AN ABSTRACT OF THE THESIS OF
Frit
Reinert for the degree of Doctor of Philosophy
presented on
Mathematics
in
Title:
June 11. 1985
The Diveraent kPlane Transform
Redacted for Privacy
Abstract approved:
i
Kennan T. Smith
The
sional
divergent kplane transform of a function f on an
real
vector space V is the function Df(a,a) =
assigns to each point a
e
V and each a
e
ndimen-
Daf(a)
which
Gk(V) the integral of f over
the translate of n(a) passing through a. Here x(a)is the nonoriented
kdimensional subspace of V associated with a and Gk(V) the Grassmann
manifold
2
Lo(0),
of
unit kvectors on V.
It is generally assumed that f
where 0 is a bounded open subset of V,
e
and that a is outside
the closure of O.
It is shown that under these conditions D f
a
e
L2 (Gk(V)), and the
adjoint is calculated. If Daf is known for infinitely many sources a,
this
determines
f uniquely,
essentially arbitrary.
while for finitely many sources
Exact and approximate inversion formulas
f
is
are
derived.
Some
formulas
independent interest.
for
integration on the
Grassmannian may have
The Divergent kPlane Transform
by
Fritz Keinert
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
Completed June 11, 1985
Commencement June 1986
APPROVED:
Redacted for Privacy
Professor of Mathematics in charge of major
Redacted for Privacy
A;2r''
Chaian
of Department of
athematics
Redacted for Privacy
Dean of Gra
te
School('
Date thesis is presented
Typed by the author
June 11, 1985
Fritz Keinert
ACKNOWLEDGEMENTS
I
would like to thank all the wonderful people in the
Mathema-
tics Department for seven enjoyable years as a student.
Special thanks go to my major professor Kennan T.
Smith,
whose
support and advice made this thesis possible, to David Finch for many
valuable hints,
and to Philip Anselone,
who supervised my
Master's
paper.
Finally,
to be,
I
would also like to thank my lovely fiancee and wife
Victoria Stevens,
necessary distraction.
and all my other friends for providing the
TABLE OF CONTENTS
Pane
I.
INTRODUCTION
1
II.
BASIC FORMULAS
5
kVectors
The Grassmann Manifold
Measure and Integration .
III.
IV.
V.
7
9
.
Riesz Potentials
17
Spherical Harmonics
22
THE TRANSFORMS P, D AND S
26
Definitions
26
Domains
27
Adjoints
33
UNIQUENESS AND NONUNIQUENESS THEOREMS
45
Uniqueness and NonUniqueness for D
45
Uniqueness and NonUniqueness for P
48
51
INVERSION FORMULAS
Exact Inversion Formulas
51
Approximate Inversion Formulas
55
Bibliography .............
.. .
.
59
THE DIVERGENT K PLANE TRANSFORM
I. INTRODUCTION
The
problem
of recovering a function on Rn,
from
n > 2,
its
integrals over kdimensional planes has been treated by many authors,
beginning with Radon in 1917 [91.
transform
of
Intuitively,
the parallel kplane
which
Pf
a measurable function on Rn is the function
assigns to each kdimensional subspace n of Rn and each point x"
the
in
subspace 70- perpendicular to n the integral of f over the trans-
late of n through x".
If k = a
1, Pf is commonly called the Radon
transform of f.
In this thesis, the following definition is used. Let V be an n
dimensional real vector space with an inner product, let Gk(V) be the
Grassmann manifold of unit kvectors on V, and for each a
e
Gk(V) let
n(a) be the nonoriented kdimensional subspace of V associated
with
a. Then
Pf(a,x")
P f(x")
a
f(x' + x") dx',
=
n(a)
where de is the kdimensional Hausdorff measure on n(a),
and x"
e
e
Gk(V)
n(a)L.
Interest
due
a
in this transform increased dramatically in the 1970's
to the introduction of Computed Tomography (CT)
and,
more
re-
2
cently, of Nuclear Magnetic Resonance Tomography (NMR). The object of
CT
and
function
NMR
is to reconstruct a two or
threedimensional
density
(see
from the values of its integrals over lines or planes
Herman LC for a summary).
In
integrals
through
CT,
not
over parallel lines,
a common source point a.
another operator D.
is
Sn-1
but rather over
lines
the
If f is a function on Rn,
of
then intuitively Daf
of f over the line passing
integral
passing
This lead to the investigation
the function on the unit sphere Sn-1 whose value at a
is
the
measure
soon turned out to be advantageous to
it
point
through
a
0 C
with
direction 0.
the divergent Xplane transform of a
In a more general setting,
function f is defined by
Df(a,a)
=
f(a + x) dx,
=
Daf(a)
n(a)
where a
V and a
Another
Gk(V).
transform related to D is the spherical kDiane trans
form S defined by
Sf(a)
f(0) de,
=
snlew n(a)
where f is a function on
and a
Gk(V).
Some
useful treatments of the parallel kplane transform can be
found in Helgason [5] and Solmon [14].
The spherical kplane
trans-
e.g., in Helgason [5]. The divergent line transform (k
form appears,
= 1) is described, e.g., in Bamaker/Smith/Solmonnagner [4].
main
The
divergent
purpose of this thesis is the
kplane
transform for arbitrary k.
connections between D,
and
P and S.
in later proofs,
Because of
of
the
many
the
a number of results on the parallel
spherical kplane transforms are included.
needed
investigation
Sometimes these
are
in other cases they are mentioned only
for
comparison with similar results for D.
Chapter
II.
contains
the background necessary for
the
later
chapters. Most of it is fairly standard, but the geometrical description
of the metric on G (V) and some of the integration formulas
on
the Grassmannian are not readily available in the literature.
Chapter
range
III.
spaces.
The
describes
some possible domains
emphasis lies with
and
squareintegrable
associated
functions
rather than with possible generalizations to differentiable functions
or
distributions.
For squareintegrable functions with compact sup-
port, the adjoint is calculated.
For the remaining chapters it is assumed that f
e
L2(0), where 0
0
is a bounded open subset of V with closed convex hull Q.
Chapter IV contains uniqueness and nonuniqueness
is
proved
that f is uniquely determined,
if
theorems.
Daf is known
for
It
any
4
infinite
set
of sources bounded away from D.
If Daf is
known
finitely many sources only and f is infinitely differentiable,
for
then
essentially nothing can be said about the behaviour of f inside 0.
Finally,
chapter
V.
contains exact and approximate
formulas to recover f from Pf or Df.
inversion
5
II. BASIC FORMULAS
1. kVectors
This
section
contains a brief review of certain aspects of k
Details and proofs can be found
vectors needed in the following.
in
many books, e.g. Greub [3].
Let
V be
a
real vector space of dimension n
with
an
inner
product. If vi,..,vk is a collection of elements of V, [vi,..,v ] is
the subspace spanned by these vectors.
The (contravariant) skewsymmetric tensors of rank k over V
a
real
A0(v)
vector space Alk(V),
where e(V) is identified with
The inner product on V induces
R by definition.
form
and
V,
an
inner
product (and thus also a norm) on Ak(V).
If
A
k+m
(V).
a
e
Alc(v).
De
If a C R = Ao(V),
Nonzero products
=
el(V),
v1A..dkvk
v1
their
anft = DA«
A.. AVk'
vj
exterior product an
is
in
aP.
V, are called kvectors. If a
is a kvector, then hail is the volume of the parallel
epiped spanned by v OO.PVke
Lemma (1.1) If a
S Hall 1101.
gonal, then
II
If the
1A..AVk and
= VI11%..AWm,
subspacesv1,..,vk
then
I Ian Pl
and [w1'' w I are ortho
m
Apl 1 = hull Ilpll.
