Explicit Inversion Formulas for the X–Ray Transform in Zn

1660-5446/10/040523-16, published online March 23, 2010

© 2010 Birkhäuser / Springer Basel AG

of Mathematics

Explicit Inversion Formulas for the X − Ray Transform in Z ⁿ

Ahmed Abouelaz

Abstract. In the present paper, we state and prove explicit inversion formulas for the X−Ray transform in the lattice Zⁿ by using certain arithmetical geometrical techniques.

Mathematics Subject Classification (2010). Primary 44A12; Secondary 05B35, 11D04.

Keywords. d−plane Radon transform, inversion formulas, dilatation of thed−planes, Smith normal form, Mobius function.

1. Introduction

Many authors contributed to the integral geometry on Crassmannian mani- folds and projective spaces see for example [1], [4–11].

We brieftly recall the definition of the classical Radond-plane transform, namelyR_c, on the Euclidean spaceRⁿ. Letd be an integer 1≤d≤n−1, n≥2.We denote byG(d, n) the Grassmann manifold consisting of all affine d-dimensional planes inRⁿ.The Radon d-plane transformR_c is defined by

R_cf(ζ) =

ζ

f(x)d_m(x), (1.1)

for all (f, ζ)∈D(Rⁿ)×G(d, n),whered_m(x) is the Euclidean measure on the d-planeζand D(Rⁿ) denotes the space of all complex valuedC^∞-functions onRⁿ with compact support.

In the case of the lattice Zⁿ, we study an analogue of (1,1) that is an average of a suitable complex valued function f on Zⁿ over the d- plane H(A, b) = {m∈Zⁿ|Am=b}, the d-plane H(A, b) is regarded as solution set of ranksof linear systemAx=bof diophantine equations, with s =n−d≥1. The matrix A is an element of the set of s×n matrices of ranks ( see section 2 for more precisions).

The goal of this paper is to reconstruct a function f of C^(Zn) from its d-plane Radon transform. WhereC^(Zn)is the space of all functionsf defined

onZⁿ with finite support. We establish also a Plancherel’s formula relatively at discrete Radon transform for the functions f ofC^(Zn).Finally, we give a inversion formula, via the theorem 3.4, for the discrete Radon transform on the latticeu1Z×. . .×u_nZwhereu1, ..., u_nare the integer numbers satisfying the conditions (1),(2) of theorem 3.4.

Our paper is organized as follows:

In section 2, we recall some properties of the discrete Grassmannian of Zⁿ and we also fix certain notation that will be used in the sequel of this paper. The main result of this section is the theorem 2.4 which shows that Rf can be inverted by the formula

f(m₁, ..., m_n) = ∞ j₁=1

...

∞ j_n=1

μ(j₁)...μ(j_n)Rf(j₁m₁, ..., j_nm_n), whereRf is a complex valued function defined onZⁿ by

Rf(m₁, ..., m_n) = ∞ k₁=1

...

∞ k_n=1

f(k₁m₁, ..., k_nm_n),

for all (m₁, ..., m_n )∈Zⁿ\ {0},withf is a function ofS(Zⁿ) (see [2] for the definition of S(Zⁿ)) such that f(0) = 0 and μ is the Mobius function (see section 2).

In section 3, we give sufficient conditions for the following identity T_(u1,...,u

n)(H(A, b)) =H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b),

where u = (u₁, ..., u_n) ∈ (N\ {0})ⁿ and ˆu_j = u₁u₂...u_j−1u_j+1...u_nfor all j = 1,2, ..., n, the function T_(u1,...,u

n) being defined by T_(u1,...,u

n)(m) = (u₁m₁, ..., u_nm_n) for allm= (m₁, ..., m_n)∈Zⁿ.Finally

T_(u1,...,u

n)(H(A, b))

=

x= (x₁, ..., x_n)∈Zⁿ|x=T_(u1,...,u

n)(m), m∈H(A, b) . More precisely, we state and prove the following result

Let (A, b) ∈ Ps,n×Z^s with A = (a_ij)

1≤i≤s 1≤j≤n

and u = (u₁, ..., u_n) ∈ (N\ {0})ⁿ.Assume thatuand the coefficients a_ij ofA = (a_ij)

1≤i≤s 1≤j≤n

satisfy the conditions:

(1)d(u_j,uˆ_j) = 1, (2)d(u_j, a_ij) = 1,

for j = 1,2, ..., n and i = 1, . . . , s, the number d(x, y) being the greatest common divisor of the integersxandy.

