AN ABSTRACT OF THE THESIS OF
DONALD C. SOLMON
(Name)
in
MATHEMATICS
(Major)
for the
DOCTOR OF PHILOSOPHY
(Degree)
presented on
April 25, 1974
(Date)
Title: THE X-RAY TRANSFORM
Signature redacted for privacy.
Abstract approved:
Kennan T. Smith
The problem of determining a function from its integrals over
0 < k < n,
k-planes,
is studied. It is shown that when k < n/2,
the x-ray transform of an arbitrary square integrable function is in a
certain LP
If
space on a fiber bundle over the Grassmann manifold
k > n/2,
the x-ray transform of a square integrable func-
tion may not exist. A characterization of the range of the transform
is given and a generalization of a theorem of Ludwig on the Radon
transform is proved. Finally a relationship between a theorem on the
null space and a particular iterative scheme that is being used to
detect brain tumors is discussed.
The X-Ray Transform
by
Donald C. Solmon
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
June 1974
APPROVED:
Signature redacted for privacy.
Professor of Mathematics
in charge of major
Signature redacted for privacy.
Chai
,-
an of Department of Mathematics
IA 4,
411,'.
Signature redacted
for privacy.
Dean of Graduate School
Date thesis is presented
Typed by Clover Redfern for
April- 25, 1974
Donald C. Solmon
ACKNOWLEDGMENT
I am indebted to Professor Kennan T Smith for suggesting the
study of the x-ray transform. Professor Smith's advice and encouragement have been invaluable throughout the research and writing of
the thesis. I also wish to thank both the mathematics faculty and
graduate students of Oregon State for numerous conversations on the
thesis topic and other mathematical areas. The openness and availability of the faculty and the comradeship of the graduate students have
made my tenure at Oregon State both profitable and enjoyable.
Professor Aldo Andreotti among the faculty and Gregory Fredricks
among the graduate students have been especially helpful.
A special expression of love and gratitude goes to my wife,
Elizabeth, who has born with my varying moods through the successes
and failures of the last four years.
Lastly, I would like to thank the typist, Mrs. Clover Redfern,
who has been very helpful in preparing the final copy.
TABLE OF CONTENTS
Chapter
Page
INTRODUCTION
1
DEFINITION AND FUNDAMENTAL PROPERTIES OF
THE X-RAY TRANSFORMATION
6
LOWER DIMENSIONAL INTEGRABILITY OF L2
FUNCTIONS
14
THE X-RAY TRANSFORM AS AN UNBOUNDED
OPERATOR
19
AN INVERSION FORMULA
26
THE SUPPORTS OF f AND Lf
29
THE RANGE OF L
40
THE NULL SPACE OF L Tr
48
BIBLIOGRAPHY
53
THE X-RAY TRANSFORM
INTRODUCTION
The problem of characterizing a function on Rn, n > 2,
means of its integrals over k-dimensional planes,
a new one.
tion f
0 < k < n,
by
is not
Radon [12] and John [8] proved that a differentiable func-
with compact support in
R'1
uniquely determined by
means of its integrals over the hyperplanes in the space. The Radon
transform is defined by
R0f(t) =
where
0E
Sn1
,
x
(x)
f()dan-1
<x,0>=t
t E R1,
and
an-1
is the (n-1)-dimensional sur-
face area measure in Rn. Considerable literature on the Radon
transform exists.
The lower dimensional problem,
has received less
k < -2,
attention. However, Helgason [6] and [7], has obtained some inter-
esting results.
In order to motivate the lower dimensional case, consider the
following situation. An object in 3-dimensional space is determined
by a density function f on the space
at the point
a function
x.
L0f
R3,
f(x) being the density
An x-ray pictkire taken in the direction
on the subspace orthogonal to
0
0
provides
whose value at a
point
x
of this subspace is the total mass along the line through
in the direction
0.
from its x-rays
Lef.
The problem is to recover a density function
x
f
If we restrict ourselves to a 2-dimensional cross section of the
3-dimensional object then the problem is exactly the Radon problem.
However, if we wish to reconstruct the object without going to cross
sections, the problem is one of integration over lines in
and the
R3
Radon theory does not apply.
The above x-ray reconstruction problem has been the object of
a great deal of recent mathematical and practical research. For
example, see [4] where a bibliography of recent results is given.
Within the past year R. Guenther and K. T. Smith [5], have detected
brain tumors in a human patient by reconstructing cross sections of
the skull from ordinary x-ray data. The research in this thesis
stems from the reconstruction problem and was suggested by K. T.
Smith.
Let us put the problem in a more abstract setting. If
direction in Rn, n >2,
space orthogonal to
0
0
is a
x is a point in the (n-1)-dimensional suband
f
is an integrable function, then the
ordinary x-ray of the object with density function f in the direction
0
at x is given by
3
L0
f(x) =
f(x+t0)dt
_oo
provided that the integral exists in the Lebesgue sense. Note that
the x-ray of a function
ES
n-1
depends on the variables
f
x E r. Thus the x-ray of
and
tangent bundle
f
with
(0, x)
is a function on the
to the sphere.