If n is a kdimensional subspace of V. an orientation
is an
6
Two bases
equivalence class of bases of n.
(111,..,v0
and (wi,..,wk)
determinethesameorientationifthemapT:n>nwhichmapsv.into
w.
An oriented subspace of V is a
has a positive determinant.
sub-
space together with an orientation.
If a =
v1
A.. A
Vk
0=
and
w1
A. AWk are kvectors,
then a is a
= [wi,..,wk) and the
positive multiple of 0 if and only if
two bases determine the same orientation. Thus, there is a onetoone
correspondence
between kvectors
of unit length
and
oriented k
dimensional subspaces of V.
each kvector a,
For
let n(a) be
the
nonoriented
subspace
associated with a. Then n(a) = n(-a).
If
p
is a unit nvector (there are two of them),
there
unique (nk)vector aL of unit length with
a
A
aL
The relationship between the associated subspaces is
n(a4)
=
n(a)L.
is
a
7
2. The Grassmann Manifold
unit kvectors in A (V) form a set Gk(V) called the
The
mann manifold of oriented kdimensional subspaces of V.
In
Grassparticu-
lar, 60(V) = (1, 1) and G1(V) = Sn-1 (the unit sphere in V).
in
It is shown
norm on A (V) makes G(V) into a metric space.
The
Whitney [17] that Gk(V) with the topology defined by this
metric
admits the structure of an k(nk)dimensional analytic manifold.
Let
a0
km (n(a0
G
(1.1),
be a fixed unit mvector,
m
The Grassmann manifold
k.
By
)4) is an (km)(nk)dimensional submanifold of Gkm (V).
the map a' > a'Aao,
its range 11(a0) = (a
a'
e
Gk_m(n(a0)4),
is an isometry, so
a'
e
0k_mor(a0)4)) is an
a
e
rk(a0) if and
a = a'A ao,
Gk(V):
(km)(nk)dimensional submanifold of Gk(V).
only
if n(a) contains n(a0) as a subspace.
The
remainder of this section is devoted to a geometric
inter-
In the following theorems,
E will
pretation of the metric on G (V).
always
n
be the orthogonal projection of V onto n(a2),
(41).
Lemma (2.1)
Let a1, a2
1
Gk(V).
a1,a2>1
Then
= Y(E),
where Y(E) is the Yacobian of E.
Proof: See Whitney [17], section 1.15.1
restricted
to
8
Remark:
space n into another and (v.)
vector
J(T)
the
is a basis of n,
Jacobian
parallelepipeds
defined as the ratio of the volumes of the
is
k-dimensional
If T is a linear transformation from one
spaimed.byrrvAand (v.1.
with lia -a
a2 e Gk(V)
Lemma (2.2) Let a1,
11
21/2, and let I
<
= J(E). Then
[2(1 - 1)]1/2.
EXP-9.1.1
=
<aval> -
2<a1 ,a2 >
From the condition
so
<a1-a ,a -a2>
lia1-a2112 =
+
<a2' a2
<
I1a1 -a2
2(1 -
=
>
a1 ,a2 >).
21/2 it follows that
<a1 ,a2 >
> 0,
<a1 ,a2 > = J by (2.1).1
21/2.
Corollary (2.3) E is non-singular if I1a1-a211
11x11 1101-
11Ex-x11
e n(a1),
Proof: For any x
such that x,x2,
.
I1a1-a2II
Gk(V) with
Theorem (2.4) Let a1,
21/2. Then
x e n(a 1
11,
choose
<
x2,..xk e
n(a1),
I1xII=
xk are mutually orthogonal. Then if J = J(R),
iExAEx2 A
..
AExkii
1
i Ex
I iX AX2 A
AXkl 1
A..
lix11
1
ilExii
<
lixl,
liRxkli by (1.1).
Exki
1,
9
Since
11Ex.II
j
< 1,
IlEx11
III !WI.
>
Ex and xEx are orthogonal, thus
I1x112
IlEx112
=
IlxEx112,
+
and
IlxEx112
=
11
1lEx112
112
11x112 2(1-3)
11x112(1-1J12)
<
11x112
=
a2112.1
3. Measure and Integration
metric
The
defined in the preceding section induces
Hausdorff area measure da on
orthogonal
Gk(V).
a
This measure is invariant
finite
under
transformations of V (since the metric is) and is usually
normalized so that the measure of the entire Grassmannian becomes
ISn-111Sn
IGk(V)I
=
Is
(V)1
=
iG1 (V)i
=
100
2,
Ils
I
..
..
ISnkl
k> 2.
Is
I
10
measurepreserving between
Gk_121(ff(a 0
> a'A ao
the map a'
dimensional measure. As an isometry,
)4)
and
ISkI
and
denotes the kdimensional unit sphere in V,
where S
its
is
r(a0).
Lemma (3.1) If h is continuous on Gk(V), then
h(a) da
=
11(0)
rk"))
5n-1
is continuous on
.
Proof: Pick any 01,
carries 0
into
1
an isometry from
If a
02
Let U be the rotation on V that
and acts as the identity on 10 ,0 14. U extends to
1
11(02)
e C1(61),
)
Sn-1.
02
2
onto rk(02).
then
a
=
1Ja
=
1
a' A
G
k-1
WO AO )-1-)
2
1
02'
SO
II
=
1%11
=
leA (el - 02)11
11
=
1
Ile -
up
II
0
0211
by (1.1).
Now
le(9 )-e(e )1
=
Jh(a) da
rk (o1
)
h(a) da
-
rk
2
)
1
11
JIh(a) -
h(lJa)1 da.
rk (
By
uniform continuity of h it is possible to find a 8
the
= Ile -0211
s whenever Ila-Uall
any s > 0 so that 111(a) - h(lia)1
for
(
6.R
Theorem (3.2)
If h is non-negative,
measurable and defined al-
most everywhere on Gic(V), then
For almost every
e e Sn-1
the restriction of h to r(0)
measurable and defined almost everywhere on
The function B(8)
rk(e).
h(a) do is measurable on
=
is
Sn1
,
and
rko»
f
1
f
h(a) da
IS---1
Gk(V)
In
Remark:
and
(b)
n-1
h(a) da de
(3.3)
ric(e)
subsequent integration formulas the analogs of
in theorem (3.2) are considered as implicit
are not stated explicitly.
formulas
and
theorem,
typical of the others,
The proof of
parts
the
of
(a)
the
present
is given in full detail. Subsequent
proofs that are basically repetitions are omitted.
Proof:
continuous
By lemma (3.1),
linear
the right-hand side of (3.3) defines
form on the space of all continuous functions
on
12
Gk(V), hence a finite regular Sorel measure on Gk(V). This measure is
obviously rotation invariant,
factor.
it is da up to a constant
so
This gives (3.3) when h
The constant is determined by setting h = 1.
is continuous.
If
the
h is the characteristic function of an open set in Gic(V),
limit
sequence
everywhere of an increasing
functions (h.l. Therefore the restriction of h to
r
of
h
continuous
(0) is measurable
forevery0.ThefunctionsH.defined by
II (0)
fh.(a) da
=
rk(9)
form
an
bounded
the
increasing
sequence of
111(0)1,
above by the constant
function
H,
which is therefore measurable.
Sn-1n-1
fH(0) dO
lim
=
h.(a) da dO
-
Sn1
,
By
Sn-1 r (e)
and
""m
Gk(V)
I5k-11
h(a) da.
Gk(V)
the
H.(0) dO
ISk-11
=
on
and converging pointwise
convergence theorem, H is integrable over
= lim
functions
continuous
h (a) da
to
dominated
13
This proves the theorem when h is the characteristic function of
an open set.
be the characteristic function of a Gs.
h
Let
everywhere
limit
of a decreasing sequence
functions of open sets.
