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Limited domain Radon transform

Arp´´ ad Kurusa

Abstract. The problem in this article is to recover a function onRnfrom its integrals known only on hyperplanes intersecting the unit ball.

1. Introduction

There is a number of papers concerning the reconstruction of a function from only a partial knowledge of the function’s Radon transform. The two most known examples are the exterior Radon transform [6] and the limited angle Radon trans- form [4].

The classical Radon transform is defined for an integrable functionf onRn by

Rf(ω, p) = Z

H(ω,p)

f(x)dxh,

whereω∈Sn−1is a unit vector,p∈R+ andRf(ω, p) is just the integral off over the hyperplaneH(ω, p) ={x∈Rn :hx, ωi=p}by the surface measure dxh on it.

In the limited angle caseRf(ω, p) is restricted inωto a subset ofSn−1. The exterior Radon transform is the restriction ofRf to the setp >1.

We define the limited domain Radon transform RLf of a function as the restriction of Rf onto the setp ≤1. In the next section we show its continuity on a weighted class L2α,β(En) of square integrable functions that are zero in a neighborhood of the origin. In Section 3 we give the null space and range of RL acting on L2α,β(En) for the odd dimensional spaces. In Section 4 we do the same for even dimensional spaces, where the injectivity ofRLonL2α,β(En) turns out.

The author thanks the Matematische Institut of the Universit¨at der Erlangen- N¨urnberg and the Department of the University of Maryland for their support and assistance while working on this paper.

AMS Subject Classification(2000): 44A12.

Supported by the Hungarian NFS, Grants No. T4427, W015452 and F016226

Math. Balkanica, 11(1997), 327–337. c A. Kurusa´

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2. Preliminarities

The limited domain Radon transform of a function f integrable on the hy- perplanes is defined in means of the formula RLf =χ[0,1)Rf, where χ[0,1) is the characteristic function of the interval [0,1).

From the well known inversion formulas [3] we know that a square integrable function is determined by the limited domain Radon transform in the unit open ball Bnif the dimensionnis odd. In even dimensions the situation is more complicated, but still there is an approximate possibility, as C. Berenstein and D. Walnut proved in [1] that to recover f to a given accuracy in a ball of radiusR >0 it is sufficient to knowRf(ω, p) only forp < R+αfor someα >0 that depends on the accuracy desired (the greater the accuracy desired, the greater αmust be).

Motivated by these observations, in the next two sections we carry out in- vestigations on the following spaces. L2α,β(En) is the Hilbert space of functions supported inEn=Rn\ ClBn equipped with the inner product

hf, giα,β = Z

En

f(x)g(x)|x|α(1− |x|−2)βdx.

Estimate on the weight shows immediately, thatL2α,β(En)⊇L2α00(En) forα≤α0 and β ≤ β0. Denote Bn ={H(ω, p) : 0 ≤p ≤1, ω ∈Sn−1}. Then the Hilbert space of functions on Bn with the inner product

hF, Giγ,δ = Z

Sn−1

Z 1 0

F(ω, p)G(ω, p)pγ(1−p2)δdpdω

is L2γ,δ(Bn). HereL2γ,δ(Bn)⊇L2γ00([0,1]) forγ0 ≤γ andδ0 ≤δ. Finally we need the Hilbert space L2γ,δ([0,1]) of functions on [0,1] with the inner product

hf, giγ,δ= Z 1

0

f(p)g(p)pγ(1−p2)δdp.

For −1 < γ ≤ 0, −1 < δ ≤ 0, the polynomials are dense in this space, and L2γ,δ([0,1])⊇L2γ00([0,1]) for γ0 ≤γ andδ0 ≤δ.

Our main tool is the spherical harmonic expansion in these spaces. Briefly, the spherical harmonics, Y`,m constitute a complete polynomial orthonormal sys- tem in the Hilbert space L2(Sn−1). If f ∈ C(Sn−1×R+) and f`,m(p) is the corresponding coefficient of Y`,m(ω) in the expansion of f(ω, p), ie. f`,m(p) = R

Sn−1f(ω, p)Y`,m(ω)dω, then the series P

`,mf`,m(p)Y`,m(ω) converges uniformly

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absolutely on compact subsets of Sn−1×Rto f(ω, p). For further references we refer to [7]. Below we shall use the expansions

g(ϕ, p) =

X

m=−∞

gm(p) exp(imϕ) and g(ω, p) =

X

`,m

g`,m(p)Y`,m(ω) in dimension two and in higher dimensions, respectively. In dimension two,ϕwill mean the angle of the respective unit vector to a fixed direction.

