# On the Exponential Radon Transform and Its Extension to Certain Functions Spaces

Authors: S. K. Q. Al-Omari; A. Kılıçman
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We investigate the exponential Radon transform on a certain function space of generalized functions. We establish certain space of generalized functions for the cited transform. The transform that is obtained is well defined. More properties of consistency, convolution, analyticity, continuity, and sufficient theorems have been established.


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## Body

## 1. Introduction

The Radon transform of a sufficiently nice functionf defined on ℝn is given by
(1)(ℛϑf)(η)≡(ℛf)(ϑ,η)≡∫ϑ⊥f(ηϑ+u)du,
where (ϑ,η)∈ℝ~n=∑n-1‍xp,∑n-1‍ is the unit sphere in ℝn, and du is the Euclidean measure on the subspace ϑ⊥ orthogonal to ϑ.Applications of the Radon transform occur in a number of areas, such as seismic signal processing, remote sensing, and system identification from output data [1, 2]. The Radon transform is extended to various spaces of distributions, rapidly decreasing and integrable Boehmians [3, 4]. More about the Radon transform is given in [5–9].The discrete Radon transform is defined by [10, 11]. The attenuated Radon transform is defined in Mikusiński et al. [12, 13]. For a uniform attenuation coefficient μ∈ℂ, the exponential Radon transform of a compactly supported real valued function f, defined on ℝ2, is given by Kurusa and Hertle [7, 8]:
(2)Tμef(ϑ,t)=∫ℝ2f(x)δ(x·ϑ-t)eμx·ϑ⊥dx,t∈ℝ,
where ϑ=(cos⁡φ,sinφ)τ is a unit vector on 𝒮1, φ∈[0,2π),ϑ⊥=(-sinφ,cos⁡φ).The exponential Radon transform constitutes a mathematical model for imaging modalities such as X-ray tomography forμ=0, single photon emission tomography for μ∈ℝ, and optical polarization tomography of trass tensor field [14]. However, if in addition μ is unknown, then one first must find μ and then find f. This is the identification problem.The exponential Radon transform, as a generalization of the Radon transform, is defined as a mapping of function spaces and is also represented in terms of Fourier transforms of its domain and range, and this leads to a characterization of the range of the transform. For more information about the exponential Radon transform, we refer to [15, 16].


## 2. General Construction of Boehmians

The minimal structure necessary for the construction of Boehmians consists of the following elements:(i)
a seta and a commutative semigroup (g,*);(ii)
an operation⊙:a×g→a such that for each x∈a and υ1,υ2,∈g,
(3)x⊙(υ1*υ2)=(x⊙υ1)⊙υ2;(iii)
a collectionΔ⊂gℕ satisfying the following:
(1)
ifx,y∈a, (υn)∈Δ, x⊙υn=y⊙υn for all n, then x=y;(2)
if(υn),(σn)∈Δ, then (υn*σn)∈Δ,Δ being the set of all delta sequences.Consider(4)𝒜={(xn,υn):xn∈a,(υn)∈Δ,xn⊙υm=xm⊙υn,∀m,n∈ℕ}.
If (xn,υn),(yn,σn)∈𝒜, xn⊙σm=ym⊙υn, ∀m,n∈ℕ, then we say (xn,υn)~(yn,σn). The relation ~ is an equivalence relation in 𝒜. The space of equivalence classes in 𝒜 is denoted by δ(a,(g,*),⊙,Δ). Elements of δ(a,(g,*),⊙,Δ) are called Boehmians.Betweena and δ(a,(g,*),⊙,Δ) there is a canonical embedding expressed as
(5)x⟶x⊙snsnasn⟶∞.The operation⊙ can be extended to δ(a,(g,*),⊙,Δ)×a by
(6)xnυn⊙t=xn⊙tυn.
In δ(a,(g,*),Δ), two types of convergence exist:(1)
a sequence(hn) in δ(a,(g,*),⊙,Δ) is said to be δ-convergent to h in δ(a,(g,*),⊙,Δ), denoted by hn→δh as n→∞, if there exists a delta sequence (υn) such that (hn⊙υn),(h⊙υn)∈a, ∀k,n∈ℕ, and (hn⊙υk)→(h⊙υk) as n→∞, in a, for every k∈ℕ;(2)
a sequence(hn) in δ(a,(g,*),⊙,Δ) is said to be Δ-convergent to h in δ(a,(g,*),⊙,Δ), denoted by hn→Δh as n→∞, if there exists a (υn)∈Δ such that (hn-h)⊙υn∈a, ∀n∈ℕ, and (hn-h)⊙υn→0 as n→∞ in a.The following theorem is equivalent to the statement ofδ-convergence.Theorem 1.
h
n
→
δ
h
(
n
→
∞
) in δ(a,(g,*),⊙,Δ) if and only if there is fn,k,fk∈a and υk∈Δ such that hn=[fn,k/υk], h=[fk/υk] and for each k∈ℕ, fn,k→fk as n→∞ in a.For further discussion see [3, 17–21].