T(Sn-1)
More generally let
be a k-dimensional subspace of Rn.
7
The x-ray of f in the direction
Tr
at the point
x"
in
Tr-L-
is
defined by
f(x'
LTrf(x") =
")dxl
It
provided that the integral exists in the Lebesgue sense.
general, once a subspace
x' and
It
is fixed, we write x = (x', x")
are the orthogonal projections of
x"
The k-dimensional subspaces of
The x-ray of
T(Gn, k)
When
f
k=
1
or
Note when
Gn, k
x
on
Tr
and
where
Tr-L.
form the Grassmann manifold
is a function Lf(Tr, x")
with base space
Here, and in
on a fiber bundle
and fibers isomorphic to Rn-k.
k = n-1, Gn,k = Sn-1.
k
n-1,
Radon transform. In fact
the x-ray transform is essentially the
4
Ref(t)
L
f(tO)
.
eJ-
Thus in this case all results on the Radon transform carry over with
appropriate change of notation.
The central problem of this thesis is the study of the relationship between a function
f
and the transformed function Lf.
Although one might expect the results to depend somewhat on the
magnitude of
k
there is a surprising difference in the results when
k < n/2. Indeed, we prove that any square integrable function. on
R
is actually integrable over almost every translate of almost
every k-space,
k < n/2,
while an example is given of a square
integrable function on Rn which is not integrable over any k-plane,
k > n/2. We also show that when k < n/2,
the x-ray transforma-
tion has a natural extension as a closed operator from a domain
Dk
C L2(R) into a certain
L2
space on the fiber bundle T(Gn,k)
Again this is shown to be impossible when k > n/2. An inversion
formula is given for all k.
A characterization of the range of the Radon transform was
given by Ludwig [10].
We do the same here for the x-ray transform,
giving necessary and sufficient conditions that a function on the fiber
bundle
T(Gn, k)
be the x-ray of a square integrable function with
support in a given compact convex set K. Our proof fails when
k
n-1,
but the result holds from the work of Ludwig. However, we
5
prove more. Indeed, we show that it is possible, in some cases, to
get inside the convex hull of the support of f from the knowledge of
the x-rays when
space of
Lif
k < n-2. Finally, a characterization of the null
is given and a few remarks are made on the relation-
ship of this theorem to the particular iterative scheme that is being
used to detect brain tumors.
6
DEFINITION AND FUNDAMENTAL PROPERTIES
OF THE X-RAY TRANSFORMATION
1.
If
is a k-dimensional subspace of
Tr
integrable function f in the direction
TT
Rrl
the x-ray of the
at the point
x"
in
is defineddefined by
(1.1)
Litf(x") =
f(x', x")dx'
it
provided that the integral exists in the Lebesgue sense. Here, and
in general, once a subspace
where
and
and
manifold
For
x = (x', x")
are the orthogonal projections of
x
on
Tr
The k-dimensional subspaces of Rn form the Gras smann
Tr-L.
T(Gn, k)
x"
is fixed, we write
it
The x-ray of f is a function on a fiber bundle
Gn,
with base space
k=1
or
k
-1, Gn,
k
Lemma (1. 2). If
k-space
f
and fibers isomorphic to
Gn, k
-
=
Sn1
is integrable on Rn,
then for each
L f is an integrable function on Tri and
f< f 11 L (R
Ll(Tr.L)
Proof.
Rn-k.
1
n
)
The proof is an immediate application of Fubini's
theorem. From (1. 1) we have
7
IIL f
f(xl , x") I dx1dx" =
<
Ilfil1(Rn)
L
L 1 (Tr-L)
Lemma (1.3). If
variable and for fixed
on
p
is a locally integrable function of one
yll E 7r1,
is integrable
p(<x", y"))LIFIf 1(x")
1T,
(1.4)
t.)
I
P(<x", y">)L f ")dx" =
p(<x, y">)f(x)dx.
Tr-L.
Proof. Fubinils theorem gives
Tr
.L.
p(<x", y">)LTrf(x")dx" =
p(<x, y">)f(xl
")dxIdx"
Tr
Tr
p(<x,y">)f(x)dx.
The Fourier transform of an integrable function on Rn is
given by
(21.)_1112
Thus if
g E L1(Tr1),
"g(
)
(2
e- <x,
f(x)dx.
Rn
where
k-n) /2
Tr
is a k-space, it is natural to define
e
-i<x,
g(x)dx,
for
E
Trj"
From Lemmas (1.2) and (1.3) we derive immediately a relationship
between the Fourier transform of f and the Fourier transform of
L f.