Then T_(u1,...,u

n)(H(A, b)) =H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b),

wherediag_n×n(ˆu₁, . . . ,uˆ_n) is the matrix (α_ij)

1≤i,j≤n given by α_ij =

⎧⎨

⎩ ˆ

u_i fori=j, i= 1,2, ..., n

0 otherwise.

In the same section we define, via the above theorem, the d-plane Radon transform in the lattice T_u1,...,un(Zⁿ) = u₁Z×. . .×u_nZ where u₁, ..., u_n are the fixed integer numbers satisfying the conditions (1),(2) of the above theorem. In the sequel of this paper, we design by C^(Zn) the space of all functions f defined on Zⁿ with finite support.

In section 4, we establish many inversion formulas for the discrete d-plane Radon transform. The essential results of this section are

1) Letf ∈ S(Zⁿ) such that f(0) = 0 and t→f(0, ...,0, t) is an even function. Then theX−Ray transformRinZⁿcan be inverted by the formula

f(m_0,1, ..., m_0,n) = 1 2

∞ j=1

μ(j)Rf(g₀+jH(I_n−1,n,0)),

where g₀= (m_0,1, ..., m_0,n −1) andS(Zⁿ) is the Schwartz space ofZⁿcon- sisting of all complex valued functionsf defined onZⁿsuch thatp_N(f)<∞ for allN∈N, wherep_N(f) = sup_m∈Zn 1 +||m||²N

|f(m)|(see, [2]).

2) Let f ∈ C^(Zn) with suppf ⊂ {x₁, . . . , x_N} (x₁, . . . , x_n ∈ Zⁿ). Then for allj > j₀=N

i=1||x_i||²,we have

f(m) =Rf(H(a_j, a_jm)),for allm∈Zⁿ, wherea_j =

1, , j, , j², . . . , , jⁿ⁻¹

∈(N\ {0})ⁿ.

2. Notations and Preliminaries

We denote byS0(Zⁿ) the subspace of the Schwartz spaceS(Zⁿ) consisting of all complex valued functionsf defined onZⁿ such thatf(0) = 0. We design byS0,+(Zⁿ) the subspace ofS0(Zⁿ) such that the functiont→f(0, ...,0, t) (t∈Z) is even.

LetNbe the set of all positive integers. Recall that the Mobius function μis defined on the setNas follows

μ(1) = 1,

μ(n) = (−1)^rifnis the product ofrdistinct prime numbers, μ(n) = 0 otherwise.

In the goal to state and prove the dilatation theorem, we shall need some notation (see section 3). Let u= (u₁, ..., u_n)∈ (N\0)ⁿ, we note by ˆu the vector ofZⁿ defined by ˆu= (ˆu₁, ...,uˆ_n) where ˆu_j =u₁ u₂....u_j−1 u_j+1...u_n, for allj= 1,2, ..., n.

Let Ps,n be the set of s×n matrices of rank s whose Smith normal form is the projection matrix I_s,n =diag_s×n(1, ...,1), i.e, thes×nmatrix whose left hand s×s block is the identity and the right-hand s×d block is 0 (see [3] for more details). Finally b is an element of Z^s . We denote by G(d, n) the discrete affine Grassmann of all discreted-planes in Zⁿ. Recall that G(d, n) = (Ps,n×Z^s)/SL_s (see[3] Theorem 2.13) where SL_k is the group of k×k matrices whose determinant is equal to ±1. In particular SL₁={±1}.ConsequentlyG(d, n) is the set of all the d-planeH(A, b) with (A, b)∈ Ps,n×Z^s.