## 3. Necessary and Sufficient Conditions

Denote byl1(𝒮1×ℝ) the space of Lebesgue complex-valued measurable functions of bounded support defined on 𝒮1×ℝ and satisfying
(7)∥|ψ((ϑ,t))|∥=∫𝒮1∫ℝ|ψ((ϑ,t))|2eμy·ϑ⊥dϑdt<∞
then Tμeψ∈l1(𝒮1×ℝ),μ∈ℝ,y∈ℝ2 being arbitrary but fixed.Byκ(ℝ2) denote the space of test functions of bounded support defined on ℝ2.LetΔ be the set of sequences (μn(x))∈κ(ℝ2) such that [3, (2.6)–(2.8)]
(8)∫ℝ2μn(x)dx=1,∫ℝ2|μn(x)|dx<M,0<M∈ℝ,supp⁡(μn(x))⟶0asn⟶∞.The convolution product between two functions is defined by the integral equation(9)(f*g)(x)=∫ℝ2f(x-y)g(y)dy,
where x∈ℝ2.Now we construct the spaceδ(l1,(κ,*),×,Δ) of Boehmians.We have the following definition.Definition 2.
Letψ∈l1(𝒮1×ℝ) and ξ∈κ(ℝ2); then we define the mapping × as
(10)(ψ×ξ)(ϑ,t)=∫ℝ2ψ(ϑ,t-y·ϑ)ξ(y)eμy·ϑ⊥dy(ψ×ξ)(ϑ,t)(ψ×ξ)(ϑ,t)===(ϑ,t)∈𝒮1×ℝ.Theorem 3.
Letψ∈l1(𝒮1×ℝ) and ξ∈κ(ℝ2); then ψ×ξ∈l1(𝒮1×ℝ).Proof.
Letψ∈l1(𝒮1×ℝ) and ξ∈κ(ℝ2). By using (10), Fubini’s theorem, and Jensen’s inequality we get
(11)∥|(ψ×ξ)(ϑ,t)|∥2=∫𝒮1∫ℝ|(ψ×ξ)(ϑ,t)|2dϑdt=∫𝒮1∫ℝ|∫ℝ2ψ(ϑ,t-y·θ)ξ(y)eμy·ϑ⊥dy|2dϑdt=∫ℝ2|ξ(y)|∫𝒮1∫ℝ|ψ(ϑ,t-y·ϑ)|2eμy·ϑ⊥dϑdtdy<M∥|ψ|∥2,
where M is a positive constant.
The proof is therefore completed.Theorem 4.
Letψn→ψ in l1(𝒮1×ℝ) and ξ∈κ(ℝ2); then ψn×ξ→ψ×ξ as n→∞.Proof of this theorem follows from Theorem3.Theorem 5.
Letψ∈l1(𝒮1×ℝ) and ξ1,ξ2∈κ(ℝ2); then one has
(12)ψ×(ξ1*ξ2)=(ψ×ξ1)×ξ2.Proof.
Let(ϑ,t)∈𝒮1×ℝ. Using (10) and (9) we write