The spherical expansions of the Radon transforms are well known [6]. Apply- ing these to the functions in L2α,β(En) we obtain

(2.1) (RLf)m(p) = 2 Z

1

fm(q)cos(marccos(p/q)) p1−p2/q2 dq for dimension two and

(2.2) (Rf)l,m(p) =|Sn−2| Cmλ(1)

Z 1

f`,m(q)qn−2Cmλ p

q 1−p2 q2

n−32

dq

for higher dimensions, where Cmλ is the Gegenbauer polynomial of degreem, λ= (n−2)/2 andp≤1.

An important consequence of these expansions is the continuity ofRL. Theorem 2.1. RL maps L2α,β(En) continuously into L2γ,δ(Bn), where α > n−2, β <1,γ >−1 andδ >

−1 if n≥3

−1/2 if n= 2.

Proof. First observe that (2.3) kRLfk2γ,δ =X

`,m

kY`,mk22 Z 1

0

(RLf)2`,m(p)pγ(1−p2)δdp.

Using (2.1), (2.2) and that |Cmλ(x)| ≤ |Cmλ(1)| for |x| ≤ 1 we can over estimate (2.3) by

c1

Z 1 0

Z 1

f`,m(q)qn−2

1−p2 q2

n−32

dq

2

pγ(1−p2)δdp,

where c1 is a suitable constant independent fromm. Forn≥3 this is less than

(2.4) c1

Z 1

f`,m(q)qn−2dq

2Z 1 0

pγ(1−p2)δdp.

Math. Balkanica, 11(1997), 327–337. c A. Kurusa´

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Forn= 2 we estimate with |1−p2/q2| ≥1−p2 and obtain

(2.5) c1

Z 1

f`,m(q)dq

2Z 1 0

pγ(1−p2)δ−1/2dp.

The second integrals in (2.4) and (2.5), respectively, explain the restrictions on γ and δgiven in the theorem.

The restrictions on α and β come from the following estimate of the first integrals in (2.4) and (2.5).

(2.6)

Z 1

f`,m(q)qn−2dq

2

=

Z 1

f`,m(q)q−1−α(1−q−2)−βqα+n−1(1−q−2)βdq

2

≤ Z

1

f`,m2 (q)qα+n−1(1−q−2)βdq×

× Z

1

q−2−2α(1−q−2)−2βqα+n−1(1−q−2)βdq.

The last two integrals need to be finite to ensure the finiteness of (2.3). The first one is finite becausef ∈Lα,β(En), the other is finite ifα > n−2 andβ <1.

Throughout this paper we shall assume thatα, β,γ andδ satisfy the condi- tions given in Theorem 2.1.

The weight 1−p2/q2(n−3)/2

is substantially different in odd and in even dimensions; it is polynomial in odd dimensions.

3. Odd dimensions

Letn= 2d+ 3 andd≥0. Thenλ=d+ 1/2 and

(3.1) (RLf)l,m(p) = |Sn−2| Cmλ(1)

Z 1

f`,m(q)qn−2Dλm p

q

dq,

where Dλm(x) =Cmλ(x)(1−x2)d.

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Lemma 3.1. Dmλ is a polynomial of the form

Dmλ(x) =

d+[m/2]

X

i=0

dλm,ixm+2d−2i,

where the coefficients dλm,i 6= 0.

The proof is a simple consequence of the formulas (8.932.1), (9.136.1) and (9.131.1) in [2].

Substituting the polynomial form of Dmλ into (3.1) we see that (RLf)l,m is also a polynomial of the form

(3.2) (RLf)l,m(p) =|Sn−2| Cmλ(1)

d+[m/2]

X

i=0

pm+2d−2idλm,i Z

1

f`,m(q)q1−m+2idq.

Let 2φ`,m(x2) = f`,m(1/x)/xn and change the variable q = 1/√

x. Then the integral in (3.2) becomes

(3.3) c`,m,i= Z

1

f`,m(q)q1−m+2idq= Z 1

0

φ`,m(x)xd−i+m/2dx

√x.

Theorem 3.2. The null space of RL in L2α,β(En) is the closure of the span of functions

gk,`,m(ω, q) =Pk(0,εm)(2q−2−1)q−nY`,m(ω),

where Pk(.,.)is the Jacobi polynomial of degree k,n−1≤α < n,−1< β≤0, k≥

m+n−1 2

and εm=

−1/2 if m even

0 if m odd.