Let A ∈Ps,n, it is clear thatA diag_n×n(ˆu₁, ...,uˆ_n) ∈ P/ s,n. Indeed for s= 1, u= (4,2,3) anda= (1,2,3), the vector

ˆ

ua= (a₁uˆ₁, a₂ˆu₂, a₃uˆ₃) = (1,2,3)

⎛

⎝ 6 0 0 0 12 0

0 0 8

⎞

⎠= (6,24,24) does not belong to the set P_1,3.

LetS(Nⁿ) be the Schwartz space associated to the setNⁿ,i.e,S(Nⁿ) is the space of all functionsf such thatp_N(f) = sup_m∈Nn 1 +||m||²N

|f(m)|

<∞,for allN ∈N

Proposition 2.1. Letf ∈ S0(Zⁿ)andm= (m₁, ..., m_n)∈Zⁿ a fixed element.

Then the mapping

k1→ ^∞

k₁=1

...

∞ k_n=1

f(k1m1, ..., k_nm_n) (2.1) is a function of S(N).

The proof of the above proposition is easy.

Proposition 2.2. [see [12]] Let ϕ ∈ S0(N). Then the transform Rϕ defined by

Rϕ(n) = ∞ k=1

ϕ(kn) for alln∈N, (2.2) can be inverted by the formula

ϕ(n) = ∞ j=1

μ(j)Rϕ(jn). (2.3)

In particular

ϕ(1) = ∞ j=1

μ(j) _∞

l=1

ϕ(lj)

. (2.4)

We define the extension of Rϕ to Zby Rϕ(m) = _∞

l=1ϕ(lm) when ϕ ∈ S0(Z).

The above proposition can be generalized toZas follows

Proposition 2.3. Letf ∈ S0(Z). Then the transform Rf defined by Rf(m) =

∞ k=1

f(km) for allm∈Z, (2.5) can be inverted by

f(m) = ∞ j=1

μ(j)Rf(mj)

= ∞ j=1

μ(j) _∞

l=1

f(lmj)

. (2.6)

Proof. Form∈Z a fixed element, define the function ϕbyϕ(x) =f(mx) for allx∈N. Applying the proposition 2.2 (formula 2.4) toϕ, we obtain

f(m) = ϕ(1)

= ∞ j=1

μ(j) _∞

l=1

ϕ(lj)

= ∞ j=1

μ(j) ∞ l=1

f(ljm), then

f(m) = ∞ j=1

μ(j)Rf(mj). (2.7)

This proves the proposition.

Recall thatS0(Zⁿ) design the subspace of the Schwartz spaceS(Zⁿ)(see [2],[3]) constitued by the functions f such that f(0) = 0. Forf ∈ S0(Zⁿ) defineRf onZⁿ by the formula

Rf(m₁, ..., m_n) = ∞ k₁=1

...

∞ k_n=1

f(k₁m₁, ..., k_nm_n). (2.8) Now, we state and prove the following theorem.

Theorem 2.4. Let f ∈ S0(Zⁿ). Then the transform R defined by (2.8) can be inverted by

f(m₁, ..., m_n) = ∞ j₁=1

...

∞ j_n=1

μ(j₁)...μ(j_n)Rf(j₁m₁, ..., j_nm_n). (2.9)

We shall make the proof of (2.9) for n = 2, the same reasonning rest valid for alln.

Proof. Forn= 2,the formula (2.8) becomes Rf(m₁, m₂) =

∞ k₁=1

∞ k₂=1

f(k₁m₁, k₂m₂)

= ∞ k₁=1

_∞

k₂=1

f(k₁m₁, k₂m₂)

Then

Rf(m₁, m₂) = ∞ k₁=1

Bf(k₁m₁, m₂), (2.10) with

Bf(k₁m₁, m₂) = ∞ k₂=1

f(k₁m₁, k₂m₂), (2.11) m2being a fixed element, the above formula (2.10) is analogue to (2.5) then