Cy
Then h
is
the
characteristic
of
The above argument proves the theorem when h
is the characteristic function of a G.
Let h be the characteristic function of a set N of measure zero.
N
characteristic
with
contained in a G8 set A of measure zero
is
function hA. From above, HA is integrable over Sn-1 and has
so HA is zero almost everywhere.
zero,
of
hA over
This shows that the integral
r (9) is zero almost everywhere,
hence that the restric-
of hA to r(0) is zero almost everywhere on
tion
integral
for
rk(o),
almost
every 0. Since N is a subset of A, this is also true for the function
h. Thus, for almost every 0, h = 0 almost everywhere on
By the regularity of da,
= A
where A is a
N,
11(0).
any measurable set E is a difference E
and NA has measure zero.
so the theorem is proved for
Bence
hE = hA
characteristic functions of measur-
able sets, hence for simple functions.
arbitrary nonnegative measurable function is the
An
limit
almost
functions.
The
everywhere of an increasing sequence
same
arguments
used above for
the
Ch.)
pointwise
of
simple
characteristic
function of an open set now yield the statement of the theorem.1
14
Corollary (3.4) If h is integrable on Gk(V),
then h
is
integ-
r (0) for almost every 0 e s'-1 .
rable over
Lemma (3.5) If f is continuous with compact support on V, then
F(a)
f(x) dx
=
n(a)
is continuous on G (v).
Proof: Pick al,
Hal-
Gk(V) with
a2
the orthogonal projection of V onto
n(a2),
211
<
21/2,
restricted
and let E be
to
n(a1
). E is
non-singular by (2.3), so
IF(a ) -
f(a+x) dx -
(a2)1
Itta)
=
I
I f(a+x)
dx
nta1)
S
I
By
I
n(a2)
I
f(a+Ex) 3(E) dx
I
n(a1)
Wit-I-WI (1-3(E))
Waft) - f(a+Ex)I dx
nta )
f(a+y) d
dx
It(Ct )
(2.4) and the uniform continuity of f the first integral can
be made arbitrarily small by taking al,
a2
sufficiently close
/
ther. (2.2) yields the same for the second integral.,
toga-
15
Corollary (3.6) If f is continuous on Sn-1, then
f(0) dO
F(a)
S
n-1
in(a)
is continuous on Gk(V).
p(Ixl) ixilk f(x/Ix1),
Proof: Define g(x)
where p(1):1) = 0
if Ixi is outside the interval [1,2], and fp(IxI)dx = 1. g is clearly
continuous, and
f(9) dO
F(a)
Sn-1
n(a)
2
f(0) de dt
1
nJp(t)
nn(a)
n-1
S
P(IxI) 1x11k f(x/IxI) dx
=
=
G(a).
w(a)
which is continuous by (3.5).11
Lemma (3.7) If g is nonnegative,
measurable and defined almost
everywhere on 5n-1, then
Isn-11
g(0) de
n-1
g(0) de do
=
ISkilIGk(V)I
G (V)
5n1mn(a)
16
g(0) de da
,nk-111G (v)1
Gk(V)
the first double integral on
Proof: By the preceding corollary,
the
defines a continuous linear form on
right
hence
a
SnInn(a)L
finite regular Borel measure on Sn-1.
space
the
C(Sn-1),
This measure is
and there is only one up to a
viously
rotation invariant,
factor.
The constant is determined by setting g = 1.
ob-
constant
The theorem is
then extended to nonnegative measurable functions as in (3.2).
The proof of the second equality is identical.1
measurable and defined almost
Lemma (3.8) If g is nonnegative,
everywhere on V, then,
ISnil
g(x) dx
-------------
=
ISkiliGk(V)1
V
I
g(x') dx' da
SI
(V)
n(a)
k
'
S
ISnklliGk(V)i
Proof:
Using
G (V)
g(x')
n(04
coordinates in V and in n(a)
polar
(3.7),
CO
g(x) dx
V
ftn
=
n
1
0
da
g(tO) dt de
and
theorem
17
CD
ISn-11
n-1
ISk-111Gk(V)I
g(tO) dt dO da
Sn-14nn(a)
k(V)
CO
ISn-11
tnk tk-1
f
ISk-111Gk(V)I
Gk
S
(V)
n-1
0
'in (a)
ISn-1I
=
,,1nk
f
ISk-111Gk(V)1
Gk(V)
g(tO) dt dO da
g(30)
dxs
da.
n(a)
The proof of the second part is identical.'
4. Riesz Potentials
The Riesz kernel Rk is defined by
Rk(x)
=
c(n,k)
1x1kn,
0 < k < n,
where
r((nk)/2)
c(n,k)
2k7n/2tlk/2)
This function comes up in various places in computed tomography.
In this thesis,
that
it appears,
e.g.,
in lemma (III.2.1), which states
18
c 8k*f(a).
Daf(a) da
Gk(V)
The following two lemmas are adapted from [11].
Lemma (4.1) If p is integrable and
(1+1x1)ap
is bounded, then
illk*p(x)1 < c (1+1x1)kn (11p111 + 11(1+12E1)11pH ).
Proof: Assume p > 0, and let M = 11(1+1x1)ap11
It will be shown first that
Rk*p(x)
10
11p11
/+
101),
then that
Rk*p(x)
1 0 11ka (11p11
+ M).
Together the two prove the lemma.
For the first estimate,
1
Rk*p(x)
p(y) dy
=
c(n,k)
V
lxy1
p(y) dy
+j
1xylka p(y) dy
19
f
,,kn
dz
+
iipll
(11p11
c
+ M).
1z1(1
Because of the first estimate, the second estimate only needs to
be proved for ixi > 1.
1
Rk*p(x)
=
Ix-yIk-n
c(n,k)
p(y)
lyklx1/2
lzylkn p(y) dy
Ix_yikn p(y) dy
+
lx1/2<lyl<21x1
ly12,21xl
s (Ix1/2)k-n 1101
ik-
/
I
ixikn
10zIkn p(Ixlz)
IzIn dz,
1/2(1z1(2
x/Izl, z =
where 0 =
Since
yilxi.
2nM,
p(Ixiz)izIn = p(Ixiz)(Ixiz)nxn
the last integral
is bounded by
2nmi
1kn
1w 113
the following lemma,
[1].
T
is the space of Bessel
It is similar to the Sobolev space cl4
,
potentials
but the functions
are
20
defined
more precisely than almost everywhere.
defined except on sets of 2kcapacity zero.
Functions in
r
are
(For any e > 0, a set of
2kcapacity 0 has (n-2k+e)dimensional measure zero).
Lemma (4.2)
If (1+
and
ixi)knf
is
e
locally
.1
L,
then Rk*f is defined almost everywhere
Moreover,
integrable.
pe
if
L1
and
(1+IxI)"p is bounded,
=
latkibf,p>1
I
f
Rksf(x) p(x) dx
V
S.
1(1101
cli(1+61)knfli
L
If
in addition f
e
L2
loc'
+ 11(1+1xl)npli
L
).
then Rk*f is defined except on a
set of 2kcapacity zero and is in
-e2k
F
loc
if k > n/2, Rk*f
is continuous, as points have positive capacity.
If
(1+Ixi)1nf f Li
and f > 0,
then
Rk*f
everywhere.
Proof:, (a)
I
j Rk*f(x) "p"(x) dx
V
JRk(xy) f(y) dy 13(x)
dx
I
is
undefined
21
if(y)I
=
f
Rk(xy) ;(x) dx Idy
I
if(y)i INk*p(y)i dy
V
c
+ 11(1+1 1)1101 .)