Proof. The fact thatgk,`,m∈L2α,β(En) follows fromα < nand−1< β. Since also n−1≤αandβ≤0 it follows thatφ`,m∈L2−1/2,0([0,1]).

According to (3.2) and (3.3) the functionRLf vanishes if and only ifc`,m,i= 0 for all`,m, and 0≤i≤d+ [m/2].

Formeven this gives that

(3.4) 0 =

Z 1 0

φ`,m(x)xjdx

√x for all 0≤j≤d+m/2.

Math. Balkanica, 11(1997), 327–337. c A. Kurusa´

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The shifted Jacobi polynomialsPk(0,−1/2)(2x−1) constitute a complete orthogonal system on [0,1] with respect to the weight 1/√

x[2(8.904)], thereforeφ`,mmust be in the closure of the span of{Pk(0,−1/2)(2x−1)}k=d+1+[m/2].

Formodd we have

(3.5) 0 =

Z 1 0

φ`,m(x)xjdx for all 0≤j≤d+ (m−1)/2.

The Jacobi polynomialsPk(0,0)(2x−1) constitute a complete orthogonal system on [0,1] [2(8.904)], soφ`,mis in the closure of the span of{Pk(0,0)(2x−1)}k=d+1+[m/2].

The results of (3.4) and (3.5) give the theorem.

In the following we determine the range of the limited domain Radon trans- form.

Theorem 3.3. RL maps the L2α,β(En)closure of the span of the functions hk,`,m(ω, q) =q−n−2kY`,m(ω), 0≤k≤d+ [m/2]

where n−1 ≤α < n and −1 < β < 0, onto the L2γ,δ(Bn) closure of the span of functions

F`,m(ω, p) =Y`,m(ω)

d+[m/2]

X

i=0

pm+2d−2ib`,m,i

continuously and bijectively.

Proof. The easy verification of hk,`,m ∈ L2α,β(En) and F`,m ∈L2γ,δ(Bn) is left to the reader.

SinceRL:L2α,β(En)→L2γ,δ(Bn) is continuous by Theorem 2.1, it takes closed set to closed set. Further it is injective on the given functions by Theorem 3.2, therefore we only have to give coefficientse0k,`,m∈Rso that

F`,m=RL(f`,mY`,m), where f`,mY`,m=

d+[m/2]

X

k=0

e0k,`,mhk,`,m.

EliminatingY`,m and reordering the summation we can search forf`,m in the form

(3.6) f`,m(q) =

d+[m/2]

X

k=0

ek,`,mq−nPk(0,εm)(2q−2−1),

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where the coefficientsek,`,m are to be determined.

In F`,m = RL(f`,mY`,m) the equality of two polynomials appears, that is equivalent to the coincidence of their corresponding coefficients. By (3.2) and (3.3) this gives

(3.7) b`,m,i= |Sn−2| Cmλ(1)dλm,i

Z 1

f`,m(q)q1−m+2idq

for 0≤i≤d+ [m/2]. Substitutingf`,m according to (3.6) this becomes a system of linear equations for ek,`,m with coefficients

ai,k,m= Z

1

q1+2i−m−nPk(0,εm)(2q−2−1)dq

=1 2

Z 1 0

xd+[m/2]−iPk(0,εm)(2x−1)xεmdx,

where 0 ≤ k ≤ d+ [m/2]. The Jacobi polynomials Pk(0,εm)(2x−1) constitute orthogonal polynomial system with respect to the weight xεm on [0,1], therefore ai,k,m= 0 fork+i > d+ [m/2] andai,d+[m/2]−i,m 6= 0 for 0≤i≤d+ [m/2]. By these properties the equations

b`,m,i= |Sn−2| Cmλ(1)dλm,i

d+[m/2]

X

k=0

ek,`,mai,k,m,

where 0≤i, k≤d+ [m/2], determine uniquelyek,`,m. This completes the proof.

4. Even dimensions

The nature of the problem changes considerable in even dimensions. On one hand we do not have easy to handle polynomials, but on the other hand just this sole inconvenience gives uniqueness on these spaces.

We have

(4.1) (RLf)l,m(p) =|Sn−2| Z

1

f`,m(q)q Dmλ(p/q) p1−p2/q2dq, where the dimensionn= 2λ+ 2 is even,λ≥0,|S0|= 2 and

(4.2) Dλm(x) =

Cmλ(x)

Cλm(1)(1−x2)λ if λ≥1 cos(marccosx) if λ= 0 is a polynomial of degree m+ 2λ.