Bf(m₁, m₂) = ∞ k₁=1

μ(k₁)Rf(k₁m₁, m₂). (2.12) But, by (2.11) we have

Bf(m1, m2) = ∞ k₂=1

f(m1, k2m2). Using the propositions 2.3 and formula 2.6, we obtain

f(m1, m2) = ∞ k₂=1

μ(k2)Bf(m1, k2m2), from (2.12), the above equality can be transformed as follows

f(m₁, m₂) = ∞ k₂=1

μ(k₂) _∞

k₁=1

μ(k₁)Rf(k₁m₁, k₂m₂)

= ∞ k₂=1

∞ k₁=1

μ(k2)μ(k1)Rf(k1m1, k2m2).

This proves the above theorem.

3. Dilatation of the discrete d -planes in Z

In this section, we shall state and prove an important result (see Theorem 3.4 below). The gool of this section is to define the d-plane in the lattice u₁Z×. . .×u_nZ, whereu₁, ..., u_nare the fixed numbers. In the followingPs,n

design the set ofs×n matrices of rank swhose Smith normal form is the projection matrix I_s,n = diag_s×n(1, ...,1), i.e, the s×n matrix whose left hand s×s block is the identity and the right-hands×dblock is 0 (see[3]

Theorem 2.13).We begin with the following proposition.

Proposition 3.1. Let (A, b)∈ Ps,n×Z^s. Then T_(u1,...,u

n)(H(A, b))⊂H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b), (3.1) for allu= (u₁, ..., u_n)∈(N\0)ⁿ.

Proof. Fory∈T_(u1,...,u

n)(H(A, b)),there exists m= (m₁, ..., m_n)∈H(A, b) such thaty= (u₁m₁, ..., u_nm_n).Showing that

y∈H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b). It is clear thatA diag_n×n(ˆu₁, ...,ˆu_n) = (a_ijuˆ_j)_1≤i≤s

1≤j≤n

,whereA= (a_ij)

1≤i≤s 1≤j≤n

. We have

(A diag_n×n(ˆu₁, ...,uˆ_n))y=

⎛

⎜⎜

⎜⎜

⎜⎜

⎝

a₁₁uˆ₁u₁m₁ + . . . + a_1nuˆ_nu_nm_n

. . . . . . .

. . . . . . .

. . . . . . .

. . . . . . .

a_s1uˆ₁u₁m₁ + . . . + a_snuˆ_nu_nm_n

⎞

⎟⎟

⎟⎟

⎟⎟

⎠ ,

(3.2) since y = (u₁m₁, ..., u_nm_n). Using the equality ˆu_ju_j = u₁u₂...u_n for all j= 1,2, ..., n,(3.2) can be transformed as follows

(A diag_n×n(ˆu₁, ...,uˆ_n))y= (u₁u₂...u_n)

⎛

⎜⎜

⎜⎝ b₁ b₂ ... b_s

⎞

⎟⎟

⎟⎠, (3.3)

with b_j = a_j1m₁+· · ·+a_jnm_n for all j = 1,2, ..., s. Since m ∈ H(A, b), it follows from (3.3) that y ∈ H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b), which

proves the proposition.

Proposition 3.2. Letu= (u₁, ..., u_n)∈(N\0)ⁿ.Then the mappingT_(u1,...,un): G(d, n)−→T_(u1,...,un)(G(d, n))is a bijection.

Proof. It suffices to prove the injectivity of this mapping. Let H(A, b) and H(A, b) be two elements ofG(d, n).Assume that

T_(u1,...,un)(H(A, b)) =T_(u1,...,un)(H(A, b)). (3.4) The equality (3.4) shows that the two systemsAx =b andAx = b have the same solutions inZⁿ, since the equalityT_(u1,...,u

n)(m) =T_(u1,...,u

n)(m) implies m = m (u_i = 0 for all i = 1,2, ..., n). By [3, Theorem 2.17] there existsQ∈SL_ssuch that A =QAand b =Qb,thenH(A, b) =H(A, b).

The proposition is proved.

