponding differential form g. The line transform of differential forms was studied in detail in Gel’fand–Gindikin–Graev [43] Ch.2 (the John transform). The Doppler transform is invariant with respect to the

Ray and Line Integral Transforms

41

gauge transformation g 7→ g + df, since all line integrals vanish for the exact form df . In the 3D case the problem is to find a non redundant variety in the data space which is sufficient for a reconstruction of dg. Vertgeim [140] has shown that dg is uniquely determined from ray integrals g (L) of g for rays L with vertices on a curve Γ of arbitrary shape that meets any plane in, at least, three points. The case of a curve at infinity was considered by Schuster [128]. Two intersection points are sufficient for reconstruction Denisjuk [27], Sharafutdinov [132]. Theorem 2.15 was published in [109]. §§2.8-9 The theory of line integrals of symmetric differential forms on Rn was developed by V. Sharafutdinov [131]. Any symmetric 2 form g has zero axial integrals if and only if it belongs to the image of the symmetric differential D which is an analog of the exterior differential d. The tensor form Vg can be explicitly reconstructed from knowledge of all line integrals of g by means of a Radon type formula [131]. Other method was used by Denisjuk [27]. According to Theorem 2.17, tensor Vg can be reconstructed from second derivatives of the ray integrals with sources on an arbitrary curve satisfying Tuy’s completeness condition. Lionheart and Sharafutdinov [77] considered the ”truncated transverse” line transform of a traceless symmetric 2-tensor fields on R3 . They have shown that a stable reconstruction is possible if the integrals are known for lines meeting one of 6 projective lines at infinity. An arbitrary symmetric 2-tensor in R3 can be reconstructed from combination of data of axial and truncated normal line transforms [113].

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Chapter 3 Factorization Method

3.1

Factorable Maps

Definition. Let F : X → Y be a proper smooth map of Riemannian manifolds of equal dimensions, Σ = {σ} be a family of smooth subvarieties of X. We say that F is factorable for this family, if for any σ ∈ Σ and arbitrary point x ∈ σ, dVY (F (x) , F (σ)) = j (x) J (σ) , dVX (x, σ) where dVX and dVY are the corresponding Riemannian volume forms. We call j : X → R+ and J : Σ → R+ Jacobian factors of the mapping F. The corresponding metric integral operators RX and RY are related by the following equation for any σ ∈ Σ and a function f on Y : Z Z RY f (F (σ)) = f dVY = J (σ) jgdVX = J (σ) RX (jg) (σ) , F (σ)

σ

where g (x) = f (F (x)) . If there is an inversion operator IX for RX |Σ, then
the operator IY (g) = j −1 IX J −1 (σ) g (σ) provides inversion for RY |F (Σ) . If F is a factorable diffeomorphism, the map F −1 is also factorable for the family F (Σ). Moreover, the transitivity property holds: if F : X → Y and G : Y → Z are factorable maps of Riemannian manifolds for the family Σ and F (Σ) , respectively, then the composition GF : X → Z is factorable for Σ with the Jacobian factors jGF (x) = jG (F (x)) jF (x) , JGF (σ) = JG (F (σ)) JF (σ) . Example 3.1. Any conformal map F is factorable for any k, and the family Σk of all k-dimensional subvarieties with the Jacobian factors jk = k/2 (F ∗ (gY ) /gX ) , Jk = 1. In particular, the geometric inversion x 7→ y = −2 −2k |x| x in E n is factorable with the factors jk (x) = |x| . Example 3.2. Let D be the unit disc in the Euclidean plane. The auto −1 2 morphism of D given by G (z) = 2z 1 + |z| is factorable for the family of circle arcs σ ⊂ D that are orthogonal to the circle ∂D (geodesics in the Poincar´e model of the hyperbolic plane). The image G (σ) of an arc σ is 43

44

Reconstruction from integral data

the chord in D that leans the arc σ (geodesics in the Beltrami-Cayley-Klein model). The Jacobian factors are
1/2 2 2 1 + r2 1 − |z| jG (z) =  , 2 , JG (σ) = r 2 1 + |z| where r is the radius of the arc σ. Example 3.3. Let e : E n → E n+1 be the injection of Euclidean spaces e (x) = (1, x), and p : E n+1 → E n be the projection p (x0 , x) = x−1 0 x. Let Λ be an invertible linear transform of E n+1 , and L = pΛe : E n \H → E n be the induced projective transformation of E defined in the complement to a hyperplane H. Proposition 3.1 The map L is factorable for all k-planes A ⊂ E and arbitrary k ≥ 1. For any x ∈ A, the equation holds 

 0 , ..., Λ ak 0 dV (L (x) , L (A)) −k−1 Λ a = |p (Λ (e (x)))| , dV (x, A) [a0 , ..., ak ]0 where a0 , ..., ak are arbitrary points in E n+1 that span A.  Here for arbitrary points b0 , ..., bk ∈ E n+1 , we denoteby b0 , ..., bk the  matrix with rows b0 , ..., bk . The positive number b0 , ..., bk 0 is defined by X  0 2 2 b , ..., bk 0 = |Bi0 ,...,ik | , 0=i0 0. p=

Proof. Choose p ∈ R and ω ∈ Sn , such that p2 > q (ω) > 0 and set
1/2 p2 − q (ω) ω0 , r= . a=− p + ω1 p + ω1 We have for x ∈ B   2 (p + ω1 ) |x − a| − r2 = # " 2 p2 − q (ω) hω 0 , x0 i |ω 0 | 2 (p + ω1 ) |x| + 2 + 2 − 2 p + ω1 (p + ω1 ) (p + ω1 )   2 2 = (p + ω1 ) |x| + 2 hω 0 , x0 i − p + ω1 = 1 − |x| (hω, yi − p) , where y = Qx. This implies  −1 h i 2 2 1 − |x| (p + ω1 ) (x − a) − r2 = hω, yi − p, 2

hence, equation hω, yi = p is equivalent to (x − a) − r2 = 0, which proves (3.1). Denote by f (x) the left-hand side and have 2

|∇y f | 1 − |x| = , |∇x f | 2r (p + ω1 ) hence, 2

|dy S| dy |∇y f | 1/2 n 1 − |x| = = (q (y) − 1) (1 + y1 ) |dx S| dx |∇x f | 2r (p + ω1 ) =

(1 + y1 )

n−1

1/2

(q (y) − 1) 1/2

(p2 − q (ω))

= j (y) J (p, ω) ,

where n−1

j (y) = (1 + y1 )

1/2

(q (y) − 1)


−1/2 , J (p, ω) = p2 − q (ω) .

46

3.2

Reconstruction from integral data

Spaces of Constant Curvature

Example 3.5. Let x = (x1 , ..., xn ) be Euclidean coordinates in a space in a subspace E n of a Euclidean space E n+1 with coordinates (x0 , x) . Consider the unit hemisphere n o 2 Sn+ = (x0 , x) : x20 + |x| = 1, x0 > 0 . and the unit ball B in E n . For any linear manifold F in E n+1 parallel to x0 -axes with the distance d < 1 from the origin, the manifold F ∩ Sn+ is a hemisphere. Let dS V and dE V be the Euclidean volume forms on F ∩ Sn+ and on F ∩ E n , respectively, and p : Sn+ → B be the projection parallel to x0 -axes. This implies that p is factorable for the family of hemispheres F ∩ Sn+ , since for any x ∈ Sn+ ,
1/2 1 − d2 dS V (x) = , y = p (x) . dE V (y) x0 The one sheet hyperboloid n o 2 Hn+ = x20 − |x| = 1, x0 > 0 , is a model of hyperbolic space. It is endowed with the metric induced by the pseudo-Euclidean metric dh s2 = −dx20 + dx21 + ... + dx2n and has constant sectional curvature −1. For the projection p : Hn+ → E n , the similar formula holds
1/2 1 + d2 dH V (x) = , y = p (x) , dE V (y) x0 where x ∈ Hn+ and dH V id the hyperbolic volume in F ∩ Hn+ . Let v = (v0 , v 0 ) be an arbitrary vector in E n+1 such that v0 ≥ |v 0 | . The projection P : Hn+ → E n = {x0 = 1} parallel to v is one-to-one and is also factorized for the family of (n − 1)-hyperboloids X ∩ H, where H is a hyperplane parallel to v. Example 3.6. The hemisphere Sn+ which is a model of the elliptic space of curvature 1. The central projection in E n+1 defines the maps γ

γ

Hn+ → E n ← Sn+ (3.2)
called gnomonic projections. We have γ Sn+ = E n and this map is invertible.
The set γ Hn+ is an open ball B of radius 1. The intersections of Sn+ , E n and Hn+ with a subspace F ⊂ E n+1 are totally geodesic submanifolds in the elliptic, Euclidean, and hyperbolic space, respectively.

Factorization Method

47

Gnomonic projection of three spaces of constant curvature. Proposition 3.3 The gnomonic projections (3.2) are factorable, namely for an arbitrary subspace F ⊂ E n+1 of dimension k + 1 we have

1/2 k+1 dVS (y0 , y) , F ∩ Sn+ = 1 + d2 (F ∩ E n ) y0 , (3.3) n dVE (x, F ∩ E )

1/2 k+1 dVH (y0 , y) , F ∩ Hn+ = 1 − d2 (F ∩ E n ) y0 , (3.4) dVE (x, F ∩ E n ) where (y0 , y) is an arbitrary point in Sn+ and Hn+ , respectively, x + γ (y) = y0−1 y ∈ E n , where the volume forms dVS , dVE , and dVH are compared, and  −(k+1)/2 2 d (G) = distE (G, 0) , y0k+1 = 1 ± |x| . Corollary 3.4 For any dimension k, 1 ≤ k < n, the metric integral transform on totally geodesic transform on hyperbolic space is equivalent to the Radon transform on the unit ball B ⊂ E n . The Radon transform in E n is equivalent to the Funk transform on the elliptic space Sn /Z2 . Proof. Let ϕ be an arbitrary function on Hn+ , which is integrable on F ∩Hn+ for any subspace F ⊂ E n of dimension k + 1. By (3.4), we have Z Z
1/2 2 ϕ (y0 , y) dVH = 1 − d (F ) f (x) dVE , F ∩Hn +

F ∩B

 −1 2 where f (x) = y0k+1 ϕ (y0 , y) and (y0 , y) = 1 − |x| (1, x) ∈ Hn+ . This function is supported in B and is integrable on F ∩B for any subspace F ⊂ E n .

48

Reconstruction from integral data

Therefore, for arbitrary dimension k, 0 ≤ k ≤ n − 1, the metric integral transform of functions ϕ on Hn+ is reduced to the Radon transform on B. In the elliptic case, we apply (3.3) and write Z Z
1/2 ϕ (y0 , y) dVS = 1 + d2 (F ) f (x) dVE F ∩Sn +

F ∩E n

for an arbitrary function ϕ on Sn+ , which is integrable on F ∩Sn+ . Here, (y0 , y) =  −1 2 1 + |x| (1, x) ∈ Sn+ , and the function f (x) = y0k+1 ϕ (y0 , y) is integrable on F ∩ E n . It follows that the Funk transform of even functions is equivalent to the Euclidean k-dimensional transform. 

3.3

Funk Transform on the Orthogonal Group

Let SO (3) be the group of orthogonal transformations of an oriented Euclidean space E 3 preserving the orientation. Let S2 be the unit sphere E 3 . Any element g ∈ SO (3) is a rotation about an axis in E 3 by an angle ϕ. The Riemannian metric dµ on the group is defined by the condition dist (g, h) = min (ϕ, 2π − ϕ) , where ϕ is the rotation angle of the element g−1 h (or hg−1 ). For arbitrary elements s,r ∈ S2 , we denote C (s,r) = {g ∈ SO (3) : gs = r}. For any g ∈ C (s,r) , this set contains elements ag and gb, where a and b are rotations around r and s, respectively. The set C (s,r) is isometric to a circle, which is a geodesic curve in the group. Conversely, any closed geodesic in SO (3) coincides with C (s,r) = C (−s, −r) for some s,r. For a function f on the group, the integral Z Mf (s,r) + f (g) dµ C(s,r)

is the geodesic transform defined on the product S2 ×S2 . It is an even function with respect to the involution (s, r) 7→ (−s, −r) . This is an analog of the classical Funk transform for functions g defined on a 2-sphere S2 ; see §5.2. The field H of quaternions is a real 4-dimensional Euclidean space with the P 2 scalar product hq, ri = Re (q ∗ r) , |q| = i qi2 . We write a quaternion q in the form q0 + q; the real number q0 and the vector q ∈ R3 are called the real and the imaginary part of q, respectively. Any unit nonreal quaternion q = q0 + q defines the rotation about the axis q/ |q| by the angle ϕ = 2 arccos q0 . Let S be the sphere of all unit quaternions. For any q ∈ S, transformation G (q) : s 7→ qsq ∗ , q ∗ + q0 − q

(3.5)

Factorization Method

49

preserves the norm and orientation, hence, G (q) ∈ SO (3) . It can be written in the form:   2 G (q) s = q02 − |q| s + 2q0 q × s + 2 hq,si q, s ∈ S2 . (3.6) The map G : S → SO (3) given by (3.5) has quadratic components and defines a surjective group homomorphism. It is two-fold, since the quaternions q and −q generate the same orthogonal transform. Let H0 be the vector space over R of all imaginary quaternions. The manifold S+ = {q ∈ S, q0 > 0} is the hemisphere with the boundary S ∩ H0 . Consider the gnomonic projection γ : S+ → 1 + H0 , q 7→

q q =1+ . q0 q0

These maps are shown in the commutative diagram S G. ↓ ∼ SO (3) S/Z2 = G↑ S+

∼ =

RP3 , ↑ π → 1 + H0

(3.7)

where Z2 = {1, −1} is a subgroup of S. This diagram shows that the manifold of SO (3) is isomorphic to the projective 3-space S/Z2 . An arbitrary circle C (s,r) can be parameterized by g = gr R (ϕ) gs−1 , where   cos ϕ − sin ϕ 0 R (ϕ) =  sin ϕ cos ϕ 0  , 0 ≤ ϕ < 2π, 0 0 1 and gs , gr are some rotation, such that gr N = r, gs N = s, where N = (0, 0, 1) . The line element dϕ on C (s,r) coincides with the metric element dg. Proposition 3.5 For any big circle CS in S, the set G (CS ) is a closed geodesic in SO (3) and vice versa. Map G is conformal with the conformity coefficient 2 with respect to the Euclidean metric on the sphere. Proof. Write the circle C in a parametric form q = q (ϕ) = r cos ϕ + s sin ϕ for some orthogonal vectors r, s ∈ S ∩ H0 . The triple (r,s, t) , t = r × s, is an orthogonal frame in H. By (3.6), the matrix of the rotation G (q) in this frame is   cos 2ϕ − sin 2ϕ 0 G (q (ϕ)) =  sin 2ϕ cos 2ϕ 0  . (3.8) 0 0 1 We see that the image C = G (CS ) is a circle in SO (3), and G (q (ϕ + π)) = G (q (ϕ)) , hence, CS covers twice C. Any big circle in S is a geodesic, and

50

Reconstruction from integral data

the image C = G (CS ) is again a geodesic. Therefore, G is conformal and the conformal coefficient equals 2, according to (3.8). Now let C be a circle in SO (3). It is a geodesic, hence, the set CS = G−1 (C) is also a geodesic, since G is conformal. This implies that CS is a big circle.  Proposition 3.6 For arbitrary s, r ∈ S2 , the circles C (s,r) and C (−s,r) are contained in orthogonal 2-planes L+ , L− in H. Proof. In the case r = s, the set C (s,s) consists of rotations about the axis through s. Circle G−1 (C (s,s)) is the set of unit quaternions q0 ± s, hence, L+ is the plane of the quaternions whose imaginary part is proportional to s. The circle C (−s,s) consists of rotations about an axis r orthogonal to s, hence, G−1 (C (−s,s)) is the set of pure imaginary quaternions orthogonal to s. The planes L+ and L− are obviously orthogonal, which proves the statement for the special case. For an arbitrary r ∈ S 2 and a rotation g, such that gs = r, we have C (±s,r) = gC (±s,s), hence, G−1 (C (±s,r)) = qG−1 (C (±s,s)) for a quaternion q, such that G (q) = g. The operator of multiplication by q is a rotation in H. Therefore, the statement holds for arbitrary r. 

3.4

Reconstruction from Non-Redundant Data

A function defined on SO (3) can be reconstructed from some 3-parameter data of circle integrals. In Chapter 2, a similar problem was considered for functions on R3 and its line integrals. Theorem 3.7 Let Γ be a closed noncontractible curve in SO (3) . Any continuous function f defined on the group can be effectively reconstructed from knowledge of integrals Mf (s, − gs) for s ∈ S2 and g ∈ Γ.. Proof. Fix g ∈ Γ and consider the function fg (a) = f (ga) . We have Mfg (s, −s) = Mf (s, − gs), and the right-hand side is known for any vector s ∈ S2 by the assumption. Therefore, we know all the integrals Z Z 1 Mfg (s, −s) = fg dµ = G∗ (fg ) ds, (3.9) 2 G−1 (C(s,−s)) C(s,−s) where ds is the Euclidean metric in S. For any s ∈ S2 , the circle G−1 (C (s, −s)) is contained in 2-sphere S ∩ H0 , and any big circle in this sphere has such form. Function G∗ (fg ) defined on the sphere S ∩ H0 is even and its big circle integral is in the right-hand side of (3.9). This function can be reconstructed by means of Funk’s formula (5.6). Therefore, fg is known on S + G (S ∩ H0 ) and f is determined on gS. We check that spheres gS cover the group SO (3) as g runs over Γ. Consider the map γΓ : Γ × S → SO (3) , (f , s) 7→ f −1 s.

Factorization Method

51

This is a continuous proper map, and the degree deg γΓ mod 2 is well defined, since Γ and S are connected manifolds. By (3.7), manifold SO (3) is isomorphic to the real projective space, hence its fundamental group is Z2 . This group is generated by an arbitrary projective line, that is, by an arbitrary circle C. Therefore, Γ is homotopically equivalent to C, which implies degγΓ = degγC . We show below that deg γC = 1, hence, deg γΓ = 1 for any Γ. This implies surjectivity of γΓ .  Lemma 3.8 We have deg γC = 1 for any circle C. Proof of Lemma. The set C of all rotations about the point N ∈ S2 is a circle in the group. For an arbitrary g ∈ SO (3) , we consider an arc joining N and g (N ) and take the middle point m of this arc. Let h ∈ S be the rotation by the angle π about m. We have hN = gN , hence, f = hg−1 is a rotation about N, which implies f ∈ C. We have f −1 h = gh−1 h = g, which means that γC (f , h) = g. The rotation h is unique if g (N ) = 6 −N, whereas equation g (N ) = −N implies that g ∈ S. In other words, each point g ∈ SO (3) \S is covered by γC only once. It follows that deg γC = 1.  Corollary 3.9 Let Γ be a closed noncontractible curve in SO (3). Any continuous function f on SO (3) can be effectively reconstructed from knowledge of integrals Mf (s, gs) for s ∈ S2 , g ∈ Γ. Proof. For a function ϕ defined on SO (3) , the pullback ϕ∗ (q) + ϕ (G (q)) is an even function on the sphere S, and the pullback C + G−1 (C) of a circle C is a big circle in the sphere. We have Z Z 1 ϕ∗ ds. ϕdµ = 2 C C By Proposition 3.6, for arbitrary s, r ∈ S, the 2-plane containing C (s, r) is orthogonal to the plane containing the circle C (s, −r). Function ψ (q) = −2 ϕ∗ (q/ |q|) |q| is equal to the extension of ϕ∗ to H ∼ = R4 as a homogeneous function of degree −2. By [108] Theorem 5.10, Z Z F4 (ψ) ds = ϕ∗ ds, C(s,−gs)

C(s,gs)

where F4 means the Fourier transform acting in R4 . This yields Z Z F4 (ψ) dµ = ϕdµ, C(s,−gs)

C(s,gs)

since ψ and F4 (ψ) are even function. By the assumption, we know the righthand side, and by Theorem 3.7, the function F4 (ψ) can be explicitly reconstructed. Function ψ is recovered by means of the inverse Fourier transform. 

52

3.5

Reconstruction from integral data

Range Conditions

Range conditions for the circle transform on SO (3) are known in two different forms. Theorem 3.10 A function M defined on S2 × S2 is equal to the circle transform Mf of a function f on SO (3) if and only if it fulfils ∆s M (s,r) = ∆r M (s,r) ,

(3.10)

where ∆ means the Laplace–Beltrami operator on S2 . This equation is apparently an analog of John’s equations §2.2. A proof based on harmonic series of f and M f can be found in [10]. We give a more geometric proof of the part “only if.” It is based on an analog of Asgeirsson’s theorem. For any point s ∈ S2 and a positive ρ, we denote by O (s,ρ), the circle in the unit sphere with the center s and radius ρ. We say that a function M defined on S2 × S2 has Asgeirsson property if Z Z M (s,q) dq = M (p,r) dp (3.11) q∈O(r,ε)

p∈O(s,ε)

for arbitrary points s,r ∈ S2 and any angle ε. We prove Theorem 3.11 The Asgeirsson property is equivalent to (3.10) for any 2smooth function M. Proof. Let t ∈ S2 and θ, ϕ be the latitude and the longitude coordinates on the sphere S2 with the pole t. The Laplace–Beltrami operator ∆=

∂ 1 ∂2 1 ∂ sin θ + sin θ ∂θ ∂θ sin2 θ ∂ϕ2

is self-adjoint with respect to the area form Ω = sin θdθdϕ on S2 . Function −1 Ep (θ, ϕ) = (2π) log tan θ/2 + const is the fundamental solution of this operator, that is, ∆Ep = δp . For an arbitrary 2-smooth function h on S2 , the Green formula for the disc with the center t of radius ε reads: Z 2π Z tan θ/2 ∆h (θ, ϕ) Ω. (3.12) h (ε, ϕ) dϕ = 2πh (t) − log tan ε/2 0 θ≤ε We apply (3.12) to a 2-smooth function M (s, q) of variable q, and to M (p, r) of variables p. We get two integrals over in the left-hand sides Z Z M (s,q) dϕ (q) , M (p,r) dϕ (p) . (3.13) O(r,ε)

O(s,ε)

Factorization Method

53

If M fulfils (3.10), then the right-hand side s of (3.12) coincide, hence, the integrals (3.13) are equal which proves (3.11). Conversely, suppose that (3.11) is fulfilled for a function M. Applying twice (3.12) we obtain Z Z tan θ/2 tan θ/2 log ∆r M (s, r) Ω = ∆s M (s, r) Ω (3.14) log tan ε/2 tan ε/2 θ≤ε θ≤ε for any s, r, and ε > 0. We have Z
1 tan θ/2 sin θdθdϕ = − πε2 + O ε3 log tan ε/2 2 θ≤ε as ε → 0. Since ∆s M and ∆r M are continuous functions, Equation (3.14) reads

π π − ε2 ∆s M (s, r) + o ε2 = − ε2 ∆r M (s, r) + o ε2 , 2 2 which implies (3.10).  Finally, we state the following: Theorem 3.12 For any bounded function f on SO (3), its circle transform M = Mf has Asgeirsson property and fulfills (3.10).  Proof. Let Rs = Rs,ω , ω ∈ S1 be the subgroup of rotations about a point s ∈ S2 . Fix an element g of the group, such that q0 = gs ∈ O (r, ε) . Then the set gRs is equal to C (s, q0 ) and Rr gRs = ∪ {C (s, q) , q ∈ O (r, ε)} , since the action of g is an isometry. On the other hand, the point p0 = g−1 r belongs to O (s, ε) , since g is an isometry and Rr g = C (p, r) . Therefore, Rr gRs = ∪ {C (p, r) , p ∈ O (s, ε)} and, consequently, ∪ {C (s, q) , q ∈ O (r, ε)} = ∪ {C (p, r) , p ∈ O (s, ε)} . Integrating f over this set, we obtain Z Z Mf (s, q) dϕ (q) = q∈O(r,ε)

Mf (t, r) dϕ (p) .

p∈O(s,ε)

This means that Mf fulfills (3.11), and (3.10) follows by Theorem 3.11. 

54

3.6

Reconstruction from integral data

Bibliographic Notes

§§3.1–2 Missing proofs can be found in [104]. Denisjuk [25] has applied the method of factorable maps for reconstructions on spaces of constant curvature. Formula (3.4) was used in [8]. §§3.3–4 Circular integrals of a function on the group SO (3) is a mathematical model for the quantitative texture analysis of polycrystalline materials by means of x-ray or neutron diffraction data [10]. The integrand is called in this context “orientation distribution function (ODF),” and the sum of integrals over a union of two orthogonal circles—“pole density function (PDF).” The latter can be obtained from x-ray diffraction experiments. The problem of texture analysis is to extract information on ODF from knowledge of PDF. Several methods of reconstruction of ODF have been known since the 1960s: a Funk-type inversion formula [81], expansion in spherical harmonics and backprojection inversion [10], and inversion by means of a singular integral operator [60]. These methods are applied to the complete 4D data, which are not technically attainable and redundant. See [106] for nonredundant reconstructions. §3.5 The consistency condition (3.10) has been known for thirty years; see [10]. Asgeirsson’s theorem [6] relates to solutions of a second order differential equation. According to Helgason’s result [58] in Chapter 6, Equation (3.10) implies the Asgeirsson property for any symmetric space X of rank one.