Math. Balkanica, 11 (1997), 327–337. c A. Kurusa´

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Lemma 4.1. In the Taylor expansion Dmλ(x)

√1−x2 =

X

i=0

dλm,ix2i+m−2[m/2],

the coefficients dλm,i are not zero. The series is convergent absolutely on (−1,1) and uniformly on [−1 +ε,1−ε]for any 1> ε >0.

Proof. Forλ >0, (8.932.1), (9.136.1) and (9.131.1) of [2] give Cmλ(x)

(1−x2)1/2−λ = Km,λF1−m

2 −λ,1 +m 2 ,1

2;x2

+xKm,λ0 F2−m

2 +λ,2 +m 2 ,3

2;x2 , whereFis the hypergeometric function [2(9.1)],Km,λ0 = 0 formeven andKm,λ= 0 formodd.

Forλ= 0 we prove cos(marccosx)

√1−x2 =KmF1−m 2 ,1 +m

2 ,1 2;x2

+xKm0 F2−m 2 ,2 +m

2 ,3 2;x2 with (8.942.1), (9.136.1) and (9.131.1) of [2], whereKm0 = 0 formeven andKm= 0 formodd.

The well known properties of the hypergeometric functions and the formulas above prove the statement.

Letbλm,i=|Sn−2|dλm,i. According to Lemma 4.1 the Taylor expansion of (4.1) is

(4.3) (RLf)l,m(p) =

X

i=0

bλm,ip2i+m−2[m/2]Z 1

f`,m(q)q2λ−2i+2[m/2]−mdq,

Let 2φ`,m(x2) = f`,m(1/x)/xn. Changing the variableq = 1/√

xthe integral in (4.3) becomes

(4.4) c`,m,i=

Z 1 0

φ`,m(x)xixεmdx.

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Theorem 4.2. The limited domain Radon transform RL is injective on L2α,β(En) if n−1≤α < nand −1< β <0.

Proof. Easy calculation shows φ`,m ∈ L2−1/2,0([0,1]). Then RLf(ω, p) = 0 for p <1 impliesc`,m,i= 0 for alli, m≥0, henceφ`,m must be zero.

Because (4.3) is an infinite series, we need more sophisticated tools to deter- mine the range ofRL.

Theorem 4.3. RL maps the L2n−1,0(En)closure of the span of the functions fk,`,m(ω, q) =q−n−2kY`,m(ω), 0≤k

onto the L2γ,δ(Bn) closure of the span of functions

Fj,`,m(ω, p) =p2j+m−2[m/2]Y`,m(ω) 0≤j continuously and bijectively.

Proof. The verification offk,`,m∈L2n−1,0(En) andFj,`,m∈L2γ,δ(Bn) is left to the reader. Since RL:L2n−1,0(En)→L2γ,δ(Bn) is continuous by Theorem 2.1, it takes closed set to closed set. Further it is injective by Theorem 4.2, therefore we only have to give coefficients e0k,`,m∈Rso that

Fj,`,m=RL(f`,mY`,m), where f`,mY`,m=

X

k=0

e0k,`,mfk,`,m.

The functions f`,m of this form can be written in the form

(4.5) f`,m(q) =

X

k=0

ek,`,mq−nPk(0,εm)(2q−2−1)

with unique coefficients ek,`,m. To find these coefficients we consider (4.5) and (4.3), where the left hand side is substituted with Fj,`,m, as a system of linear equations with infinite dimensional matrix of entries

ai,k,m= Z

1

q−2−2i+2[m/2]−mPk(0,εm)(2q−2−1)dq

= 1 2

Z 1 0

xiPk(0,εm)(2x−1)xεmdx.

Math. Balkanica, 11(1997), 327–337. c A. Kurusa´

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By the orthogonality of Pk(0,εm) these are zero fork > i. Fori≥k

ai,k,m= (k!)2 i

k k

Y

h=0

1 h+i+ 1 +εm

by [2(7.391.3)]. In virtue of the Stirling formula for the factorials we find the (not sharp) inequality

(4.6) 16< lim

i→∞

ai,i,m i 4e

i.

The system of equations

(4.7) δi,j=

0 ifi6=j 1 ifi=j =bλm,i

i

X

k=0

ek,`,mai,k,m, i, j≥0

obtained from (4.3), by substitutingFj,`,minto the left hand side and substituting f`,m into the right hand side according to (4.5), determine the coefficientsek,`,m.