Remark 3.3. Let (u₁, ..., u_n)∈(N\0)ⁿ and(A, b)∈ Ps,n×Z^s. Then H

Adiag_n×n(ˆu1, ...,uˆ_n),(u1...u_n)b

is not included inT_(u1,...,un)(H(A, b)).

For the simplest case n = 2, we can take a= (1,1) andu= (2,4). In this caseuˆ= (4,2)and the setT_(2,4)(H((1,1),0)) becomes

T_(2,4)(H((1,1),0)) ={(−2t,4t)|t∈Z}. On the other hand, we have

H((2,1),0) = H(ˆua,0)

= {(t,−2t)|t∈Z}. Then (1,−2)∈H((2,1),0), but(1,−2)∈ {/ (−2t,4t)|t∈Z}. In the general case, we can construct an element of the set

H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b), which does not belong toT_(u1,...,u

n)(H(A, b)).

Now, we shall seek arithmetical conditions betweenA ∈ Ps,n andu= (u1, ..., u_n)∈(N\0)ⁿ for following identity

T_(u1,...,u

n)(H(A, b)) =H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b). The answer to this question is given by the following theorem.

Theorem 3.4. Let(A, b)∈ Ps,n×Z^s,andu= (u₁, ..., u_n)∈(N\{0})ⁿ.Assume that uand the coefficientsa_ij of A= (a_ij)

1≤i≤s 1≤j≤n

satisfy the conditions:

(1) d(u_i,uˆ_i) = 1 for alli= 1,2, ..., n

(2) d(u_j, a_ij) = 1for allj = 1,2, ..., nandi= 1,2, ..., s,

withd(x, y)denotes the greatest common divisor of the integersxandy.Then T_(u1,...,un)(H(A, b)) =H(A diag_n×n(ˆu1, ...,uˆ_n),(u1...u_n)b), (3.5) where

diag_n×n(ˆu₁, ...,uˆ_n) = (α_ij)

1≤i≤s 1≤j≤n

=

⎧⎨

⎩

α_ii = ˆu_i for i= 1,2, ..., n

0 otherwise.

Proof. It is clear that A diag_n×n(ˆu1, ...,uˆ_n) = (a_ijuˆ_j)

1≤i≤s 1≤j≤n

, where A = (a_ij)

1≤i≤s 1≤j≤n

∈ Ps,n.By taking

m= (m₁, ..., m₂)∈H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b), we obtain

⎛

⎜⎜

⎜⎜

⎜⎜

⎝

a11uˆ1 . . . . . . . a1nuˆn

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

a_s1uˆ₁ . . . . . . . a_snuˆ_n

⎞

⎟⎟

⎟⎟

⎟⎟

⎠

⎛

⎜⎜

⎜⎝ m₁ m₂ ... m_n

⎞

⎟⎟

⎟⎠= (u₁u₂...u_n)

⎛

⎜⎝ b₁

... b_s

⎞

⎟⎠.

(3.6)

The equality (3.6) can be written

⎧⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎩

a11uˆ1m1+· · ·+a_1nuˆ_nm_n = (u1...u_n)b1 (I1) .

. .

a_s1uˆ₁m₁+· · ·+a_snuˆ_nm_n = (u₁...u_n)b_s (I_s)

(3.7)

Now, consider the equation (I₁).We have

a₁₁uˆ₁m₁= (u₁u₂...u_n)b₁−a₁₂uˆ₂m₂− · · · −a_1nuˆ_nm_n, then

a11uˆ1m1=u1(ˆu1b1−a12

ˆ u₂

u₁m2− · · · −a_1nuˆ_n

u₁m_n), (3.8) which implies

a₁₁uˆ₁m₁=u₁β₁, where

β1= (ˆu1b1−a12

ˆ u₂ u1

m2− · · · −a1nuˆ_n u1

mn)∈Z,

because u⁻_i¹(ˆu_j)∈Z wheni =j, for all (i, j)∈ {1,· · ·, n}² . The equality (3.8) implies that the integeru₁devidesa₁₁uˆ₁m₁,from the Gauss theorem and (1),(2) of the theorem 3.4, we haveu₁|m₁thenm₁=u₁α₁withα₁∈Z. Now, we consider the equation (I2),applying the same reasonning as above we obtainm2=u2α2withα2∈Zand so on we obtainm= (u1α1, ..., u_nα_n), that is m = (m1, ..., m_n) = T_(u1,...,un)(α) where α= (α1, ..., α_n). Replace m_j =u_jα_j(j= 1,2, ..., n) in the system (3.7).Since ˆu_iu_i = (u1u2...u_n) for alli= 1,2..., n,the formula (3.7) becomes