Chapter 4 General Method of Reconstruction

4.1

Geometric Integral Transforms

Let X, Σ be manifolds of dimension n > 1, Z be a smooth hypersurface in X × Σ, such that the natural projections pX : Z → X, and pΣ : Z → Σ have rank n. It follows that Z (σ) + p−1 Σ (σ) is a smooth hypersurface in X for any σ ∈ Σ and Z (x) = p−1 X (x) is a smooth hypersurface in Σ. A real smooth function Φ defined on a neighborhood of Z , such that Z = Φ−1 (0) and dΦ 6= 0 on Z will be called generating function. These conditions imply that dx Φ|Z = − dσ Φ|Z 6= 0. Suppose that (i) map DX : Z × R+ → T0∗ (X) + {(x, ξ) ∈ T ∗ (X) , ξ 6= 0} is a diffeomorphism, where DX (x, σ, t) = (x, tdx Φ (x, σ)) . This condition implies that pX is a proper map and for any x ∈ X, Z (x) is diffeomorphic to a (n − 1)-sphere that meets each ray {ξ = tξ0 , t > 0} in one point. Proposition 4.1 The map DX is a local diffeomorphism if and only if detJx,σ (Φ) 6= 0 on Z, where   ∂2Φ ∂2Φ ∂Φ ... ∂x1 ∂σ1 ∂x1 ∂σn 1   ∂x ... ...  ...  2 2   ∂Φ ∂ Φ ∂ Φ Jx,σ (Φ) =  ∂x ... ∂xn ∂σn  , ∂xn ∂σ1 n     ∂Φ ∂Φ 0 ... ∂σ1 ∂σn and x1 , ..., xn ; σ1 , ..., σn are local coordinates on X and Σ, respectively. Conversely, if DX is a local diffeomorphism, then det Jx,σ (Φ) = 6 0 for arbitrary coordinate systems. Proof. Consider the map ∆ : X × Σ × R+ → T0∗ (X) × R, where ∆ (x, σ, t) = (x, tdx Φ (x, σ) , Φ (x, σ)) = (DX (x, σ, t) , Φ (x, σ)) . We have det ∂∆/∂ (x, σ, t) = tn det Jx,σ (Φ) by a direct calculation. On the 55

56

Reconstruction from integral data

other hand, we have ∆ (x, σ, t) = (DX (x, σ, t) , 0) for (x, σ) ∈ Z, and the condition det J (Φ) (x0 , σ0 , 1) 6= 0 implies dx Φ (x0 , σ0 ) = 6 0. Take a coordinate system, say (x1 , x0 ; σ) , x0 = (x2 , ..., xn ) on a neighborhood U of (x0 , σ0 ), such that ∂Φ/∂x1 6= 0. There exists a smooth function ξ (x0 ; σ), such that x1 = ξ (x0 ; σ) in Z ∩ U , and we have ∂Φ ∂DX ∂∆ (x, σ, t) = det . det ∂ (x, σ, t) x1 =ξ ∂x1 ∂ (x0 , σ, t) This implies that det ∂DX /∂ (x0 , σ, t) 6= 0, that is, DX is a local diffeomorphism. These arguments are invertible.  Remark Condition det Jx,σ (Φ) 6= 0 can be written in the form J (Φ) 6= 0 ∧n that does not depend on coordinate systems, where J (Φ) = dx Φ∧(dx dσ Φ) ∧ dσ Φ. Let dX be an odd volume form on X. For an arbitrary function f ∈ C00 (X), the integral Z Z 1 MΦ f (σ) = δ (Φ (x, σ)) f (x) dX + lim f dX ε→0 2ε x:|Φ(x,σ)|≤ε X is defined for almost any σ ∈ Σ. We call MΦ f geometric integral transform of f. The function MΦ f is supported by the set L = pΣ p−1 X (suppf ), which is also compact, since pX is a proper map. If dX is equal to the volume form dg X of a Riemannian metric g, then Z f (x) dg S MΦ f (σ) = , |d x Φ (x, σ)|g Z(σ) where dg S is the volume form on Z (σ) . If X is oriented by an even volume form dX, then for any σ ∈ Σ, the quotient dX/dx Φ is an even volume form that defines an orientation in Z (σ), and we have Z f (x) dX MΦ f (σ) = . Z(σ) dx Φ (x, σ) The backprojection operator M∗Φ g (x)

1 = lim ε→0 2ε

Z

Z gdΣ =

σ:|Φ(x,σ)|≤ε

g Z(x)

dΣ dσ Φ

is defined for any x ∈ X and any continuous function g on Σ, since Z (x) is compact and is oriented by the quotient form dΣ/dσ Φ. Note that for a function g with compact support, the backprojection may not have compact support if pΣ is not proper.

General Method of Reconstruction

4.2

57

Reconstruction

Definition. Let Φ be a generated function defined on X × Σ. Points x, y ∈ X are called conjugate for Φ, if x = 6 y, Φ (x, σ) = Φ (y, σ) and dσ Φ (x, σ) ∧ dσ Φ (y, σ) = 0 for some σ ∈ Σ. Condition (ii): There are no conjugate points.

Conjugate points of big circles in a sphere. Let dΣ be a volume form on Σ. If Φ is n + 1-smooth with respect to σ, then by Proposition 7.17, the singular integral Z 1 dΣ Qn (x, y) = n Z(y) (Φ (x, σ) − i0) dσ Φ (y, σ) is well defined for any x, y ∈ X, y = 6 x and is a continuous function. We have Z 1 dΣ Re Qn (x, y) = , n d Φ (x, σ) σΦ Z(y) Z π dΣ Im Qn (x, y) = δ (n−1) (Φ (x, σ)) . (n − 1)! Z(y) dσ Φ Theorem 4.2 Suppose that dX is the Riemannian volume form for a Riemannian metric g on X, and Φ is an n + 1-smooth generating function Φ satisfying conditions (i),(ii), and (iii): Re in Qn (x, y) = 0 for any x, y ∈ X such that x = 6 y.

(4.1)

58

Reconstruction from integral data

Then for any odd n, an arbitrary function f ∈ C0n−1 (X) can be reconstructed from integral data of MΦ f by Z 1 f (x) = n−1 δ (n−1) (Φ (x, σ)) MΦ f (σ) dΣ, (4.2) 2j Dn (x) Σ and for any even n and any f ∈ C0n (X) , by Z MΦ f (σ) dΣ (n − 1)! , f (x) = n j Dn (x) Σ Φ (x, σ)n where Dn (x) =

1

Z

|Sn−1 |

Z(x)

−n

|dx Φ (x, σ)|g

dΣ . dσ Φ

(4.3)

(4.4)

The integrals (4.2) and (4.3) converge to f uniformly on any compact set K ⊂ X. For generating functions satisfying (i) and (ii), condition (iii) is necessary for (4.2–4.3) to hold. Proof. Let τ1 , ..., τn be some smooth tangent fields on X whose linear envelope is equal Tx (X) at any point x ∈ X. Introduce the norm kf k = max sup τ i1 ...τnin f (x) K,m

i1 +...+in ≤m x∈K

1

for m-smooth functions on X. By (i), there exists a smooth tangent field t and a smooth function t0 on X such t (Φ) + t0 Φ = 1. Equations (4.2–4.3) can be written in the form Z 1 f (x) = n−1 δ0 (Φ (x, ·)) gn−1 , (4.5) 2j Dn (x) Σ Z (n − 1)! f (x) = − n (− log |Φ (x, ·)| t (gn−1 ) + t0 gn−1 ) , (4.6) j Dn (x) Σ for odd and even n, respectively. Here, gn−1 = Tn−1 (MΦ (f ) dΣ) and Tn−1 = (t + t0 ) (t + 2t0 ) ... (t + (n − 1) t0 ) . Lemma 4.3 The geometric integral transform MΦ defines an operator C0k (X) → C k (Σ) that is bounded in the sense that for arbitrary compact K in X and compact Λ in Σ there exists a constant CK,Λ , such that kMΦ f kΛ,k ≤ CK,Λ kf kK,k

(4.7)

for functions f with support in K. Proof of Lemma. We abbreviate notation MΦ to M. If the function f is continuous and has compact support, then Mf is also continuous. By Proposition 7.10,   dF dX ∇σ Mf = −M ∇σ Φ , F =f (4.8) dΦ dΦ

General Method of Reconstruction

59

and

dF df ∧ dX/dΦ + f d (dX/dΦ) = . dΦ dΦ Calculate the right-hand by means of Proposition A.8: df ∧ dX/dΦ f1 dX dF dX = , = (f1 + χf ) , dΦ dΦ dΦ dΦ where f1 =

h∇f, ∇x Φig 2 |∇x Φ|g

, χ=

d (dX/dΦ) . dX

By (4.8), ∇σ Mf = −M ((f1 + χf ) ∇σ Φ) . It follows that Mf is 1-smooth if so is f , and so on. If f is k-smooth, then Mf is also k-smooth, and (4.7) holds for any compact set Λ.  Corollary 4.4 The right-hand side of (4.5) and (4.6) defines a bounded operator C0n−1 (X) → C 0 (X), respectively, C0n (X) → C 0 (X) . Proof. The integral of (4.5) is taken over the compact manifold Z (x) of the continuous density gn−1 that is continuous by Lemma 4.3, since f ∈ C0n−1 (X) . The integrand in (4.6) is a locally integrable density by loc. cit., since f ∈ C0n (X). The support of the density is contained in the compact set pΣ p−1 X (suppf ), which implies the convergence of (4.6).  For even n and any function f ∈ C0n (X) , we have for even n by Fubini’s Theorem Z Mf (σ) dΣ Re In f (x) = n Σ Φ (x, σ) Z Z dΣ f (y) dY = n Σ Φ (x, σ) Z(σ) dy Φ (y, σ) Z Z 1 dΣ f (y) dY = n X Z(y) Φ (x, σ) dσ Φ (y, σ) Z = Re Qn (x, y) f (y) dY, (4.9) X

since Z = ∪ {Z (σ) , σ ∈ Σ} , where (Y, dY ) is a copy of (X, dX) and dY dΣ ∧ dΣ = ∧ dY dy Φ (y, σ) dσ Φ (y, σ) in Z. If n is odd, we have by (A.4) Z π Im In f (x) = δ (n−1) (Φ (x, σ)) Mf (σ) dY (n − 1)! Z(x) Z =− Im Qn (x, y) f (y) dY, Y

(4.10)

60

Reconstruction from integral data

that is, Im Qn is the kernel of −In . In both cases, the kernel vanishes off the diagonal by (iii). By Corollary 4.4, the diagonal term has the form ± Re in Qn (x, y) = A (x, D) δx (y) ,

(4.11)

where δx means the delta-function and A : C n (X) → C 0 (X) is a differential operator. To complete the proof, we show that A is an operator of zero degree, that is, A = A (x) is a function.

4.2.1

Continuation of the Proof

Fix a point x ∈ X and choose a coordinate system  ξ = ξ (y) = (ξ1 , ..., ξn ) on 2 a neighborhood U of x, such that gij = δij + O |ξ| and ξ (x) = 0. Take a smooth function e on R with support in [−1, 1], such that e (0) = 1, and set  2 eε (x, y) = e |ξ| /ε2 for x, y ∈ X and ε > 0. Set Z Eε (x) =

eε (x, y) Qn (x, y) dY, Y

where (Y, dY ) is a copy of (X, dX) . Consider the map ω : Z ∩ U → U × Sn−1 , (y, σ) 7→ (y, ω (y, σ)) , where Φ (y, σ) = 0, ω (y, σ) +

∇ξ Φ (y, σ) . |∇ξ Φ (y, σ)|

(4.12)

By (i), this map is a diffeomorphism if U is sufficiently small. The inverse map σ = σ (y, ω) is n-smooth, since so is ω (y, σ) . Set ϕ (x, y; ω) + Φ (x, σ (y, ω)) . We have dΣ/dσ Φ (y, σ) = λ (y, ω) Ω (ω) for a smooth function λ on U × Sn−1 . Therefore, Z Z eε (x, y) dΣ Eε (x) = dY n d Φ (Φ (x, σ) − i0) σ (y, σ) Y Z(y) Z Z eε (x, y) λ (y, ω) = Ω n dY n−1 Y (ϕ (x, y; ω) − i0) ZS = Eε (x, ω) Ω,

(4.13)

Sn−1

where

Z Eε (x, ω) = Y

eε (x, y) λ (y, ω) dY n . (ϕ (x, y; ω) − i0)

Lemma 4.5 For even n, we have Re Eε (x, ω) →

jn λ (x, ω) , (n − 1)! |Sn−1 | |dx Φ (x, σ)|n

(4.14)

General Method of Reconstruction

61

and for odd n, Im Eε (x, ω) →

2πjn−1 λ (x, ω) , |Sn−1 | |dx Φ (x, σ)|n

(4.15)

as ε → 0. Proof of Lemma 4.5. We have ϕ (x, x; ω) = 0 and ∇ξ Φ (y, σ) +

∂Φ (y, σ) ∇ξ σ = 0, σ = σ (y, ω) , ∂σ

since Φ (y, σ (y, ω)) = 0. It follows that ∇ξ ϕ (x, y; ω) =

∂Φ (x, σ) ∇ξ σ ∂σ

= −∇ξ Φ (y, σ) +

∂ (Φ (x, σ) − Φ (y, σ)) ∇ξ σ, ∂σ

which implies ∇ξ ϕ (x, y; ω) = − |∇ξ Φ (y, σ)| ω (y, σ) + ψ (x, y, σ) ,

(4.16)

where ω is as in (4.12) and ψ = ∇σ ϕ∇ξ σ. Derivatives of ψ up to order n are equal to O (|ξ|) for small |ξ| . Let T be a smooth tangent field T on X × U that fulfils T (ϕ (x, y; ω)) = 1 and has the form T = − h|∇ξ Φ (y, σ)| ω + τ, ∇ξ i , where the vector τ and its derivatives up to order n are equal to O (|ξ|) . We have
T ϕk = kϕk−1 , k ∈ Z, T (log |ϕ|) = ϕ−1 . Integrating by parts, n gives Z ∗ 1 T (eε (x, y, σ) λ (y, σ)) dY (4.17) Eε (x, ω) = n−1 (n − 1) (ϕ (x, y; ω) − i0) Z −1 = ... = log (ϕ (x, y; ω) − i0) eε,n (x, y, σ) dY, (n − 1)! where T ∗ = −T + div T is adjoint to T and  n λ ω ∗ n , ∇ξ eε,n = (T ) (eε λ) = n eε + .... n |∇ξ Φ| ε |∇ξ Φ|
n−1 The omitted terms are O ε1−n and αn = (−1) / (n − 1)!. Note that  n  n   ω ω 2 , ∇ eε (|ξ| /ε) = ε−n , ∇η eε |η| + ..., |∇ξ Φ| |∇ξ Φ| where η = ε−1 ξ.

62

Reconstruction from integral data

Even n We have by (4.16)   2 ϕ (x, y; ω) = − h∇ξ Φ (x, σ) , ξi + O |ξ| , since ϕ (x, x, ω) = 0. This yields Z Z log |ϕ| eε,n (ξ) dY = log |hω, ηi| eε,n dY Z + (log ε |∇ξ Φ (x, σ)|) eε,n dY + O (ε) ,

(4.18)

where η = ε−1 ξ. The second term vanishes, since eε,n−1 has compact support and integrating by parts yields Z Z Z eε,n dY = T ∗ (eε,n−1 ) dY = T (1) eε,n−1 dY = 0. We substitute dY = (1 + O (|ξ|)) dξ in (4.18) and have Z Z log |ϕ| eε,n dY = log |ϕ| eε,n dξ + O (ε) n 
ω , ∇η λ (x, ω) e ρ2 dη + O (ε log ε) . |∇ξ Φ| The first term in the right-hand side equals Z
λ (x, ω) n log |hω, ηi| hω, ∇η i e ρ2 dη. n |∇ξ Φ (x, σ)| |η|≤1  Z  = log |hω, ηi|

By (4.17), it coincides with the limit of − (n − 1)! Re Eε (x, ω) . Substitute η = ρθ, θ ∈ Sn−1 , ρ > 0 and integrate by parts n − 1 times with respect to ρ : Z
n log |hω, ηi| hω, ∇η i e ρ2 dη
Z Z 1 ρn−2 ∂e ρ2 = − (n − 2)! Ω dρ n−2 ∂ρ Sn−1 0 hω, ηi
Z Z 1 ∂e ρ2 Ω (θ) = − (n − 2)! dρ. n−2 ∂ρ Sn−1 hω, θi 0 The first integral in the right-hand side is equal −in π Sn−2 by Example 7.18. The second one equals −1. It follows that Z


n log |hω, ηi| hω, ∇η i e ρ2 dη = in (n − 2)!π Sn−2 =

jn |Sn−1 |

,

which implies lim Re Eε (x, ω) =

ε→0

jn λ (x, ω) . (n − 1)! |Sn−1 | |∇ξ Φ (x, σ (x, ω))|n

Now (4.14) follows, since |∇ξ Φ (x, σ)| = |dΦ (x, σ)|g by the choice of the coordinates ξ.

General Method of Reconstruction

4.2.2

63

Odd n

By (4.17), Z Im Eε (x, ω) = =

1 n (n − 1)! |∇ξ Φ|

θ (−ϕ (x, y; ω)) eε,n (x, y; σ) dξ Z


n hω, ∇η i e ρ2 dη + O (ε) ,

hω,ηi≤0

where θ (t) = 1 for t > 0 and θ (t) = 0 for t < 0. Integrating by parts backward gives the equation Z Z n−2 1 m−1 ∂ m e n m−1 s hω, ∇η i edη = −2 (n − 2)!! S ds, (4.19) ∂sm 0 hω,ηi≤0 where m = (n − 1) /2. Integrating by parts m − 1 times in the right-hand side, m we get the quantity (−1) (m − 1)!. Therefore, (4.19) equals 2πjn−1 m (−1) 2m−1 (m − 1)! (n − 2)!!π Sn−2 = n−1 . |S | Finally, lim Im Eε (x, ω) =

ε→0

2πjn−1 λ (x, ω) , |Sn−1 | |dx Φ (x, σ (x, ω))|ng

which implies (4.15) and completes the proof of Lemma 4.5.  Let p = p (ξ) be an arbitrary polynomial that vanishes for ξ = 0. The proof of Lemma 4.5 can be repeated with the function eε replaced by peε and the factor µ = pλ instead of λ. The limits (4.14) and (4.15) are equal to zero. By (4.13), this implies that A (x, D) δx p = 0 where A is as in (4.11). It follows that A is an operator of order zero, that is, a function. This function is continuous, since of Lemma 4.4.

4.2.3

Calculation the Coefficient in (4.11)

By (4.13), integrating (4.14), we get jn (n − 1)! |Sn−1 |

Z

jn = (n − 1)! |Sn−1 |

Z

Re Eε (x) →

Sn−1

Z(x)

λ (x, ω) nΩ |dx Φ (x, σ)|g 1 dΣ n |dx Φ (x, σ)|g dσ Φ (x, σ)

n

=

j Dn (x) , (n − 1)!

as ε → 0, hence, A (x) = lim Re Eε (x) =

jn Dn (x) (n − 1)!

64

Reconstruction from integral data

for even n. This and (4.9) imply (4.3). Integrating (4.15), we prove that for odd n, 2πjn−1 Im Eε (x) → Dn (x) (n − 1)! and A (x) = lim Im Eε (x) =

2πjn−1 Dn (x) . (n − 1)!

This and (4.10) yield (4.2). This completes the proof of Theorem 4.2. 

4.3

Integral Transforms with Weights

Constructions of §§4.1, 4.2 can be generalized by inserting arbitrary weights in the geometric integral transform. Let Φ be a generating function on X ×Σ, dX be an odd volume form on X, and w = w (x, σ) ∈ C 0 (X × Σ) be a weight function. For an arbitrary bounded function f on X with compact support, the integral Z MΦ,w f (σ) + w (x, σ) δ (Φ (x, σ)) f (x) dX converges for almost all σ ∈ Σ. We call MΦ,w weighted geometric transform. For a weighted function u ∈ C 0 (X × Σ) , we define the weighted filtered backprojection M∗Φ,u . If u ∈ C n−1 (X × Σ) , the operator Z π NΦ,u g (x) + u (x, σ) δ (n−1) (Φ (x, σ)) g (σ) dΣ (4.20) (n − 1)! Σ is defined for odd n and g ∈ C0n−1 (Σ) . The integral Z u (x, σ) g (σ) dΣ NΦ,u g (x) + n Φ (x, σ) Σ

(4.21)

is well defined for even n and g ∈ C0n (Σ) . Theorem 4.2 is generalized for the weighted geometric operators: Theorem 4.6 For arbitrary generating function Φ ∈ C n (X × Σ) satisfying (i), (ii), and smooth weight functions w, u ∈ C n (X × Σ) fulfilling condition (iii’): Re in Qw,u (x, y) = 0 where Z u (x, σ) w (y, σ) dΣ Qw,u (x, y) = , n Z(y) (Φ (x, σ) − i0) dσ Φ (y, σ) any function f ∈ C0n−1 (X) or f ∈ C0n (X) can be reconstructed from its weight transform by NΦ,u MΦ,w (f ) = Dw,u f,

General Method of Reconstruction where Dw,u (x) = n−1

and cn = 2j

1 cn |Sn−1 |

Z σ∈Z(x)

65

u (x, σ) w (x, σ) Ω 6= 0 n |∇x Φ (x, σ)|

for odd n and cn = jn for even n, respectively.

For a proof, we repeat arguments of Theorem 4.2 with obvious modifications.

4.4

Resolved Generating Functions

We call a generating function Φ defined on X × Σ resolved, if Σ = R × Sn−1 and Φ (x; p, ω) = q (x, ω) − p, p ∈ R, ω ∈ Sn−1 for a 2-smooth function q on X × Sn−1 . The product dΣ = dp ∧ Ω will be taken as the basic volume form on Σ. Note that integral transform MΦ f of a function f ∈ C00 (X) can be defined as a derivative of sublevel integral Z SΦ f (p, ω) = f dX, q(x,ω)≤p

since ∂SΦ (f ) /∂p = MΦ (f ) . For a resolved functions Φ, Theorem 4.2 reads in a more simple way. We n−1 ∧n−1 have J (Φ) = (−1) dx q ∧ (dx dω q) ∧ dp, and the key integral looks as Z Ω 6 y. Qn (x, y) = − n, x = q(y,ω)=p (q (x, ω) − q (y, ω) − i0) Theorem 4.7 Let Φ = q − p, q ∈ C n (X × Sn ) be a resolved generating function satisfying (i),(ii), and Re in Qn (x, y) = 0 for x, y ∈ X, x = 6 y. If n is even, an arbitrary f ∈ C0n (X) is reconstructed from MΦ f by Z Z ∞ (n − 1)! MΦ f (p, ω) dp Ω f (x) = n j Dn (x) Sn−1 −∞ (q (x, ω) − p)n Z Z ∞  n−1 1 ∂ Ω = n MΦ f (p, ω) . j Dn (x) Sn−1 −∞ ∂p q (x, ω) − p For odd n, we have 1 f (x) = n−1 2j Dn (x)

Z Sn−1



∂ ∂p

n−1

MΦ f (p, ω)

Ω, p=q(x,ω)

66

Reconstruction from integral data

where Dn (x) =

Z

1 |Sn−1 |

Z(x)

Ω n, |dx q (x, ω)|g

and Z (x) is oriented by the form Ω. The integrals uniformly converge on any compact set in X. For resolvent generating functions, there is a simple consistency condition: Proposition 4.8 Suppose that q (x, ω) is a trigonometric polynomial of ω of order m for any x ∈ X. Then for an arbitrary f ∈ C00 (X) and integer k ≥ 0, the function Z pk MΦ f (p, ω) dp

µk (ω) = R

is a trigonometric polynomial of order ≤ km. Proof. By Fubini’s, Z µk (ω) = R

pk

Z f p=q(x,ω)

dX dp = dq

Z

q k f dX,

X

where q k is a trigonometric polynomial of order km. 

4.5

Analysis of Convergence

We state here a more detailed version of Theorems 4.2, 4.6, and 4.7 in terms of Sobolev spaces of functions defined on manifolds [136]. For a manifold X with a volume form dX and a compact set K ⊂ X, the Sobolev space of order α of function supported in K is denoted Hα (K) ; the norm in this space will be denoted k·kα . In particular, for α = 0, we have H0 (K) = L2 (K) and Z kf k0 =

2

|f | dX

1/2 .

K

The union of spaces L2 (K) for all compact sets K ⊂ X is denoted by L2 (X)comp . A sequence of functions ϕi converges to ϕ in L2 (X)comp if ∪suppϕi is contained in a compact set K ⊂ X and kϕ − ϕi k0 → 0 as i → ∞. Notation L2 (X)loc is used for the space of functions f on X such that product φf has a finite L2 (X)-norm for any bounded function φ on X with compact support.R A sequence of functions fi ∈ L2 (X)loc converges to f if kϕ (f − fi )k0 → 0 for any such function φ. An operator A : L2 (X)comp → L2 (Y )loc is called continuous if it maps any convergent sequence to convergent one. We denote ν = (n − 1) /2.

General Method of Reconstruction

67

Theorem 4.9 I. For any generating function Φ satisfying (i) and (ii), any smooth weights w, u, arbitrary compact set K ⊂ X, and α ≥ 0, the estimate kMΦ,w f kα+ν ≤ Cα,K kf kα

(4.22)

holds for an arbitrary function f ∈ H0 (K) , where MΦ,w f is supported by Λ + pΣ p−1 X (K) and Cα,K is a constant. For an arbitrary compact set Λ ⊂ Σ, a number α and a test function φ on X, the similar inequality holds

φM∗Φ,u g ≤ Cα,Λ kgkα (4.23) α+ν for any function g supported in Λ. II. The operator NΦ,u defined by (4.21–4.20) is continuous bounded from Hν (Σ)comp to L2 (X)loc . Proof. We give a sketch of the proof and skip subscripts w and u to simplify the formulas. The first statement was proven in [105] based on a result of [59]. To prove II for odd n, we apply Proposition A.19 and write Z (n − 1)! NΦ g = δ (Φ (x, σ)) Tn−1 (g) (σ) dΣ π Σ = M∗Φ (Tn−1 (g)) , where Tn−1 is a differential operator of order n − 1. Operator Tn−1 acting from Hν (Λ) to H−ν (Λ) is bounded for any compact set Λ ⊂ Σ. By (4.22), operator MΦ is bounded from L2 (K) to Hν (Λ) where Λ = pΣ p−1 X (K) , and by (4.23) M∗Φ is a continuous operator from H−ν (L)comp to L2 (X)loc . This completes the proof for odd n. For even n, we have by (A.33) §A.5 (n − 1)! NΦ g (x) = GΦ (Tn−1 (g)) , π where

Z GΦ h (x) =

h (σ) dΣ. Φ (x, σ)

It is sufficient to show that GΦ is a continuous operator from H−ν (L)comp to L2 (X)loc . Suppose first that Φ = q − p is a resolved generating function and dΣ = λdp ∧ Ω for a function λ in Σ. We have then Z Z Z Z dΣ dt dp GΦ h (x) = h (σ) = λh (σ) Ω d Φ t s −p σ Φ=t q(x;ω)=s = M∗Φ (Hp7→s h) (x) , where s = p + t and H is the Hilbert operator. This operator is bounded in arbitrary Sobolev space Hα (K), where K ⊂ R × Y, and arbitrary manifold Y. The composition M∗Φ H is a continuous operator from H−ν (L)comp to L2 (X)loc .