Lemma 1.5). For each k-space
(LTrf)
Proof.
") = (2Tr)
(
Tr
ki2Af(t")
and integrable function
for
f,
t" E 7r.
The proof is immediate if one takes
p(t) = e
-it
Lemma (1. 2).
If
V
is any
dimensional subspace of Rn and
(k+1)
is an arbitrary point in Rn,
-
then
E
i
for some k-space
contained in V. Lemma (1.5) tells how to compute
L f. Since
Tr
A
f(
)
from
is uniquely determined by its Fourier trans-
f E L1 (Rn )
form,
f
7 C V.
Thus we have proved the following result,
is uniquely determined by its x-rays in the directions
the submainfold of Gn,k
Gn, k(V)
being
consisting of the k-spaces contained in V.
Corollary (1.6). For any (k+1) dimensional subspace
Rn,
Tr
an integrable function f on
V
is uniquely determined by
its x-rays in the directions Gil, k(V).
Note that if k =
sphere
Sn-i .
1,
then
Gn,k(V)
of
is a great circle on the
9
The preceding result can be improved upon when f has
compact support. First a well known algebraic result is needed.
Lemma (1.7). The vector space of homogeneous polynomials of
degree m in
is spanned by those polynomials of the form
R1.1
<a, x>
Proof-
The monomials x a
al
=
an
.. xn
xl
with
=M
i=1
form a basis for the homogeneous polynomials of degree m. A
linear functional A on this vector space can be represented by
aaxa ) =
A(
am
b
aaa,
b0 EC.
lal=rn
Now
<a, x>
=
(
al ...
a
al
a)a ...a n
n
l
n
a
a
1
xl
xn
Ial =m
where
(a
If
A
1.
.
a
)
=
al
m!
!.
an!
vanishes on all polynomials of the form <a, x>
then
10
al
an
ba(a1... an)a1 ...an =0.
But this is a polynomial in a. Thus
ba(al... an) =
0
for all
a.
ba =0 for all a.
So
Lemma (1.8). A nonzero homogeneous polynomial of degree
m in
Rn
can vanish identically on at most m distinct hyper-
planes.
Proof. The proof is by induction on m. If m = 0,
then the
result is trivial. Suppose that the result has been established for
m<I
0.
J_
and that
±0.,
Pi vanishes on the hyperplanes 01, ., 01+1,
j. Without loss of generality we may assume
01 = el'
i
the direction of the
x1
axis. Then PI (x) = 0
whenever
xi = 0.
Thus
PI (x) = x1Q1 -1(x)
with
Q1
homogeneous of degree 1-1. But Q1 vanishes on
)2 -
0\1 01_L for
it follows that
i
= 2, ...,/+1. Since
vanishes on
Q -1
hypothesis implies
Q
For any integer
tive integers
=
1, .
Q
-1
0.,
vanishes on a closed set
and the induction
i>2
a- 0.
k>0
1 k)
let Nk be the set of k-tuples of posisuch that
<
2<
k.
11
Definition (1.9).
such that any
of the
(k+1)
,
are linearly independent. For each
().
let
Nk,
E
0.1 be a sequence of directions in Rn
Let
be the k-space generated by
el
.
The set
:j
{Tr
Nk}
E
is called a fundamental system
1
of k-spaces for RII
Theorem (1.10). If
has com act susoort, then
f E L1(
L f coming from any fundamental
is uniquely determined by the
Tr
system of k-spaces for Rn.
Proof. Suppose
Tr, j
E
J
for a fundamental set of k-spaces
=0
L
Nk. Formula (1.4) gives for
">
( 1 . 11 )
JTr
where P
y II
E
Cx, y >f(x)dx = P (y")
f(x ")dx" =
Tr
is the homogeneous polynomial defined by the right hand
side of (1.11). From the assumption P
Let
I
E
Nk-1.
For each
is contained in
IT ,
j
Thus
Tri. C
J
hyperplane in
Tri
1
P m vanishes on
J
7 ,1_
such that j = (I, jk)
Nk
E
vanishes on
771
1
.
Each
Tr-L.
for all
the space
is a distinct
3
by construction of the fundamental system. Since
Tr-!.-
J
that P m vanishes on
for all
TrI,
I
i
I
E
E
Nk
it follows from Lemma (J..8)
Nk-l Proceeding inductively it
easily follows from the construction of the fundamental system of
k-spaces that Pm vanishes on the hyperplanes
Oi
where the
Oi
12
are the directions generating the fundamental system as in Definition
(1. 9).
Now Lemma (1.8) implies Pm is identically zero for all
while (1.11) and Lemma (1.7) imply f
nomials. Since
f
is orthogonal to all poly-
has compact support,
Note that if k = 1,
f
0
almost everywhere.
a fundamental system is contained in any
infinite set of directions with distinct entries. Thus if f has compact support,
is uniquely determined by any infinite set of
f
ordinary x-rays.