⎧⎪

⎨

⎪⎩

a11α1+· · ·+a_1nα_n=b1

...

a_s1α1+· · ·+a_snα_n=b_s, that is

A

⎛

⎜⎝ α₁

... α_n

⎞

⎟⎠=b= (b₁, ..., b_s).

Then α = (α₁, α₂, ..., α_n) ∈ H(A, b) and m = (u₁α₁, ..., u_nα_n) ∈ T_(u1,...,un)(H(A, b)). From proposition 3.1, we obtain

T_(u1,...,u

n)(H(A, b)) =H(A diag_n×n(ˆu₁, ...,uˆ_n),(u₁...u_n)b).

This proves the theorem.

In the sequel, we denote byA₀= (α_ij)

1≤i≤s 1≤j≤n

the matrix defined by

⎧⎨

⎩

α_1j= 1 for all j= 1,2, ..., n α_ii= 1 for all i= 1,2, ..., s α_ij= 0 otherwise.

Corollary 3.5. Let u= (u₁, ..., u_n) ∈(N\ {0})ⁿ. Assume that d(u_i,uˆ_i) = 1 for alli= 1,2, ..., n.Then

(1) (ˆu₁, ...,uˆ_n)∈ P1,n

(2)T_(u1,...,un)(H(A0, b) =H(A0diag_n×n(ˆu1, ...,uˆ_n),(u1...u_n)b) ; (3)T(β,1,...,1)(H(A₀, b) = H(A₀ diag_n×n(1, β, ..., β), βb), for all β ∈ N\ {0}.

Proof. First, the matrixA₀is an element ofPs,n,indeed A₀=I_s,n.V₀where V₀ = (β_ij)

1≤i≤n 1≤j≤n

∈ SL_n with β_ii = 1 for all i = 1,2, ..., n, β_1j = 1 for j = 1, ..., n and β_ij = 0 otherwise. Showing that (ˆu₁, ...,uˆ_n) ∈ P1,n when d(u_i,uˆ_i) = 1 for all i = 1,2, ..., n. By Bezout equality there exist v_i and w_i(i= 1,2..., n) with (v_i, w_i)∈Z² such that

u_iw_i+v_iuˆ_i= 1 for alli= 1,2..., n.

Then n

i=1

(u_iw_i+v_iˆu_i) = 1. (3.9) Prove that there existβ₁, β₂, ..., β_n belonging toZ such thatβ₁uˆ₁+β₂uˆ₂+

· · ·+β_nuˆ_n = 1. For this, we develop the formula (3.9) to obtain λ(u₁u₂...u_n) +

n i=1

μ_iuˆ_i= 1, (3.10) whereλ, μ_i∈Zfor alli= 1,2, ..., n.By the equality (u₁u₂...u_n) =u₁uˆ₁,the formula (3.10) becomes

β₁uˆ₁+ n i=2

η_iuˆ_i= 1, withβ₁, η_i∈Zfor alli= 2, ..., n.

The assertions (2),(3) are the simple consequences of the theorem (3.4) and

this proves the corollary.

4. Inversion Formulas for the discrete X− Ray Transform in Z

We begin by proving some results which will be useful in the sequel of this paper.

Proposition 4.1. Let(A, b)∈ Ps,n×Z^sandm₀∈Zⁿ.Then for allj∈N\ {0}, we have

m₀+jH(A, b)⊂m₀+H(A, jb)⊂H(A, jb+Am₀). (4.1) Proof. Let m ∈ jH(A, b) this implies that m =jm with Am = b, hence Am = jAm = jb. Let t ∈ m0+H(A, jb) it follows that A(t−m0) = jb thusAt=jb+Am0.The proposition is proved.