68

Reconstruction from integral data

For an arbitrary generating function Φ satisfying (i–ii), an arbitrary point (x0 , σ0 ) ∈ X × Σ, we can choose a neighborhood X0 × Σ0 and a chart (τ, σ 0 ) , σ 0 = (σ2 , ..., σn ) on Σ0 , such that ∂Φ/∂τ = 6 0 on X0 × Σ0 . It follows that Φ (x, σ) = (τ − φ (x, σ 0 )) ρ (x, σ) for some smooth functions φ and ρ 6= 0. We have dΣ = λdτ ∧ dσ 0 on X0 × Σ0 for a smooth function λ. The binomial Ψ (x, σ) = τ − φ (x, σ 0 ) is a resolved generating function on X0 × Σ0 satisfying (i–ii). Therefore, continuity of GΦ is reduced to continuity of eGΨ for functions h supported in Σ0 and arbitrary smooth function e supported in X0 , since the factors ρ and λ can be absorbed by the weights w and u. The factor e and the restriction for h can be removed by appropriate partitions of the unity. 

4.6

Wave Front of Integral Transform

Proposition 4.10 Let Φ be a smooth generating function defined on X × Σ, with the zero set Z, such that dx Φ 6= 0 and dσ Φ = 6 0 on Z, and dX be an odd volume form on X. The integral transform MΦ is extended to a continuous operator acting on distributions D0 (X)comp → D0 (Σ) and to a transform of generalized functions K0 (X)comp → K0 (Σ) . Proof. The extensions are defined in terms of the backprojection transform M∗Φ acting on test functions and test densities: MΦ (u) (ψ) = u (M∗Φ (ψ)) , u ∈ D0 (X)comp , ψ ∈ D0 (Σ) ,

(4.24)

MΦ (f ) (ρ) = f (M∗Φ (ρ)) , f ∈ K0 (X)comp , ϕ ∈ K (Σ) .

(4.25)

The transform M∗Φ

Z (ψ) (x) = Z(x)

ψdΣ dσ Φ

(4.26)

is well defined for any test function ψ, since pX is proper and the integral is infinitely differential, according to Lemma 4.3. It follows that the action of a distribution u on M∗Φ (ψ) has sense, and, moreover, it defines a continuous functional of the argument ψ, which is denoted MΦ (u) in (4.24). To define the action of MΦ on generalized functions f , we need to check that the backprojection ! Z ρ ∗ MΦ (ρ) (x) = dX Z(x) dσ Φ is a continuous operation on test densities ρ. We write ρ = ψdΣ, where ψ is

General Method of Reconstruction

69

test function density on Σ and obtain the integral like that in (4.26). This gives a sense to (4.25) agree. It is easy to see that (4.24) and (4.25) in the sense that MΦ (f dX) = MΦ (f ) dΣ.  Definition. A distribution (generalized function) f on a manifold X is called (microlocally) smooth at a point (x0 , ξ0 ) ∈ T0∗ (X) , if there exists a test function (density) 6 0 and for any q ≥ 0,   α on X, such that α (x0 ) = P −q F (αf ) (ξ) = O |ξ| as |ξ| → ∞, and covector ξ = ξi dxi belongs to a conic neighborhood of ξ0 . Here, the Fourier transform F (αf ) is defined in terms of the local coordinate system x1 , .., xn . The set of smooth points is always open in T0∗ (X) . The wave front of f denoted W F (f ) is the complement in T0∗ (X) of the set of points where f is smooth. This set is closed and is the union of local cones W Fx (f ) ⊂ T0∗ (X) . This definition does not depend on the choice of local coordinates. Example. If f ∈ C ∞ (X) , then W F (f ) = ∅. Let ϕ ∈ C ∞ (X) and dϕ = 6 0. For any function h on R which is smooth on an open set U, we have W F (h (ϕ)) ⊂ {(x, ξ) : ϕ (x) ∈ R\U, ξkdx ϕ} . Note that W F (αf ) ⊂ W F (f ) for any smooth function α.

4.6.1

Transformation of the Wave Front

For a generalized function f, the wave front of MΦ (f ) can be described in terms of the wave front of f. Proposition 4.11 Under conditions of Proposition 4.10 for an arbitrary generalized function f with compact support in X, W F (Mf ) ⊂ ∪ {(σ, η) ∈ T0∗ (Σ) : ∃x : (x, σ) ∈ Z, ηkdσ Φ (x, σ) ,

(4.27)

(x, dx Φ (x, σ)) ∈ W F (f )} . Proof. Take an arbitrary point (x0 , σ0 ) ∈ X × Σ, its coordinate neighborhood U = Ux0 × Uσ0 , a test function α on Ux0 and a test function β on Uσ0 . We shall show that there exists such a neighborhood U of (x0 , σ0 ) that any point (σ, η) ∈ T ∗ (Σ) is not contained in W F (M (αf )) if ∀x : (x, σ) ∈ Z ∩ U : ηkdσ Φ (x, σ) , (x, dx Φ (x, σ)) ∈ / W F (f ) . Let σ = (σ1 , ..., σn ) be a coordinate system on Uσ0 . The integral Z Z I0 (η) + dξ β (σ) exp (j (hη, σi + hξ, xi)) γ0 (ξ) φ (ξ) dZ Rn

Z

is defined for η ∈ Rn , where φ (ξ) = F (αf ) and dZ = dX ∧ dσ/dx Φ. Here γ0 is a bounded homogeneous function of degree 0 on Rnξ , such that γ0 = 0 on an

70

Reconstruction from integral data

open cone W , such that W F (αf ) ⊂ Ux0 × W and γ0 = 1 on a neighborhood of the set D + {(x, ξ) : ξ k dx Φ (x, σ) , (x, σ) ∈ U } . Note that sets Ux0 × W and D are separated in the sphere |ξ| = 1 if Ux0 is sufficiently small. There exists a smooth tangent field T0 on X × Σ × Rnξ × Rnη that is parallel to X × Σ and fulfills T0 (Φ) = 0 on Z and T hη, σi = |η|   at any point (x, σ, ξ, η) ∈ U × Rnξ \W × Rnη . We have Z

Z

dξ Rn Z =−

T0 β (σ) exp (j (hη, σi + hξ, xi)) γ0 φdZ Z dξ div T0 β (σ) exp (j (hη, σi + hξ, xi)) γ0 φZ,

Z

Rn

Z

where div T0 = T0 (dZ) /dZ (see §1.4) is a smooth function of (x, σ). Further, T0 (β exp (j (hη, σi + hξ, xi))) = (T0 (β) + jβ (|η| + T0 hξ, xi)) exp (j (hη, σi + hξ, xi)) . Comparing the last two equation yields Z Z |η| I0 (η) = dξ exp (j (hη, σi + hξ, xi)) tγ0 φdZ, Rn

(4.28)

Z

in a neighborhood of σ0 , where t = −T0 hξ, xi + j−1 (div T0 − T0 (β)) . Obviously, |t| ≤ C (|ξ| + 1) for a constant C. The factor γ0 φ on (4.28) decreases fast as |ξ| → ∞, since γ0 = 0 on W. The same is true for the product tγ0 φ, hence, the integral converges and product |η| I0 (η) is bounded on Rn . −2 Applying this transformation to (4.28), we find that |I0 (η)| ≤ C (|η| + 1) −2 and so on. Finally, we have |I0 | ≤ Cq (|η| + 1) for q = 2, 3, ... Consider integral Z Z Eε (η) + β exp (j hη, σi) exp (j hξ, xi) eε γ1 φdξdZ, Z

Rn

  2 where γ1 = 1−γ0 and eε = eε (ξ) = exp −ε |ξ| . The inner integral may not converge if ε = 0 and the product γ1 φ grows as ξ → ∞. We rearrange it to get a convergent integral. For this, we choose a tangent field T1 on X ×Σ×Rnξ ×Rnη , which is parallel to X × Σ, such that T1 (Φ) = 0, T1 hη, σi = 0, T1 hξ, xi = |ξ| + 1

(4.29)

General Method of Reconstruction

71

at any point (x, σ, ξ, η) ∈ U × suppγ1 × Rnη . It is possible, because η is not collinear to dσ Φ (x, σ) for ξ ∈ suppγ1 , since suppγ1 ⊂ W is disjoint with the closure of D. By (4.29), we obtain Z Z eε γ1 φ Eε (η) = β exp (j hη, σi) T1 (exp (j hξ, xi)) dξdZ j (|ξ| + 1) Z Rn Z Z =− (T1 (β) + β div T1 ) Z

Rn

× exp (j (hη, σi + hξ, xi))

eε γ1 φ dξdZ. j (|ξ| + 1)

(4.30)

We apply this partial integration r times to reach an integral like that in −r the right-hand side of (4.30), with the factor (|ξ| + 1) instead of eε (ξ) . If r is sufficiently large, the integral absolutely converges for ε = 0, hence, Eε → E0 as ε → 0 and E0 (η) is bounded. We repeat this transformation once −1 more and get an extra factor (|ξ| + 1) . Then we integrate by parts with the field T0 which brings the growing factor T0 hξ, xi , but the integral is still absolutely convergent, and an equation like (4.28) holds. This implies that the product |η| E0 (η) is also bounded, and so on. Finally, we get estimate −q |E0 (η)| ≤ Cq (|η| + 1) for any q ≥ 0. The same is true for the sum I = E0 + I0 , and we have Z  Z I (η) = exp (j hη, σi) β (σ) exp (j hξ, xi) φ (ξ) dξ dZ Rn ZZ dX ∧ dσ = exp (j hη, σi) β (σ) α (x) f (x) dx Φ Z ! Z Z dX = exp (j hη, σi) β (σ) αf dσ dΦ Σ Z(σ) Z = exp (j hη, σi) β (σ) M (αf ) (σ) . Σ

It follows that any point (σ, η) in a neighborhood of (σ0 , η0 ) does not Pbelong to W F (M (αf )) . We can find test functions α1 , ...αN on X, such that αi (x) = 1 on suppf and (σ0 , η0 ) does not belong to W F (αi f ) for each i. Therefore, this point is not contained in W F (f ) ⊂ ∪i W F (αi f ) .  Conclusion. According to condition (i) of §4.1, the map D−1 X is well defined as well as the composition ∗ ∗ DΣ D−1 X : T0 (X) → T0 (Σ) ,

where DΣ D−1 X (x, ξ) = (σ, η) , ξ = tdx Φ (x, σ), and η = tdσ Φ (x, σ) for some t > 0. Condition (i) guarantees that for any x ∈ X and any covector ξ at x, there exists a point σ, such that (x, σ) ∈ Z and dx Φ (x, σ) k ξ. If (x, ξ) is a smooth point of a function f, then by Proposition 4.11, (σ, dσ Φ (x, σ)) is a smooth point of MΦ f. If (ii) is also satisfied, the contribution of the point x cannot be cancelled by another point y ∈ X.

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Reconstruction from integral data

Proposition 4.12 For any ∞-smooth generating function Φ satisfying conditions (i,ii), we have W F (MΦ f ) = DΣ D−1 X (W F (f )) . Proof. If Φ is a resolved generating function, a parametrix exists for the operator MΦ [112]. Proposition 4.11 holds also for the parametrix that gives the opposite inclusion (4.27). For an arbitrary generating function satisfying (i,ii), a parametrix can be constructed using the method of Theorem 4.9. We omit here the details.  Remark. In other words, the set W F (f ) can be recovered from W F (MΦ f ) by pure geometrical arguments without checking (iii). Conversely, if the function MΦ f is not known on a part of Σ, the corresponding part of W F (f ) is lost.

4.7

Bibliographic Notes

§4.1 Guillemin [49] defined a generalized Radon transform R for an arbitrary double fibration and treated it as a Fourier integral operator. He has shown that R∗ R is an elliptic pseudodifferential operator under the “Bolker condition.” More details are given in Quinto [120]. A parametrix for a class of generalized Radon transforms was constructed by Beylkin [11] in terms of the Fourier integral operators. Popov [117] used a perturbation method for a class of operators close to the classical Radon transform. Geodesic x-ray transform was studied by Pestov and Uhlmann [115]. In [112], a method close to this section is used for construction of a parametrix for a general class of geometric integral transforms M. §4.3 Theorem 4.2 is new. §4.6 Proposition 4.9 is a corollary of H¨ormander’s estimate of the Fourier integral operators [59]; see details in [105]. Natterer [83] gave a proof for the Radon transform and the similar estimate for the line transform without using Fourier integral operator techniques. §4.7 Notion of the wave front is due to H¨ormander. Proposition 4.14 is an analog of H¨ ormander’s result for Fourier integral operator [59].

Chapter 5 Applications to Classical Geometries

5.1

Minkowski–Funk Transform

Let X be a manifold with a Riemannian metric g. For any submanifold S ⊂ X, the restriction of the metric generates an odd volume form dg S defined on S, which is called Riemannian volume form. For any bounded function f on X with compact support and any closed submanifold S, the integral Z Rg f (S) = f dg S S

will be called metric integral transform of f . A submanifold S of X is called totally geodesic if it is connected, and any geodesic Γ in X tangent to S lies entirely in S. Any complete, simply connected Riemannian manifold X of constant sectional curvature has many geodesic submanifolds. For any point x ∈ X and any subspace V ⊂ Tx (X) , there is a (unique) geodesic submanifold S through x, such that Tx (S) = V. Any Euclidean space E has zero sectional curvature, and geodesic submanifolds are linear subvarieties. The geodesic integral transform is just the usual Radon transform on E. The classical reconstruction can be obtained by means of Theorem 4.7 applied for
the generating function Φ (x; p, ω) = hω, xi − p defined on E × R × Sn−1 . See Chapter 1 for more inversion formulas. The unit sphere Sn in a Euclidean space E n+1 is supplied with the metric e induced from E n+1 . This metric has constant sectional curvature −1. The sphere where antipodal points x and −x are identified is called elliptic space. In other terms, Xe + Sn /Z2 . The elliptic space is orientable if and only if n is odd. A function f on Xe can be considered as an even function on Sn and vice versa. Any geodesic manifold is the image of a big sphere Sk ⊂ Sn , k < n. The metric integral transform defined on big spheres of dimension n − 1 in Sn is called the Minkowski–Funk transform on n-dimensional elliptic space Z 1 f de S, σ ∈ Sn . Re f (σ) = 2 hσ,xi=0,x∈Sn It vanishes for any odd function f and is invertible for even functions, that is, for functions defined on the elliptic space Xe . 73

74

Reconstruction from integral data

Proposition 5.1 The metric integral transform Re coincides with the geometric transform Me Z Ω Me f (σ) = f (x) , σ ∈ Σ = Sn hσ, dxi Z(σ) generated by the function Φe + hσ, xi , where Ω is the Euclidean volume form on Sn , and Z (σ) = {x ∈ Xe , hσ, xi = 0} . 2

Proof. We have |hσ, dxi|e = |σ| = 1, by Proposition A.12, the form dX/ hσ, dxi = 0 is equal to the Euclidean volume form on Z (σ), hence, Me f = Re f .  Note that Me f (−σ) = Me f (σ) , hence, the Minkwoski–Funk transform is well defined on the dual elliptic space Σ = Sn /Z2 . Let dΣ be the Euclidean volume form on the sphere Σ. Theorem 5.2 Any even function f ∈ C n (Sn ) can be reconstructed from Re f (σ) by Z dΣ (n − 1)! Re f (σ) (5.1) f (x) = n n j hσ, xi Sn /2 for even n and by f (x) =

1

Z

2jn−1

δ (n−1) (hσ, xi) Re f (σ) dΣ

(5.2)

Sn /2.

for odd n. Here, the elliptic space Σ is modelled by an arbitrary hemisphere Sn /2. Proof. It is sufficient to apply Theorem 4.2 to the function Φe defined in Xe × Σe . 

5.1.1

Funk’s Method

Formulas (5.1–5.2) can be simplified by splitting the variables σ. Following Funk’s method, we fix a point x ∈ Sn and consider the integral of the geometric transform Mf over (n − 1)-sphere Sθ = {σ ∈ Sn , hx, σi = sin θ} of radius cos θ : Z dΣ 1 Mf (σ) , Ff (x, θ) = n−1 |S | cosn−1 θ hσ,xi=sin θ hx, dσi where dΣ/ hx, dσi equals the Euclidean volume form on Sθ and 0 ≤ θ ≤ π/2. By Fubini’s Theorem, integral (5.1) can be written as follows Z Z π/2 Z dΣ dθ dΣ Mf (σ) = Mf (σ) n n sin θ hσ,xi=sin θ hx, dσi hx, σi Sn /2 0 = Sn−1

Z 0

π/2

cosn−1 θdθ Ff (x, θ) . sinn θ

(5.3)

Applications to Classical Geometries

75

Corollary 5.3 For even n, π/2

Z

dθ sin θ

f (x) = cn 0 n/2

where cn = 4/ (−4π)

 n−1 ∂ cos θ cos θ Ff (x, θ) , ∂θ

(5.4)

Γ (n/2) Γ (n) , and for odd n,

n−1  ∂ Ff (x, θ)|θ=0 , cos θ f (x) = cn cos θ ∂θ n−1/2

where cn = 4/ (−4π)

(5.5)

Γ (n/2) Γ (n) .

Proof. We apply Proposition A.17 for function f = sin θ on a circle. We have t (f ) + t0 f = 1, where t = cos θ∂/∂θ and t0 = sin θ. By (A.19–A.20), the singular integral in the right-hand side of (5.3) is equal to Z (n − 1)! 0

Z

π/2

= 0

Z

π/2

= 0

π/2

dθ cosn−1 θFf sinn θ


dθ (t + t0 ) (t + 2t0 ) ... (t + (n − 1) t0 ) cosn−1 θFf dθ sin θ  n−1 dθ ∂ cos θ cos θ Ff dθ. sin θ ∂θ

The last equation can be proven by induction on n.  For n = 2, we apply (5.4) and integrate by parts: 1 f (x) = − π

Z

1 π Z

Z

π/2

0

dθ sin θ



 ∂ cos θ cos θ Ff (x, θ) dθ ∂θ

π/2

1 ∂ Ff (x, θ) dθ sin θ ∂θ   1 ∂ + sin θ + cos θ Ff (x, θ) dθ π 0 ∂θ Z π 1 π/2 ∂ dθ 1 =− Ff (θ) + Ff , π 0 ∂θ sin θ π 2 =−

0 π/2

which implies f (x) = −

1 π

Z 0

π/2

 π dFf (θ) dθ 1 + Ff x, . sin θ π 2

This is the original formula of P. Funk [36].

(5.6)

76

5.2

Reconstruction from integral data

Nongeodesic Hyperplane Sections of a Sphere

Let B ⊂ E n+1 be the unit ball and b be a closed ball, such that b b B. If a is the center and r is the radius of b, then for any σ ∈ Σ + Sn the hyperplane H (ha, σi + r, σ) is tangent to b and meets ∂B = Sn along an (n − 1)-sphere Sb,σ . The family of spheres Sb,σ is generated by the function Φb (x, σ) = hx − a, σi − r. The geometric integral transform Mb generated by Φb is Z Ω , Mb f (σ) = f Zb (σ) hdx, σi where Zb (σ) = ∂B ∩ H (ha, σi + r, σ) . Theorem 5.4 Let t ∈ Σ and X = {x ∈ ∂B : hx − a, ti > r}. For any odd n ≥ 3, an arbitrary function f ∈ C0n−1 (X) can be recovered from Mb f (σ) , σ ∈ Σ by Z 1 f (x) = n−1 δ (n−1) (hx − a, σi − r) Mb f (σ) dΣ. (5.7) 2j Dn (x) Sn For even n, any f ∈ C0n (X) is reconstructed by Z Mb f (σ) (n − 1)! dΣ, f (x) = n j Dn (x) Sn (hx − a, σi − r)n where  Dn (x) =

2

|x − a| − r2

(n−2)/2

n−1

|x − a|

(5.8)

.

(5.9)

Proof. If a = 0, r = 0, hypersurfaces Zb (σ) are big n-spheres in ∂B, and the statement follows from Theorem 5.2. In the general case, we shall apply Theorem 4.2. It is easy to see that Φb fulfils (i) and (ii). To verify (iii), we have to show that the integral Z 1 Ω (σ) n Qn (x, y) = Re i n hy − a, dσi (Φ (x, σ) − i0) b Zb (y) vanishes for arbitrary x, y ∈ X, y 6= x. We check that the linear function Φb (x, σ) = hx − a, σi − r has a zero in Zb (y) = {σ; Φb (y, σ) = 0}. Take the 2-plane P that contains x, y and a. The line L through y and x does not touch b, since x, y ∈ X; see the following picture.

Applications to Classical Geometries

77

Let L1 , L2 ⊂ P be the rays started from y that are tangent to the circle P ∩ b, and n1 , n2 be the exterior unit normal vectors to this circle at the points of tangency. The set Zb (y) is a sphere of dimension n − 1 > 0, hence, it is connected. The function hx − y, σi is continuous on this set, and its values hx − y, n1 i and hx − y, n2 i have different signs. Therefore, this function has a zero σ0 . It follows Φb (x, σ0 ) = Φb (y, σ0 ) + hx − y, σ0 i = 0. By Theorem 7.20, and Qn (x, y) = 0 for any n ≥ 2. By (4.4), we have |dx Φb (x, σ)|e = |σ| = 1 and Z 1 Ω (σ) Dn (x) = n−1 . |S | Zb (x) hx − a, dσi  1/2 2 We have |hx − a, dσi|e = |x − a| − r2 on the sphere Zb (x) and, by Proposition A.12, Ω (σ) = hx − a, dσi

Ω0 (σ) 2

|x − a| − r2

1/2 ,

where Ω0 is the Euclidean volume form on this sphere. This yields Z Z Ω (σ) 1 = Ω0 (σ) 1/2 2 Zb (x) hx − a, dσi Z (x) b |x − a| − r2 (n−1)/2 n−1  S 1 − (r/ |x − a|)2 , =  1/2 2 |x − a| − r2

78

Reconstruction from integral data

 1/2 2 since the radius of the sphere Zb (x) equals 1 − (r/ |x − a|) . This implies (5.9).  Remark. The Euclidean metric transform for the family of spheres Sb,σ is related to the geometric transform  −1/2 2 Mb f (σ) = 1 − (ha, σi + r) Rf (σ) . Therefore, f can be reconstructed from the Euclidean integrals Rf over spheres Sb,σ.

5.3

Totally Geodesic Transform in Hyperbolic Spaces

The hyperbolic space of dimension n is modelled by the upper sheet of the hyperboloid n o 2 Xh = x = (x0 , x0 ) ∈ E n+1 : [x, x] = 1, x0 > 0 , 2

where [x, y] + x0 y0 − hx0 , y 0 i , x0 = (x1 , ..., xn ) , y 0 = .... The hyperbolic metric h is induced by the pseudo-metric h (dx) = −dx20 + dx21 + ... + dx2n = − [dx, dx] in E n+1 . For a point a = (a0 , a0 ) ∈ E n+1 , the generating function Φa (x, σ) = [σ, x − a] is defined on Xh × Σh , where  Σh + σ ∈ E n+1 , [σ, σ] = −1 . The metric dΣh is induced from the pseudo-metric − [dσ, dσ] . For any σ ∈ Σh , the set Za (σ) = {x ∈ Xh , [σ, x − a] = 0} is one sheet hyperboloid of dimension n − 1. Consider the geometric transform generated by Φa Z dXh Ma f (σ) = f (x) , [σ, dx] Za (σ) where dXh = dx0 ∧ ... ∧ dxn / [x, dx] is the hyperbolic volume form in Xh . 1/2 Denote |x|h = [x, x] if x ∈ E n+1 , [x, x] ≥ 0. Theorem 5.5 Suppose that
[a, a] ≥ 0 and a0 ≤ 0. An arbitrary function f on Xh , such that f = O x−n at infinity can be recovered from Ma f by 0 Z 1 f (x) = n−1 |x − a|h δ (n−1) ([σ, x − a]) Ma f (σ) dΣh 2j Σh for odd n and by Z dΣh f (x) = (n − 1)!j−n |x − a|h Ma f (σ) (5.10) n [σ, x − a] Σh Z |x − a|h dΣh n−1 −n =j hx − a, ∇σ i Ma f (σ) n [σ, x − a] [x − a, x − a] Σh for even n.