(11p11
If(y)1 (1+1x1)kndx
Li
V
The rest of (a) now follows by choosing p to be the characteristic function of a bounded measurable set.
This is proved in
Ill.
lyxl / 1y1 + Ix!
/ 1
+ Ixl,
/ ly1(1+1x1),
so lyxl < max(1,1y1)
(1+1x1),
Rk*f(y)
if
ly1 j1,
if lyl > 1,
and
=
V
>
min(l.iy lkn)
j
V
which is infinite for all y.N
(1+ixi)k-11 f(x) dx,
22
S. Spherical Harmonics
of this section consists of a brief review of some
Most
facts
about
needed
A more detailed treatment can
spherical harmonics.
be
found e.g. in Seeley [10].
Let
and
let
Pm
H
be the space of homogeneous polynomials of degree in on V,
geneous polynomials of degree m.
sion
Hm
(2m+n-2)(m+n-3)1/(m!(n-2)!).
svherical harmonics
gf
n-1
harmonic
homo-
is a real vector space of
dimen-
consist of the restrictions to S
functions in H
The
in
are
called
degree m.
In spherical coordinates the Laplace operator on V is given by
a
2
n-1
8r2
where
8
1
ar
r2
---
A
r
A' is the Laplace operator on
5n-1
Hm
functions of A with eigenvalue m(m+n-2), and
the H.
every f e
gonal direct sum of
basis of H, then
f(0)
Thus, if
2
L (S
=
For fixed in and each e
such that
m,
m,j
e
L2(Sn-1)
1 <
j
is the
ortho-
dim(H ), form a
) can be uniquely written as
a
m
Hm
n-1
(hj I,
is the space of eigen
h
.(0).
m,j
j
n-1
S
there is a unique function eel
e
23
h(0) Z:(0) de
=
h(e)
(5.1)
sn-1
any h
for
is
e
Z:(0) depends only on the angle between e and 0 and
II.
called the zonal harmonic
odd about the hyperplane
al
order in with pole 1.
Z: is even
or
depending on whether in is even or odd.
Lemma (5.2) If in is even and a 6 G (eA), then
0.
ZI:(0) de
sn-1
Proof:
described
from
e Sn-1,
For fixed e
by coordinates (r,o),
an arbitrary point 0
e Sn-1
where r is the (geodesic)
can
be
distance
0 to e and a is the intersection of the great circle through
0
and e with the (n-2)sphere perpendicular to e. In these coordinates,
=
a2
3
+
sin2 r A",
(n-1) cot r
ar
8r
where A" is the Laplace operator on the (n-2)sphere
perpendicular
to e.
Since
Zm depends only on r in these coordinates,
Zm
satisfies
e
is not the
the differential equation
h" + (n-1) cot r
' = m(m+n-2)12.
Zm is even about r = n/2, so (Zm)1(n/2) = 0. Since
24
zero
solution,
Zm(n/2) 0 0.
since Zm is constant on 5n-1
Theorem (5.3)
The lemma follows directly from
this,
n(a).111
If H is a subspace of Hm invariant under orthogo-
nal transformations, then either H = (0) or H = H.
Proof:
i
Define
over
a non-zero h 6 H and a point e with
Choose
O.
as the function whose value at (r,a) is the average
all points with distance r from e.
transformations of h,
it must be a non-zero multiple of Z:.
(orthogonal complement in II).
orthogonal
transformations,
it
of
h
As an average of orthogonal
the new function remains in H.
zero because the value at e is unchanged.
HJ.
h(e)
and it is
not
Since R depends only on r,
e
By (5.1), g(e) = 0 for any g
Since H4 is also invariant
follows that g must be
under
identically
zero, so H4 =
The following theorem is standard for bounded operators.
Theorem (5.4)
domain
n-1
on S
Let T be a closed linear
operator
on the space L2(Sn-1) of complex square-integrable
with
dense
functions
If T commutes with orthogonal transformations, then each
Hm
is in the domain of T, and on H, T is a multiple of the identity.
Proof:
means
To say that T commutes with
orthogonal
transformations
that whenever U is a unitary operator arising from an orthogo-
nal transformation of the sphere, so
UT CTU.
(5.5)
25
This implies that the intersection of the domain of T with Hm is
invariant, hence is either 0 or
Suppose
the latter,
Hm.
and let Tml be the restriction of T to Ems
followed by the projection on H. The null space and range are invariant,
either
so
onto,
in
which
case m and I must be equal,
different dimensions.
2
L
Because
multiple
of
one-to-one
is identically zero or it is
ml
as different
and
have
Em
It follows that TO! )m, since the Hm span
T must
the invariance of eigenspaces of T,
be
the identity on H. This proves the theorem when
of
a
the
domain of T has a non-zero intersection with H .
By
(5.5),
**
(TU)
2 U T .
(UT)
2
(ru)
TU has
dense domain,
*
Because U is bounded,
from which it
*
(UT)
**
= T U
**
that
follows
**
Therefore, T U =
*
*
2 U T . Replacement of U by U gives UT c.T U, hence
UTTcTTU.
This
identity
has
implies
that each projection E in the resolution
for the self-adjoint operator T T commutes with U.
been proved for bounded operators,
identity
on
implies that H
each H ,
of
By what
E must be a multiple of
therefore either 0 or 1
on
each
Hm,
the
the
which
is in the domain of T T, hence in the domain of TI
26
III. The Transforms P. D and S
1. Definitions
f be a measurable function on V.
Let
If a is a
kvector.
the
parallel kplane transform of f is defined by
Pf(a,x")
=
=
Paf(x")
n" e
f f(xl+x")
n(a)
n(a)
whenever
The kvector a is called the
the Lebesgue integral exists.
xray direction.
For a fixed point x Q V, the divergent tDiane transform of f is
defined by
Df(a,x)
=
Dxf(a)
=
f
f(x+x') dz.
n(a)
whenever the Lebesgue integral exists.
nu
The point x is called the
g-
source.
The transforms Pa and Dx are related by
(1.1)
Dxf(a) = Paf(Em4x),
where Ex is the orthogonal projection of x onto n(a) = n(a)4.
If
f
is a measurable function on S
transform of f is defined by
n-1
,
the spherical kplane
27
f(0) de
=
Sf(a)
S
n-1
n(a)
whenever the Lebesgue integral exists.
2. Domains
Let 0 be a bounded open set in V with closure 0. C(Q) and
are,
respectively,
functions
defined
the
spaces of continuous and square
on V and zero outside Q.
The fiber
L2(n)
0
integrable
bundle
T =
T(Gk(V)) is defined by
=
((dor"): a
e
e
ek(V).
nta) ).
T has a natural measure g so that
f dg
5
Gk(V)
In
denoted
the
by c.
f(a.x")
dx" da.
n(a)-1.
most constants will
be
The value of a need not be the same from one line
to
following theorems and proofs,
the next.
Lemma (2.1) If f is nonnegative,
measurable and defined almost
everywhere on V, then Daf is measurable, and
28
15k-11
Daf(a) da
1Gk(V)1
=
Rk*f(a)
c(n,k) -18n-11
Gk(V)
where Rk = c(n,k) 11kn is the Riesz kernel defined in section 11.4.
Proof: By (11.3.5), Daf is continuous if f is continuous and has
compact support.
The proof that Daf is measurable proceeds like
the
proof of theorem (11.3.2).