What remains to prove is only that the function defined by (4.5) with these coefficients is in L2n−1,0(En), that is equivalent to

(4.8)

Z 1

X

k=0

ek,`,mPk(0,εm)(2q−2−1)

!2

q−2dq <∞.

Using Cauchy’s inequality we can over estimate the left hand side by

(4.9)

X

k=0

(k+ 1)2e2k,`,m

X

k=0

Z 1

Pk(0,εm)(2q−2−1) k+ 1

!2 q−2dq.

First we estimate the second multiplier in (4.9). With a change in the variable we get

(4.10) Z

1

Pk(0,εm)(2q−2−1)2

q−2dq=

√2 4

Z 1

−1

Pk(0,εm)(x)2

(1 +x)−1/2dx.

For m even, the right hand side evaluates explicitly 1/(4k+ 1) by [2(7.391.1)].

For m odd,Pk(0,εm)(x) =Ck1/2(x) by [2(8.962.4)], which is less thanCk1/2(1) = 1.

Therefore the second multiplier in (4.9) is less then P

k=0(k+ 1)−22/6.

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To estimate the first multiplier in (4.9) we introduce for N > j the N- dimensional vectors

δN = (bλm,j)−10,j, . . . , δi,j, . . . , δN,j)

eN = (e0,`,m,2e1,`,m, . . . ,(k+ 1)ek,`,m, . . . ,(N+ 1)eN,`,m)

and theN×N quadratic matrixAN = (ak+1i,k,m)Ni,k=0. With these notations, (4.7) can be rewritten asδN =ANeN and therefore

(4.11)

N

X

k=0

(k+ 1)2e2k,`,m =keNk22≤ kδNk22· k AN−1

k22.

Obviously kδNk22 = (bλm,j)−2 does not depend on N. Since the matrix AN is triangular so is its inverse

AN−1

, and the diagonal elements in AN−1

are (i+ 1)/ai,i,m, that are also the eigenvalues of

AN−1

. In virtue of (4.6) the estimate

k AN−1

k22≤ max

0≤i≤M

i+ 1 ai,i,m

is valid for M big enough. Together with (4.11) this completes the proof.

As a consequence we obtain the following.

Theorem 4.4. The range of RL is dense in L2γ,δ(Bn), where −1 < γ ≤ 0 and 0> δ >

−1 if λ >0

−1/2 if λ= 0.

Proof. According to Theorem 4.3 it is enough to prove thathF, Fj,`,miγ,δ = 0 for allj, `, m≥0 impliesF = 0 forF ∈L2γ,δ(Bn).

Let Φ(ω, p2) =F(ω, p). ThenF∈L2γ,δ(Bn) implies Φ∈L2(γ−1)/2,δ(Bn), hence

0 =hF, Fj,`,miγ,δ = Z 1

0

Φ`,m(p)pjpγ−12 +m2−[m2](1−p)δdp for allj≥0 implies Φ`,m≡0 for all`, m≥0, i.e. F = 0.

Note that we used the special valuesα=n−1 andβ = 0 in Theorem 4.3 only to simplify the estimate of (4.10) that would need tedious calculations in general.

Math. Balkanica, 11(1997), 327–337. c A. Kurusa´

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However, the results can be extended to the spaces L2α,β forn−1 ≤α < n and

−1< β≤0.

References

[1] C. A. BERENSTEIN and D. WALNUT, Local inversion of th Radon transform in even dimensions using wavelets, Proceeding of the Conference 75 years of Radon transform, International Press (1994), 45–69.

[2] I. S. GRADSHTEYN and I. M. RYZHIK,Table of integrals, series, and products, Aca- demic Press, New York, 1980.

[3] S. HELGASON,The Radon transform, Birkh¨auser, Boston, 1980.

[4] W. R. MADYCH, Summability and approximate reconstruction from Radon trans- form data, Cont. Math., 113(1990), 189–219.

[5] E. T. QUINTO, The invertibility of rotation invariant transforms, J. Math. Anal.

Appl., 91(1983), 510–522.

[6] E. T. QUINTO, Singular value decompositions and inversion methods for the exterior Radon transform and a spherical transform, J. Math. Anal. Appl., 95 (1983), 437–448.

[7] R. SEELEY, Spherical harmonics, Amer. Math. Monthly, 73(1966), 115–121.

A. K´ URUSA, Bolyai Institute, Szeged University, Aradi v´ertan´uk tere 1., H-6720 Szeged, Hungary; e-mail: kurusa@math.u-szeged.hu

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