Applications to Classical Geometries

79

Proof. If a = 0, the hypersurfaces Z (σ) are totally geodesic in Xh , and [x, x] = 1. The factorization method and (3.4) reduce the problem to the Radon transform. In the general case, we have [x − a, x − a] > 0 for x ∈ Xh , since of −a0 ≥ |a0 | . The assumptions guarantee that Φa fulfils (i). Condition (ii) is easy to check. To verify (iii),we have to show that the integral Z 1 dΣh Qn (x, y) = Re in n d Φ ([σ, x − a] − i0) σ a (y, σ) Za (y) vanishes for arbitrary x, y ∈ Xh , y = 6 x. The set Za (y) is the intersection of Σh and the hyperplane  La (y) + σ ∈ E n+1 , [σ, y − a] = 0 . ∗

The normal vector (y − a) = (y0 − a0 , σ 0 − y 0 ) fulfills  ∗ ∗ (y − a) , (y − a) = [y − a, y − a] > 0, which implies that Za (y) is an ellipsoid of dimension n − 1. Lemma 5.6 We have Ωn−1 dΣh = , dσ Φa (y, σ) |y − a|h where Ω is the volume form of a sphere in a linear coordinate system on Zh (y) . Proof. We have dΣh = dσ0 ∧ ... ∧ dσn / [σ, dσ] and dΣh dσ0 ∧ ... ∧ dσn = . dσ Φa (y, σ) [y − a, dσ] ∧ [σ, dσ] Take a linear map A preserving the form [y, y], such that det A = 1 and y − a = (|y − a|h , 0) . Apply A to variable y − a and σ. The brackets and the numerator do not change, and we find [y − a, dσ] = |y − a|h dσ0 . This yields 1 dΣh dσ1 ∧ ... ∧ dσn = dσ Φa (y, σ) |y − a|h [σ 0 , dσ 0 ] and Lemma follows.  Finally, we have Qn (x, y) = Re in

Z Sn

Z Z(x)

Ω n. ([σ, x − a] − i0)

Equations [σ, y − a] = [σ, x − a] = 0 define a (n − 2)-dimensional ellipsoid Z (x)∩Z (y), since y = 6 x. In the case n = 2, this is a two points set. Therefore,

80

Reconstruction from integral data

the denominator [σ, x − a] has a zero in the sphere. By Theorem 7.20, and Qn (x, y) = 0 for all n ≥ 2, hence, we can apply Theorem 4.2. The corresponding denominator is Z 1 1 dΣh Dn (x) = n−1 . n |S | Za (x) |∇x Φa | dσ Φa (x, σ) It can be calculated in terms of the coordinate system as in the above Lemma. We have |∇x Φa | = |σ 0 | = 1, since 0 = [x − a, σ] = |x − a|h σ0 , which yields σ0 = 0. Therefore, by the Lemma, Z 1 −1 Dn (x) = n−1 Ω (σ 0 ) = |x − a|h , |S | |x − a|h |σ0 |=1 which completes the proof of Theorem 5.5. The singular integral in (5.10) is regularized by means of Proposition A.16, with t = −1 [x − a, x − a] hx − a, ∇σ i , t0 = 0. 

5.3.1

Alternative Formulas

1. Formulas like (5.4) and (5.5) hold for the hyperbolic case. 2. One more method can be applied for the hyperbolic n-space realized as the open unit ball B ⊂ E n , n ≥ 2, with the conformal metric dg s =  −1 2 ds. The resolved generating function Φ = q − p is defined on 1 − |x| B × Σ, where Σ = Sn−1 × (−1, 1) and q (x, ω) =

2 hω, xi 2

|x| + 1

.

For any (p, ω) ∈ Σ, we have Z (p, ω) = B∩O, where O is a sphere (or a plane for p = 0) orthogonal to ∂B. This intersection is a totally geodesic manifold in B. Conversely, any totally geodesic manifold H ⊂ B is equal to Z (p, ω) for a point (p, ω) ∈ Σ. The function Φ satisfies (i) and (ii). We  have q (x, ω) − q (y, ω) = 2 2 a (x, y) hω, zi where a = 6 0 and z = |y| + 1 x − |x| + 1 y. It is easy to see that z = 6 0 if x 6= y ∈ B. Therefore, by loc. cit., (iii) is also fulfilled. According to Theorem 4.7, for even n, Z Z Z π ∂ n−1 Mh f (p, ω) dp 1 p  f (x) = n 0 Ω , 2 j Dn (x) ω∈Sn−1 0 p |x| + 1 − 2 hω, xi where Dn0

(x) =

1 |Sn−1 |

Z Sn−1

Ω 

2

|x| + 1

2

2

n/2 .

− 4 hω, xi

The corresponding reconstruction formula holds also for any odd n.

Applications to Classical Geometries

81

By §5.1, the geometric transform Mh is equivalent to the metric transform:
1/2 Mh ϕ (p, ω) , Rh f (p, ω) = 2 1 − p2 where

5.4

 1−n  −1 2 2 ϕ (x) = 1 − |x| |x| + 1 f (x) .

Horospherical Transform

Consider the generating function Ψ (x, σ) = [x, σ] − 1, x ∈ Xh \0, σ ∈ Σ0 + {σ : [σ, σ] = 0, σ 6= 0} . Hypersurfaces Z (σ) = {x : [x, σ] − 1 = 0} are horospheres in the hyperbolic space Xh .

Geodesics and horocycles. Proposition 5.7 If n is odd, any bounded function f on Xh decreasing sufficiently fast at infinity can be reconstructed from its horospherical integrals by Z dh Σ 1 f (x) = n ∂ n−1 Mh f (σ) (5.11) j Σ0 p [σ, x] − 1 for even n and by f (x) = for odd n.

1 2jn−1

Z Z(σ)

δ (n−1) ([σ, x] − 1) Mh f (σ) dh Σ

(5.12)

82

Reconstruction from integral data

Proof. It is easy to check that Ψ fulfils (i) and (ii). For any x ∈ Xh , the set Z (x) is an ellipsoid. For arbitrary x, y ∈ Xh \0, y = 6 x, function Ψ (y, ·) is linear and changes its sign in Z (x) . By Theorem 7.20, and (iii) is fulfilled and Theorem 4.2 can be applied. The dominator Dn is constant by the same reasons as in Proposition 5.5. Calculate Dn (x0 ) for x0 = (1, 0) . This point is contained in Z (σ) for σ = (1, σ 0 ) and arbitrary σ 0 , such that |σ 0 | = 1. We have ∇x Ψ (x0 , σ) = (1, −σ 0 ) , and the restriction of the gradient to the hyperplane T tangent to Xh at x0 satisfies ∇Ψ (x0 , σ)|T = |σ 0 | = 1. We find Dn (x0 ) = 1 by calculations like that in Theorem 5.2. 

5.4.1

Parallel Spheres in a Hyperbolic Space

Let X = E n \Sn−1 , where E n is a Euclidean space and Σ = R × Sn−1 . Fix a number ε, 0 ≤ ε ≤ 1, and consider the resolved generating function qε (x, ω)−p on X × Σ, where ε − hω, xi n−1 qε (x, ω) = 2 . (5.13) 2 , ω ∈S 1 − |x| For any p ∈ R and ω ∈ Sn−1 , the set Z (p, ω) = {x : qε (x, ω) = p} is a sphere, such that the sphere Z (p, ω) ∩ Sn−1 is contained in the hyperplane, H (ω) = {hω, xi − ε = 0}, which does not depend on p. Remark. Let B be the open unit ball in E n endowed with the hyperbolic metric. For any ω ∈ Sn−1 and real p and q, the hyperbolic distance between a point x ∈ Z (p, ω) ∩ B and the sphere Z (q, ω) ∩ B is the same for all x, that is, these spheres are parallel in B. For a proof, we show that any curve in X orthogonal to all hypersurfaces Z (p, ω) ∩ B, p ∈ R is a geodesic. If n = 2, the sphere H (ω) ∩ ∂B consists of two points, say, a, b, and any Z (p, ω) ∩ B is a circular arc through a and b. For any real λ, the set A (λ) + {x : |x − a| = λ |x − b|} is a circle orthogonal to all Z (p, ω) (Apolloneus circle). It is also orthogonal to the unit circle ∂B, since Z (p, ω) → ∂X as p → ∞. Therefore, A (λ) is a geodesic. An arbitrary point in B is contained in A (λ) for some λ. If n > 2, we take a plane P in Rn that contains points 0, ω and apply the above arguments to the disc P ∩ B. We check that Φε fulfills conditions of Theorem 4.7 for arbitrary ε, 0 ≤ ε ≤ 1 inside the ball B, as well as in the complement X\B. Checking (i) and (ii) is a routine. To verify (iii), we write qε (x, ω) − qε (y, ω)   −1    2 2 2 2 = 2 1 − |x| 1 − |y| hω, r (x, y)i + ε |x| − |y| .     2 2 where r (x, y) + y 1 − |x| −x 1 − |y| . It is easy to check that |r (x, y)| > 2 2 6 y ∈ B by squaring both sides. Dividing both |x| − |y| for arbitrary x =

Applications to Classical Geometries 2

83

2

sides by |x| |y| , we prove this inequality for x = 6 y ∈ X\B. It follows that qε (x, ω) − qε (y, ω) = hω, a (x, y)i + b (x, y) , −n

where |b| < |a| . The integral of Re in (qε (x, ω) − qε (y, ω) − i0) vanishes by loc. cit. and Theorem 4.7 works. Write the reconstruction in terms of the hyperbolic integral transform. Because of the factorization −1
1/2 2 |∇qε (x, ω)| = 1 − |x| p2 − 2εp + 1 , the metric transform Re is related to the geometric transform Mε generated by Φε : Z f (x) Mε f (p, ω) = de S Z(p,ω) |∇qε (x, ω)|
−1/2 = p2 − 2εp + 1 Re f1 (p, ω) ,   2 where f1 (x) = 1 − |x| f (x) . Corollary 5.8 For any function f with compact support in the unit ball, a reconstruction is given by Z (n − 1)! f (x) = Ω (5.14) 2 jn Dn (x) 1 − |x| Sn−1 Z Re f (p, ω) dp × 1/2 n R (p2 − 2εp + 1) (qε (x, ω) − p) for even n and by f (x) =

1

Z

2 2jn−1 Dn (x) 1 − |x|

× ∂pn−1

1/2 2 (p − 2εp + 1)

Ω

(5.15)

Sn−1

Re f (p, ω)

p=qε (x,ω)

for odd n, where qε is as in (5.13). For a proof, we apply Theorem 4.7 to Φε .  Euclidean transform Re can be replaced by the hyperbolic metric transform Rg in (5.14–5.15), changing the dominator Dn0 to Dn00 (x) =  n−1  n−1 2 2 Dn0 (x) 1 − |x| , since de S = 1 − |x| dg S for (n − 1)-surfaces S.  −1 2 for In the case n = 2, ε = 0 we have q0 (x; p, ω) = −2 hω, xi 1 − |x| x ∈ B and 2 2 Z 1  dω 1 − |x| 2 D2 (x) = 1 − |x|  2 = 2. 2π 1 + |x| S1 4 hω, xi2 + 1 − |x|2

84

Reconstruction from integral data


−1/2 For p = tan ϕ, we have dp = cos−2 ϕdϕ, p2 + 1 = cos ϕ, and (5.14) yields Z 2π Z π/2 2 1 + |x| g (p, ω) dϕdω 1 f (x) = − 2  2    2 . (5.16) 2π 2 2 −π/2 0 1 − |x| tan ϕ |x| − 1 + 2 hω, xi

5.4.2

Alternative Inversion

The hyperbolic space is modelled by the open unit ball B and horospheres as spheres in B tangent to ∂B. We take the generating function Φ1 = q1 − p as above, with ε = 1, where q1 (x, ω) = 2

1 − hω, xi 1 − |x|

2

.

The sets Z (p, ω) are horospheres; they are also equidistant for any fixed ω. Reconstruction formulas can be obtained from (5.14–5.15), taking ε = 0 and replacing Dn0 by Dn00 . We can also use the reconstructions (5.14–5.15) for the family of spheres S ⊂ E n \B tangent to ∂B.

5.5

Hyperboloids

Consider the family of hyperboloids Z (p, ω) in an euclidean space E n given by q (x; ω) = p,, where 2

2

q (x; ω) = |x| − 2 hx, ωi , ω ∈ Sn−1 , p ∈ R. They are obtained from the hyperboloid −x21 + x22 + ... + x2n = p by means of rotations in E n . This family is generated by the function Φ (x; p, ω) = q (x, ω) − p. The geometric integral transform MΦ is related to the metric transform R by Z MΦ (2 |x| f ) (p, ω) = f dS = Rf (p, ω) , (5.17) |x|2 −2hx,ωi2 =p

since |∇x q| = 2 |x| . Theorem 5.9 An arbitrary even function f with compact support in E n \0 can be reconstructed from the Euclidean integrals Rf by Z ∞ n−1 Z (2 |x|) dp Ω f (x) = ∂pn−1 Rf (p, ω) . (5.18) n j q (x, ω) − p Sn−1 /2 −∞ for any even n ≥ 2 and by n−1

f (x) =

(2 |x|) 2jn−1

Z Sn−1 /2

∂pn−1 Rf (p, ω)|p=q(x,ω) Ω

(5.19)

Applications to Classical Geometries

85

for odd n, where Sn−1 /2 denotes an arbitrary hemisphere in Sn−1 . Proof. The function f is well defined on X + (E n \0) /Z2 . The euclidean volume forms dx and dp∧Ω generate volume forms in X and Σ + R×Sn−1 /Z2 . We will check conditions of Theorem 4.7 for the function Φ defined on X × Σ. The function det J (Φ) does not vanish for n = 2, x = 6 0 and vanishes only if x is orthogonal to ω for n > 2. It follows that det J (Φ) 6= 0 on Z and condition (i) is fulfilled (and does not hold in R × Sn−1 ). Condition (ii) is violated in E n if and only if x = −y. It holds on X. To check (iii), we write Z Ω n Qn (x, y) = Re i n, Sn−1 (q (x, y; ω) − i0) where 2

2

2

2

q (x, y; ω) = |x| − |y| − 2 hx, ωi + 2 hy, ωi . Lemma 5.10 Polynomial q is oscillatory in the sense of §7.6. Proof. Suppose that |x| ≥ |y| and take the point ξ = x/ |x| as a pole in the sphere Sn−1 . Let P be an arbitrary plane through ξ and −ξ and S + P ∩Sn−1 . 2 2 2 The inequality 2 hy, ξi < |x| + |y| implies q (x, y; ±ξ) < 0. Let η ∈ S be a 2 2 2 point orthogonal to x. We have q (x, y; ±η) = |x| + 2 hy, ηi − |y| > 0, except for the case hy, ηi = 0, |y| = |x|. It follows that q has four simple zeros on the circle S or two multiple zeros in the last case.  By this Lemma and Theorem A.20, the kernel Qn vanishes for x 6= y. For odd n, we write Ω (ω) = sinn−2 θdθ ∧ Ω0 (υ) , where ω = (θ, υ) ∈ Sn−1 and Ω0 is the volume form on Sn−2 (see A.6). We have Z Sn−1

Ω n = (q (ω) − i0)

Z Sn−2

Ω0 (υ)

Z 0

π

sinn−2 θ n dθ. (q (θ, υ) − i0)

Take the sum of contributions of the right-hand side at an arbitrary point (θ, υ) and the point (τ, −υ) where τ = π − θ sinn−2 θ sinn−2 τ 0 0 dθ ∧ Ω (υ) + n n dτ ∧ Ω (−υ) . (q (θ, υ) − i0) (q (τ, −υ) − i0) These points are opposite in the sphere and q (τ, −υ) = q (θ, υ) since q is even. We have sin τ = sin θ, dτ = −dθ and Ω0 (υ) = Ω0 (−υ) , since n − 1 is even. Therefore the sum of contributions vanishes, which implies vanishing of the integral and yields again Qn = 0.  Now (5.18) and (5.19) follow from Theorem 4.7 and (5.17). 

86

5.6

Reconstruction from integral data

Cormack’s Curves

Cormack’s α-curves Aα (p, θ) are given by rα cos (α (ϕ − θ)) = p, |ϕ − θ| < π/2α in a plane with polar coordinates (r, ϕ) ; p > 0 and θ are parameters. The curves with α = 1/2 are parabolas with the focus at the origin. They have self-intersections if α < 1/2. The corresponding metric integral transform is defined by integrals Z Rα f (p, θ) =

f ds, Aα (p,θ)

where ds is the Euclidean metric. It is well defined for functions f on the plane with compact support. Cormack’s method of inversion is based on the relation between harmonic decompositions of functions f and Rα f in variables ϕ and θ, respectively. The integral equation is explicitly solved for each frequency, which gives the reconstruction [20],[21]. If α = 1/k for a natural k, an inversion can be given in filtered backprojection format. For any angle θ, the conformal map zθ (w) = wk : Cθ → C\0 is defined on the half-plane Cθ + {w : |argw − θ/k| < π/2} . The image of line u cos θ/k + v sin θ/k = p, p > 0 is equal to the curve A1/k (p, θ), written in coordinates u = Re w, v = Im w, since u cos θ/k + v sin
θ/k = |w| cos ((ϕ − θ) /k) . For a function f = f (z) , the function f wk is Zk k−1 periodic in w-plane and ds = k |w| |dw|. This yields Z R1/k f (p, θ) = f (z) ds A1/k (p,θ)

Z g (w) |dw| = Rg (p, θ/k) ,

= u cos θ/k+v sin θ/k=p


k−1 where g (w) = k |w| f wk and R means the Radon transform on w-plane. Function g can be reconstructed by inverting the Radon transform. The reconstruction of f is unambiguous, since g (p, θ + 2π/k) = g (p, θ) is Zk -periodic in angular variable.
Corollary 5.11 For any function f ∈ C 2 E 2 with compact support, we have
f wk = −

Z

1 2π 2

|w|

k−1

0

π

Z

∞

−∞

∂p R1/k f (p, θ) dpdθ , u cos θ/k + v sin θ/k − p

where we set R1/k f (p, θ) = R1/k f (−p, −θ) for p < 0.

(5.20)

Applications to Classical Geometries

87

Confocal parabolas.

5.6.1

Generalization

Consider the family of plane curves defined by s (r) cos (θ − ϕ) = p, |θ − ϕ| < π/2, where (r, ϕ) are polar coordinates of X + E 2 \0 and p, θ are parameters. Generating function Φ = s (r) cos (θ − ϕ) − p is defined for x ∈ X, p ∈ R, 0 ≤ θ < 2π. Suppose that ss0 = 6 0 and have det J (Φ) = r−1 s (r) s0 (r) 6= 0. For arbitrary conjugate points x, y ∈ X, we have s (r) cos (θ − ϕ) = s (ρ) cos (θ − ϕ) , s (r) sin (θ − ϕ) = s (ρ) sin (θ − ψ) , where (ρ, ψ) are polar coordinates of y. This yields tan (θ − ϕ) = tan (θ − ψ) and ϕ = ψ, since of the limitation |ϕ − θ| < π/2α. Equation Φ (x; p, θ) = Φ (y; p, θ) implies x = y, which proves (ii). Next, we have Z 2π dθ Q2 (x, y) = 2. (s (ρ) cos (θ − ψ) − s (r) cos (θ − ϕ)) 0 The second order trigonometric polynomial s (ρ) cos (θ − ψ) − s (r) cos (θ − ϕ) has four real zeros, if s (ρ) = 6 s (r) or ψ 6= ϕ. By Theorem A.20 Q2 (x, y) = 0 for y 6= x, which implies (iii). By Theorem 4.7, any function f on X with compact support is reconstructed by Z 2π Z ∞ 1 Mf (p, θ) dpdθ f (x) = − 2 (5.21) 2, 4π D2 (x) 0 −∞ (s (r) cos (θ − ϕ) − p) where D2 (x) =

1 2π

Z 0

2π

s02

(r) cos2

dθ . (θ − ϕ) + r−2 s2 (r) sin2 (θ − ϕ)

88

Reconstruction from integral data

The Euclidean integrals Rf can be used for reconstruction if |∇Φ (x; p, ω)| = m (x) µ (p, ω) for some functions m and µ. In this case, R (f /m) = µMf for arbitrary f. This factorization is possible if s (r) = ±rs0 (r), which implies s = cr or s = c/r. The corresponding curves are lines or circles through
−1 the origin. Two more cases of factorization are s± (r) = 2r 1 ± r2 , where

2 2 1/2 |∇Φ± | = 2 1 ± r 1∓p , and (5.21) yields a reconstruction from data of Rf . In these cases, the reconstruction coincides with (5.16) §5.3 or reduces to this formula by means of the geometric inversion, which is a conformal map. For the family of curves   s r1/k cos ((θ − ϕ) /k) = p, the reconstruction problem is reduced to the previous one if k is a natural number. It is sufficient to apply the conformal map z = wk .

5.7

Confocal Paraboloids

The family of paraboloids of revolution in Euclidean space E n with the focus at the origin is given by the generating function Φ (x; p, ω) = q (x, ω) − p, q (x, ω) = |x| − hx, ωi defined on the manifold X × Σ, X = E n \0, Σ = R+ × Sn−1 , where ω ∈ Sn−1 and p > 0. The Euclidean volume form dx is used for X. We consider two types of Euclidean integrals over rotation paraboloids Z Z f Rf (p, ω) = f dS, Mf (p, ω) = dS, Z(p,ω) Z(p,ω) |∇q| where dS is the Euclidean volume form on the paraboloid Z (p, ω) = {q (x, ω) = p} , p > 0. We have ∇q (x, ω) = ξ −ω, ξ = x/ |x| , and |∇q (x, ω)| = −1/2 1/2 |x| (2q) , which yields   −1/2 1/2 Mf (p, ω) = (2p) R |x| f (p, ω) . Both the integral transforms can be inverted according to:
Theorem 5.12 For an arbitrary function f ∈ C02 E 2 , the following reconstruction holds

f (x) = − =−

1/2

4π 2 |x|

Z

1 4π 2

2π

Z

1

|x|

0 2π

0

Z Ω

0

∞

Z Ω

0

∞

 ∂  1/2 dp p Rf (p, ω) ∂p |x| − hx, ωi − p

∂ dp (pMf (p, ω)) . ∂p |x| − hx, ωi − p

(5.22)

Applications to Classical Geometries
For any f ∈ C02 E 3 and any x ∈ E 3 \0, we have Z 1 ∂ 2 1/2 Ω f (x) = − p Rf (p, ω) 2 2 1/2 ∂p p=|x|−hx,ωi 2 S (4π) |x| Z 1 ∂2 =− pMf (p, ω)|p=|x|−hx,ωi Ω. 2 (4π) |x| S2 ∂p2

89

(5.23)

Remark. For the case n = 2, reconstructions (5.22) and (5.20) are based on different generating functions. The first formula (5.23) is similar to Cormack’s reconstruction

r1/2 f (x) = −



Z

1 2

(4π)

S2

∂ r ∂r

2

g (p, η) dη, p3/2 p=r(1+hξ,ηi)/2

(5.24)

where r = |x| and, apparently, g (2p, ω) = Rf (p, ω) . However, integral (5.24) looks divergent near the point η = −ξ. On the other hand, the correct model for the “scattered light filtrated on travel times” is the integral Mf . It is different from the Euclidean surface integral Rf, since |∇q| is not a constant. Proof. Let n ≥ 2 be arbitrary. The manifold Σ is equipped with the volume form dp ∧ Ω. The function Φ does not fulfill (i), since the function J (Φ) = const hξ − ω, ωi dp ∧ Ω vanishes for ω = ξ + x/ |x| . Let Z0 = {(x; p, ω) ∈ Z, ω = ξ} . The map DX : Z × R+ → T0∗ (X) is not defined on Z0 , since dx q vanishes. The image of DΦ is U + {(x, θ) , hθ, xi 6= 0} . To fix the situation, we introduce the weight 2 function w (x, ω) + |∇x q (x, ω)| and have Z Mw f (p, ω) = f |∇x q (x, ω)| dS Z(p,ω) 1/2

Mw f = (2p)

    −1/2 −1 R |x| f = 2pM |x| f ,

where M = M1 . We prove (5.23) by applying Theorem 4.6 to the weight functions w (ε) and u (ε) , where w (ε) = u (ε) = 0 on ε-neighborhood Zε of the set Z0 and w (ε) = w, u (ε) = 1 otherwise. Note that (ii) is fulfilled, since the equation dω q (x, ω) = −x. According to Theorem 4.6, we have Nu(ε) Mw(ε) f = Dε f + Qε f, since Φ satisfies (i) on suppw (ε) ∩ suppu (ε) . Here, Qε is the operator with the kernel Z u (ε; x, ω) w (ε; y, ω) Ω Qn,ε (x, y) = Re in n S2 (q (x, ω) − q (y, ω) − i0) and Dε (x) =

1 |Sn−1 |

Z Sn−1

u (ε; x, ω) w (ε; x, ω) Ω . n |∇q (x, ω)|

90

Reconstruction from integral data

Obviously, Dn,ε (x) → Dn,w (x) for x 6= 0, where Dn,uw (x) =

1 |Sn−1 |

Z Sn−1

1 uwΩ n = n−1 |S | |∇x q|

Z Sn−1

Ω n−2

|∇x q|

= 2n−2 ,

since |∇x q| = 2 sin ψ/2, where ψ is the angle between θ and ω. Lemma 5.13 We have Qn,ε (x, y) → 0 for arbitrary x = 6 y ∈ E n \0 as ε → 0, where n = 2, 3. Proof of Lemma. The difference q (x, ω) − q (y, ω) = (|x| − |y| − hx − y, ωi) /2

(5.25)

is a linear function of ω for arbitrary x 6= y. This function has a zero ω ∈ Sn−1 , since ||x| − |y|| ≤ |x − y| . Weight function uw = 2 (1 − hη, ωi) , η = y/ |y| is linear in ω, which implies that Qw (x, y) = 0 by Theorems A.20. Now it is sufficient to check that Qn,ε → Qn as ε → 0. For this, we show that (5.25) does not vanish at ω = η. We have q (x, η) − q (y, η) = (|x| − |y| − hx − y, ηi) /2 = (|x| |y| − hy, xi) /2 |y| > 0, since y 6= x and x 6= 0, y = 6 0. This completes the proof for n = 3. The case n = 2 is handled in a similar way.  To complete the proof of Theorem, it is sufficient to check that the integrals (5.22) and (5.23) converge in the classical sense.
Lemma 5.14 For any f ∈ C 2 E 3 , we have ∂p2 Mf (p, ω) = O (1)

(5.26)

as p → 0 uniformly with respect to ω. Proof of Lemma. Variables r = |x| , θ and ϕ are spherical coordinates in R3 , where θ = θ (y, ω) , 0 ≤ θ < π is the spherical distance between a point y ∈ S2 and ω, and ϕ is rotation angle about ω and Z (p, ω). Equation p = (x − hx, ωi) /2 = r sin2 θ/2 defines the surface Z (p, ω), and we have
1/2 dS = dr2 + r2 dθ2 r sin θds for the Euclidean area dS of Z (p, ω) where ds is the length element of a circle 1/2 r = const in this surface. We have ds = r sin θdϕ and dS = 2 (pr) drdϕ. This yields Z −1/2 Mf (p, ω) = p r1/2 f (r, θ, ϕ) dS Z ∞ Z 2π =2 f (r, θ, ϕ) dϕrdr. p

0

Applications to Classical Geometries

91

For a function h ∈ C 2 (P ) in a plane P, any circle integral can be represented in the form Z h (y) dφ = 2πh (0) + h2 (s) s2 , |y|=s

where the remainder fulfills h2 = O (1) . Applying this equation to f in a plane P orthogonal to ω for s = r sin θ, we obtain Z 2π 2 f (r, θ, ϕ) dϕ = 2πf (r cos θω) + (r sin θ) ψ (r, θ, ω) , 0 2

where ψ = O (1) uniformly. We have r cos θ = r − 2p, (r sin θ) = 4p (r − p) and integrate against rdr : Z ∞ Z 2π 2 f (r, θ, ϕ) dϕrdr p

Z

0 ∞

(πf ((r − 2p) ω) + 2p (r − p) ψ (r, θ, ω)) rdr.