(13)(ψ×(ξ1*ξ2))(ϑ,t)=∫ℝ2ψ(ϑ,t-y·ϑ)(ξ1*ξ2)(y)eμy·ϑ⊥dy=∫ℝ2ψ(ϑ,t-y·ϑ)(∫ℝ2ξ1(y-x)ξ2(x)dx)eμy·ϑ⊥dy=∫ℝ2(∫ℝ2ψ(ϑ,t-y·ϑ)ξ1(y-x)eμy·ϑ⊥dy)ξ2(x)dx.
The substitutiony-x=z, y,x,z∈ℝ2, implies
(14)(ψ×(ξ1*ξ2))(ϑ,t)=∫ℝ2(∫ℝ2ψ(ϑ,t-(z+x)·ϑ)ξ1(z)eμ(z+x)ϑ⊥dz)×ξ2(x)dx=∫ℝ2(∫ℝ2ψ(ϑ,(t-x·ϑ)-z·ϑ)ξ1(z)eμx·ϑ⊥dz)×ξ2(x)eμx·ϑ⊥dx=∫ℝ2(ψ×ξ1)(ϑ,t-x·ϑ)ξ2(x)eμx·ϑ⊥dx=((ψ×ξ1)×ξ2)(ϑ,t).
This completes the proof of the theorem.Theorem 6.
Letψ1,ψ2∈l1(𝒮1×ℝ) and ξ∈κ(ℝ2); then
(15)(ψ1+ψ2)×ξ=ψ1×ξ+ψ2×ξ2.Theorem 7.
Letα∈ℂ, ψ∈l1(𝒮1×ℝ) and ξ∈κ(ℝ2); then
(16)α(ψ×ξ)=(αψ)×ξ.Proof of Theorems6 and 7 follows from simple integration. Detailed proof is thus avoided.Theorem 8.
Letψ∈l1(𝒮1×ℝ) and (μn)∈Δ; then ψ×μn→ψ as n→∞.Proof.
Letψ∈l1(𝒮1×ℝ). Since κ(ℝ×ℝ) is dense in l1(𝒮1×ℝ), we can choose α∈κ(ℝ×ℝ) such that
(17)∥|ψ-α|∥<ϵ,ϵ>0.
From the analysis applied for proving Theorem3 and by applying (17) we get
(18)∥|(ψ-α)×μn|∥≤M∥|ψ-α|∥<Mϵ.
Also, for each fixed (ϑ,t)∈𝒮1×ℝ define
(19)g(y)=α(ϑ,t-y·ϑ)eμy·ϑ⊥,y∈ℝ2;
then g(y)∈κ(ℝ2) and hence g(y) uniformly continuous on ℝ2. Thus, there is δ>0 such that |g(y)-g(x)|<ϵ whenever |y-x|≤δ.
Moreoversupp⁡α(ϑ,t)⊆[a,b]×𝕜,𝕜⊆ℝ-{0} implies α(ϑ,t)=0, ∀(ϑ,t)∉[a-δ,b+δ]×𝕜.
Hence, by (8) and the fact that |(g(y)-g(0))|<ϵ, by Jensen’s inequality, we have
(20)∥|(α×μn-α)(ϑ,t)|∥2=∫𝒮1∫ℝ|(α×μn-α)(ϑ,t)|2dϑdt=∫𝒮1∫ℝ|∫ℝ2(α(ϑ,t-y·ϑ)μn(y)eμy·ϑ⊥-α(ϑ,t)μn(y)α(ϑ,t-y·ϑ)μn(y)eμy·ϑ⊥)dy∫ℝ2(α(ϑ,t-y.ϑ)μn(y)eμy.ϑ⊥|2dϑdt=∫𝒮1∫ℝ|∫ℝ2(g(y)-g(0))μn(y)dy|2dϑdt≤∫𝕜∫a-δb+δ∫ℝ2|(g(y)-g(0))|2|μn(y)|dydϑdt≤ϵ2∫𝕜∫a-δb+δ∫ℝ2|μn(y)|dydϑdt=ϵ2Mm(𝕜)(b-a+2δ),