Now
Daf(a) da
f
=
Gk(V)
n(a)
Gk(V)
ix,in-k ix'
J
f(a+z9 dz' da
k-
f(a+x') dx' da
n(a)
Gk(V)
15k-11 1G (V)1
Isn-ll
lxi
f(a+z) dx
V
by corollary (11.3.8).1
Theorem (2.2)
(a)
If
f
e
LP(B),
0
1
sources a outside
IID
< g < =,
5,
then D f
a
e
LP(Gk(V))
and for such a,
fit
<
LP(Gkon)
c
11f11Lg(n).
for
all
29
(b)
If
e Li
(1+Izl)k-1tf
e L2i
loc
and f
then for all sources
Daf(a) is defined
a set of 2kcapacity zero,
outside
a
for
almost every a, and Daf is integrable on Gk(V). If k > n/2,
this is true for any a.
Proof: (a) For
p(
IDaf(a)IP
co,
=
f
I
f(a+x) dx IP
n(a)
(
.1
If(a+z)IP dx
Xn(a+x) (ix) Pici
n(a)
n(a)
If(a+x)IP
c
dz
= cDIIP(a),
n(a)
where X
is the characteristic function of 0 and 1/p +
IID fir
a
LP(Gk(V))
S
IDaf(a)
I
Gk(V)
c
j
DalfIP(a) da
elfIP(a).
G (V)
=
Iaxlka IfIP(x) dx
c
V
c
[dist(a,0)]ka IlflIP
LP(o)
.
= 1. Thus
30
where dist(a,0) is the distance from a to 0.
The case p =
is easy to check.
(b) follows from (2.1) and (11.4.2).1
Theorem (2.3)
If f
e
LP(o).
Gk(V), Pf
p < c*,
1
0
,
e L(T),
IPf I I
for all a
c
<
Li,
11f11gm)
then Pf is defined almost everywhere
thus for almost every
on T and is locally integrable,
Gk(V), Pf is defined for almost everywhere on n(a)4.
is locally integrable.
Proof:
(a) For
p<
co,
1Pcif(e)IP
=
I
f(x"ft ) di'
n(a)
c
e
Lg(0)
0
e
)
c 11f11
LP(T)
(1+Ix1)kaf
0
LP(n(a)4)
a
If
LP(n(a)
and
IIP fll.1
I
e
then P f
a
j If(x"+e)IP de
n(a)
as in the proof of (2.2)(a). Thus
IP
a
e
and
31
iPaf(x")1P dx"
x(a)4
I
n(a)4
if(x"+x9IP dx' dx"
=
c lifilP
Lg(n)
w(a)
To get the result for P, integrate over Gk(V). Again, the case p
=
is easy.
(b)
is proved in Solmon [15]..
Theorem (2.4) If f
e LP(sn-1).
1
1
then Sf
I)
and
HUH
c
lifil
Lp(sn-1)
.
LP(Gk(V))
Proof: For p
ISf(a)IP
f(e) de IP
=
n-1
S
If(0)1
n(a)
elf+,
w(a)
where c = i5k-11P/q, 1/p + 1/q = 1, so
ISf(a)iP da
IlsfII"
LP(Gk(V))
G (V)
e
LPtek(v)),
32
j
S
Gk(V)
=
C
n-1
If(0)1P dO
n w(a)
lifliP
1,P(Sn-1)
by theorem (II.3.7).11
Theorem (2.5) Let S
be defined by
S*f(0)
f(a) da.
=
rk(e)
If f
e
k
CV))
P
=, then S*f
Hefn
e
LP(S/1-/),
and
c lifil
Lp(G (V))
Lp(sn-1
Proof: Almost identical to (2.4), with obvious adaptations.1
The various possible domains can be summarized as follows:
Pa:
P:
Da:
S:
1,13(0)
->
LP(n(a)-L)
1.10)(0)
LP(T)
L(Q)
LP(Gk(V)),
LP(Sn-1
0
->
LINGk(V))
a outside
5,
33
bounded
are
operators
for all 1
p
=.
Pa and P
also
can
be
regarded as unbounded operators on L2(V) with domain
;6k
=
(f
e
L2(V): (1-1-1x1)k-nf
e
L1(V)).
3. Adioints
Lemma (3.1) Let n be a fixed k-dimensional subspace of V, n4 its
orthogonal complement.
If g is non-negative and measurable on n4 and
has support in Ban n4, then
I
=
g(E70.0) dO
ISk1I
-2
f
g(x) (1 - ixi2
(k-2)/2
dx.
sn-1
Proof:
Bn(1) = Bn.
first
Let
Bn(r)
be the ball of radius r around
Fix an arbitrary (k-1)-dimensional subspace n
writing
the integral over Sn-1 as one over
n' +
origin,
the
n4
of n.
and
applying Fubini's theorem,
g(Ex) (1 - Idx
=
g(E7140) de
(n'+1/4) A Bn
sn-1
Ix"!)-1/2
g(x1)
n Bn
ty2 -
B(y)
a" de,
By
then
34
where y2 = 1
1x'12.
Setting x" = te,
the inner integral in polar
coordinates becomes
t2)-1/2 dt.
tk-2 (y2
1Sk-2
0
With the substitutions $ = t/y = cos v this equals
(/
sk-2 (1
x,12)(k-2)/2 1 k 21
2)_1I2 ds
0
n/2
(1
lx'
j
)2)(k_2)12 i5k-2,
1
cos
k-2
v
v
0
=
rl
(k."1. ) /2)
r. (1/2)
1Sk-2
X,
(1
2 r(k/2)
1Sk-1
=
1
(1
1e12)(k-2)/2.11
2
Lemma (3.2) Let p be nonnegative and measurable on [0,1],
n
a
nk-1 (1t 2 (k-2)/2
dt
fixed kdimensional subspace of V. Then
1
p(1%40 ) de
n1
=
1Sk-11 15nk
p(t)
0
Proof: Using (3.1) and polar cordinates on
n4.
35
p(1%401)
dO
n-1
=
p(lx1)
ISk-11
(1
1
12)(k-2)/2 dx
Bn A n-t-
=
t''
1Sk-11 ISnk-1
p(t) (1
t2
(k-2)/2
dt.1
0
Lemma (3.3)
Let p be nonnegative and measurable on [OM,
fixed point on Sn-1. Then
p(lEco.01) da
Gk(V)
IGk(V)I 1Sk-11
i
11
1Sn-11
f
(k
(1t)2)/2
dt.
p(t)
0
Proof: The first integral is clearly independent of 0, so
f01%1.0 )
p(lEa40 ) da
G (V)
k(V)
1
p(lEco.01) dO da
isn-11
Gk(V)
da de
36
IG (V)I
Ip(lEcri.01)
d0,
ISnlI sn-1
since the inner integral is independent of a."
Corollary (3.4)
d
iEa4Olk
Gk
=
IG (0.)1
'Ski
2
(V)
Proof: By (3.3),
0
f
J
1E
az.
(I)I
isk-11 Isn-k -11
(1-
Oik-11+1 da
Isn-11
,
k(V)
dt
0
Gk (V)
I
2 (k 2)/2
n/2
iSn-E-41
I
cos os
Isn-11
0
lGk(V)I iSk-li is k-11
le-11
ISki 1G (04)1
s ds
r(ia2)
r(1n)
2 n(k+1)/2)
after simplification.'
2
For fixed x
e
V, let T be defined by T a = xA a/ I Ix A al , when-
ever x An # 0. If F: I
Y and g is a function on Y, then Pig = goF.
37
Lemma (3.5)
If
h is continuous on rk (0) '
p
nonnegative
and
measurable on [0,1], and
1
p(lEa40i) da
p(t)
(1t2)(k-3)12 dt
=
C
<
m,
Gk-1(V)
then
!ric(e)!
h(a) da
h(T0
G
k-1
rk(6)
a) p(IEa40i) da.