=4 p

Applying the operator ∂p2 to the right-hand side, we obtain a bounded function of ω. This implies (5.26). 

5.8

Cassini Ovals and Ovaloids

Cassini ovals with foci ±1 are defined by equation 2

2

|x − (1, 0)| |x − (−1, 0)| ≡ x21 + x22


2


+ 1 − 2 x21 − x22 = p,

where x = (x1 , x2 ) and p > 0 is a parameter. In the case p = 1, this is the Bernoulli lemniscate. The family of Cassini ovals with foci in opposite points on the unit circle are given by generating function Φ (x; p, ω) = q (x, ω)−p = 0, where 4 2 q (x, ω) = |x| − 2 |x| cos 2 (ϕ − ω) + 1, and ϕ is the polar angle
of a point x ∈ B + {0 < |x| < 1} and 0 ≤ ω 0.

92

Reconstruction from integral data

Cassini ovals and the disc of reconstruction. Proposition 5.15 Any even function f ∈ C02 (B) can be reconstructed from the Euclidean integrals Rf by Z πZ ∞ Rf (p, ω) dpdω 1   f (x) = − 2 . 2 (q (x, ω) − p) 0 0 16π 2 |x| 1 − |x| Proof. Any even function f is well defined on the manifold X = B/Z2 , where the generator of the group Z2 acts by x 7→ −x. The function Φ fulfills (i) and (ii) on B/Z2 . We have Z π dϕ Re Q2 (x, y) = Re 2 = 0, 0 (p4 (x, y; ω) − i0) where p4 (x, y; ω) + q (x, ω) − q (y, ω) 4

4

2

2

= |x| − |y| − 2 |x| cos 2 (ϕ − ω) + 2 |y| cos 2 (ψ − ω) and ψ is the polar angle of y. The trigonometric polynomial p4 (ω) is π-periodic and has 4 zeros on the unit circle 0 ≤ ω < 2π if x, y ∈ X, x = 6 y. Suppose that |y| < |x| . It is easy to see that for ω = ϕ,    2 2 2 2 −p4 (ω) ≥ 2 − |x| − |y| |x| − |y| > 0, 2

2

that is, p4 (ω) < 0. For ω = ϕ + π/2, we have p4 (ω) ≥ |x| − |y| > 0, hence, p changes sign 4 times in S1 . The case |y| < |x| is symmetric, in the

Applications to Classical Geometries

93

case |y| = |x| the statement is obvious, since ψ 6= ϕ. We apply Proposition A.21 4.7 for Φ and take into account that Mf = R (f / |∇q|) =  and Theorem   2 R f /2 |x| 1 − |x| .

5.8.1

Polynomial Ovals

A polynomial oval is given by equation |p (x + iy)| = const > 0, where p (z) is a holomorphic polynomial on R2 . For p (z) = z k − 1, this is a Cassini oval of order k. The ovals of order 3 look as follows.

Polynomial ovals. Proposition 5.16 For any natural k, an arbitrary function f ∈ C02 (B) invariant with respect to rotations by 2π/k about the origin can be reconstructed form Euclidean integrals Rf over Cassini ovals of order k by Z 2π/k k Rf (p, ω) dω   f (x) = − 2. k−1 k (q (x; ω) − p) 0 8kπ 2 |x| 1 − |x| Proof. The resolved generating function 2k

Φk = qk − p, qk (x, ω) = |x|

k

− 2 |x| cos (k (ϕ − ω)) + 1

is well defined on X = B/Zk , satisfies (i) and (ii), and we have |∇x q| =  k−1 k 2k |x| 1 − |x| . The corresponding kernel Z Re Qk (x, y) = Re 0

2π/k

dϕ

2,

(p2k (x, y; ω) − i0)

94

Reconstruction from integral data

where 2k

p2k (ω) = |x|

− |y|

2k

k

k

− 2 |x| cos k (ϕ − ω) + 2 |y| cos k (ψ − ω) ,

vanishes for any x, y ∈ X, y = 6 x, since the trigonometric polynomial p2k has 2k zeros on the circle. 

5.8.2

Cassini Ovaloids 4

Take the generating function Φ = q − p on E n , where q (x, ω) = |x| − 2 2 |x| cos (2θ (x, ω)) + 1, x ∈ E n , ω ∈ Sn−1 , and θ (x, ω) = arccos hx/ |x| , ωi is the angle between x and ω. Hypersurface q (x, ω) = p looks like a barbel. Proposition 5.17 For any even n, an arbitrary even function f supported in the ball Bn + {x ∈ E n , 0 < |x| < 1} can be reconstructed form Euclidean integrals Rf over Cassini ovaloids by Z Z ∞ Rf (p, ω) dp 1   (5.27) f (x) = − 2 Ω. 2 n−1 2 (q (x, ω) − p) S /2 0 16π |x| 1 − |x| Proof. We take X = Bn /Z2 and
follow the arguments of the case n = 2 above. We have |∇q| = 4r r2 − 1 and Z 2π dϕ Re Qn (x, y) = Re 2, (p4 (x, y; ω) − i0) 0 where 4

4

2

2

p4 (x, y; ω) = |x| − |y| − 2 |x| cos 2θ (x, ω) + 2 |y| cos 2θ (y, ω) . Lemma 5.18 Polynomial p oscillatory in the sense of §A.6 for arbitrary y 6= x ∈ Bn .   4 4 2 2 Proof. Suppose that |x| ≥ |y|. We have |x| − |y| < 2 |x| − |y| , since x, y, ∈ Bn . Let P be an arbitrary plane that contains the points ξ = x/ |x| and −ξ. Check that p4 has four zeros ω in the circle S = P ∩ Sn−1 . We have −2 −2 2 cos 2θ (y, ξ) = 2 |x| |y| hy, xi − 1 and 4

4

−2

2

2

p (±ξ) = |x| − |y| − 2 |x| + 4 |x| hy, xi − 2 |y|   4 4 2 2 < |x| − |y| − 2 |x| − |y| < 0.

2

Let ω ∈ S be a point orthogonal to x. We have 4

4

2

2

p (±ω) = |x| − |y| + 2 |x| + 2 |y| cos 2θ (y, ±ω)   4 4 2 2 > |x| − |y| + 2 |x| − |y| > 0, which implies the statement.  By Theorem 7.20, and we conclude that Re Qn (x, y) = 0 for y 6= x ∈ X, and (5.27) follows from Theorem 4.7. 

Applications to Classical Geometries

5.9

95

Bibliography Notes

§§5.1 The classical formula for reconstruction on 2-sphere from data of big circle integrals is due to Funk [36]. The Minkowski–Funk transform was generalized by Helgason [53] for totally geodesic submanifolds in hyperbolic spaces. He found formulas for reconstruction of even functions on Sn from data of integrals over big k-spheres for even k. His reconstruction has Backprojection filtration structure p (∆) M∗ M, where p is a polynomial of degree k/2 and ∆ is a Laplace–Beltrami operator. Semjanisty˘ı [130] generalized this result for odd k = n − 1 in a form close to (5.1). For odd k = n − 1, Helgason’s formula [57] is analogous to that of Funk. Rubin [125] obtained inversion of the Funk transform for the same case in BPF format. Theorem 5.4 is not known, to my best knowledge. Reconstructions (5.1–5.2) were obtained in [130]. §5.2 Formulas (5.7–5.8) for a = 0 can be found in Gel’fand–Gindikin–Graev [43] Chapter 5. In the case of a = 6 0, the reconstructions are new. §5.3 Inversion formulas for the totally geodesic transform of even and odd dimensions k on a hyperbolic space were obtained by Helgason [53], [54], [58]. Helgason’s formulas have BPF structure. For odd k, a similar formula in BPF formal was obtained by Berenstein and Tarabusi [8]. A FBP reconstruction for the hyperbolic disc was obtained in Lissianoi and Ponomarev [78]. Reconstructions for Z (σ) = H ∩Xh for some families of hyperplanes H parallel to a vector v, [v, v] ≥ 0 are done in [43]. Formulas (5.11–5.12) for horospherical transform were obtained in Gel’fand–Graev–Vilenkin [39]; see also [43]. Reconstructions of BPF type are mentioned in Helgason [55]; see also Berenstein-Tarabusi [9]. Reconstructions given in Theorem 5.5 are apparently new. §5.4 The method of Compton scattering was proposed by Norton [93] for reconstruction of electron density on a body. It is based on counting of the scattered photons of a specific energy Eω emitted from a point source S in the plane collimated by the detector D. Another acquisition geometry was proposed by Nguyen and Truong [97],[138], where both source and detector rotate around a fixed point being always in opposite positions. In this geometry, the detector counts are collected from families of equidistant circular arcs in the hyperbolic unit disc. The authors solved the exterior problem: To determine a function from line integrals, both are given in the complement to a disc in a plane. The FBP-formula (5.16) was obtained in [107], [108], and §5.3 (5.16). Corollary 5.8 is a generalization of this result for n-dimensional case. See [97] and [138] for further results. §5.5 The result is new. No reconstruction for a family of ellipses or ellipsoids is known so far. Romanov [124] stated uniqueness of a function with known integrals over a family of ellipses with a focus in the origin and another focus moving along a line. The case of rotation ellipsoids is also studied. §5.6 Pioneering papers of A. Cormack [19] were devoted to medical application. Cormack developed a method of inversion from integrals over plane

96

Reconstruction from integral data

curves of special types (α and β). He found an integral relation between harmonic decomposition of a function f and its curve integrals and inverted it [20],[21]. Nguyen and Troung [97] applied Cormack’s method for the family of arcs in a disc leaning two points (family Φ+ ). They studied the generalizations of Cormack curves and described the cases of factorization of |∇Φ| [138]. §5.7 In [22], Cormack proposed a method for determination of distribution of a scattered material in Rn , n = 2, 3 from a point radiation source by filtering the scattered light on travel times [23]. The mathematical problem is to reconstruct a function on a plane from data of integrals over a family of confocal parabolas or confocal rotation paraboloids. Cormack’s reconstruction is (5.24). The reconstruction from data of Mf is closer to the problem. §5.8 Results are new.

Chapter 6 Applications to the Spherical Mean Transform

6.1

Oscillatory Sets

Definition. Let p be a real polynomial on Rn of degree m > 0 and Z its zero set. We call p and Z oscillatory with respect to a point a ∈ Rn \Z, if p has m simple zeros in L for almost any line L ⊂ Rn through a. Such a point a is called hyperbolic. We call a hyperbolic cavity of an oscillatory polynomial p any maximal connected set H of hyperbolic points a. This definition is close to that of hyperbolic polynomials in the sense of Petrovski˘ı [116]. Let p be a homogeneous polynomial on Rn of degree m. It is hyperbolic with respect to a vector θ ∈ Rn , if p (θ) = 6 0, and the equation p (tθ + η) = 0 has only real zeros t = tk (η) , k = 1, ..., m for any η ∈ Rn . By [37], for any polynomial p, the set H of hyperbolic vectors of p is open (may be empty), and any connected component Γ of H is a convex cone called cone of hyperbolicity of p. If Γ is a cone of hyperbolicity, then so is the cone −Γ. Proposition 6.1 Any hyperbolic polynomial p in Petrovski˘ı’s sense is oscillatory, and any cone of hyperbolicity of p is a hyperbolic cavity. Proof. Let Γ be a cone of hyperbolicity for p. Take arbitrary θ ∈ Γ and a vector η ∈ Rn \0 and consider the line L = {θ + tη, t ∈ R} . We have p (θ + tη) = tm p (sθ + η) , where s = 1/t and t = 6 0. The polynomial . r (s) = p (sθ + η) has m real zeros, since p is hyperbolic with respect to θ. Therefore, p (θ + tη) has m real zeros for any η 6= 0, hence, p is oscillatory with respect to θ, that is, Γ is a hyperbolic cavity.  There many nonhyperbolic and nonhomogeneous oscillatory polynomials. For any oscillatory set Z with a hyperbolic cavity H and any affine subspace A in Rn , the set Z ∩ A is oscillatory with the cavity H ∩ A if this set is not empty. From any oscillatory polynomial, one can construct a hyperbolic one: Proposition 6.2 For an arbitrary oscillatory polynomial p of degree m and
an arbitrary hyperbolic point a of p,the polynomial Pa (τ, ξ) = τ m p τ −1 ξ + a on Rn+1 is homogeneous and hyperbolic with respect to θ = (1, 0, ..., 0) . Proof. We have Pa (θ) = p (a) = 6 0, and for an arbitrary point η = (τ, ξ) ∈ 97

98

Reconstruction from integral data

Rn+1 ,

m

P (tθ + η) = P (t + τ, ξ) = (t + τ ) p (sξ + a) , −1

where s = (t + τ ) . The polynomial p (sξ + a) has m real roots s1 , ..., sm . The numbers tk = s−1 k − τ are real roots of P (tθ + η) , k = 1, ...., m.  The following result describes geometry of oscillatory sets Z. A point x ∈ Z is called regular if dp (x) = 6 0. An oscillatory set Z is compact if and only if the polynomial p is elliptic, that is, the principal part pm does not vanish on Rn \0. Theorem 6.3 Let p be a polynomial on Rn , n > 1 of degree m, with the compact zero set Z, which is hyperbolic with respect to a point a. We have Z = Z1 ∪ ... ∪ Zµ , where Z1 , ..., Zµ , µ = m/2 are ovals (homeomorphic images of a sphere). They are nested in the sense that a set of regular points of Zi is contained in the interior of Zi+1 for i = 1, ..., µ. The oval Z1 is convex, and its interior H is the unique hyperbolic cavity. Proof. Let a be a hyperbolic point of p and Sn−1 be the unit sphere in Rn with the center at the origin. For any ω ∈ Sn−1 , we numerate zero points of p by xk (ω) = a + tk (ω) ω, k = −σ (ω) ..., −1, 1, ..., τ (ω), counting with multiplicity in such a way that t−σ(ω) ≤ ... ≤ t−2 ≤ t−1 < 0 < t1 ≤ t2 ≤ ... ≤ tτ (ω) . Fix an arbitrary ω0 ∈ S and arbitrary real numbers u < v, such that p (a + tω0 ) = 6 0 for t = u, v. Let D be the disc in the complex plane that contains the real segment [u, v] as a diameter. The holomorphic function f (t) = p (a + tω0 ) does not vanish on ∂D, since p is oscillatory and f (t) has only real zeros. By the Rouche theorem, the number of zeros t = tk (ω) of p (a + tω), such that u ≤ tk (ω) ≤ v is constant for all ω in a neighborhood U of ω0 , since the p (a + tω) has no complex zeros. It follows that the number of zeros of p (a + tω) is equal to m for all ω, since Z is compact. This implies σ (ω) + τ (ω) = m for any ω. Therefore, we have σ = τ = µ + m/2 and t−k (ω) = tk (−ω) , k = 1, ..., µ, since the sphere Sn−1 is connected. For any k = 1, ...,  µ, function xk (ω) = a + tk (ω) ω is defined and continuous, hence, Zk = x = xk (ω) , ω ∈ Sn−1 is an oval. We have Z = ∪µ1 Zk , and the variety Zk ∩ Zj has dimension < n − 1 for any j = 6 k. Therefore, sets Zk are nested, and a belongs to the interior H of the oval Z1 . Now we check that any b ∈ H is a hyperbolic point. For any ω ∈ Sn−1 , the line L (b, ω) = {x = b + tω, t ∈ R} intercepts each closed oval Z1 , ..., Zµ at least two times. These 2µ points are different for almost all ω, since the intersection Zk ∩ Zj has dimension < n − 1 for k = 6 j. The total number of intersections is ≥ 2µ = m. It is equal to m, since p is a polynomial of degree m. This shows that b is a hyperbolic point. It follows that H is the set of all hyperbolic points. Let us check prove that H is convex. If it is not the case, then there exists a sequence of lines Lj → L, such that L touch a concave point b ∈ ∂H, but

Applications to the Spherical Mean Transform

99

Lj does not meet the set U \Z for a neighborhood U of b. By the Rouche theorem, the number of zeros of p in L\U is the same as in Lj \U , which is equal to m. The point b ∈ Z makes the number of zeros of p in L larger m, which is impossible. If there is a hyperbolic point b ∈ Rn \H, there exists a line L through b that is tangent to ∂H. Then again, the number of zeros of p changes for lines close to L, which causes a contradiction.  Examples 6.1. Any polynomial p of one variable of order m is oscillatory if and only if all the zeros are real. It has m + 1 cavities. The same is true for the polynomial p (l (x)) in Rn , where l is an arbitrary linear function on Rn . 6.2. Two-sheet hyperboloid is oscillatory with two hyperbolic cavities, whereas any ellipsoid, elliptic paraboloid, elliptic, and parabolic cylinder is oscillatory with only one hyperbolic cavity. One-sheet hyperboloid, elliptic, and parabolic hyperboloids Z are not oscillatory, since the curvature form of Z is determined.
3 6.3. The zero set Z of the oscillatory polynomial p = x2 + y 2 −
2
12 x2 + y 2 + 7x2 y 2 + 30 x2 + y 2 − 20 is compact, since the principal part
3 of p is the elliptic polynomial x2 + y 2 . The set Z is regular and oscillatory.

Oscillatory set of degree 6. 6.4. A Cartesian leaf given by x3 + y 3 − 3xy = 0 is oscillatory, with only one hyperbolic cavity. It is a unicursal curve parametrized by ξ1 =
−1
−1 3t 1 + t3 an interval, ξ2 = 3t2 1 + t3 . 6.5. Half-ellipsoid. Let p2 be a second order polynomial with positive principle part. The set p3 = 0, where p3 = x1 p2 is the union of an ellipsoid and a

100

Reconstruction from integral data

hyperplane. If the set H = {p2 (x) < 0, x1 > 0} is not empty, it is a hyperbolic cavity for p3 . We can approximate p3 by a polynomial p˜3 close to p3 with a ˜ The set Z ˜ consists of an oval Z1 = ∂H and an regular oscillatory zero set Z. unbounded component Z2 6.6. A “triangle” hypotrochoid is given implicitly by an elliptic oscillatory
2
polynomial p4 (x, y) = 4 x2 + y 2 − 4x3 + 12xy 2 − 27 x2 + y 2 + 27 = 0. It is oscillatory of degree 4.

Hypotrochoid (thick), a regularization (thin). 6.7. Pentagon trochoid Z is given parametrically by x = 2 cos 3t + 5 cos 2t, y = 2 sin 3t − 5 sin 2t, etc.

Applications to the Spherical Mean Transform

101

Pentagon trochoid. 6.8. The surface of normals of the system of crystal optics is given by the equation 3 X ξi2 = 1, 2 i=1 |ξ| − σi where σ1 , σ2 , σ3 are positive constants. It can be written in the form p4 (ξ) = 0, where p4 (ξ) is an elliptic oscillatory polynomial of degree 4. The polynomial
P (τ, ξ) = τ 4 p4 τ −1 ξ of forth degree in space-time R4 is the symbol of the hyperbolic system of crystal optics. Equation p4 (ξ) = 0 defines a compact oscillatory surface Z in R3 , which consists of two ovals that meet one another at four conic singular points, where dp4 = 0 but the form d2 p4 is not degenerate.

6.2

Reconstruction

The spherical integral transform on an Euclidean space E n Z Rf (r, ξ)) = f (x) dS, ξ ∈ Z |x−ξ|=r

restricted to an arbitrary oscillatory set Z can be inverted by the universal filtered backprojection formula: Theorem 6.4 Let H ⊂ E n be the hyperbolic cavity of an elliptic oscillatory

102

Reconstruction from integral data

polynomial p on E n , n ≥ 2, with the regular zero set Z, q be an arbitrary polynomial of degree < deg p. An arbitrary function f ∈ C n−1 (E n ) with support in H can be reconstructed from its spherical integrals by n−1 Z  qdξ 1 ∂ Rf (r, ξ) 1−n p (x) f (x) = j (6.1) q (x) Z r ∂r r r=|x−ξ| dp for odd n, and by Z Z ∞ dr2 p (x) q (x) Z 0 |x − ξ|2 − r2  n−1 1 ∂ Rf (r, ξ) qdξ × r ∂r r dp

f (x) = 2j−n

(6.2)

for even n if f ∈ C n (E n ) . We assume that pq > 0 on H and Z is oriented by the inward normal field with respect to H. Proof. We apply Theorem 4.7 for the generating function Φ (x; s, ξ) = 2 |x − ξ| − s defined on H × Σ, Σ = R × Z. We have n−1

J (Φ) = (−1)

 ∧n−1 2 2 dx |x − ξ| ∧ dx dξ |x − ξ| ∧ ds ∧n−1

= ±2n n!dV ∧ (x − ξ) ∧ (dξ1 , ..., dξn )

∧ ds,

where dV = dx1 ∧ ... ∧ dxn is the volume form on E n . The product ∧n−1 (x − ξ) ∧ (dξ1 , ..., dξn ) vanishes only if vector x − ξ is tangent to Z. This is impossible, since x is a hyperbolic point and Z is regular. This implies (i). To prove (ii), we suppose that y = 6 x are conjugate points in H, which means 2

2

2

|x − ξ| = |y − ξ| , dξ |x − ξ| = dξ |y − ξ|

2

(6.3)

for some ξ ∈ Z. The first Equation (6.3) implies that |x − ξ| = |y − ξ| . It follows that the line L through h = 1/2 (x + y) and ξ is orthogonal to x−y. By the second Equation (6.3), we have hx − y, dξi = 0, hence, x − y is orthogonal to Z at ξ. Therefore, L is tangent to Z at ξ, which is impossible, since H is convex and no line through h ∈ H is tangent to Z. Consider an integral Z q (ξ) dξ Θn (x, y) = Re in , n Z (ϕ (x, y; ξ) − i0) dp (ξ) where 2

2

2

ϕ (x, y; ξ) + |y − ξ| − |x − ξ| = 2 hy − x, ξi + |y| − |x| = hz, ξ − hi , z = 2 (y − x) .

2

(6.4)

Lemma 6.5 Equation Re in Θn (x, y) = 0 holds for arbitrary x and y = 6 x.

