http://www.acta.hu/
A characterization of the radon transform
and its dual on Euclidean space
Árpád Kurusa
Abstract. In this paper we present a characterization of the Radon transform without any restriction on its range. We also consider the boomerang
transform B [5], which is essentially the dual of the Radon transform.
A. Hertle [3] gave a characterization of the Radon transform by investigating its behavior under rotations, dilations and translations and making some restrictions on its range. We present an alternative characterization that needs no
restriction on the range.
By definition
Z
R(f )(ω,p) =
Z
f (x)dx,
hx,ωi=p
B(h)(x) =
h(hω, xiω)dω,
n−1
Sx
where ω ∈ S n−1 , 0 ≤ p ∈ R, dx and dω is the surface measure, h and f are good
enough functions on Rn (for example continuous functions with compact support)
and
Sxn−1 = ω ∈ S n−1 : hω, xi ≥ 0 .
One can easily verify (see [4]), that the boomerang transform has the following
properties:
(1) B(α1 f1 + α2 f2 ) = α1 B(f1 ) + α2 B(f2 )
(2) B(f ◦ U ) = B(f ) ◦ U
(3) B(f (δx))(y) = B(f )(δy)
(4) B(f )(y+z) = B f x hx,x+zi
hx,xi
(y)
(5) If fk → f locally uniformly on Rn then B(fk )(x) → B(f )(x) for any x ∈ Rn ,
where α1 , α2 ∈ R, f, f1 , f2 , fk are functions on Rn , U is an orthogonal transformation, 0 < δ ∈ R and x, y ∈ Rn .
Acta Sci. Math., 54 (1990), 273–276.
c Á. Kurusa
http://www.math.u-szeged.hu/tagok/kurusa
2
Á. KURUSA
Theorem 1. If a nonzero function-transformation β has properties (1) — (5) then
β(1)
β = B(1)
B on C(Rn \{0}).
Proof. Let fi (x) = |x|i . The following simple equations prove, that β(fi ) = fi ·c0i ,
where the constants c0i depends only on i ∈ N.
β(fi )(δx) = β(fi (δy))(x) = δ i β(fi )(x)
β(fi )(U x) = β(fi (U y))(x) = β(fi )(x)
This implies that if i is even
c0i (t + |x|)i = β(fi )(x+(tx/|x|)) = β(fi (y + tey hey , ex i))(x) =
=
i
X
i
j
ti−j β(|y|j hex , ey ii−j )(x) ,
j=0
x
and t ∈ R. Since this is an equality of two polynomials, we have
where ex = |x|
obtained for even i that
c0i fj (x) = β(|y|j hex , ey ii−j )(x)
(∗)
Now this equation and the condition (4) imply
β(he, ey ii )(e0 ) = β(he, ey ii )(e+e0 ) = c0i ,
where e, e0 are unit vectors. Let ci = B(fi )/fi . The integration of our last equation
over Sen−1
gives
0
Z
Z
c0i B(1) =
β(he, ey ii )(e0 ) de = β
he, ey ii de
= β(ci )(e0 ) = ci β(1).
Sen−1
0
Sen−1
0
(e0 )
This way, we have derived the following equation for any even i:
β(1)
B(fi ) =
B(fi ).
B(1)
Furthermore if i is even
β(fi (x + y)) = β(hx + y, x + yii/2 )
X i/2 =
2m |x|2l+m β(|y|2k+m hex , ey im ),
k, l, m
k,l,m
where k + l + m = i/2 and k, l, m ∈ N. Taking (∗) into account this gives rise to
(by calculating B(fi (x + y)))
β(1)
B(fi (x + y)) =
B(fi (x + y)) .
B(1)
Since the translates of radial polynomials are locally uniformly dense in C(Rn )
this completes the proof.