Remark. No function f is determined by any finite set of
x-rays. Indeed if 01, ..., 0
is a finite set of directions,
oo
any open set in Rn, 4 E Co (0),
P(
)
C2
is
and we define
= <,(31>
0m>
then
[P(D)qi]A(
)
=
Thus Lemma (1.5) shows that if g = P(D)kii,
LO. g = 0,
The set of all functions
i = 1,
g
..
then
, m.
P(D)Lii
with
Do
if,
E
C0 (0) is obviously
infinite dimensional. Thus the set of functions with compact support
in any open set
C2 C Rn having the same x-rays in a finite number
13
of directions is infinite dimensional.
Also if the assumption of compact support is dropped in
Theorem (1.10), the theorem fails even for analytic functions.
Indeed let
I:0
E Cx0(Rn)
with
exists an open set G C Sn - 1
is empty. If
ip
supp (I:1C B(b, 1),
is the inverse Fourier transform of
kp
Logi
0
There
1°
such that for all 0 E G,
analytic and Lemma (1.5) implies
L0
b
B(b, 1)
0
(1),
ip
is
for all 0 E G. Thus
vanishes in a continuum of directions. In fact by choosing
sufficiently large the x-rays of
ip
b
can be made to vanish in all
directions outside of an arbitrarily small neighborhood of b-Lrm SnI
This is the best that can be said about the vanishing of the x-rays of an
integrable function since the Fourier transform of a non-zero function
in
Ll(Rn)
is continuous and thus nonzero on some open set.
14
2. LOWER DIMENSIONAL INTEGRABILITY OF Lz FUNCTIONS
The symbol
L2(T(Gn,k))
will be used to designate the
measurable functions on the fiber bundle T(Gn,k)
which satisfy
sg(ff' XIT)dx"dp. < co
G
where
la
n, k
is the finite measure on
invariant under orthogonal
Gil, k
transformations [11], and normalized so that
I-L(Gn,k) =
Isn-k-11
the bars denoting the appropriate area measures on the spheres.
The symbol L
be used when the x-ray is acting in the
direction of a fixed k-space
be denoted by
Tr.
The total x-ray transformation will
L.
Lemma (2.1). If g is nonnegative and measurable on the
sphere
Sn - 1 ,
then
Gn, k
Proof.
g(0)d0
g(co)doidp.
-1
Tr
Sn - 1
The integral on the left defines a continuous linear
form on the space C(Sn-1),
(hence a measure on 5n-1 ). This
15
form is obviously finite and rotation invariant and there is only one
such up to a constant factor, namely the integral on the right. The
normalization of
is chosen to make the constant
[1,
1.
Once the
formula is established for continuous functions it extends immediately
to nonnegative measurable functions by the standard arguments of
measure theory.
Lemma (2.2). If
is nonne ative and measurable on Rn,
then
Ix
G
n, k
I
g(x")dx"dp. =g(x)dx
Rn
Tr
Proof. Using polar coordinates in the integral over
Tr-1T
gives
x" k g(x")dx"dp,
,51
n, k
r n-1 g(re")d0"dp.dr.
=.C'S
0
Gn, k
Sn-1
The result follows from Lemma (2.1).
In terms of Fourier transforms the operator
(2.3)
(Af)A(
)
If()
A
is defined by
16
To avoid cumbersome notation the same symbol is used irrespective
of the space Rn,
or subspace of Rn in which
Theorem (2.4). The ma
from L2(Rn)
into
(2Tr
-k/2 Ak/2L)
A
acts.
is an isometry
L2(T(Gn,
Proof. It is enough to establish the result on a dense subset of
L2(Rn)
.
then clearly A(L f)
f E Cm(Rn)
If
0
The definition of
The
Tr
is in
(2.3), the Parseval relation on
A,
L2().
Tr-L,
Lemma
(1.5), and the last lemma give
k/2
SG
=
S
L
LTr
f(x")12dx"dp.
n, k
kA
1,"1 If(,")1
(27)k
2
"dI-1 = (2'7
11
112 2
L (R
Gn, k
n
)
The space of bounded, measurable, functions which vanish out00
side a compact subset of Rn is denoted by
L0 (Rn )
The next theorem, which is a rather surprising one, is part of a
recent paper done jointly with K. T. Smith. See [13] for complete
details of the proof.
Theorem (2.5). For
in on
k
and
n
k < n/2 there is a constant
such that if f E L2(Rn
c
(de end-
then for almost ever
17
Tr EGk'
n,
is defined almost everywhere on
LTrf
by an absolutely
Tr-I-
convergent integral and
(Z. 6)
S
ii
Lirf 11
2q
L (TT
Gn, k
c2 110
dp.