For every j ∈ N\ {0}, the subset m₀+jH(A, b) of H(A, jb+Am₀) will be called a sub-d-plane of the d-plane H(A, jb+Am₀). In the sequel, we reconstruct f(m₀) from Rf(m₀+jH(A, b)), restriction of the discrete d-plane Radon transform to the sub-d-planem₀+jH(A, b) (see the above Proposition).

In this section, we establish an explicit inversion formula for the dis- creteX-Ray transform R.More precisely, we state and prove the following inversion theorem.

Theorem 4.2. Let m₀= (m_0,1, ..., m_0,n)∈Zⁿ andf ∈ S0,+(Zⁿ).Then f(m_0,1, ..., m_0,n) =1

2 ∞ j=1

μ(j)Rf(g₀+jH(I_n−1,n,0)), whereg₀= (m_0,1, ..., m_0,n−1).

Proof. Consider the transformationRdefined by Rf(α) =

∞ j=1

f(0, ...,0, jα), for allα∈Z. (4.2) By Proposition 2.3 (see formulas (2.5),(2.6)), we have

f(0, ...,0,1) = ∞ j=1

μ(j)Rf(j). (4.3)

But

Rf(jH(I_n−1,n,0)) =

m_n∈Z

f(0, ...,0, jm_n), then

Rf(jH(I_n−1,n,0)) = 2 ∞ m_n=1

f(0, ...,0, jm_n), (4.4) since the function m_n −→ f(0, ...,0, m_n) is even and f(0) = 0. Then the equality (4.4) can be transformed as follows

Rf(jH(I_n−1,n,0)) = 2Rf(j), see formulas (4.4) and (4.2).From (4.3),it follows that

f(0, ...,0,1) = 1 2

∞ j=1

μ(j)Rf(jH(I_n−1,n,0)). (4.5) Now, replacing f by _g0f (where_g0f(x) =f(x+g0)) in the equality (4.5), we obtain

f(m_0,1, ..., m_0,n) =1 2

∞ j=1

μ(j)Rf(g0+jH(I_n−1,n,0)),

which proves the theorem.

Now, we shall give an other inversion formula for the discrete d-plane Radon transform. More precisely, we state and prove the following theorem.

Theorem 4.3. Letf ∈ S(Zⁿ)andm₀∈Zⁿ.Then the discreted-plane Radon transform R can be inverted as follows

f(m₀) =

b∈Z^s

Rf(H(A, b))−

b∈Z^s\{0}

Rf(H(I_s,n, b+ (m_0,1, ..., m_0,s)))

−Rf

H(I_s,n,0)^∗+m₀

, (4.6)

for allA∈ Ps,n, whereH(I_s,n,0)^∗=H(I_s,n,0)\ {0}.

Proof. Recall thatS(Zⁿ) is the Schwartz space ofZⁿ (see [2]). It clear that

m∈Zⁿ

f(m) =

(m₁,...,m_s)∈Z^s\{0},(m_s+1,...,m_n)∈Z^d

f(m₁, ..., m_s, m_s+1, ..., m_n)

+f(0) +

(m_s+1,...,m_n)∈Z^d\{0}

f(0, ...,0, m_s+1, ..., m_n). (4.7) In the right hand side of the equality (4.7), the first series can be writ- ten as

b∈Z^s\{0}Rf(H(I_s,n, b)) while the second one can be transformed as Rf(H(I_s,n,0)^∗).As for the series of the left hand side of (4.7),it is equal to

b∈Z^sRf(H(A, b),for all A∈ Ps,n (see [2, Corollary 3.10]). Consequently the equality (4.7) becomes

f(0) =

b∈Z^s

Rf(H(A, b))−

b∈Z^s\{0}

Rf(H(I_s,n, b)−Rf

H(I_s,n,0)^∗ . (4.8) Replacing f by _m0f, where _m0f(m) = f(m+m0), and using the propo- sitions 3.2, 3.8 of [3] (see also [2]), we obtain (4.6) and the theorem is

proved.