(V)
Proof: Let
L(h)
=
h(T0
G
k-1
Ta is undefined for a
Gkl(V)
of
a) p(iE4 8 ) da.
a
(V)
rk-1(9),
but this is a submanifold
of
L
is
smaller dimension and therefore has measure zero.
clearly a continuous linear form on C(1*(0)).
Let U be any rotation in V with U0 = 0,
from Gk-1(V) into itself also be called U.
and let the induced map
Then T011 = UT01 thus U T9
**
= T0 U ,
so
L(U h)
T h(Ua)
0
=
Gk-1
(V)
p(iEa401)
da
38
T h(Ua) P (lEue.01) da
=
L(h).
Gk 1(V)
This
shows that L is rotationinvariant and thus equal
up to a constant.
integral on the left,
to
the
This constant is determined
by setting h = 1.1
If h is continuous on
Corollary (3.6)
r (0),
then
2
=
h(a) da
h(Tea) 1E0Ikn da.
sk-1
rk(e)
G
(V)
k-1
Theorem (3.7)
(a)
Let f
e
L2(0),
0
e
h
L2 (Gk(V)), and let a be a point outside
Then
Daf(a)
h(a) da
=
H*f(a),
Gk(V)
where
H(x)
=
ixikn
h(a) da
(3.8)
39
(b)
If
considered as an operator on L2(0) and
is
Da
0
point outside
5,
Dah(x)
the adjoint operator
f
lxalkn
=
Da
a
is
a
is given by
h(a) da
rk(1-11E!T)
for h
e L2(Gk(V))
and x
e n.
Proof: (a) By (3.6), the formula given for H is equivalent to
2
H(x)
Eazx
h(Txa
Iskli
1k
da.
Gk-1(V)
This latter form will be used in the proof.
Assume
first
that both f and h are continuous and
proceed
induction.
For k = 1,
ri(x/IxI) = (x/Ixl,
H(x)
=
(h(x/Ix1) +
x/IxI),
thus
h(x/IxI)) I
I n,
a formula in Hamaker/Smith/Solmon/Wagner [4].
Assume now that the result is true for Gk(V).
Daf(a) h(a) da
Ck+1(V)
f(a+e) dx' da
h(a)
010.1(V)
n(a)
by
40
1
f(a+e)
h(a)
mle,001.00,10
k+1
da de
n(a)
(0)
1
'ski
sn-1
f(a+x"+t0) dt dx" da' dO
f
h(a'n 8)
n(e)
Gk(04)
-40
1
h(a'AO)
j
'ski sn-1
a+x") dx" da' de
P0
f(EOA
n(a9
Gk(84)
1
h(a'AO) D
P f(a') du' dO
E04a 0
J
ISkI
n-1
Gk(84)
f(Es-e)
P
dx' dO
by induction, where
H
(e)
0
2
h(Tx,a"A8)
k-11
'ski
k -n+1
iEa,IAe
k-1(OA)
Thus the last integral in the sequence above is equal to
CO
H(e)
f(E0 a-x'-t0) dt
4
de
CO
f(s-x'-t0) dt dx' de
8(x')
.041:0
da".
41
H0 (E04
1
n-1
x) f(ax) dx dO
f*H(a),
=
V
S
where
H
=
11(x)
94
0
=xa"A
dO
2
ISk
'Ski
I
Now T
E
x
11"4
S
a"
multiples of Eej.x
more, E
=
T (a" AO),
Au" AO =
E(a,
ow
,
x Aa"A 0 and have length
positive
Further-
1.
so the integral equals
h(Tx(a"
0))
4x1
1E.a 0)kn+1
da" dO,
t
n_i
S
xikn+1 da" dO
since both sides are
2
isk-11
E
a 9 L 04
Gk-1(04.)
0
=
E
1E ,
(9)
Gk-1(04)
2
h(T a) 1Eal,xikn+1 da dO
---------isk-11
Isk1
ç(e)
2
IE
xfkn+1
h(Txa)
!Sk i
Gk(V)
Now if (fj) is a sequence of functions in C20(0) converging in L2
to a function f
e L2(n) .
in L2 by theorem (2.2)(a).
then Daf,
D
a
f.2(G (V)) and D f. > Df
a
a
42
Likewise,
)
if
is a sequence of continuous functions conver-
(11i
ging in L2 to a function h
L2(Sn-1)
e
then by theorem (2.5), S*h
L2(Gk(V)),
e
and S*h. -> S*h in L2.
Choose r so that 0 is contained inside the open ball B of radius
r around a, and let H' be the restriction of H to B. Then
2(k-n))
2
Lo(B)
IIS*012
'
lisahll
dt
L (Sn-1)
L2(Sn-1)
0
and likewise for the H
This shows that H'
e
-> H' in L2. Since H*f(a)
L20(B) and that H
= H"f(a)andH.*f(02211.'*f(a),
the integrals on both
sides
(3.4) converge, and by Cauchy-Schwartz,
'Jaya) hi(a) da
->
f
Daf(a) h(a) da,
Gk(V)
Gk(V)
H'*f.(a)
->
H'*f(a).
(b) is now an easy consequence of (0.11
Theorem (3.9)
(a)
If f
e L2(Sn-1),
h
e
L2(Ok(V))D then
Sf(a) h(a) da
Gk(V)
f(0)
n-1rk(0)
h(a) da.
of
43
its adjoint
If S is considered as an operator on L2(S
(b)
is given by
S
S h(0)
h(a) da.
=
ric(0)
Proof:
Define g(x) = izilk f(x/Izi) for 1 < lxi<
(a)
2
zero otherwise. Then
II
112
en-1 t2(1k)
Ilf11 22
L
L(V)
ts n-1
dt
)
1
lifil2
L2(Sn-1).
so g e
L2 (V). Since
2
Dog(a)
=
j
g(z) dx
tk-1 t1k
=
f(0) de dt
n(a)
n(a)
=
Sf(a),
it follows that
Sf(a) h(a) da
Dog(a) h(a) da
=
Gk(V)
Gk(V)
2
H(z) g(x) dz
=
V
=
tn-1 tkn t1k
H(-0) f(0) dO
n-1
and
44
h(a) da de,
f(0)
sn-1
rk(e)
since 11(0) = 11(-0).
follows directly from (a) and the known mapping
(b)
properties
of S and S
The following theorem and corollary are proved in Solmon [15].
If P is considered as an operator on q)(V),
Theorem (3.10)
formal adjoint
Pit
its
is given by
P#g(
g(a,Eaz) da,
j
=
)
Gk(V)
e
where g
For
locally
every g
e
P0 g is defined almost everywhere
L2 (T),
square integrable.
*
joint
L2(T) and Eal. is the orthogonal projection onto n(a)L.
P
Moreover,
and
is
g is in the domain of the ad-
0
of P if and only if P g is globally square
integrable,
in
which case F g = P* g.
Corollary (3.11)
If f is nonnegative,
measurable and
almost everywhere in V, then
isn-11
Rk*
c(n.k)
1s'1
1G (V)i
where Rk is the Riesz kernel defined in section (II.4).
defined
45
IV. UNIQUENESS AND NON-UNIQUENESS THEOREMS
1. Unioueness and Non-Unioueness for D
0
Let
convex
hull
be
a bounded open set in V with closure
O.
C0
and L2(0) are the spaces of
(0)
0
5
and
closed
continuous
square integrable functions defined on V and zero outside 0,
and
respec-
tively.