Applications to the Spherical Mean Transform

103

Proof. Let S be the unit sphere in E n and ω ∈ S, t > 0 are spherical coordinates in E n with the pole h. We write ξ = h+tω and have p = p0t dt+dω p. This yields dξ = tn−1 dt ∧ Ω = tn−1

dp ∧ Ω, p0t

tn−1 dξ = 0 Ω, t > 0. dp pt

(6.5)

2

We denote ω± = ω ± iεz/ |z| for a small number ε > 0. For even n, we have by (6.4) 1 n ϕ (x, y; ξ)   1 1 1 = lim + n 2 ε→0 ϕ (x, y; h + tω+ )n ϕ (x, y; h + tω− )   1 1 1 = n lim + , n 2t ε→0 hz, ω+ in hz, ω− i which yields Z

1 qdξ n Z ϕ (x, y; ξ) dp  Z  1 1 qdξ 1 = lim n + n n ε→0+ 2t 2dp hz, ω+ i hz, ω− i Z Z 1 qdξ 1 = n t Z hz, ωin 2dp

Re Θn (x, y) =

(6.6)

as ε → 0 for any t 6= 0. Applying the push down operation over the central projection Z → S acting by ξ = h + tω 7→ ω to the integral (6.6), we get by (6.5) Z X µ q (ξk (ω)) Ω Re Θn (x, y) = n, 0 (ξ (ω)) t (ω) p hz, ωi k k S t k=1

where ξk (ω) = h + tk (ω) ω, k = −µ, ..., µ are zeros of p (ξ), such that t−µ < .. < t−1 < 0 < t1 < ... < tµ , as in Theorem 6.3. Let S+ be a hemisphere in S. The sum of contributions of opposite points ω ∈ S+ and −ω equals µ X k=1

=

µ

X q (ξk (−ω)) Ω q (ξk (ω)) Ω − 0 tk (ω) pt (ξk (ω)) tk (−ω) p0t (ξk (−ω)) k=1

µ X k=−µ

q (ξk (ω)) Ω . tk (ω) p0t (ξk (ω))

(6.7)

The opposite sign appears in the first line, since ∂/∂t0 = −∂/∂t for t0 = −t. It disappears in the second line, because of tk (−ω) = −t−k (ω) . The expression

104

Reconstruction from integral data

(6.7) is equal to the sum of residues of the form ρ (t, ω) =

q (tω) dt p (tω) t

at the points t = tk , −µ ≤ k ≤ µ. Integrating this form over the circle of radius R > maxS |tµ (ω)| and applying the Residue theorem gives Z µ X 1 ρ (t, ω) = res0 ρ (t, ω) + restk ρ (t, ω) 2πi |t|=R k=−µ

=

q (h) + p (h)

µ X k=−µ

q (ξk ) tk p0t (ξk )

= −res∞ ρ (t, ω) = 0. Here, ξk ∈ Z|k| ,
k = −µ, ..., µ and the residue at infinity vanishes, since ρ (t, ω) = O t−2 . It follows that µ X k=−µ

q (h) q (ξk ) =− , tk p0t (ξk ) p (h)

(6.8)

and we come up with the equation Z Z Ω Ω q (h) q (h) ± Re Θn (x, y) = = . p (h) S+ hz, ωin 2p (h) S hz, ωin The right-hand side vanishes by Theorem 7.20, and hence, Re Θn (x, y) = 0. For odd n, we argue in the similar way. For t > 0 and h + tω ∈ Z, we have 1 n (ϕ (x, y; ξ) − i0)   1 1 1 − = lim n 2i ε→0 ϕ (x, y; h + tω− )n ϕ (x, y; h + tω+ )   1 1 1 = n lim n − n ε→0 2i |t| (hz, ω− i − i0) (hz, ω+ i + i0) 1 1 = n Im n. |t| (hz, ωi − i0)

Im

(6.9)

For t < 0, −n

Im ϕ (h + tω± )

−n

= Im (t hz, ωi ± itε)

−n

= |t| −n

−n

Im (hz, ωi ∓ iε)

,

and the factor |t| appears in (6.9) instead of t . Moving integration from Z to S yields Z 1 qdξ 2iΘn (x, y) = 2i Im n dp (ϕ − i0) Z Z µ X q (ξk ) Ω = Im n. |tk | p0 (ξk ) S hz, ω − i0i k=1

Applications to the Spherical Mean Transform

105

Compare contributions of opposite points ω and −ω. Let Σ+ (ω) denote the sum in the right-hand side and Σ− (ω) be the similar sum taken for k = −1, ..., −µ. We have Σ+ (−ω) = −Σ− (ω), since |tk (−ω)| = |t−k (ω)| and p0 (h − tk (−ω) ω) = −p0 (h + t−k (ω) ω) , since ∂/∂t0 = −∂/∂t for t0 = −t. The second factor is the same: −n −n −n Im hz, −ω − i0i = − Im hz, ω + i0i = Im hz, ω − i0i . Omitting this factor, the sum of the contribution is equal to Σ+ Ω (ω) + Σ− Ω (−ω) = ΣΩ (ω) , where Σ it the sum taken for all k. Here, we take into account that equation Ω (−ω) = −Ω (ω) holds for any even dimension n − 1. After integration of the sphere, both terms give the same quantity. This yields finally µ X

Z 2iΘn (x, y) = ± Im

S+ k=−µ

=±

q (h) Im 2p (h)

Ω q (ξk ) 0 tk p (ξk ) hz, ω − i0in

Z Sn−1

Ω n. (hz, ωi − i0)

(6.10)

The right-hand side vanishes according to Theorem A.20. This, together with (6.10), implies Im Θn (x, y) = 0 and completes the proof of Lemma 6.5.  Theorem 6.4 now follows from Theorem 4.7 applied to the generating function Φ as above. We change the variable s = r2 and take into account that |∇Φ| = 2 |x − ξ| = 2r and
−1 Mf r2 , x = (2r) Rf (r, x) in loc. cit. To complete the proof, we only need to calculate the dominator Z 1 1 qdξ Dn (x) = n n−1 2 |S | Z |ξ − x|n dp for x ∈ H. By (6.5), for any point ξ ∈ Z, q (ξ) dξ q (ξ) Ω = , n |ξ − x| p0t (ξ) |ξ − x| dp where p0t (ξ) = h∇ |ξ − x| , ∇pi . We can write ξ = x + tk (ω) ω for a unique ω ∈ Sn−1 and k, 1 ≤ k ≤ µ. This yields Dn (x) =

1 n 2 |Sn−1 |

Z

µ X

Sn−1 k=1

q (ξk ) Ω , |tk (ω)| p0t (ξk )

106

Reconstruction from integral data

where p0t (ξ−k ) = −p0t (ξk ) . The sum of contributions of points ω ∈ S+ and −ω equals µ X k=1

µ

X q (ξk ) Ω q (ξ−k ) Ω + 0 |tk (ω)| pt (ξk ) |tk (−ω)| p0t (ξ−k ) k=1

µ µ X X q (ξk ) Ω sgnk q (ξk ) Ω = . = |tk (ω)| p0t (ξk ) tk (ω) p0t (ξk ) k=−µ

k=−µ

The factor sgnk appears, since ∂/∂t0 = −∂/∂t for t0 = −t. It cancels, since tk (−ω) = −t−k (ω) for all k = 1. By (6.8), the right-hand side sum is equal to −q (x) /p (x) Ω, and the integration over Z is reduced to the integration of this form over S+ . The orientation of the hemisphere is opposite to the standard orientation, since the choice of the orientation in Z. This yields Z q (x) q (x) Ω = n+1 Dn (x) = n , 2 p (x) |Sn−1 | S+ 2 p (x) and (6.1–6.2) follow, since 2r∂/∂s = ∂/∂r. 

6.3

Examples

Ellipsoid. An arbitrary ellipsoid Z is the zero set of an elliptic second order polynomial p, such that H = {p > 0} is the hyperbolic cavity. Any function f supported in H can be reconstructed by the formula n−1 Z  1 ∂ Rf (λ, ξ) qdξ 1−n p (x) f (x) = j (6.11) q (x) Z r ∂r r r=|x−ξ| dp for odd n, and by −n p (x)

f (x) = 2j

q (x)

Z Z

qdξ dp

Z 0

∞

dr2 2

|x − ξ| −

 r2

1 ∂ r ∂r

n−1

Rf (r, ξ) r

for even n, from data of spherical integrals Rf (r, ξ) , ξ ∈ Z.

(6.12)

Applications to the Spherical Mean Transform

107

Circles centered on an ellipse. Any elliptic paraboloid Γ is a noncompact oscillatory set with only one hyperbolic cavity H. Formulas (6.12) and (6.11) are valid for the polynomial p that defines Z and is positive in H. However, the integral over Z is improper and ought to be regularized by the method (6.13) below. Hypotrochoid. The “triangle” hypotrochoid Z is given by the elliptic oscillatory polynomial p(x1 , x2 ) as in Example 6.6. The curve has 3 selfintersection points. It can be approximated by the nonsingular oscillatory
curve Zε defined by the elliptic polynomial p + εq, where q = 4 x21 + x22 − 9. Theorem 6.4 holds for Zε and any small ε ≤ 0. x1 (ϕ) = 2 cos 2ϕ − cos ϕ, x2 (ϕ) = 2 sin 2ϕ + sin ϕ, 0 ≤ ϕ < 2π. On the other hand, it can be approximated by the nonsingular curves Zε =
{x : p + εq = 0}, where q = 4 x21 + x22 − 9 and ε > 0. Theorem 6.4 holds for any curve Zε with small ε ≥ 0. Half-plane. Oscillatory polynomial p (x) = x1 defines the hyperplane Z = {x1 = 0} and the cavity H = {x1 > 0} in E n . Theorem 6.4 does not apply to this polynomial directly, since it is not elliptic and the map DX : Z × R+ → T0∗ (X) is neither proper nor surjective. The image of DX misses the set S (X) of points (x, θ) ∈ T0∗ (X), with θn = 0. To give a meaning to (6.12–6.11), these integrals must be regularized. Proposition 6.6 In the case p = x1 , q = 1, integrals (6.12–6.11) can be regularized for any function f ∈ C0n−1 (H) and f ∈ C0n (H), respectively. 2

Proof. The generating function Φ (x; ξ, s) = |x − ξ| − s chosen in §6.2 fulfills ts (Φ) = 1, where ts = −∂/∂s. Taking t0 = 0, we can apply (4.6) and

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Reconstruction from integral data

get Z (n − 1)! 2 f (x) = n log |x − ξ| − s gn , j Dn (x) Σ Z   1 2 f (x) = n−1 δ0 |x − ξ| − s gn−1 , 2j Dn (x) Σ

(6.13) (6.14)

where gk = tks (M (f ) dΣ) = tks M (f ) dΣ, k = n − 1, n, since div ts = 0. By Propositions A.11 and A.9, we have   d f dX = −M (tx (f )) , ts (M (f )) = −M dΦ dΦ where tx =

1 2 |x − ξ|

X 2

(xi − ξi )

∂ . ∂xi

(6.15)


Finally, gk = M tkx (f ) dΣ for any natural k and
M tkx (f ) ≤ Ck (|ξ| + 1)−k−1 , where the constant Ck depends also on the diameter of suppf. Here, we use n−1 (6.15) and equation |∇x Φ| = 2 |x − ξ| . For odd n, condition  f ∈ C0 (X)
−n (f ) is continuous and equals O |ξ| as |ξ| → implies that function M tn−1 x n o 2 ∞. Therefore, gn−1 is integrable over Z (x) = (ξ, s) : |x − ξ| = s for any   −n−1 x ∈ H and (6.14) converges. For even n, we have M(tnx (f )) = O |ξ| , which guarantees convergence of (6.13).  Remark. Cutting off a neighborhood of infinity in the improper integrals over Σ causes error in the reconstruction whose wave front contains a neighborhood of the set of covectors missed in Im DX . The same arguments can be applied to arbitrary nonelliptic oscillatory polynomials. We skip details here.

6.3.1

Andersson’s Formula

Compare the reconstructions with the simple looking formula [4]: n−1 fˆ (ξ) = const |ξ1 | |ξ| gˆ (σ, |ξ|) ,

(6.16)

written in terms of the Fourier transform fˆ (ξ) = Fx7→ξ f (x) , gˆ (σ, ρ) = Fx0 7→σ Fr7→ρ g (x0 , r) . Here, f is a function with compact support in E n that vanishes for x1 < 0, g (x0 , r) = r1−n Rf (x0 , r), and g (x0 , r) is considered as a function in E n−1 ×E n , that is, Z F|y|7→ρ g (x0 , ρ) = exp (−i hξ, yi) g (x0 , |y|) dy, ρ = |ξ| . Rn

Applications to the Spherical Mean Transform

109

The last formula is n-dimensional Fourier transform of a function depending only on the radial coordinate. Here, the last step of inversion fˆ 7→ f causes a problem, since the right-hand side of (6.16) is in general unbounded as ξ → ∞even if f has compact support.

6.4

Time Reversal Structure

Let E n be a Euclidean space, n ≥ 2; consider the Cauchy problem for the wave equation on the space-time R × E n   2 ∂ − ∆ u = 0, (6.17) ∂t2 u (x, 0) = f (x) , u0t (x, 0) = 0 for a function f on E n . Let En+1 (t, x) be the forward propagator for (1.6). Theorem 6.7 Let p be an elliptic oscillatory polynomial in E n with the cavity H and q an arbitrary polynomial of degree < deg p that is positive in H. Formulas (6.1–6.2) are equivalent to the time reversal method for (6.17) acting on the following steps: (i) transmission (forward propagation) of a function f supported in H to the mirror manifold R × Z: Z d f 7→ u (ξ, t) = En+1 (ξ − x, t) f (x) dx, ξ ∈ Z, dt H (ii) filtration v = Fu +

4 u, t

(iii) retransmission by time reversion Z Z ∞ d qdξ g (x) = En+1 (x − ξ, t) v (ξ, t) dt , dt dp Z 0 and (iv) normalization f (x) =

p (x) g (x) . q (x)

110

Reconstruction from integral data

Geometry of the time reversal. Proof. Denote ωZ = qdξ/dp. For odd n = 2m + 1 ≥ 3, the forward propagators is  m−1   1 ∂ 2 2 En+1 (x, t) = θ (t) δ t − |x| . 2π m ∂t2 The solution of (6.17) equals ∂ 1 u (ξ, t) = w (ξ, t) , w (ξ, t) = ∂t 4π m



∂ ∂t2

m

Rf (ξ, t) , t

where Rf (ξ, t) is the integral of f over the sphere |x − ξ| = t. Denote τ = t2 , 2 σ = |x − ξ| , and have by (6.1)  n−1 Z f (x) 2n−1 ∂ Rf (ξ, t) − = n−1 ωZ p (x) j ∂t2 t Z t=|x−ξ| m

= (−1)

4 πm

Z



∂ ∂t2

Z

∞

ωZ Z

Z

m

w (ξ, t) 

t=|x−ξ| m

4 ∂ ωZ δ (τ − σ) w (ξ, t) dτ πm Z ∂τ 0 Z Z ∞  m−1 2 ∂ 1 ∂ =− m ωZ δ (τ − σ) w (ξ, t) dτ π ∂τ t ∂t Z 0  Z Z  4 1 = 4 ωZ En+1 (t, x − ξ) u (ξ, t) dt = E∗ Ef t t Z R m

= (−1)

Applications to the Spherical Mean Transform

111

is the retransmission of v = 4u/t = Fu. For even n = 2m, by solving the Cauchy problem by means of the propagator  m−1 ∂ θ (t − |x|) 1 En+1 (x, t) =  1/2 , 2π m ∂t2 2 t2 − |x| we obtain u (ξ, t) = ∂w (ξ, t) /∂t, where Z w (ξ, t) = En+1 (ξ − x, t) f (x) dx

(6.18)

H

1 4π m

Z

1 = 4π m

Z

=

∞

−∞



∂ ∂τ

∞

−∞

m−1

dρ 1/2

(τ − ρ)+

dρ

Sf (ρ, ξ) 1/2 (τ − ρ)+  m−1 ∂ Sf (ρ, ξ) . ∂ρ

Here, we denote τ = t2 , Sf (σ, ξ) +

Rf (r, ξ) 2 , σ = |x − ξ| r

and Rf (r, ξ) = 0 for r < 0. Solving (6.18) by Abel’s method, we find 

∂ ∂σ

m−1 Sf (σ, ξ) = 4π

m−1

d dρ

Z

w (ξ, t) 1/2

(σ − τ )+

dτ.

By (6.12), m−1

f (x) (−1) = n−1 n p (x) 2 π

m−1

=

(−1) πn

Z Z

Z ωZ Z

n−1 1 ∂ Sf (ρ, ξ) 2 |x − ξ| − r2 r ∂r 0 !  n−1 Z ∞ ∂ 1 dρ Sf (ρ, ξ) , π n 0 σ − ρ ∂ρ

Z ωZ

∞

dr2



(6.19)

112

Reconstruction from integral data

where ρ = r2 . The inner integral is calculated by means of Proposition A.7 and (6.18). Integrating by parts yields 1 πn =

∞

Z 0

4

dρ σ−ρ Z ∞



∂ ∂ρ

n−1

dρ σ−ρ

Sf (ρ, ξ) 

∂ ∂ρ

m+1 Z

w (ξ, t)

dτ 1/2 (ρ − τ )+  m Z Z ∂ 4 ∂ 1 dρ = m+1 w (ξ, t) dτ 1/2 π ∂τ ∂τ σ − ρ R (ρ − τ )+  m Z ∂ 4 ∂ dτ w (ξ, t) =− m 1/2 π ∂τ ∂τ R (τ − η)+  m Z 2 1 ∂ dτ = m u (ξ, t) 1/2 π t ∂τ R (τ − σ)+  m Z 2 1 ∂ dt = m u (ξ, t)  1/2 π t ∂τ 2 R t2 − |x − ξ| Z ∂ 1 =4 u (ξ, t) En+1 (−t, x − ξ) dt. t ∂t π m+1

0

This, together with (1.32), gives Z Z f (x) 4 ∂ = ωZ u (ξ, t) En+1 (x − ξ, −t) dt, p (x) t ∂t Z where again v (ξ, t) =

6.5

4 u (ξ, t) .  t

Boundary Isometry for Waves in a Cavity

Formulas (6.1–6.2) imply Corollary 6.8 Let p be an elliptic oscillatory polynomial in Rn , n ≥ 1 and q be an arbitrary polynomial of degree < deg p. For an arbitrary function f supported in H, Z Z ∞ Z 2 dt q (ξ) dξ 2 q (x) dx 4 |u (ξ, t)| = |f (x)| . (6.20) t dp (ξ) p (x) Z 0 H The set Z is endowed with the orientation defined by the inward normals with respect to H.

Applications to the Spherical Mean Transform

113

Proof. In the case n = 1, the zero set Z of p consists of real simple zeros ξk , k = 1, ..., m, which divide the real axis in m + 1 intervals. Each of the intervals is a hyperbolic cavity. Fix a cavity H = (a, b), where a ≥ −∞, b ≤ ∞. The value of the form qdξ/dp at a point ξk ∈ Z is equal to q (ξk ) εk , where εk is the orientation of Z, that is, εk = 1 if ξk < a and εk = −1 if ξk > b. For a function f supported in H, the solution of the initial problem (6.17) is equal to u (x, t) = 1/2 (f (x + t) + f (x − t)) . According to the geometry of characteristic of the d’Alembert equation, we have Z Z 2 X εk q (ξk ) Z ∞ |u (ξk , t)|2 qdξ ∞ |u (ξ, t)| dt = 4 dt 4 t p0 (ξk ) 0 t 0 Z dp ξk b a ξk 1, with X + {x ∈ E n : x1 > 0} . The function 2

2

Φ (x, σ) + |x − σ| − r2 = |x0 − ξ| + x21 − 2rx1 , defined on X × Σ, generates the family of spheres in X tangent to the hyperplane P = {x : x1 = 0}. Here, x0 , ξ ∈ P, r > 0, and manifold Σ = R+ × P is supplied with the volume form drdξ.

Spheres tangent to a hyperplane. Theorem 6.10 For any odd n and any f ∈ C0n−1 (X) , the equation holds Z Z ∞   2n dr 2 f (x) = n−1 x21 δ (n−1) x21 + |x0 − ξ| − 2rx1 Rf (r, ξ) dξ. (6.22) j r P 0 For even n and f ∈ C0n (X) , we have Z Z ∞ (n − 1)!2n+1 2 Rf (r, ξ) dr  n dξ. f (x) = x 1 2 jn r 2 0 P 0 x + |x − ξ| − 2rx 1

(6.23)

1

Proof. We have |det J (Φ)| = 2n+1 rx1 6= 0, which implies that DΦ is a local diffeomorphism. It is not proper and surjective, since any covector (x, η) ∈ T0∗ (X) with η = (η1 , 0, ..., 0) , η1 > 0 does not belong to Im DΦ . It is easy to check that there is no conjugate points. Condition (iii) of Theorem 4.7 is fulfilled according to

116

Reconstruction from integral data

Lemma 6.11 Qn (x, y) = 0 for y = 6 x.  Proof . Forany point (x, σ), such that Φ (x; r, ξ) = 0, we have r = 2 x21 + |x0 − ξ| /2x1 . Therefore, for any x ∈ Z (y) , 2

Φ (x; r, ξ) = x21 + |x0 − ξ| − 2rx1 = ϕ (x, y; ξ) /y1 , 2

2

2

ϕ (x, y; ξ) + y1 |x| − x1 |y| + 2 hx1 y − y1 x, ξi − (x1 − y1 ) |ξ| , which yields Qn (x, y) =

y1n

n

Z

Re i

P

dξ n, (ϕ (x, y; ξ) − i0)

(6.24)

−1

since dΣ/d  σ Φ = (2y  1 ) dξ in Z (y) . Suppose that y1 6= x1 and write ϕ = 2 (y1 − x1 ) |η| − ρ , where 2

η=ξ−

x1 y − y1 x x1 y1 |x − y| , ρ= 2 > 0. y1 − x1 (x1 − y1 )

By (6.24), Qn (x, y) = where the integral Z  P

y1n (y1 − x1 )

dη 2

|η| − ρ − i0

n

2n

Z

dη

Re i

n = Sn−2

P

Z



2

|η| − ρ − i0

∞

0

(t2

n ,

tn−2 dt n − ρ − i0)

is calculated by terms of spherical coordinates (t, ψ) on Rn−1 . For even n, we have Z ∞ Z ∞ 1 tn−2 dt tn−2 dt = Re Re n n, 2 2 (t2 − ρ − i0) −∞ (t − ρ − i0) 0 and the right-hand side vanishes by Proposition A.21. For odd n ≥ 3, we change variable s = t2 and get Z ∞ Z ∞ (n−3)/2 tn−2 dt 1 s ds Im = Im n 2 − ρ − i0)n 2 (t (s − ρ − i0) 0 0 = πRess=ρ

s(n−3)/2 ds n = 0, (s − ρ)

which implies again Qn (x, y) = 0. In the case y1 = x1 , equation Qn (x, y) = 0 holds by continuity.  The dominator Dn (x) can be calculated by means of (4.7). Set Z (x) is parametrized by ξ ∈ P. We have 2

dΣ dξ x2 + (x0 − ξ) =− , |∇x Φ (x, σ)| = 2r = 1 dr Φ 2x1 x1

Applications to the Spherical Mean Transform

117

in Z (x) and Dn (x) =

6.7.1

1 2x1 |Sn−1 |

xn1 dξ

Z P



2

x21 + (x0 − ξ)

n =

1 2n+1 x21

.

Regularization of Improper Integrals

To complete the proof of the theorem, we need to regularize integrals (6.23– −1 6.22). For this, we argue as in §6.3 and take the field tr = − (2x1 ) ∂/∂r fulfills tr (Φ) = 1 and set t0 = 0. By (4.6), we obtain reconstruction in a regularized form Z (n − 1)! 2 f (x) = n log x21 + |x0 − ξ| − 2rx1 gn , (6.25) j Dn (x) Σ Z   j1−n 2 f (x) = δ0 x21 + |x0 − ξ| − 2rx1 gn−1 , (6.26) 2Dn (x) Σ where gk = tkr (M (f ) dΣ) = tkr M (f ) dΣ, k = n − 1, n, since div tr = 0. By Propositions A.11 and A.9, we have   d f dX tr (M (f )) = −M = −M (tx (f )) , dΦ dΦ where tx =

1 2 |x − ξ|

X 2

(xi − ξi )

∂ . ∂xi

Finally, for any natural k,
gk = M tkx (f ) dΣ.
For odd n and any function f ∈ C0n−1 (X) , function M tkx (f ) is continuous  −n and equals O |ξ| as |ξ| → ∞. This guarantees convergence of (6.26). For  
−n−1 , hence, (6.25) even n and f ∈ C0n (X) , we have M tkx (f ) = O |ξ| absolutely converges.

6.8

Summary of Spherical Mean Transform

For a continuous function f on a Euclidean space E n with compact support, the integral Z Rf (S) = f dS S

118

Reconstruction from integral data

is well defined on the variety Λ of all spheres S and hyperplanes in E n . Equation 2 a0 |x| + 2 ha, xi + b = 0 2

defines a not empty hypersurface if |a| − a0 b > 0. It is a sphere if a0 = 6 0 and a hyperplane if a0 = 0. The vector (a0 , a, b)n can be considered a point o 2 in the projective space Pn+1 , and the set Λ + |a| − a0 b > 0 in Pn+1 is bounded by a quadratic cone. The problem is to reconstruct a function f defined on a subset X ⊂ E n+1 from knowledge of integrals Rf (S) over spheres (hyperplanes) S ∈ Σ, where Σ is a n-dimensional subvariety of Λ. For the following varieties X and Σ, a FBP reconstruction method is known: I. X = E n+1 , Σ is the variety of all hyperplanes, R is the Radon transform; see Chapter 1. II. X = Sn , Σb is the variety of spheres S = Sn ∩ H, where H is a hyperplane in E n+1 tangent to a ball b b B and B is the unit ball (R is the Minkowski–Funk transform if b is the center of B); §5.1–2. III. X = B, Σ is the set of subvarieties B∩S or S\B, where S is a sphere or a hyperplane orthogonal to ∂B (totally geodesic hypersurfaces in hyperbolic space); §5.3. IV. X = B, Σr is the family of surfaces B ∩ S or S\B, where S is a sphere, such that the intersection S ∩ ∂B has radius r (equidistant spheres; see §5.4). V. X = B, Σ is the variety of spheres S ⊂ B or S ⊂ E n \B tangent to ∂B (horospheres); §5.4. VI. X is a half-space, ΣH is the variety of spheres tangent to a hyperplane H; §5.6. VII. X = H is a hyperbolic cavity of an oscillatory hypersurface Z ⊂ Rn , Σ is the set of spheres with centers in Z; §6.4–5. VIII. X = SO (3), Σ is the set of geodesics (circles) in X; §3.4. IX. X is a half-disc, Σ is a family of circular arcs in plane orthogonal to the diameter in a half-disc [104]. X is a 3D half-ball, Σ is the family of hemispheres in X orthogonal to the equatorial plane; see Example 3.5.

Applications to the Spherical Mean Transform

119

Families of spheres in a ball admitting FBP reconstruction.

6.9

Bibliography Notes

§6.1 Oscillatory polynomials appeared in analysis of hyperbolic differential equations by Petrovski˘ı [116] and Garding [37]. Arnold [5] applied the property of hyperbolic polynomials for construction of a distribution of electric charge that generates the zero electrical field in an open set. §§6.2–3 Reconstruction problem for the family spheres centered on a hypersurface Z is motivated by applications to the SAR (synthetic aperture radar) technic [18],[90] and to velocity inversion methods in geophysics [124],[12] and in photo-thermo-acoustic tomography [72]. For a hyperplane Z, reconstructions were given by Fawcett [31] and Andersson [4]. They developed a method based on relating the Fourier transform of a function and its integral data. However, application of Andersson’s formulas meets difficulty, since the result is interpreted as a singular function. J. Klein [71] gave a detailed analysis of convergence in Andersson’s method. Papers [26], [89] are also devoted to this problem. Reconstruction problem for Fawcett-Andersson’s geometry was also addressed in Gel’fand-Graev-Gindikin [43] Chapter 5.2. Explicit reconstructions of FBP type for spheres Z are given in Finch et al. [33] (odd dimensions) and in [34] (even dimensions). Xu and Wang [141] used solutions of the wave equation in the frequency domain for inversion of the photoacoustic transform for spheres, cylinders, and planes Z. An alternative

120

Reconstruction from integral data

method for FBP reconstructions was used by Kunyanski [74]. Linh [76] found some consistency conditions for the integral data for spherical center set Z. Kunyanski [75] studied reconstruction with center set that is the boundary of a square or of a cube. A reconstruction for arbitrary ellipsoids Z in R3 was done in Natterer [87] and in Palamodov [108] for ellipsoids of arbitrary dimension; see also Salman [127] and Haltmeier [51]. Haltmeier et al. [50] proposed a method of photoacoustic 3D tomography based on data of “integrated” line or plane arrays of detectors instead of point detectors. These data are modelled by data integrals of spherical means. See a survey of related analytic reconstructions in Elbau-Scherzer-Schulze [29]. §6.4 Results of this section were published in [111]. §6.5 The isometry identity of this kind were obtained by A. L. Bukhgeim and Kardakov [16] (odd n) and Narayanan and Rakesh [82] (even n) for p (x) = x1 , g = 1. It looks like Equation (6.20), however, the factor 4 is absent. Nevertheless, (6.20) agrees with the results of [16] and [82], since the authors extend f to an even function of x1 . §6.6 Agranovsky, Kuchment, and Quinto [1] showed that the totality of the orthogonality conditions like (b), together with the moment conditions, describe the range of the spherical mean operator if the central set is a sphere of arbitrary dimension n. Simultaneously, Finch and Rakesh [35] has shown that for odd n the moment conditions can be dropped. This fact was checked for all n in [2]. Theorem 6.5, giving only necessary condition, was proven in a slightly different form in [1]. Moment conditions of a different kind are proposed in [105]. §6.7 For the family of spheres tangent a hyperplane, Gel’fand-GraevGindikin [43] applied for the reconstruction the method of “kappa”-operator. The technical problem is a homotopy between some noncompact cycles in Λ. For the family Σ (Y ) of tangent spheres to an arbitrary compact manifold Y ⊂ Rn , Goncharov [45] applied a modified method of “kappa-operator”. The reconstruction problem was considered also in [104] as the Goursat type problem for the Darboux equation with data on Σ (Y ). In both methods, the difficulty of a reconstruction is to give sense to an integral with the singular kernel.