Acta Sci. Math., 54 (1990), 273–276.
c Á. Kurusa
http://www.math.u-szeged.hu/tagok/kurusa
A characterization of the radon transform
3
Theorem 2. If a nonzero function-transformation % has the following properties
then % = cR, for some c ∈ R on Cc (Rn ), the compactly supported continuous
functions.
(1’) %(α1 f1 + α2 f2 ) = α1 %(f1 ) + α2 %(f2 )
(2’) %(f ◦ U ) = %(f ) ◦ U
(3’) %(f (δx))(ω,r) = δ 1−n %(f )(ω,δr)
(4’) %(f (x + y))(ω,r) = %(f )(ω,r+hω,yi)
(5’) If fk → f locally uniformly on Rn then %(fk ) → %(f ) (with common support)
The notations are the same as in Theorem 1.
Proof. Let fˆ = B(%(f )). Then we have
(a) (α1 f1 + α2 f2 )ˆ = α1 fˆ1 + α2 fˆ2
(b) (f ◦ U )ˆ = fˆ ◦ U
(c) (f (δx))ˆ (y) = δ 1−n fˆ(δy)
(d) (f (x + y))ˆ (z) = fˆ(z + y)
(e) If fk → f locally uniformly on Rn then fˆk → fˆ (with common support)
It is well known [1], that if a transform satisfies (a), (d) and (e) then there is a
function g on Rn such that
Z
fˆ(x) =
f (y)g(x − y) dy.
Rn
A short calculation with (b) and (c) gives that
fˆ(x) =
Z
f (y)
Rn
d
dy ,
|x − y|
where d is a suitable constant. This implies by [2,pp.104] that fˆ(x) = cB(R(f ))
for some nonzero constant c. Thus we have for any function f that
B(%(f ) − cR(f )) = 0.
From Theorem 1.3 of [4] we know the invertibility of the boomerang transform
on the set of radial continuous functions. So if we prove that the continuity of f
implies the continuity of %(f ) then we shall obtain that % is equal to cR on the
continuous radial functions. Since the translates of radial continuous functions are
locally uniformly dense in Cc (Rn ) this will imply the theorem.
Thus to finish the proof we only have to prove the continuity of %(f ) for any
continuous radial function f . For any ε > 0 we have
Acta Sci. Math., 54 (1990), 273–276.
c Á. Kurusa
http://www.math.u-szeged.hu/tagok/kurusa
4
Á. KURUSA
%(f )(r+ε) − %(f )(r) = %
=%
r+ε
r
n−1 !
r+ε
f
x
r
r+ε
r
!
n−1 r+ε
f
x − f (x)
r
− %(f )(r)
(r)
(r)
If ε tends to zero then
r+ε
r
n−1 r+ε
x − f (x) → 0
f
r
uniformly on supp(f ). This just means that %(f ) is a continuous function.
Theorem 1 (resp. Theorem 2) is valid for any other function space in which
C(Rn \{0}) (resp. Cc (Rn )) is dense.
The author would like to thank L. Fehér for making valuable suggestions on
the form of this article.
References
[1] W. F. DONOGHUE, Distributions and Fourier Transforms, Academic Press, Inc.,
1969.
[2] S. HELGASON, Groups and geometric analysis (Pure and applied mathematics), Academic Press, Inc., 1984.
[3] A. HERTLE, A characterization of Fourier and Radon transforms on Euclidean space,
Trans. Amer. Math. Soc., 273 (1982), 595–608.
[4] Á. KURUSA, New Radon inversion formulas, Acta Math. Hung., 60(1-2) (1992),
283–290.
[5] Z. I. SZABÓ, Hilbert’s Fourth Problem I., Adv. in Math., 59 (1986), 185–301.
Á. KURUSA, Bolyai Institute, Aradi vértanúk tere 1., 6720 Szeged, Hungary;
[email protected]
Acta Sci. Math., 54 (1990), 273–276.
e-mail:
c Á. Kurusa
http://www.math.u-szeged.hu/tagok/kurusa