)
2
L2(R)
q-
Proof. It is sufficient to prove inequality (2.6) for
2(n-k)
n-2k
f E Lc); (Rn)
Indeed if this has been done and f E L2(Rn), f > 0,
we can choose
n
fn E L 000(R n)
an increasing sequence of functions
and with
converging to
fn
f.
with f> 0,
fn
Inequality (2. 6) for the fn
and the
monotone convergence theorem give the desired result for nonnegative
f.
The result follows immediately for all f E L2(Rn).
Assume that f E Lco(Rn). It is clear that
0
oo
Lf E L 0(Tr-L)C 2L2() so that Af E L 2
,
According to Lemma
Tr
(2. 2)
STri. IV!
kA2
fl dVdt1 <
Gn,k
and it follows that for almost every
E
Fix any such
Tr.
f.(
Since
II) =
VI I
g(V)
TT
L2(71')
.
18
and since both
f
and
are in
g
the theorem in [13] on
L2(7-1-)
the Riesz transform gives that
Lf=-R k/2 a
(2.7)
Tr
-
R
k/2 L f
/21 A1
Tr
where
R ah(x) = C(n, a) S
Rn
Now, Sobolev's inequality in the space Rm
R ag
q
Cg
In the present case we have
0 < a < n.
la-nh(Y)dY,
a = k/2
1
1
for
2
[14], asserts that
q
=
2
-
a
m > O.
and m = n-k,
so we get from
(2.7) and Sobolev's inequality that
q-
< C Aki2L fil ,
11
TT
L.TrfLq(71.)
Squaring and integrating over
LL(Tri-)
2(n-k)
n-2k
and making use of Theorem
Gn,k
(2.4) gives the desired result.
Remark (2.8).
f(x) =
The function
lx1-n/2,(loglxh
0, otherwise
-1
>2
is square integrable on Rn but is not integrable over any plane of
dimension >n/2.
19
3.
THE X-RAY TRANSFORM AS AN UNBOUNDED OPERATOR
We shall study the x-ra,y transformation as an unbounded
operator from L2(Rn)
L2(T)
into
L2
(T(Gn,k)). (We will usually write
in place of L(T(Gn,k)) .) The natural domain would appear
to be the set
(3.1)
Dk
{f
E
L2 (Rn )
k11L Tr
c1P- <
f112,G
Gn,
}
L (Tr)
k < n/2,
This turns out to be entirely satisfactory for
but not for
k > /2. In the latter case, for example, we have not been able to
decide whether the operator is even closable with domain Dk.
Lemma (3.2).
(3.3)
fE L1
If
111422
<
L (T)
Proof.
Since
f
E
rTh
2 )1(0
L1
L2,
then f
L2
E
Dk
11f1122)
11-1111121
it follows that
and
.
E
L2
and an easy calculation shows that
(3.4)
11
10
-k /2A 2
f11< 1Sn-11 11/112,)
1111122
L
L
On the other hand, Parseval's relation on
Lemma (2.2) with g = 10 -k/2A give
Tr-L,
Lemma (1. 5), and
20
Lf1122
L (T)
ddi.
sc
(27)k SI
=
Gn,
-k /2A 2
f
L2
k
= (2Tr)
11
1
The result follows from (3. 4), Parseval's relation on Rn and the
fact that
co
11/f11
11f11
L
L
Let us define
2
(3.5)
Dk = {f E L: I
Tr
whenever
fED
Proof.
CC)°
7n
E
in
and
0
(R
n
D
and for almost
k
If
/2A
f
a. e.
k > n/2,
1T,
on
then
D
k
D
.
k
0 < k < n and assume that
such that fn
A
f
in
L
2
and
fE
1
1
Dk.
-k12"
fn
Choose
10
f
A
be the inverse Fourier transforms of fn
respectively. Since Afn E COoo (R n ) , it follows that
L2. Let
il
Let
(27)k
A
)
.
Tr
(L f=
(3. 7)
2
-k /2A
f E L2}
k < n/2 then Dk =
Theorem (3. 6). If
every k-space
I
fn E 1,4 (Rn),
fn
and
f
(the space of rapidly decreasing functions of Schwartz).
Lemma (1.5) and Lemma (2.2) give
21
f )A_ (270k/2/p 2
S
11(.1-Tr
I
n, k
= (2Tr)
L
2
_k A 0\ 2
s
(2Tr)nn
111'n-71122
L (Tr
Gn, k
But the last integral converges to
)
as
0
so, choosing a
co,
subsequence if necessary, we have for almost every
LTrf n
(3. 8)
where
iT
a. e.
g
Tr
is defined by
g
(3.9)
= (2Tr)
k/2Af
on
Tr
TT -I-
Theorem (2.5) shows that for almost every
However when k < n/2,
Tr,
Tr,
and again a suitable subsequence,
LTrfn LTrf
a. e.
It follows that for almost every
=Lf
(3.10)
Tr
Hence
fE
a. e.