Appendix A

A.1

Distributions and Generalized Functions

Definition. Let X be an open set in Rn , a function ϕ on X is called k-smooth for some natural k or k = ∞ if derivatives Di ϕ are continuous on X for |i| ≤ k, where i i Di + (∂/∂x1 ) 1 ... (∂/∂xn ) n , |i| = i1 + ... + in for a coordinate system x1 , ..., xn on Rn . The space of k-smooth function in X is denoted C k (X) ; notation C0k (X) will be used for the subspace of functions with compact support. Elements of the space D (X) = C0∞ (X) are called test functions in X. This is a vector space over the field C of complex numbers. A sequence ϕk , k = 1, 2, ... converges to ϕ in the space D (X) , if (i) ∪ supp ϕk ⊂ K for a compact set K b X and (ii) Di (ϕk − ϕ) ⇒ 0 uniformly in K for arbitrary multiindex i. A functional u : D (X) → C is called continuous, if u (ϕk ) → u (ϕ) for any convergent sequence ϕk → ϕ. Any linear continuous functional on D (X) is called distribution on X. Any distribution u on X satisfies the inequality |u (ϕ)| ≤ CK kϕkK,m

(A.1)

for any compact set K ⊂ X and some integer m = m (K) , where kϕkK,m + max max Di ϕ (x) . x∈K |i|≤m

Conversely, any linear functional u on D (X) that possesses this property is a distribution. The product dx = dx1 ∧ ... ∧ dxn is a volume form on X (see A.3). A test density ρ on X is a smooth density, with compact support, that is, ρ = ϕdx, where ϕ is a test function on X. The space of test densities with compact supports in X is denoted K (X) . A sequence of test densities ρk converges to ρ if . the sequence of test functions ϕk = ρk /dx converges to ϕ = ρ/dx in D (X). Any linear continuous functional f on the space K (X) is called generalized function on X (sometimes we will call f simply function). For a function v, the distribution u = vdx is defined by u (ϕ) + v (ϕdx) . This is one-to-one correspondence as far as the volume form dx is fixed. Nevertheless, we distinguish 121

122

Reconstruction from integral data

functions and distributions, since derivations, coordinate transformations, pull back and push forward operations act on them differently. Example A.1. A function g : X → C is locally integrable, if its restriction to an arbitrary compact subset K ⊂ X is integrable in sense of the Riemann or the Lebesgue theory. Any locally integrable function g on X defines a generalized function [g] and a distribution [gdx] by Z Z [g] (ρ) = [gdx] (ϕ) = gϕdx = gρ, (A.2) X

X

where ρ = ϕdx ∈ K (X) . The functional [g] equals zero if and only if g vanishes almost everywhere. For any smooth volume form ∆ on X (see §7.3) and a generalized function u, the product u∆ is well defined as a distribution on X. The support of a distribution u (function) defined on X is the smallest closed subset G ⊂ X, such that u (ϕ) = 0 (u (ρ) = 0) for an arbitrary test function ϕ (test density ρ) supported in X\G. The support is denoted suppu. Example A.2. Dirac delta-function was the first clearly nonclassical object. The usual definition is confusing. To avoid the confusion, we use the following splitted definition. The delta-distribution ∆a ∈ D0 (X) at a point a ∈ X is continuous functional ∆a (ϕ) = ϕ (a) on the space of test functions ϕ ∈ D (X) . The support of ∆a is the one-point set {a} . If a volume density dX on X is fixed, then the delta-function at a point a ∈ X is defined as the functional δa on K (X), such that δa (ϕdX) = ϕ(a). We can write δa = ∆a /dX and ∆a = δa dX. Proposition A.1 [Schwartz theorem] Any distribution u supported by a point a ∈ X has the form u = A (D) ∆a , where A is a differential operator. If u satisfies (A.1) for a compact neighborhood K of a, then degA ≤ m. For a proof, see [129]. A family of distributions (generalized functions) uλ defined for λ in an open set Λ ⊂ C is called holomorphic if the function uλ (ϕ) is holomorphic on Λ for any test function (density) ϕ. Proposition A.2 For any λ ∈ C, the limits Z lim (x ± εı)λ ϕ(x)dx ε→+0

exist for arbitrary ϕdx ∈ K(R). They are generalized functions on R depending morphologically on λ. These functions are denoted (x ± 0ı)λ . Proof. The function (x + εı)λ is holomorphic with respect λ on the whole complex plane for any x and any ε = 6 0. For an arbitrary natural k, integrating by parts k times yields Z Z k (−1) (x ± εı)λ ϕ(x)dx = (x ± εı)λ+k ϕ(k) (x)dx. (A.3) (λ + 1) ... (λ + k)

Appendix

123

The function (x ± εı)λ+k is a continuous for ε ∈ R and Re λ > −k. Therefore, the right-hand side of (A.3) is also a continuous function of ε and of λ except for λ = −1, −2, ..., 1 − k, Re λ > −k. The left-hand side is an entire function of λ for ε = 6 0. By the maximum principle, it is continuous for ε → 0 for all λ, Re λ > −k, and the limit is a holomorphic function. This is true for all λ, since k is arbitrary and the functional (x ± 0ı)λ is an entire function of the parameter λ ∈ C.  Proposition A.3 We have jδ0 = (x − 0ı)

−1

− (x + 0ı)

−1

(A.4)

and Z (x + 0ı)

−1

Z ρ+

(x − 0ı)

−1

Z ρ = 2 lim

ε→0

|x|>ε

ρ , ρ ∈ K (R) . x

(A.5)

Proof. Set λ = −1 in (A.3), integrate by parts, and take the limit Z Z (x ± 0ı)−1 ϕ(x)dx = − log(x ± 0ı)ϕ0 (x)dx. We have log(x − 0ı) − log(x + 0ı) = jθ (−x) , where θ (t) = 1 for t > 0 and θ (t) = 0 for t < 0. Therefore, Z

(x − 0ı)−1 ϕ(x)dx −

Z

(x + 0ı)−1 ϕ(x)dx = j

Z

0

ϕ0 (x) dx = jϕ (0) ,

−∞

which proves (A.4). The sum equals Z Z (x − 0ı)−1 ϕ(x)dx + (x + 0ı)−1 ϕ(x)dx Z ∞ Z dx 0 ϕ.  =2 log (|x|) ϕ (x) dx = 2 lim ε→0 −∞ |x|>ε x The limit in the right-hand side is the distribution denoted by [dx/x]. Taking k-th derivative of (A.4), we obtain   (k) −k−1 −k−1 jδ0 (x) = k! (x − i0) − (x + i0) . (A.6) Example A.3. Let x+ = max (x, 0) , x− = max (−x, 0) . Functions xλ± are locally integrable for any λ, Re λ < −1, and we have (x ± 0ı)λ = xλ+ + e±λπı xλ− .

124

Reconstruction from integral data

Inverting these relations, yields for λ 6= −1, −2, ... xλ+ =

eλπı (x − 0ı)λ − e−λπı (x + 0ı)λ , 2i sin λπ

(A.7)

and similarly for xλ− . The numerator and the denominator are entire functions of λ. It follows that xλ± has meromorphic continuation to a family with first order poles λ = −1, −2, ....

A.1.1

Operations

Linear operations, multiplication by a smooth function a, and partial derivatives are defined for distributions u and for generalized functions v as follows: au (ϕ) = u (aϕ) , av (ρ) = v (aρ) ,   ∂ϕ , i = 1, ..., n. ∂xi

∂u (ϕ) = −u ∂xi

These definitions are consistent with the rule (A.2). In particular, we have ∂/∂x(x ± 0ı)λ = λ(x ± 0ı)λ−1 for all λ. See A.3 for more operations on distributions and arbitrary manifold X.

A.1.2

Tempered Distributions

Let V be a real vector space of dimension n. The Schwartz space L (V ) of test functions on a vector  space V consists of smooth functions ϕ on V , such −q i that D ϕ = O |x| at infinity for any q ≥ 0 and any multiindex i = n (i1 , ..., in ) ∈ Z . This property does not depend on the choice of a linear coordinate system x = (x1 , ..., xn ) on V. This space is the basis for Schwartz’s theory of the Fourier transform of distributions. A sequence of functions ϕi converges to ϕ in L (V ) if kϕi − ϕkk → 0 as i → ∞ for any natural k, where k kϕkk = max sup (|x| + 1) Di ϕ(x) |i|≤k

x

is a norm on the Schwartz space. The dual space L0 (V ) is the space of all linear continuous functions on L(V ). An element of this space is called tempered distribution in V . The functions (x ± 0ı)λ are tempered for all λ. It is easy to check by means of (A.3). The Schwartz test density on V is, by definition, a product ρ = ϕdx, where ϕ is a Schwartz function. We denote by Ld (V ) the space of Schwartz densities on V. An arbitrary element of the dual space L0d (V ) is called tempered function on V . For an arbitrary tempered function v, the product u = vdx is a tempered distribution, and any tempered distribution can be written in this form.

Appendix

125

A sequence of distributions uk (or of tempered functions uk ) has a weak limit if uk (ϕ) → w (ϕ) for any test function ϕ (density) where w (ϕ) ∈ C. Example A.4. Euler-Riesz kernels. For an arbitrary complex λ, Re λ > 0, Hλ =

xλ−1 + dx Γ (λ)

is a locally integrable density. This follows from (A.7), since (x ± 0ı)λ are tempered functions. Proposition A.4 The family H λ has holomorphic continuation into the whole complex plane C. It satisfies the differential equation d λ+1 H = Hλ dx

(A.8)

for all λ, and we have (k)

H −k = δ0 dx +

dk δ0 dx, k = 0, 1, 2, .... dxk

(A.9)

Proof . The family H λ has holomorphic continuation for λ = −1, −2, ..., since numerator and denominator have poles of order 1. For Re λ > 0, we have Z 1 d λ+1 H (ϕ) = − xλ ϕ (x) dx dx Γ (λ + 1) Z 1 = xλ−1 ϕ (x) dx = H λ (ϕ) , Γ (λ) where we integrate by parts and take into account that xλ−1 ϕ (x) vanishes at x = 0 and at x = ∞. It follows that Equation (A.8) holds for Re λ > 0. Both parts are entire functions of λ, hence, by uniqueness of analytic continuation, it holds for all λ. To calculate H 0 , we take a Schwartz function e that coincides with exp (−x) for x > 0. We have for Re λ > 0 H λ (ϕ) = H λ (ϕ − ϕ (0) e) + ϕ (0) H λ (e) , Z ∞ 1 H λ (e) = xλ−1 exp (−x) dx = 1, Γ (λ) 0 Z 1 H λ (ϕ − ϕ (0) e) = xλ−1 (ϕ (x) − ϕ (0) e (x)) dx. Γ (λ) The last integral has analytic continuation for Re λ > −1, since the function ϕ (x) − ϕ (0) e (x) vanishes at x = 0. On the other hand, the denominator Γ has a pole at λ = 0, hence, the right-hand sides vanish at λ = 0. Therefore, H λ (ϕ) → ϕ (0), as λ → 0, that is, H 0 = δ0 dx. Equations (A.9) with k = 1, 2, ... follow from (A.8). 

126

Reconstruction from integral data

Example A.5. For Euclidean coordinates (x1 , x2 ) on a plane, we have t

div div

x

∂ x1 ∂ x2 + = 2πδ0 , ∂x1 |x|2 ∂x2 |x|2

2

=

2

= 0,

|x| t ∗ x |x|

where the column t x is transposed to the row x = (x1 , x2 ) and x∗ = (−x2 , x1 ) . For a proof, we apply the left-hand side to an arbitrary test density ϕdx on E and use polar coordinates (r, θ) : Z t x hx, grad ϕi div 2 (ϕdx) = − dx 2 |x| |x| Z 2π Z ∞ ∂ϕ (r, θ) =− dθ dr = 2πϕ (0) . ∂r 0 0 The second equation can be proven in the same way.

A.2

Fourier Transform

Let ρ be a density on a vector space V. One can write ρ = f dx for a function on V and a volume form dx + dx1 ∧ ... ∧ dxn invariant with respect to shifts in V. The form dx + dx1 ∧ ... ∧ dxn fulfills this condition if x = (x1 , ..., xn ) is a linear coordinate system on V. Suppose that the density is integrable, that is, integral Z Z |ρ| ≡ |f | dx V

V

converges. The Fourier transform of ρ is given by the integral Z F (f dx) (ξ) = fˆ(ξ) = f (x) exp(−j hξ, xi)dx, where hξ, xi + ξ1 x1 + ... + ξn xn . The real variables ξi , i = 1, ..., n form the dual coordinate system on the dual vector space V ∗ . This makes the operator F : ρ 7→ F (ρ) independent on the choice of the coordinate system on V . The adjoint Fourier integral can be written in the similar form Z F ∗ (γ)(x) = g(ξ) exp(j hξ, xi)dξ, where γ = gdξ is an integrable density in V ∗ and dξ = dξ1 ∧ ... ∧ dξn .

Appendix

A.2.1

127

Basic Properties

I. For any integrable function f in V , Z |F (f dx) | ≤

|f | dx. V

II. F ∗ is inverse to F in the sense that for any integrable function f in V , such that fˆ is integrable in V ∗ , F ∗ (F (f dx)dξ) = f. III. Under the same conditions, Z 2 Z ˆ 2 f dξ = |f | dx. This equation allows to extend the Fourier transform to the isometry in the spaces of square integrable functions. IV. If f is an integrable function and possesses integrable derivative ∂f /∂xi for some i, 1 ≤ i ≤ n, then d ∂f = jξi fˆ(ξ). ∂xi V. If f and xi f are integrable functions, then the function F (f ) has a continuous derivative in ξi and ∂ fb d = −jx if . ∂ξi Notation. We denote by Lf (V ) and Ld (V ) the Schwartz spaces of test functions, respectively, of test densities on V. Recall that ϕdx ∈ Ld (V ) if ϕ ∈ Lf (V ) . Proposition A.5 If a volume form dx on V is chosen, the Fourier transform acts continuously on the Schwartz spaces: F : Ld (V ) → L(V ∗ ). A proof can be done by means of properties I, IV, and V.  Definition. The Fourier–Schwartz transform of tempered distributions is the dual operator F 0 : L0 (V ∗ ) → L0d (V ) defined by F 0 (v)(ϕdx) + vF (ϕdx) for v ∈ L0d (V ∗ ). Any functional v is a tempered distribution, and its image F 0 (v) is a tempered function. The same construction applied to F ∗ makes an operator F ∗0 , which is inverse to F 0 .

128

Reconstruction from integral data

Proposition A.6 The Fourier–Schwartz transform matches the classical Fourier transform and makes the commutative diagram F 0 : L0 (V ) → L0d (V ∗ ) ∪ ∪ F : Ld (V ) → L (V ∗ ) . Proof. An arbitrary test density ρ =R ϕdx defines a tempered distribution [ρ] by means of integration [ρ] (f ) = f ρ, f ∈ L(V ). Then F 0 ([ρ]) is the tempered function on V ∗ defined by F 0 ([ρ]) (σ) = [ρ] (F (σ)) for any test density σ in V ∗ . We obtain [ρ] (F (σ)) = F ([ρ]) (gdξ) by changing the order of integrations. This implies commutativity of the above diagram.  Later we make no difference between F and F 0 . Examples A.5. F (δ0 dx) = [1], A.6. F ([dx/πx]) = −πisgnξ, k k A.7. F ((x + 0ı)−1−k dx) = −jξ+ , F ((x − 0ı)−1−k dx) = jξ− for k = 0, 1, 2, ...,

−λ −λ λ λ A.8. F H λ = (jξ +
0) P , F H− = F (H ) = (−jξ + 0) for λ ∈ C, P A.9. F δk (the Poisson formula), k∈Z δk dx = A.10. F (exp(−2πq(x))dx) = | det q|−1/2 exp(−2πq ∗ (ξ)), where X q(x) = 1/2 q ij xi xj , Re q (x) ≥ 0, det q = det{q ij },  −1 P . and q ∗ = 1/2 qij ξ i ξ j where {qij } = q ij . See A.4 for more formulas.

A.2.2

Convolution

Let u = f dx and v = gdx be locally integrable densities on V. The convolution  Z Z u∗v = f (x − y) g (y) dy dx (A.10) is well defined as a density, provided the interior integral converges in mean for any x ∈ V. It is the case if the densities u and v are integrable and we have Z Z Z Z Z |u ∗ v| = f (x − y) g (y) dy dx ≤ |f | dx |g| dx. Moreover, v ∗ u = u ∗ v and the equation F (u ∗ v) = F (u) F (v)

(A.11)

Appendix

129

holds, where F (u) and F (v) are continuous functions. Example A.11. Equation µ λ+µ λ H+ ∗ H+ = H+

(A.12)

holds for arbitrary λ, µ ∈ C. Here, the convolution is defined, since the interior integral in (A.10) is always supported in the segment [0, x] . In the case Re λ > 0, Re µ > 0, it can be calculated by Euler’s method, which implies (A.12). This equation holds for all λ and µ, since both sides are entire functions of these arguments. In particular, H −1 ∗ H 1/2 ∗ H 1/2 = H 0 = δ0 , that is, Z 0 Z y g (y) dy f (t) d √ √ = πf (x) , g (y) = dt. dx −∞ x − y y−t −∞ This is the inversion formula for the Abel transform A = π 1/2 H 1/2 . A convolution can be defined and fulfills (A.11) for some pairs of nonintegrable densities. A reasonable construction of convolution of distributions is equivalent to the problem of consistent definition of the product F (u) F (v) of generalized functions. This problem has, however, no satisfactory solution in general. The Hilbert transform of a function f with compact support is defined pointwise as the principal value integral Z Z 1 ∞ f (y) dy dy Hf (x) = = lim . f (y) ε→0 π −∞ x−y |x−y|>ε x − y It is equal to the convolution with the distribution [dx/πx]; see (A.5). By Example A.6, c (ξ) = −ı sgn ξ fb(ξ). Hf

(A.13)

It follows that H is extended to a unitary operator on L2 (R), such that H2 = −I, where I is the identity operator. Proposition A.7 We have Z ∞ −∞

dρ (τ −

1/2 ρ) ρ+

=

π 1/2 τ−

,

(A.14)

where the left-hand side is the principal value integral. Proof. The integral can be defined as convolution Z

∞

dσ −∞

dρ 1/2

(τ − ρ) ρ+

=

dρ dρ ∗ ρ ρ1/2 +

! (τ )

of distributions, neither of which is integrable, but the second one is locally

130

Reconstruction from integral data

integrable. To calculate this convolution, we regularize the first factor by con
volution with the test density σε = exp −επρ2 dρ, ε > 0. The convolution Z ∞    dρ 2 dρ exp −ε |τ − ρ| Rε + ∗ σε = dτ ρ ρ −∞   −1 is smooth and decreases as O |τ | at infinity. The integral Z 0 dτ dρ Sε + Rε ∗ 1/2 = Rε (τ − ρ) 1/2 ρ −∞ τ− is the regularized convolution (A.14). It is an improper Riemann integral with singular points ρ = 0, ∞ that converges for any τ. Therefore, this convolution is well defined by (A.10). The Fourier transform of this convolution is equal to     dρ dρ dρ ∗ σε ∗ 1/2 F Rε ∗ 1/2 = F ρ ρ ρ       dρ dρ 1/2 1/2 =π F F (σε ) F F H , ρ ρ1/2 where by (A.13)     πξ 2 dρ Fρ7→ξ = πisgnξ, F (σε ) = exp − ρ ε     dρ −1/2 Fρ7→ξ = F π 1/2 H 1/2 = π 1/2 (jξ + 0) ρ1/2   −1/2 −1/2 = 2−1/2 i−1/2 ξ+ + i1/2 ξ− ,   −1/2 −1/2 F (Sε ) = F (σε ) π2−1/2 i1/2 ξ+ + i−1/2 ξ−   1/2 = F (σε ) πF π 1/2 H− ! ! dσ dσ = F (σε ) πF → πF 1/2 1/2 σ− σ− as ε → 0. It follows that the regularized convolution Sε weakly converges to the right-hand side of (A.14). 

A.3 A.3.1

Manifolds and Differential Forms Manifolds

A k-smooth manifold X of dimension n is a topological space together with a set of charts (U, µ) , where U is an open set in X and µ : U → µ (U ) ⊂ Rn is

Appendix

131

a homeomorphism, such that the connecting map νµ−1 is k-smooth (in its domain of definition) for any pair (µ, ν) of charts. If x1 , ..., xn are coordinates on Rn , the functions xi (µ (p)) , i = 1, ..., n are local coordinates of a point p ∈ U. An orientation of a k-smooth manifold X for k ≥ 1 is a choice of a set A of charts (atlas) that cover all X, such that the Jacobian det ∂y/∂x is positive for any two charts in A. An object is called smooth if it is ∞-smooth. Example A.12. Let V be a vector space of dimension n. The set of subspaces L ⊂ V of dimension 1 has the structure of manifold of dimension n−1; it is denoted P (V ) and called the projective space of V . For a coordinate system x1 , ..., xn on V , the set Uj of lines L, such that xj = 6 0 on L is open in P (V ) . The coordinates x1 , ..., xj−1 , xj+1 , ..., xn of the intersection point L ∩ {xj = 1} form a local coordinate system νj on Uj , j = 1, ..., n. The connecting map equals νj νk−1 (x) = (x1 /xj , ..., 1/xj , ..., xn /xj ) and is real analytic for any j = 6 k. Real projective space Pn−1 = P (V n ) is orientable for even n and nonorientable for odd n. Example A.13. The set L (V ) of all straight lines in V is a manifold of dimension 2n − 2. Let x1 , ..., xn be linear coordinates on V. Take a hyperplane H0 = {x : x1 = 0} and consider the set U1 of lines L that are not parallel to H0 . Let (0, y) and (1, z) be the points of intersection of H0 and H1 = {x1 = 1} with L. The map µ1 : U1 → R2n−2 L 7→ (y, z) is a chart, and similar maps µk : Uk → R2n−2 , k = 2, ..., n are other charts on the manifold L (V ) , since the connecting maps µi µ−1 j are given by regular rational functions. This manifold is orientable for odd n and nonorientable for even n. Let X and Y be smooth manifolds. A continuous map f : X → Y is called k-smooth map of manifolds if for any charts (U, µ) on X and (V, ν) on Y the restriction fU V : U ∩ f −1 (V ) → V of f is k-smooth, with respect to the respective local coordinates on U and V. The map f has rank k if rank {∂y/∂x} = k for any chart y on Y and any chart x on X. A k-smooth map f : X → Y is called k-diffeomorphism, if there exists a k-smooth map f −1 : Y → X that is two-side inverse to f. A subset Y ⊂ X is called submanifold if for any point y ∈ Y there exists a neighborhood U with coordinates x1 , ..., xn on X and smooth functions ϕ1 , ..., ϕk on U such that rank

∂ (ϕ1 , ..., ϕk ) =k ∂ (x1 , ..., xn )

and Y ∩ U = {x ∈ X : ϕ1 (x) = ... = ϕk (x) = 0} . The set Y has structure of smooth manifold of dimension n − k. By the rank condition, one can choose a coordinate system (y1 , ..., yn ) on a neighborhood U 0 ⊂ U of y, such that yi = ϕi (x) for i = 1, ..., k. The functions (yk+1 , ..., yn ) restricted to Y form a local coordinate system on Y ∩ U 0 . It is easy to check compatibility of these coordinate systems on Y. Number k is called codimension of Y. A submanifold of codimension 1 is called hypersurface. The inclusion Y → X is a map of manifolds. Let G be a group that acts by continuous transformation of a manifold X.

132

Reconstruction from integral data

For a point x, the set Gx + {gx, g ∈ G} is called orbit of G. The action is called discrete, if for any point x ∈ X there exists a neighborhood U , such that gU is disjoint with U for any g ∈ G different from the unity element e. Consider the set Y of all orbits in X. There is a natural map p : X → Y, p (x) = Gx and a natural topology, such that a subset V ⊂ Y is open if the set p−1 (V ) is open in X. The topological space of all orbits is called quotient manifold and denoted X/G. It has a structure of manifold if the action of G is discrete. It is easy to see in the following examples. Examples 7.14. For any natural m, the cyclic group Zm acting on X = R2 \0 generated by rotation by the angle 2π/m. This action is a discrete, and the quotient X/Zm is diffeomorphic to X. Another example: The generator of group Z2 acts on a sphere Sn by the antipodal map x 7→ −x. The quotient manifold Sn /Z2 is the projective n-space.

A.3.2

Tangent Vectors and Covectors

Let X be a manifold and x ∈ X. Any linear functional t defined on the space of 1-smooth functions f : X → R satisfying the equation t (f g) = t (f ) g (x) + f (x) t (g) is called tangent vector in X at x. For any local coordinate P system x1 , ..., xn on X, any tangent vector can be written in the form t = ti ∂/∂xi , ti ∈ R. The set Tx of all tangent vectors at x is called tangent space at x. A vector t is tangent to a submanifold Y ⊂ X if t (f ) = 0 for any function f ∈ C 1 (X) that vanishes on Y. A family of tangent vectors t = t (x) defined at any point x ∈ X is called tangent field . It is smooth if the function t (f ) is smooth for any smooth function f. For any point p ∈ X the functional f 7→ t (f ) (p) is a tangent vector at p. The vector space Tx∗ of linear functionals on Tx is called cotangent space ∗ at x ∈ X. Any element P i of Ti x is called covector. Any covector can be written in the form v = v dxi , v ∈ R.