where m(𝕜) is the Lebesgue measure of 𝕜.
Hence, using (17), (18), and (20) we, for large values of n, get
(21)∥|ψ×μn-ψ|∥≤∥|(ψ-α)×μn|∥∥|ψ×μn-ψ|∥≤+∥|α×μn-α|∥+∥|α-ψ|∥∥|ψ×μn-ψ|∥<Mϵ+ϵ2Mm(𝕜)(b-a+2δ)+ϵ∥|ψ×μn-ψ|∥=(M+ϵ2Mm(𝕜)(b-a+2δ)+1)ϵ.
Hence ψ×μn→ψ as n→∞.
The Boehmian spaceδ(l1,(κ,*),×,Δ) is constructed.
The sum and multiplication by a scalar of two Boehmians are naturally defined in the respective ways:(22)[(fn)(ϕn)]+[(gn)(τn)]=[(fn×τn)+(gn×ϕn)(ϕn*τn)],η[(fn)(ϕn)]=[(ηfn)(ϕn)],η being complex number.
The operations× and the derivative are defined by
(23)[(fn)(ϕn)]×[(gn)(τn)]=[(fn×gn)(ϵn*τn)],𝒟α[(fn)(ϕn)]=[(𝒟αfn)(ϕn)].
Betweenl1(𝒮1×ℝ) and δ(l1,(κ,*),×,Δ) the canonical embedding admits
(24)f⟶[(f×ϕn)(ϕn)]asn⟶∞.
The operation× can be extended to δ(l1,(κ,*),×,Δ)×l1(𝒮1×ℝ) by
(25)[(fn)(ϕn)]×f=[(fn×f)(ϕn)].
Byδ(l1,(κ,*),*,Δ) denote the corresponding Boehmian space obtained from l1(𝒮1×ℝ), κ(ℝ2), Δ and the product *.Theorem 9.
Letψ∈l1(𝒮1×ℝ) and ξ∈κ(ℝ2); then
(26)Tμe(ψ*ξ)(ϑ,t)=(Tμeψ×ξ)(ϑ,t),(ϑ,t)∈𝒮1×ℝ.Proof.
Let(ϑ,t)∈(𝒮1×ℝ). By employing (2) for (9) we get
(27)Tμe(ψ*ξ)(ϑ,t)=∫ℝ2(ψ*ξ)(x)δ(x·ϑ-t)eμx·ϑ⊥dx=∫ℝ2(∫ℝ2ψ(x-y)ξ(y)dy)δ(x·ϑ-t)eμx·ϑ⊥dx=∫ℝ2(∫ℝ2ψ(x-y)δ(x·ϑ-t)eμx·ϑ⊥dx)ξ(y)dy.
The substitutionx-y=z implies x=y+z and dx=dz.
Thus we get(28)Tμe(ψ*ξ)(ϑ,t)=∫ℝ2∫ℝ2ψ(z)δ(z·ϑ-(t-yϑ))eμz·ϑ⊥eμy·ϑ⊥dzξ(y)dy=∫ℝ2∫ℝ2Tμeψ(t-y·ϑ)eμy·ϑ⊥ξ(y)dy=(Tμeψ×ξ)(ϑ,t).
This completes the proof of the theorem.