In this section, we shall establish inversion formulas for the Radon trans- form inZⁿ.We begin by proving the following theorem

Theorem 4.4. Let f be a function ofC^(Zn) such that suppf ⊂ {x₁, . . . , x_N}, withx₁, . . . , x_N ∈Zⁿ. Then for allj > j₀=N

i=1||x_i||²,we have

Rf(H(a_j, a_jm)) =f(m), for allm∈Zⁿ, (4.9) wherea_j =

1, , j, j². . . , jⁿ⁻¹

∈ P (see [2])

Proof. Let m₀ and m be two elements of Zⁿ, with m = (m₁, . . . , m_n) and m0 = (m0,1, . . . , m0,n). Given f ∈ C^(Zn), we have by definition of Radon transform (see [2])

Rf(H(aj, ajm0)) =

m∈Zⁿ,a_j(m₀−m)=0

f(m),for allj∈N\ {0}, Show that Rf(H(a_j, a_jm₀)) = f(m₀), for all j > j₀ = N

i=1||x_i||². The equalitya_j(m₀−m) = 0 implies that

(m₁−m_0,1) +j(m₂−m_0,2) +· · ·+jⁿ⁻¹(m_n−m_0,n) = 0 (I)

The equation (I) gives (m₁−m_0,1) =−j

(m₂−m_0,2) +· · ·+jⁿ⁻²(m_n−m_0,n) ,

then j | (m₁−m_0,1), thus |m₁−m_0,1| ≥ j if m₁ = m_0,1. It follows that j0 =N

i=1||x_i||² < j≤ ||m||²+||m0||², where m andm0 belong to the set {x1, . . . , x_N}, which is absurd. We deduce thatm1=m_0,1, and therefore the equation (I) becomes

(m₂−m_0,2) +j(m₃−m_0,3) +· · ·+jⁿ⁻²(m_n−m_0,n) = 0. (I) Applying the same reasoning as in the case of the equation (I) to (I), we obtain m_i = m_0,i, for i = 2, . . . , n. And this completes the proof of the

theorem.

Now, letϕ_N0(j) be a sequence defined by

ϕ_N0(j) =

⎧⎪

⎪⎨

⎪⎪

⎩

0 ifj≤N₀ exp (−j)

_∞

k=N₀+1exp (−k) =ψ(j) ifj≥N0+ 1,

(4.10)

whereN₀ is a fixed number. The above theorem can be written as follows Theorem 4.5. Let f be a function ofC^(Zn) such that suppf ⊂ {x₁, . . . , x_N}, withx₁, . . . , x_N ∈Zⁿ andj₀=N

i=1||x_i||² . Then the discrete Radon trans- form inZⁿ can be inverted by the following formula

j∈N

ϕ_2j0(j)Rf(H(a_j, a_jm)) =f(m), for allm∈Zⁿ, (4.11) Proof. Letf be a function ofC^(Zn) such that suppf ⊂ {x₁, . . . , x_N},with x₁, . . . , x_N ∈ Zⁿ.The expression

j∈Nϕ_2j0(j)Rf(H(a_j, a_jm)),(m∈Zⁿ) can be written as follows

j∈N

ϕ_2j0(j)Rf(H(a_j, a_jm)) =

j≤2j₀

ϕ_2j0(j)Rf(H(a_j, a_jm))

+ ∞ j=2j₀+1

ϕ_2j0(j)Rf(H(a_j, a_jm)). (4.12) From (4.9) and (4.10),the formula (4.12) becomes

j∈N

ϕ_2j0(j)Rf(H(a_j, a_jm)) = f(m)

⎛

⎝ ^∞

j=2j₀+1

ϕ_2j0(j)

⎞

⎠

= f(m), since_∞

j=2j₀+1ϕ2j₀(j) = 1.The theorem is proved We deduce, from the above theorem, the Plancherel formula for the discrete Radon transform inZⁿ.