For k
0, the operator DI: is defined by
CO
whenever
The
the integral exists.
proof of the following theorem is analogous to the proof of
Theorem (1.1)
side
5,
f(a+t0) ItIk dt,
=
El f(0)
a
e LPG),
If f
then Dkf
a
e
Lp(Sn-1) for all a out-
and for such a,
c
IIDkfIl
a
Theorem (1.2)
G (V), then
If
f
e
2
Lo (0),
lIfIl
q(0)
LP(Sa-1)
a
is a point outside
a
and
a
e
46
1
Daf(a)
Proof:
S Dk
=
a
2
-1
f(a).
By using polar coordinates in n(a),
f
=
Daf(a)
f(a+x) dx
n(a)
CO
k-1
f(a+t0) t
n ff(a)
dt dO
0
1
=
f(a+t0)
lk-1 dt de
2
Sn-1
71(a)
1
S Dk-1f(a).111
2
a
Theorem (1.3) S is one-to-one on even functions in L2(Sn-1).
Proof: By (11.5.4), S acts as a multiple of the identity on each
H. To show that S is one-to-one on the set of even L2-functions, it
remains
even.
to
show that this multiple is not zero for the
For this,
Hm
with
m
it suffices to exhibit one function in each such if
that does not get mapped into zero by S.
By (11.5.2),
the functions
in have this property, where e is arbitrary.1
Remark: Clearly, S maps all odd functions into zero.
47
Lemma (1.4)
If f 6 L(0),
= 0 for almost every a
e
Gk(V),
if a is a point outside E and D f(a)
a
then Dtif(0) = 0 for almost every 0
S.
n-1
Proof: Immediate from (1.2) and (1.3).N
Theorem (1.5)
If f
bounded away from 0,
Gk(V) and a
2
e
Lo
of
points
then f is determined uniquely by Daf(a) for a
e
A.
Proof: By lemma (1.4),
uniquely by
almost
(0) and A is an infinite set
Dk1a
e Sn-1
f(0) for 9
identical
Solmon/Wagner
it suffices to show that f is determined
and a 6 A.
to the proof of theorem
[4],
The proof of this
(5.1)
in
Hamaker/Smith/
theorem (3.19) in that paper in
with
is
place
of
(3.2).N
Theorem (1.6) Let
outside
0,
f0 e
let A be a finite set of
L1(0),
sources
let K be a compact set in the interior of the support of
f0, and let g be any integrable function on K.
Then
there
is a function f
with Daf = D f for all a
a 0
e
e
L1(0) with the same shape as
0
A, and f = g on K.
Proof: Theorem (6.15) in Leahy/Smith/Solmon [7] shows that there
is an f so that Daf(0) = Daf0(0) for all 0
e Sn-1
The theorem then follows immediately from (1.2).N
and for all
k > 0.
48
2. Unioneness and NonUnicueness for P
A
The Fourier transform f of a function f is defined by
A
=
f(C)
(2n)
n/2
f(x)
ei(x>
dx.
V
is given by
The Fourier transform of P
(P f)A(
a
The
following
"
=
(20
k/2
4" e
f(4"),
(2.1)
n(a)4.
theorems are easy generalizations
of
(5.1)
in
Smith/Solmon/Wagner [13].
1
Theorem(2.2)Letfe(V)
Lo and let (
of unit kvectors.
1'
a
2'
..I
be a collection
If the subspaces n(aj)4 of V are not contained in
a proper algebraic variety, then f is uniquely determined by Pa f.
i
Proof:
Since
f has compact support,
its Fourier
transform
f
extends to an entire function on Cn with a Taylor expansion
'(4)
=
m=.0
where
pm(4) is a homogeneous polynomial of degree m.
A
If P
a.
f = 0,
then by (3.1),
f vanishes on n(a.)-1. For any
Ie
49
CO
f(tt')
(4')
tm p
=
=
0
for all t,
m=0
from which it follows that pm = 0 on n(aj)-L. Since no nonzero pm can
vanish on all the n(a
p
= 0, so f = 0 and finally f = 0.1
m
Theorem (2.3) Let fo Q C(V),
kvectors.
If
( 1, a2,
a collection of unit
of the subspaces 3T(a3) J- of V are contained in
all
if IC is any compact set in the
proper algebraic variety on V,
a
inte-
and if fl is any function in C(V), there
rior of the support of fo,
is a function f C(V) so that
on 1,
f1
=
Pajf
supp f
for all j,
Pa f0
C. supp fo.
Proof.LetQbeapolynomialthatvanishesonallff(a.)-L.
theorem
constant
of
EhrenpreisMalgrange on the existence
coefficient
existence of functions
pe
u0
C(V)
so
and
u1
=
the
in Cm(V) so that
=
k = 0,1.
that p = 1 in a neighborhood of IC
vanishes outside the support of fo. Now let
vk
solutions
partial differential equations guarantees
Q(D)uk
Choose
of
The
Q(D)(puk),
k = 0.1.
and
p
50
The last two formulas show that
vk
= fk in a neighborhood of
K,
and vk = 0 outside the support of fo.
By (2.1),
(Pa.v k)4(41)
=
(2w)ki2Ct(i41)(puk)4(49
for
0
4'
e
n(a
thus
Pa.vk =
Finally,
0
definef=f0
for k = 0,1 and all j.
0
+v1. Thenf=f1 inaneighborhood
of K, supp fd:supp fo and P_ f = P f- for all j.
aj u
aij
Remark:
The
theorem automatically applies
finite set (take Q(t) = <t,41><t,t2>..(t,t/i>
0).
if (al,..,aN) is
whealse
t.
j
a
nta.), t. A
J
J
51
V. INVERSION FORMULAS
1. Exact Inversion Formulas
Throughout this chapter, let C be the constant
ISn-k-11
(V)i
=
iSk-11i0k
_____...._L__
=
(2u) k
in-li
in-li
c(n,k)
10k(V)1 .
-,
11kn
x
Thee Riesz kernel Rk = c(n,k)
0 < k < n,
was defined in
section (II.4). The constant c(n,k) is chosen so that
A
Rk(c)
(2)-n/2 141-k
A
where Rk is the Fourier transform of Rk, defined in section (IV.2).
The operator
Ais
defined by
(AO^ (t)
It iD*(4),
=
so that formally
Ak(Rk*f)
Rk*Akf
=
=
f.
From (IV.2.1) it follows that formally
PA f
=
#Pf,
This, together with (III.3.11), shows that
=
C
*
Alc
(PPf)
=
C
*
k
P PA f
=
C*AkFf.
52
Here and in general the Fourier transform and the operator Aact
on each x()4, if f is a function defined on T.
Theorem (1.1)
(a)
e
If f
141
L2(V) and
0
e
f
then for almost every x
L2 (V),
v,
Dxf(a) du
C
=
f(x)
Gk(V)
=
Paf(Ee.x) dx,
C
Gk(V)
(b)
e
If f
2
L (V) and igi
ki2A
f
e
L2 (V), then for almost every x
e
v,
f(x)
=
C
j
gc.Paf(Ecti.x) dx.
G (V)
Remark:
the condition
k < n/2,
If
141k? e
L2(V)
in
(a)
is
automatically satisfied.
Proof:
(a)
Both formulas are equivalent to
f
By
(11.4.2),
Rk*f
=
/i (Rk*f).
is defined almost
everywhere.
condition guarantees that Rk*f is in L2(V), and
The
second
53
ok Rk"A(t
so f
=A
14 ik
t
1'4)
7(t),
=
in L2(V) and therefore almost everywhere.
(Itk*f)
(b) By (11.3.8),
IgIknif II
II
11k 612(g)
=
2
dg
L (V)
V
1412k
f
c
I
Gk(V)
so
II
e
f
(4) dg da
n(a)
L2(n(a)-L) for almost every a
which means
Gk(V),
e
fih)af is welldefined for almost every a, and Ahlf
If g
e
that
L2(T).