A.3.3

Differential Forms

Let V be a vector space with a coordinate system x = (x1 , ..., xn ) . A differential form a on U of degree p (p-form) can be written as the sum X α= ai1 ,...,ip (x) dxi1 ∧ ... ∧ dxip . (A.15) i1 0, ω ∈ S (E), and u = f dx for a homogeneous function f of degree α − n. We can write f (x) = rα−n s (ω) for a function s on S(E) and u = rα−1 dr ∧ sΩ, where sΩ equals the restriction of e x u to S (E). By separating variables, we obtain Z Z ∞ u (ϕ) = sψΩ, ψ (ω) + ϕ (rω) rα−1 dr (A.23) S(E)

0

for an arbitrary test function ϕ on E\0.
Distribution uε = ϕε u decreases fast at infinity, where ϕε (x) = exp −εr2 , ε > 0, and uε → u as ε → +0. Function u bε is smooth and u bε (ξ) = v (ψε,ξ ) , where Z ∞
ψε,ξ (ω) + exp −jr hξ, ωi − εr2 rα−1 dr. 0

By Proposition 7.1, integral (A.23) has meromorphic continuation for α ∈ C with the poles at α = 6 0, −1, −2, .... We have −jr hξ, ωi − εr2 = −jr (hξ, ωi − iε/2π) . Therefore, −α

Z

ψε,ξ (ω) → γα + (j (hξ, ωi − i0))

∞

exp (−t) tα−1 dt

0

= Γ (α) (j (hξ, ωi − i0))

−α

,

as ε → 0. Integrating against Ω yields Z −α u ˆε (ξ) → Γ (α) j−α (hξ, ωi − i0) e x u, S(E)

which completes the proof of (A.22). 

Appendix

141

Proposition A.14 Let q : V → R be a positive quadratic form: q(x) = P 1/2 q ij xi xj . Then ! −λ λ−n/2 (2πq ∗ (ξ)) (2πq (x)) dx = F , (A.24) Γ (λ) Γ (n/2 − λ)  P . where q ∗ = 1/2 qij ξ i ξ j and matrix {qij } is inverse to q ij . . 2 Proof. Let q (x) = 1/2 |x| . Distribution u = q λ−n/2 dx is homogeneous of degree 2λ. By Theorem A.13, we have Z −2λ λ−n/2 (hξ, xi − i0) (x) Ω (x) . F (u) (ξ) = Γ(2λ)j−2λ q G

The integral is a function of |ξ| , hence, we can take ξ = (|ξ| , 0, ..., 0) . Choose the cycle G = {x : x1 = ±1} and get F (u) (ξ) = Γ(2λ)j−2λ  Z −2λ −2λ × (|ξ| − i0) + (− |ξ| − i0) −2λ+n/2

= Γ (λ) Γ−1 (n/2 − λ) (2π)

Rn−1 −2λ

|ξ|



 2 λ

1 + |x0 |

dx0

,

where we applied the reflection and duplication formulas for the Gamma function. This proves (A.24) for the radial function q. An arbitrary positive quadratic function q can be made radial after a linear transformation A on V. The dual transform A0 is applied to the dual   quadratic form.
 2 2 1/2 Example A.17. Distribution K = δ x3 − x1 + x2 dx on R3 is homogeneous of degree 2, and its Fourier transform can be found by Theorem A.13 Z e x K (x) ˆ (ξ) = j−2 K 2. G (hξ, xi − 0i) We take the integration cycle G = {x : x3 = 1} and have e x K (x)|G = x3


1/2  ∂ x K (x)|H = δ 1 − x21 + x22 dx1 dx2 . ∂x3

This yields ˆ (ξ) = j−2 K

Z

= j−2

Z

dx1 dx2

1

x21 +x22 =1 2π

2

(ξ1 x1 + ξ2 x2 + ξ3 − 0i) d (x21 + x22 )1/2 dϕ 2

(ξ1 cos ϕ + ξ2 sin ϕ + ξ3 − 0i) 0
−3/2 1 =− |ξ3 | ξ32 − ξ12 − ξ22 if ξ32 > ξ12 + ξ22 . 2π

142

Reconstruction from integral data

ˆ (ξ) has holomorphic continuation into the half-plane Im ξ3 < 0 for Function K any ξ1 , ξ2 , since K is supported in {x : x3 > 0} . The right-hand side also has holomorphic continuation into the cone ξ32 < ξ12 + ξ22 , if we bypass the singular
1/2 − iε with small ε > 0. This yields cone setting ζ3 = ± ξ12 + ξ22
ˆ (ξ) = i ξ3 ξ 2 + ξ 2 − ξ 2 −3/2 if ξ 2 < ξ 2 + ξ 2 . K 3 1 2 1 2 3 2π

A.4.1

Singular Degrees

Proposition A.15 Any homogeneous distribution u on V \0 of degree α = −k, k = 0, 1, 2, ..., such that Z q exu=0 (A.25) S(E)

for arbitrary homogeneous polynomial q of degree k can be extended to V as a homogeneous distribution. Its Fourier transform is equal to k

u b(ξ) = −

(−j) k!

Z

k

hξ, xi log (2π hξ, xi − i0) e x u (x) + p (ξ) ,

(A.26)

Sn−1

where the first term is a homogeneous function, and p is a homogeneous polynomial of degree k. Proof. First, we check that Z

k

k

hξ, xi log (2π hξ, xi − i0) e x u (x)

v (ξ) + S(E)

is a homogeneous function of degree k. We have for any t > 0 Z k v(tξ) = (t hξ, xi) log (2π htξ, xi − i0) e x u Z k = tk hξ, xi (log (2π hξ, xi − i0) + log t) e x u Z k = tk v(ξ) + tk log t hξ, xi e x u. The last integral vanishes by (A.25), hence, v is a homogeneous. ε For an arbitrary ε ∈ R, the product uε + |x| u is a homogeneous distribution on V \0 of degree ε − k. It admits a unique extension to a homogeneous distribution on E if k − ε is not an integer ≥ 0. According to Theorem A.13, the Fourier transform of uε is equal to Z k−ε u bε (ξ) = Γ (ε − k) (2π hξ, xi − i0) e x u, S(E)

Appendix

143

since uε = u on S (E) . By (A.25), we can write Z h i k−ε k u bε (ξ) = Γ (ε − k) (2π hξ, xi − i0) − (2π hξ, xi) e x u. S(E)

This product has an indeterminacy ∞ · 0, as ε → 0. We apply the formula k −1 Γ (ε − k) ≈ (−1) (k!ε) for the first factor and the Lagrange formula for the second: k−ε

(2π hξ, xi − 0)

k

k

k

− (2π hξ, xi) ≈ −εjk hξ, xi log (2π hξ, xi − i0) .

This yields Z lim u bε (ξ) = ck

ε→0

k

hξ, xi log (2π hξ, xi − i0) e x u = ck v (ξ) ,

(A.27)

S(E)

k

where ck = − (−j) /k!. The right-hand side coincides with the first term of (A.26). Distribution vdξ is homogeneous of degree k + n, and by Proposition A.13 function, F −1 (vdξ) is homogeneous of degree −k − n. Distribution Uε + F −1 (b uε dξ) dx coincides with uε and tends to f + F −1 (vdξ) dx as ε → 0, since of (A.27). On the other hand, uε → u in the space of distributions on V \0. It follows that u = f on V \0, that is, f is an extension of u. This yields fˆ = v, which gives the first term in (A.26). Any other extension of u to E is equal to f + w, where w is a homogeneous distribution supported by the origin. Therefore, a polynomial term p = F (w) appears in (A.26). 

A.5 A.5.1

Principal Value of Integrals Singular Functions

Let X be a manifold and f be a real smooth function on X. We say that a zero x of f is simple if df (x) = 6 0. For a natural k, we consider the singular integral " # Z ρ ρ Ik (ρ) = + lim , (A.28) k k ε→+0 (f + i0) X (f + iε) where ρ is an arbitrary smooth density on X with compact support. Proposition A.16 If all roots of a function f are simple and the density ρ is k-smooth, then limit (A.28) exists and is a generalized function on X. Proof. We argue by induction on k ≥ 1. Choose a vector field t, a smooth function t0 on X, such that t (f ) + t0 f = 1. In the case k = 1, we integrate by

144

Reconstruction from integral data

parts and apply (A.20) for the locally integrable function u = log (f + i0) = limε→0+ log (f + iε) : Z Z (t (f ) + t0 f ) ρ I1 (ρ) = = t (log (f + i0)) ρ + t0 ρ f + i0 Z = (− log (f + i0) t (ρ) + t0 ρ) . If k > 1, we have by (A.19) Z (t (f ) + t0 f ) ρ Ik+1 (ρ) = k+1 (f + i0) ! Z 1 1 =− t ρ + Ik (t0 ρ) k k (f + i0)   Z 1 t (ρ) t (ρ) + Ik (t0 ρ) = Ik + t0 ρ . = k k k (f + i0) We apply this transform to t (ρ) and so on. Finally, 1 I1 (Tk ρ) k! Z 1 (− log (f + i0) t (Tk ρ) + t0 Tk ρ) , = k!

Ik+1 (ρ) =

(A.29)

where Tk ρ = (t + t0 ) (t + 2t0 ) (t + 3t0 ) ... (t + kt0 ) ρ.  The limit " # Z ρ ρ = lim k k ε→+0 (f − i0) X (f − iε) exists under the same assumptions; the sum " # " #   1 ρ 1 ρ ρ + + fk 2 (f + i0)k 2 (f − i0)k is called principal value of the integral of ρ/f k . If ρ is a real density, we have   Z Z ρ ρ ρ = Re = Re . k k k f (f − i0) X (f + i0)

A.5.2

Delta-Functions and Derivatives

The pull back construction of §7.3 can be applied to an arbitrary function −k f : X → Y = R, with simple roots and to generalized functions (t ± i0) on R. Indeed, by the assumption, f fulfills the rank condition for all points −k x, t = f (x) = 0, and (t ± i0)  is a smooth function for all t 6= 0. Therefore, −k the pull back f ∗ (t ± i0) is well defined. It is easy to see that it coincides

Appendix

145

with Ik and I¯k . The pull back of the delta-function δ0 (t) is also well defined. It is usually denoted δ0 (f ) as if it were a composition of functions. It can be represented as Z 1 δ0 (f ) (ρ) + lim ρ ε→0 2ε |f |≤ε or as

Z δ0 (f ) (ρ) = X0

ρ . df (k)

(k)

Proposition A.17 For any natural k, the pull back δ0 (f ) of δ0 can be calculated by Z (k) δ0 (f ) (ρ) = d (... (d (ρ/df )) /df.../df ) ,

∈ K0 (R)

f −1 (0)

where the dominator df appears k + 1 times. In particular, Z  Z Z 1 d (ρ/df ) = lim 2 − ρ . δ00 (f ) (ρ) = − ε→0 ε df 0≤f ≤ε −ε≤f ≤0 X0 (k)

Functionals δ0 (f ) can be determined in terms of differential operators applied to δ0 (f ) . Proposition A.18 Let t be a smooth vector field and t0 be a smooth function on X, such that t (f ) + t0 f = 1. For any k ≥ 1, we have (k)

δ0 (f ) (ρ) = δ0 (f ) (Tk ρ) , ρ ∈ K (X) ,

(A.30)

where Tk ρ = (t + t0 ) (t + 2t0 ) (t + 3t0 ) ... (t + kt0 ) ρ. Proof. By applying the pull back operation to (A.6), we obtain   (k) −k−1 −k−1 jδ0 (f ) = k! (f − i0) − (f + i0) . −k−1

From (A.29), we find k! (f ± i0)

−1

(k)

jδ0 (f ) (ρ) = (f − i0)

−1

(ρ) = (f ± i0)

(Tk ρ) , hence, −1

(Tk ρ) − (f + i0)

(Tk ρ) .

Taking k = 0 in (A.31) and applying to the density Tk ρ : −1

(f − i0)

−1

(Tk ρ) − (f + i0)

This equation and (A.32) yield (A.30). 

(A.31)

(Tk ρ) = jδ0 (f ) (Tk ρ) .

(A.32)

146

A.6

Reconstruction from integral data

Integrals of Rational Functions

The function t (ϕ) =

m X

ak cos kϕ + bk sin kϕ

k=0

is called trigonometric polynomial of degree m if am = 6 0 or bm 6= 0. Any trigonometric polynomial is 2π-periodic and has holomorphic continuation into the cylinder C/2πZ. It has at most 2m zeros on the cylinder. We say that a trigonometric polynomial t of degree m is oscillatory if it is real and has 2m real roots. Proposition A.19 Let t (ϕ) be an oscillatory polynomial of degree m > 0 and s (ϕ) be a real trigonometric polynomials of degree < pm for some natural p. Then Z 2π s (ϕ) (A.33) Re p dϕ = 0. (t (ϕ) ± i0) 0 Proof. Suppose that all roots of t are simple. Let a1 < a2 < ... < a2m be all roots of the polynomial ∂t/∂ϕ on the circle R/2πZ. The sign σk = sgn∂t/∂ϕ is constant on the interval (ak , ak+1 ) for k = 1, ..., 2m, where α2m+1 = α1 . We have Im t (ϕ + iσk ε) > 0 for any k and small ε, and Z ak+1 Z αk+1 s (ϕ) s (ϕ) dϕ = p dϕ. p (ϕ + iσ 0) t (t (ϕ) + i0) k ak αk Take the sum for k = 1, ..., 2m and obtain  Z ak+1 X Z ak+1 s (ϕ) s (ϕ) dϕ + dϕ tp (ϕ + iσk 0) tp (ϕ − iσk 0) ak ak k Z 2π Z 2π s (ϕ) s (ϕ) = p dϕ + p dϕ. (t (ϕ) + i0) (t (ϕ) − i0) 0 0

(A.34)

The left-hand side can be rearranged, separating the terms with positive and negative signs ±σk . Summating, we get  Z ak+1 X Z ak+1 s (ϕ) s (ϕ) dϕ + dϕ tp (ϕ + i0) tp (ϕ − i0) ak ak k Z 2π Z 2π = r (ϕ + i0) dϕ + r (ϕ − i0) dϕ, (A.35) 0

0

where r + st−p . By Cauchy’s theorem, we can replace r (ϕ ± i0) dϕ in the right-hand side by r (ζ) dζ, where ζ = ϕ ± iη and η > 0 runs on the cylinder C/2πZ. Function r (ζ) is meromorphic and has no poles for η = 6 0

Appendix

147

due to the assumption and |r (ζ)| → 0 as η → ∞, since deg s < pm. It follows that the right-hand side of (A.35) vanishes. Therefore, the right-hand side of (A.34) equals zero, which implies (A.33).

Transformation of integration cycle. In the general case, we can approximate any polynomial t with real roots by polynomials t˜ with real simple roots. The Equation (A.33) holds for r˜ = s/t˜p , hence, it is true for r.  Example A.19. For any even n and unit vector θ, we have Z Ω n/2 n−2 , π S n−2 = (−1) ω∈Sn−1 hω, θi where Ω is the volume form on Sn−1 . Integrating over spheres hω, θi = cos ϕ, 0 ≤ ϕ < π yields the quantity
Z Z n−2 π sinn−2 ϕdϕ n−2 π 1 − cos2 n/2−1 ϕdϕ S = S cosn−2 ϕ cosn−2 ϕ 0 0 Z π p (cos ϕ) n/2 n−2 = (−1) π S + dϕ, n−2 ϕ cos 0 where p is a polynomial of degree n − 2. The last term vanishes by Proposition A.19. Definition. Let t be a polynomial of degree m on an Euclidean space V n . We say that t is oscillatory on the unit sphere Sn−1 with respect to a point a if t (a) = 6 0 and for any 2-subspace P in V n through a, t is oscillatory polynomial of degree m on P ∩ S (V ) . A linear polynomial p is oscillarory on S n−1 , if and only if vanishes on a n − 2 sphere.

148

Reconstruction from integral data

Theorem A.20 Let t be an oscillatory polynomial on Sn−1 of degree m, with respect to a point a and s be an arbitrary real polynomial of degree ≤ pm−n+1 for some natural p. Then for even n, Z s (ω) Ω Re (A.36) p = 0. Sn−1 (t (ω) + i0) If n is odd and m = 1, then Z Im Sn−1

s (ω) Ω p = 0. (t (ω) + i0)

(A.37)

Proof. Let n be even and θ ∈ S1 , υ ∈ Sn−2 be the polar coordinates on S with the pole at a. We have Ω (ω) = sinn−2 θdθ ∧ Ω0 (υ) , where Ω0 is the volume form on Sn−2 . This yields Z Z Z 2π s (ω) Ω 1 sinn−2 θ s (θ, υ) 0 Ω (υ) p = p dθ. 2 Sn−2 (t (θ, υ) + i0) S(E) (t (ω) + i0) 0 n−1

The numerator sinn−2 θ s (θ, υ) is for any υ a trigonometric polynomial on S1 of degree deg s + n − 2 ≤ pm − n + 1 + n − 2 < pm. The inner integral is pure imaginary by Proposition A.19, since t (θ, υ) is oscillatory in θ for each υ. This yields (A.36). If n is odd, we have t (ω) = hω, bi − c for some b ∈ Sn−1 and real c, |c| < 1, since t is oscillatory. By averaging the integral over spheres hω, bi = const, we obtain Z Z n−2 π r (ω) sinn−2 θdθ s (ω) Ω S = p p, Sn−1 (t (ω) + i0) 0 (cos θ − c + i0) where r is a real polynomial depending only on cos θ = hω, bi . This yields Z Z π s (ω) Ω r (cos θ) sinn−2 θdθ Im Im p = p (cos θ − c + i0) Sn−1 (t (ω) + i0) 0
(n−3)/2 r (ζ) 1 − ζ 2 dζ r sinn−3 θd cosθ = π Resα = −π Resa , p p (cos θ − c) (ζ − γ) where γ = arccos c. The residue vanishes, since deg r + n − 3 ≤ p − 2. This proves (A.37). 

A.7

Interpolation of Bandlimited Functions

Definition. A function f ∈ L2 (R) is called a-bandlimited function for some number a > 0, if supp fˆ ⊂ [−a, a]. Any bandlimited function can be interpolated on an arbitrary finite interval:

Appendix

149

Proposition A.21 For arbitrary positive a and b, any a-bandlimited function φ ∈ L2 (R) can be interpolated on (−b, b) as follows:   p  p  Z ∞ Z −b sin A y 2 − b2 φ(ζ) = exp α b2 − ζ 2 − φ(y)dy, (A.38) π (y − ζ) b −∞ p where the branch is defined by Re b2 − ζ 2 > 0 and A = 2πa. Proof. Denote Γ+ = [b, ∞), Γ− = (−∞, b], Γ = Γ+ ∪ (−Γ− ) , where −Γ− means the opposite orientation on Γ− . The form   p exp −A b2 − ζ 2 φ (z) dz ψ (z) = j (z − ζ) is well defined and meromorphic on C\Γ. It is absolutely integrable on each side of Γ, since of the Paley-Wiener theorem. Let Γ± (ε) be ε-neighborhood of Γ. By the Residue theorem, for any point ζ ∈ C\Γ+ (ε) ∪ Γ− (ε) Z   p ψ = − exp −A b2 − ζ 2 φ (ζ) , Γ(ε)

. where Γ (ε) = ∂Γ+ (ε) − ∂Γ− (ε) is an oriented cycle. For arbitrary ζ ∈ Γ and two close points ζ± ∈ Γ (ε), such that ± Im ζ± > 0, Re ζ± > 0, and ζ+ + ζ− = 2ζ, we have     q q  p  2 2 − exp −A b2 − ζ− ≈ 2ı sin A ζ 2 − b2 . exp A b2 − ζ+ In the case Re ζ± < 0, we obtain the negative of this quantity Therefore, √
Z Z sin A z 2 − b2 φ(z)dz. − α→ π (z − ζ) γ(ε) Γ as ε → 0.  Corollary A.22 If φ is a-bandlimited function and b > 0, then Z Z 2   p 2 2 2 |φ (x)| dx, exp −A b − x φ (x) dx ≤ |x|≤b

|x|≥b

where A = 2πa. Proof. Equation (A.38) implies  p  Z sin A y 2 − b2 Z √ χ (y) dy 2 2 φ (y) dy = e−A b −x φ (x) = π (y − x) π (x − y) Γ

(A.39)

150

Reconstruction from integral data   p for |x| ≤ b, where χ (y) = ∓ sin A y 2 − b2 φ (y) for y ∈ Γ± and χ (y) = 0 otherwise. The right-hand side equals the Hilbert transform of χ. This transform is unitary, which yields Z Z Z 2 −A√b2 −x2 2 2 φ (x) dx = |χ| dx ≤ |φ| dx.  e q(x)0

Example A.20. Inequality (A.39) is close to the optimal for an estimate of interpolation of bandlimited functions. The function √
cos A x2 − b2 φa,b (x) = 2 x2 − b2 − (π/2A) is a-bandlimited. It is easy to check that (A.39) fails if φ = φa,b and A is replaced by any smaller number A0 < A in the left-hand side.

A.8

Cauchy Integral Equation on Interval

Consider the Cauchy type equation Z b f (y) dy = g (x) , a ≤ x ≤ b x−y a

(A.40)

on an interval [a, b] where g ∈ L2 (a, b) . Theorem A.23 The general solution of (A.40) has the form Z b 1/2 1 C p (y) g (y) f (x) = − 1/2 dy + 1/2 , x−y πp (x) a p (x)

(A.41)

where p (x) = (b − x) (x − a) and C is the constant, such that the right-hand side belongs to L2 (a, b) . Remark. The function p−1/2 do not belong to L2 (a, b) . Proof. If f is holomorphic function in C\ [a, b] , we replace the left-hand side of (A.40) by one half of the integral over an arbitrary cycle γ in C\ [a, b] that runs one time about this segment in the positive direction. Lemma A.24 Function f (x) = Cp−1/2 (x) is the general solution of (A.40), with g = 0. Proof. Let q be a branch of the function p1/2 , which is univalent and holomorphic on C\ [a, b] . The boundary values q± on the interval fulfils q− = −q, hence, Z b Z dy 1 dy =− , 2 γ q (y) (x − y) a q (y) (x − y)

Appendix

151

where γ is a clockwise contour in C\ [a, b] around the interval. We can move the contour to ∞ by Cauchy’s theorem. The integral tends to zero, since |q (z)| ≥ |z| − c for some constant c, hence, the integral vanishes. It follows that f = q −1 is a solution of the homogeneous Equation (A.40). Conversely, the Cauchy integral Z 1 b f (y) dy h (x) = j a x−y admits the unique holomorphic   continuation h (z) to C\ [a, b] . It allows the −1/2 uniform estimate O |p (z)| . By Equation (A.4) A.1, h+ +h− = 2j−1 g = 0 on [a, b] , where h± are boundary values of h on [a, b] . This yields h− = −h+ , hence, the product qh is holomorphic and bounded on C. By Liouville’s theorem, qh = d for a constant
d and h = d/q. According to (A.4), we have −1 −1 jf = h− − h+ = d q− − q+ = ±2dp−1/2 , that is, f = Dp−1/2 , where −1 D = ±2dj .  Lemma A.25 We have Z Z b −1/2 1 b p1/2 (t) dt p (t) dt = 1, =0 π a (τ − t) (t − σ) a (t − σ) (τ − t) for arbitrary a, b and σ, τ ∈ [a, b] . Proof. The left-hand side is equal to the residue of the corresponding form at infinity.  Proof of Theorem. It follows that the second term in (A.41) can be included in the first one, with g˜ = g − C/π. Apply the Poincar´e–Bertrand commutation formula for the segment Γ = [a, b] and the function ϕ (t, τ ) = p−1/2 (t) p1/2 (τ ) g˜ (τ ) defined for t, τ ∈ Γ = (a, b) : Z Z Z Z dt ϕ (t, τ ) dτ ϕ (t, τ ) dt − dτ (A.42) t − σ τ − t (t − σ) (τ − t) Γ Γ Γ Γ = −π 2 ϕ (σ, σ) , a ≤ σ ≤ b. We have ϕ (σ, σ) = g 0 (σ) and
Z b Z b Z b −1/2 p (t) H p1/2 g˜ dt dt ϕ (t, τ ) dτ =π τ −t t−σ a t−σ a a    2 −1/2 =π H p H p1/2 g˜ , where H is the Hilbert transform and function p1/2 g˜ is extended by 0 outside
of (a, b) . The right-hand side is well defined in L2 (a, b) , since p−1/2 H p1/2 g˜ ∈ L2 (a, b) . By Lemma A.25, Z b Z b ϕ (t, τ ) dt dτ (t − σ) (τ − t) a a Z b Z b −1/2 p (t) dt p1/2 (τ ) g˜ (τ ) dτ = = 0, a a (t − σ) (τ − t)

152

Reconstruction from integral data

since the second factor vanishes. Now (A.42) reads    H p−1/2 H p1/2 g˜ = −˜ g. The last equation implies that   f + −π −1 p−1/2 H p1/2 g˜ = −

1 πp1/2

Z a

b

p1/2 g C dy + 1/2 x−y p

is a solution to (A.41).  Remark. The constant C is uniquely defined, since the function p−1/2 does not belong L2 (a, b).

A.9

Bibliographic Notes

§§7.1–2 See [135] for elementary introduction to the theory of distributions and Fourier transform. The classical book of L. Schwartz [129] gives much more. §7.4 The Herglotz–Petrovski˘ı formula [116] for fundamental solutions of hyperbolic equations is based on a formula like (A.22); see also [63]. Proposition 7.15 is published in [110]. §7.5 Methods of regularization of divergent integrals go back to Cauchy, Hadamard, and M. Riesz; see [129]. §7.6 Proposition A.19 is taken from [108]. §7.7 For Proposition 7.25, see [114]. Another method of interpolation of functions of two variables was proposed in [142]. §7.8 The result of this section is published in [80].

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