## 4. The Exponential Radon Transform of Boehmians

Definition 10.
Letβn=[ψn/μn]∈δ(l1,(κ,*),*,Δ); then we define its exponential Radon transform as the mapping
(29)Tμe→[ψnμn]=[Tμeψnμn]
in the space δ(l1,(κ,*),×,Δ).
Definition10 is well defined by Theorem 9.
To show that (29) is well defined, let [ψn/μn],[ξn/ϵn]∈δ(l1,(κ,*),*,Δ) and [ψn/μn]=[ξn/ϵn]; then
(30)ψn*ϵm=ξm*μn.
Employing Tμe for (30) and using Theorem 9 imply that
(31)Tμeψn×ϵm=Tμeξm×μn.
From (31) we see that Tμeψn/μn~Tμeξn/ϵn in the sense of δ(l1,(κ,*),×,Δ).
This completes the proof of the theorem.Theorem 11.
Letβ1,β2∈δ(l1,(κ,*),*,Δ); then Tμe→(β1*β1)=Tμe→β1×β2.Proof.
Assume the requirements of the theorem are satisfied for someβ1,β2∈δ(l1,(κ,*),*,Δ); then there are (fn),(κn)∈l1(𝒮1×ℝ) and (φn),(ϕn)∈Δ such that β1=[(fn)/(φn)] and β2=[(κn)/(ϕn)]. Therefore, we write
(32)Tμe→(β1*β2)=Tμe→([(fn)*(κn)(φn)*(ϕn)])=[Tμe((fn)*(κn))(φn)*(ϕn)]=[(Tμefn)×(κn)(φn)×(ϕn)]=[(Tμefn)(φn)]×[(κn)(ϕn)].
Thus we get thatTμe→(β1*β2)=Tμe→(β1)×β2.
This completes the proof.Theorem 12.
T
μ
e
→ defines a linear mapping from δ(l1,(κ,*),*,Δ) into δ(l1,(κ,*),×,Δ).The proof is straightforward.Definition 13.
Letβ∈δ(l1,(κ,*),×,Δ) be such that β=[(Tμefn)/(ϕn)]. Then we define the inverse transform of 𝒯μe as
(33)(Tμe→)-1[(Tμefn)(ϕn)]=[(fn)(ϕn)]
for each (ϕn)∈Δ.Theorem 14.
T
μ
e
→ defines an isomorphism from δ(l1,(κ,*),*,Δ) onto δ(l1,(κ,*),×,Δ).Proof.
Assume that[(Tμefn)/(ϕn)]=[(Tμegn)/(ψn)] in δ(l1,(κ,*),×,Δ). Using (29) and Theorem 9 we get Tμefn×ψm=Tμegm×ϕn. Once again, Theorem 9 implies Tμe(fn*ψm)=Tμe(gm*ϕn). Hence fn*ψm=gm*ϕn. Therefore, [(fn)/(ϕn)]=[(gn)/(ψn)].
Now, let[(Tμefn)/(ϕn)]∈δ(l1,(κ,*),×,Δ); then Tμefn×ϕm=Tμefm×ϕn, ∀m,n∈ℕ. Theorem 9 leads to Tμe(fn*ϕm)=Tμe(fm*ϕn). Hence [(fn)/(ϕn)]∈δ(l1,(κ,*),*,Δ) is the Boehmian that satisfies Tμe→[(fn)/(ϕn)]=[(Tμefn)/(ϕn)].
This completes the proof of the theorem.Theorem 15.
Let[(fn)/(ϕn)]∈δ(l1,(κ,*),*,Δ) and ϕ∈κ(ℝ2); then
(34)Tμe→([(fn)(ϕn)]*ϕ)=[(Tμefn)(ϕn)]×ϕ.Proof.
Applying Definition10 for each [(fn)/(ϕn)]∈δ(l1,(κ,*),×,Δ) and ϕ∈κ(ℝ2) yields
(35)(Tμe→)([(fn)(ϕn)]*ϕ)=[(Tμe)((fn)*ϕ)(ϕn)].
By Theorem9 we get
(36)(Tμe→)([(fn)(ϕn)]*ϕ)=[(fn)×ϕ(ϕn)]=[(fn)(ϕn)]×ϕ.
This completes the proof of the theorem.Theorem 16.
The mappingsTμe→ and (Tμe→)-1 are continuous with respect to δ and Δ convergence.Proof.
First of all, we show thatTμe→ and (Tμe→)-1 are continuous with respect to δ convergence.
Letβn→δβ in δ(l1,(κ,*),*,Δ) as n→∞; then we show that Tμe→βn→Tμe→β as n→∞. By virtue of Theorem 1 we can find fn,k and fk in l1(ℝ2) such that βn=[fn,k/ϕk] and β=[fk/ϕk] such that fn,k→fk as n→∞ for every k∈ℕ.