Tri
IT
on
Tr
Moreover (3.9) and (3.10) show that (3.7) is valid.
Dk.
Conversely suppose
f (x) = e
in
_,-,21x I 2
'
A
consequently f
fE
Dk,
k < n/2,
It is easy to see that
f(x).
f
in
and define
f
f
in
L
2
and
L.2 Lemma (2.2) implies (for a suitable
22
sequence of
pis)
(3.11)
uirn
that
0
p
for
=0
I
Tr -I-
P
a. e.
I
and Theorem (2.6) gives (for a suitable sequence of
L fL f
IT p
for
in
p's)
a. e.
that
IF,
TT
and hence that
Lf
LTif p
(3.12)
where
T(rr-L)
in
for
T(iri)
a. e.
denotes the tempered distributions in
the Fourier transform is a topological isomorphism on
E
fp
it,
Since
Tr-L.
err
)
and
Lemma (1.5) and (3.12) give
L1,
(3. 13)
(2Tr)
k /ZI\
f
(L f)A
in
(
Tr)
a. e.
for
Tr.
From this and (3.11) it follows that
(LTrf)A
for any
it
(27)k
/2A
f
a. e.
on
for which (3.11) and (3.13) hold. Since
(2.2) shows that
case
=
ki /2A
1
1
f E L2 ,
fE
Dk, Lemma
and the theorem is proved for the
k < n/2.
To prove the second part of the theorem, we construct a
23
function
such that
g E L2
1
1-k12',..gA
E
LZ
but
g
is not Lebesgue
integrable over any plane of dimension k > n/2. Assume that
k >n/2 and let
be the function in Remark (2.8). Define
f(x)
0
by
f(x),
(3. 14)
g0
(x)
,
mks run through the integers.
where the
line,
..
g0
2mk < xk < 2mk +1, k = 1,
otherwise
(For example, on the
on alternate intervals between integers.) Let
=0
en
,n
be the unit vectors along the axes, and put
g
(
)
g1(x)
= gk-1(x+ek)
One may easily check that
is not Lebesgue integrable over any
gn
k-plane. Moreover
= (e
-
(e
-1)go(
)
which gives the desired result since
(e
-1)...(e n-1)10 -11.
is bounded.
Theorem (3.15). The x-ra transform with domain
admits a closure L with domain Dk (=Dk for
co
Co
k < n/2 and
Dk
24
k > n/2).
for
Proof. First we show that the x-ray transformation admits a
closure. Suppose that fn
in
oo
E
in
0
fn
CO
L(T). We must show that g = 0.
L (T)
2
r
Trj_
Now
A
lim
II
E
Gn, k
lk y
E
>0
A
fn
SiGn, k
00
G
>0
n
it,Tiki/f\
I.'
In
es.
I
>
n, k
,
is fixed and we let n
is arbitrary and
g
E
L2(T)
made arbitrarily small. Thus g = 0.
2
dt"di.t.
71-
Gn,k
E
have been used.
Lemma (2. 2) gives
co ,
A
Ig
Since
2
dVdP,
If
where Lemma (1.5) and Parseval's relation on
E
Lfn
urnLirfnli 22
L
n
If
and
n-00 G n, k
(2T)k 1im
E
L2(Rn)
,
the right-hand side can be
g
25
Let
Dk
be the domain of L . If f
E
D
then there exists
k'
00
such that fn
in L2(Rn) while Lfn -in L2(T). The proof of the first part of the theorem shows that for
a sequence
fn
almost every
E
CO
Tr,
(and an appropriate subsequence if necessary),
k/2"
(2Tr)
fa. e.
n
Tr
on
71
But we also know that for an appropriate sequence of n,
on R
a. e.
fn
It follows that
(3.16)
gT, =
Since g E L2(T) ,
2
-k/21\
f L (R11).
E
2
k /2A
a,. e. on
Tr
for
a. e.
it follows from Lemma (2.2) and (3.16) that
Thus
Finally suppose h
f
E
E
Dk.
the first part of the proof of Theorem
Dk,
h in
(3.6) shows that there exists hn
(Rn) such that hn
Thus it suffices to show
L2(Rn) and
Lhn converges in L2(T)
E
that
fC
Ll Rn)
(Rn) C -I5k
(Rn),
But this is immediate from (3.3). Indeed, if
simply choose
L2(Rn),
e. ,
00
fn
in
E
C0
L1
such that
(Rn)and
in
fn
f
L2(Rn).
in
26
4. AN INVERSION FORMULA
From Theorem (Z. 4) we know that
isometry from L2 (Rn )
into
L2 (T)
=
L2
(2Tr)
-k/2 (k/2A L)
(T(Gn,k))
is an
and it follows
that
(27)k
[(Ak/2 L) (Ak/2 L)]f = f
where
(A
k /2
L)
*
denotes the adjoint of
if
f E L2(Rn)
(Ak/2 L). In this chapter
we investigate the adjoint of the x-ray transform,
L,
inversion formula which shows how to recover
from Lf or Ef
whenever
f E Dk .