L2(V), then
0
=
(11,RLPf> 2
L (T)
<(P0A,(RcTI)A>
L (T)
(Pag)A(g) (AkPDA(g) dg da
n(a).4.
Gk(V)
(20k
f
G (V)
g(g) T(t) at
T(g) dg da
1(g)
V
n(n)1'
=
C
AA
=
, f >2
C < ,f>
.
2
L (V)
L (V)
This shows that A Pf is in the domain of P
=
c P AkPf
and that
54
almost everywhere.11
inversion formula for D is not quite satisfactory
in
this
form. In any practical application, the point x is located inside the
support
of f,
port of f.
whereas Daf is measured for points a outside the sup-
It is desirable to rewrite the formula as an
integration
over a set of points surrounding the object.
Let A be a sphere of radius r around 0,
Lemma (1.2)
not on A, and let g be non-negative and measurable on
jg(e) de
=
f g(T!=!T)
in-xl-n
x a
Sn-1.
point
Then
(a,a-x>1
A
Sn-1
Proof: See Leahy/Smith/Solmon (71.1
Theorem (1.3)
If
L2(0) and A is a sphere of radius r
f
rounding 0, then for almost every x
f(x)
e n,
IGk(V)I
DaDaf(x)
*
r c(n,k)
ISn-11
1(a,a-x>I da.
A
12=211 By (11.3.3),
1
Dxf(a) da de
Daf(a) da
I5k-11
Gk(V)
11_1
rk(e)
sur-
55
ISk-11
r
Df(a) da
laxln <a,ax>I da
Daf(a) da
laxln 1<a,ax>I
e , ax
A
1
r ISk-11
da
ax
A
1
-------r
I
1Sk-11
Daf((ax)/lax1)1 =
a a
A
Danaf(x) = 1/2 laxln CDaf(ax/laxl)
Special Case: For k = 1,
+
D D f(x) 1<a,ax>I da.111
laxln
Daf(ax/lax1).
(1.3) reduces to for
mula (3.11) in Leahy/Smith/Solmon [71:
1
f(x)
=
I.
-------r c(n,l)
Daf(TiFiT
axln 1<a,ax>I
da.
A
2. Approximate Inversion Formulas
In
practice,
none
of the formulas presented in the
preceding
section is suitable for numerical inversion of D or P, because of the
presence
of A.
reconstruction
function.
The usual way to resolve this problem is to
of
e*f instead of f,
where e is an
seek
a
approximate 8
56
Also.
the integral of f is zero,
unless
integrable at 4 = 0 for k
1
1401
is not square-
so it is not obvious how to
n/2,
inter-
pret formula (1.1)(a) in this case.
Lemma (2.1)
e
If e, f
0
A(e*f)
Proof:
L2(V), then
L2(V), 141ke
EAk(e*f)]
=
(4)
/1!te*f.
=
141k (e*f)A(4)
(20n/2 itik 1Nt) i(t)
=
(20n/2
(Ae)A(4) if(4)
Lemma (2.2) If e, f
e
L2(V), then
0
P(0f)
a
Proof:
(#e*6^(4).1
=
Pa(esf)(x")
P e*P f.
Q
e*f(x"+e) de
=
n(a)
=5
n(a)
=J
S
e(y) f(x"+e-y) dy dx'
V
n(a)
e(y"+y9 f(x"+x'- "-y') dy' dy" de
n(a)4
n(a)
') de dy' dy"
e(y"4-74)
n(a)4
n(a)
n(a)
57
j f(x"+e-y") dx' dy' dy"
Je(y"479
n(a)-1
n(a)
n(a)
If
n(a)4
e(Y".179 Paf(x"-y") dy' dy"
n(a)
Pae(y") Paf(x"-y") dy"
P e*P f(x").111
a
a
n(a)4
Theorem (2.3) If e,
every x
e
e
f
L2(V),
0
1
1k/2 A
i4i
e
e
L2 (V).
then for almost
V.
e*f(x)
k*Paf(Ea4x) dr,
=
Gk(V)
where
= Cae.
Proof;
By (1.1)(b), (2.1) and (2.2),
e*f
C
=
C
P*/6'(e*f)
ille(Pe*Pf)
P* (Ak Pe*Pf)
P*(k*P0.111
Lemma (2.4) If f, g are non-negative and measurable with support
in
and A is the sphere of
the ball of radius r around 0,
around 0, then for almost every a
e
Gk(V),
radius
r
58
Paf(a") g(a") da"
x(a)L
1
D f(a) g(EaLa) (1
r
a
nk+1 isk-11
1E4a12)(2k)/2
a
da.
A
Proof: Immediate from (III.1.1) and (III.3.1).1
a) this yields
For g(a) = k(EaLx
Theorem (2.5) If e,
every x
2
f
eo(0),
k
e e
141/2A
L2 (V),
then for almost
ea
e*f(x)
1
=
Daf(a) k(Eat,(x a)) (1-1Ea4aj2)(2h)/2 da da.
nk+1 isk-11
Gk(V)
A
where k is the same function as in (2.3).
59
BIBLIOGRAPHY
Aronszajn, K.
Completion, Univ.
Report 10, 1954.
N.
Smith,
of Kansas,
T.
Functional Spaces and Functional
of Mathematics, Technical
Dept.
Christ, Estimates for the kplane Transform, Indiana University Math. Journal, vol. 33, no. 6 (1984), 891-910.
M.
W. Greub, Multilinear Algebra, 2nd ed., Springer 1978.
C.
Hamaker, K. T. Smith, D. C. Solmon, S. L. Wagner, The DiverRocky Mtn. Journal of Math. 10
gent Beam XRay Transform,
(1980), 253-283.
S. Helgason, The Radon Transform, Birkhauser 1980.
Herman,
Press, 1980.
G.
T.
Image Reconstruction from Projections, Academic
Smith and D. C. Solmon, Uniqueness, Non
uniqueness and Inversion in the XRay and Radon Problems, Internat. Symposium on Illposed Problems. Univ. of Delaware. Newark.
DE, 1979.
J.
V.
Leahy,
K.
T.
Oberlin and E. M. Stein, Mapping Properties of the Radon
M.
Transform, Indiana University Journal, vol. 31, no. 5(1982),
641-650.
D.
Radon, Uber die Bestimmung von Funktionen durch ihre lutegralwerte langs gewisser Mannigfaltigkeiten, Her. Verb. Sachs.
Akad. Wiss. Leipzig, Math.Nat. Kl. 69 (1917), 262-277.
Y.
R. T. Seeley, Spherical Harmonics, Am. Math. Monthly 73(1966) #4
part II, 115-121.
K. T. Smith and F. Keinert, Mathematical Foundations of Computed
Tomography, to appear.
T. Smith and D. C. Solmon, Lowerdimensional Integrability of
L functions, JMAA 51(1975), 539-549.
K2
T. Smith, D. C. Solmon, S. L. Wagner, Practical and Mathematical Aspects of the Problem of Reconstructing Objects from
Radiographs, Bull. Am. Math. Soc., vol. 83 #6 (1977), 1227-1270.
K.
D. C. Solmon, The XRay Transform, JMAA 56(1976), 61-83.
Solmon, A
71(1979), 351-358.
D.
C.
Note
on kplane Integral
Transforms.
JMAA
60
R. S. Strichartz, LPEstimates for Radon Transforms in Euclidean
and NonEuclidean Spaces, Duke Math. Journal, vol. 48 No. 4 (Dec
1981), 699-727.
Whitney,
Press, 1957.
H.
Geometric Integration Theory, Princeton University