Employing the continuity condition ofTμe transform implies Tμefn,k→Tμefk as n→∞ in the space l1(𝒮1×ℝ).
Thus,(37)[Tμefn,kϕk]⟶[Tμefkϕk]asn⟶∞
in δ(l1,(κ,*),×,Δ).
To prove the second part, letgn→δg in δ(l1,(κ,*),×,Δ) as n→∞. Then, once again, by Theorem 1, gn=[Tμefn,k/ϕk] and g=[Tμefk/ϕk] and Tμefn,k→Tμefk as n→∞. Hence fn,k→fk in δ(l1,(κ,*),*,Δ) as n→∞. That is, [fn,k/ϕk]→[fk/ϕk] as n→∞. Using (33) we get
(38)(Tμe→)-1[Tμefn,kϕk]⟶(Tμe→)-1[Tμefkϕk]asn⟶∞.
Now, we establish continuity ofTμe→ and (Tμe→)-1 with respect to Δ convergence.
Letβn,β∈δ(l1,(κ,*),*,Δ) be such that βn→Δβ as n→∞. Then, by Theorem 1 we can find that fn∈l1(𝒮1×ℝ) and (ϕn)∈Δ such that (βn-β)*ϕn=[((fn)*ϕk)/ϕk] and fn→0 as n→∞. Employing (29) we get
(39)Tμe→((βn-β)*ϕn)=[Tμe((fn)*ϕk)ϕk].
Hence, we have Tμe→((βn-β)*ϕn)=[((Tμefn)×ϕk)/ϕk]=Tμefn→0 as n→∞ in l1(𝒮1×ℝ).
Therefore(40)Tμe→((βn-β)*ϕn)=(Tμe→βn-Tμe→β)×ϕn⟹asn⟶∞.
Hence, Tμe→βn→ΔTμe→β as n→∞.
Finally, letgn→Δg in δ(l1,(κ,*),*,Δ) as n→∞; then we find Tμefk∈l1(𝒮1×ℝ) such that (gn-g)×ϕn=[(Tμefk×ϕk)/ϕk] and Tμefk→0 as n→∞ for some (ϕn)∈Δ.
Now, using (33), we obtain
(41)(Tμe→)-1((gn-g)×ϕn)=[(Tμe)-1(Tμefk×ϕk)ϕk].
Theorem 9 implies
(42)(Tμe→)-1((gn-g)×ϕn)=[(fn)*ϕkϕk]=fn⟶0∈l1(𝒮1×ℝ)asn⟶∞.
Thus
(43)(Tμe→)-1((gn-g)×ϕn)=((Tμe→)-1gn-(Tμe→)-1g)*ϕn→0∈l1(𝒮1×ℝ)asn⟶∞.
From this we find that(Tμe→)-1gn→Δ(Tμe→)-1g∈δ(l1,(κ,*),*,Δ) as n→∞.
This completes the proof of the theorem.Theorem 17.
The transformTμe→ is consistent with Tμe→:l1(𝒮1×ℝ)→l1(𝒮1×ℝ).Proof.
For everyf∈l1(𝒮1×ℝ), let β∈δ(l1,(κ,*),*,Δ) be its representative; then ∀n∈ℕ, (φn)∈Δ, β=[(f*(φn))/(φn)]. For all n∈ℕ it is clear that (φn) is independent of the representative.
We have(44)Tμe→(β)=Tμe→([f*(φn)(φn)])=[Tμe(f*(φn))(φn)]=[Tμef×(φn)(φn)]
which is the representative of Tμef in the space l1(𝒮1×ℝ).
Hence the proof is completed.Theorem 18.
The necessary and sufficient condition for[(gn)/(ψn)]∈δ(l1,(κ,*),×,Δ) to be in the range of Tμe→ is that gn belongs to range of Tμe for every n∈ℕ.Proof.
Let[(gn)/(ψn)] be in the range of Tμe→; then of course gn belongs to the range of Tμe,∀n∈ℕ.
To establish the converse, letgn be in the range of  Tμe, ∀n∈ℕ. Then there is fn∈l1(𝒮1×ℝ) such that Tμefn=gn,n∈ℕ.
Since[(gn)/(ψn)]∈δ(l1,(κ,*),×,Δ) we get gn×ψm=gm×ψn, ∀m,n∈ℕ.
Therefore, Theorem9 yields
(45)Tμe(fn*φn)=Tμe(fm*φn),∀m,n∈ℕ,
where fn∈l1(𝒮1×ℝ) and φn∈Δ, ∀n∈ℕ.
Thusfn*φm=fm*φn, m,n∈ℕ. Hence,
(46)[(fn)(φn)]∈δ(l1,(κ,*),*,Δ),Tμe→([(fn)(φn)])=[(gn)(ψn)].
The theorem is therefore completely proved.



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