2
Lemma (4.1). If
L
f
and give an
,
is in the domain of
then
provided that
/(x)
g(Tr,x")dy. E L2 (Rn )-
Gn,k
Moreover the above formula defines the adjoint operator, i.e.
= Lg.
Proof. By Theorem (3.15) it is enough to compute the adjoint
of the restriction of
Lf E
oo
L0
(T)
L
to
00
Co
(Rn). If
oo
f E Co ,
then
and the use of Fubin.Ps theorem is justified in the
27
following calculations. If
tion of
x
on
g E L2(T)
and
P
is the projec-
(x)
then
Tr4",
oo >
L f(x")g(Tr, x")dx"di.i.
Gn, k
SS
n, k
f(x` x")g(Tr, x")dxIdx"dp.
Tr
1T
Sf(x)g(Tr,
P
Rn
(x))dxdp.
Tr"1"
Gn, k
=f(x)(
Rn
f712dx
P (x )dp.
g
g(,
_L
dx
)(Tr,
5G
Tr
k
.
1R.1.1
Since the range of L
must be contained in L2(Rn)
the result
follows.
Theorem (4. 2). (Inversion Formula)
k < n/2,
or L.f = g, k >n/2,
(2TT)
(Note that the
eNs.
If
fED
k
and
Lf = g,
then
-kAk/2 L A k/2 g= f.
Ak/2
on the left is acting in Rn while the
on the right is acting on each fiber in T(Gn, k)
)
Ak2/
28
Proof.
Let
f
and
satisfy the hypotheses of the
g
theorem. First we show that Ak /2g
hEC
(Rn
0
).
is in the domain of
L. Let
Parseval's relation, Theorem (3.6), (3.16), and Lemma
(2.2) give that
= <(Lh ) A ,(A /2 g) A>
L-(T)
Lz (T)
<Lh, Ak/2 g>
2
k"
= (2w) <h, IV 1 k/24f>
L2 (T)
=
(21T)k<i\l,
-1(/21.\>
1
L2
(Rn
)
kAA
= (21r) <h, u>
L2(Rn)
where
(4.3)
Since
-1c/2/\
f
u=
4"2 n
f EU
E
L
(R
).
k'
A
transform of u,
Thus if
u
Parseval's relation on Rn gives that
<Lh, Ak /2g>
2
=
(27)k<h,
L (T)
for every
Ak /2g
hE
is the inverse Fourier
00
Co
(Etn ).
/2g =
)
According to Theorem (3.15) it follows that
is in the domain of
L*Ak
u>
L2
L
and that
(27)k u.
The theorem now follows from the definition of
A and (4.3).
29
THE SUPPORTS OF f AND Lf
5.
If
is a subset of Rn the support function of A
A
is
defined by
(5. 1)
SA(
) = sup <
y>
yEA
The support function is convex, homogeneous of degree
lower semicontinuous, with values in
(-00,00].
and
1,
The value
+00
may
be obtained when A is unbounded.
Until further notice
convex subset of
Rn.
Lemma (5.2). If
then
x+Tr
intersects
(5.3)
<x,(4> <
Proof. If
some
yE
Tr.
will denote an n-dimensional, compact,
K
is a k-dimensional subspace of Rn,
Tr
if and only if
K
SK
x+Tr
But if
for all
(co)
CO
E
intersects K then
E
Tr-I-
is in
(x+y)
K
for
then
<x, co> = <x+y, co> < SK(w).
Conversely suppose (5.3) holds. Let
projections of K and
x
on
Tr-L.
K"
Now for all
<x" , 0j> = <X, W> < SK(W) =SKI, (w)
and
u)
x"
E
be the
30
Thus
tains
x"
is contained in each halfspace in
Since
K"
is closed and convex,
K"
intersects K and hence
x" + Tr
which con-
is in K". Thud
x"
intersects
x + Tr
The following is an immediate consequence of Lemma (5.2).
Corollary (5.4). If f is integrable and has support in
then
LTrf(x") = 0
<x", w> >
whenever
for someJ-
(w)
w
E
7T
.
More interestingly the converse of Corollary (5.4) holds when
k < n-2.
Theorem (5.5). Let f
k-space
Tr,
LTrf(x") = 0
does not intersect
X" +IT
outside
K.
E
Ll
and
k < n-
for almost every
K,
then
f
x"
E
.
If for every
Tr -L-
such that
vanishes almost ever where
The result fails when k = n-l.
Proof. Let H be a supporting hyperplane of K and let
H+
{x: <x, e> >t0}
be the open half space determined by H dis-
joint from K. Define
f(x), XEH
h(x) =
,
otherwise.
31
If
Tr
LTrh(x") = 0
for if <x", 0> <
other hand if
t'
then
<x", 0> > to,
L h(x)
k < n-2,
for
a. e.
x" + 7
x"
E
does not intersect
H+.
On the
then (x" +Tr) C H. and
LTrf(x") = 0
Tr
Since
it is clear that
is a k-dimensional subspace in
for
a. e. x"
E
H
rTh
Tr-L
Corollary (1. 6) shows that h vanishes almost
everywhere. Thus f vanishes almost everywhere in H+. Since
H was an arbitrary supporting hyperplane of
K,
the result fol-
n-1
it is more con-
lows.
To see that the theorem fails when
k
venient to use the Radon transform notation. Recall that the Radon
transform of an integrable function f is defined by
Ref(t) =
where
0
E
Sn-1, t
E
<x, 0>=t
R1,
f(x)dan- 1 (x)
and
an-1
is the
n-1
dimensional
Lebesgue measure. Notice that
Ref(t) =
LerLf(t0).
Thus it suffices to construct f
E
Ll(Rn)
such that f does not have
co12-n
compact support but Ref(t) = 0,
Let
h
for
1,
log Ix
1
for
be a
when
x
C
t
n > 3. When n =
> 1. Let
A
for all
< 00
0.
which is equal to
function on Rn
let h be equal to
,
be obtained from h by differen-
f
tiation in such a manner that f
pact support. If
<r
32
is integrable but does not have com-
denotes the Laplacian, then
Af = Ah = 0,
>1.
X
Moreover
t >1,
(Ref)" = R (46f) = 0,
primes denoting differentiation with respect to t. But since Ref
0
is an integrable function of t,
as
0
Ref
t
Thus
00.
jt >1,
Ref(t) = 0,
and the theorem is proved.
Remark (5.6). A function
a neighborhood of a point
hood
in
9
Rn
ofin
Tr°
g(rr, x)
(Tro, xo)
Gn, k
on
T(Gn, k)
vanishes in
if there exists an open neighbor-
and an open neighborhood
such that
g(Tr, x) = 0
for all
(Tr, x)
E
(g
X
U)
T(Gri,
U
of
x
0
33
The function
vanishes in an -neighborhood of
g
above can be chosen to contain the ball
vanishes in a neighborhood,
A C T(Gn, k)
(E
B(xo,
if
(70, xo)
E
As usual we
E
U
f
sa:,r
-neighborhood), of a subset
if it vanishes in a neighborhood,
(E
-neighborhood),
each point of A.
If
is a subset of Rn and 6)
U
is a subset of Gn, k, we
define
u=
Note
U
X E CU r1
(Pxu)
T(Gn, k)
consists of the pairs
(Tr, x)
with
E9
ir
and
j-)
The idea for the proof of the next theorem comes from Lemma
(3.1) in [10]. The importance of the theorem is that it shows that it is
possible in some cases to get inside the convex hull of the support of
f
from a knowledge of its x-rays in dimensions
Theorem (5.7). Let
Tro
k < n-2.
be a k-space k < n-1,
an open, connected, unbounded subset of
Tr-L0 .
If
f
and
co
E
C
(Rn)
V
be
and
0
Lf
vanishes in a neighborhood of
Tr
0
[E) V
then f
vanishes on
Tro
Proof. Without loss of generality assume that
Choose neighborhoods
N1,
Nk
of
Tro = [e1,
el, ..., ek such that
ek].
34
whenever w = (w1
wk) E NiX
in a subspace of dimension k. Now if
s
00
L f(y) =
then Trw = [w1,
xNk,
yE
,
define
Rn,
00
f(y+tiwi+. .+tkwk)dti...dtk
-00
Note
1
(5. 8)
Lwf(Y) =
Wk
L n wf(Y")
Lf vanishes in a neighborhood of Tro ar V,
Since
vanishes in a neighborhood of each point
Lwf
yE
w1
+V),
(1T
1
=
1
al
a
a
)(
a
w
a
k
goo
wki.
-00
00 a
00
-00
ki
21
al
.
a
)
\ wil
=
1.
If w.w. )
1
(L f(y))
Iw=(e
1,
w
E N.
with
and
1
-00
a = (al,
ak)
1
. .
f(y+tlw1+...+tkwk t ...dtk
(y+tiel++tkek)dti...dtk
I
and
I al =
i=1
vary in
, ek)
)
-00
ak
f
. tk ax a
t1
.
.
where
Letting
wk)
a
2
/a
w
a
1
(y, w1
is fixed, then
+V
TrO
b.e. E N..
w.
(5.8) shows that
Tro + V
we have
w=(ei,
ek)
a
as
I
I
35
tkkf(y+tlel
t1
c:- CO
a
a
a1
00
. +tkek)dt1...dt
.
= O.
Thus
(5.9)
s:
Pa(Y)
a
a
is a polynomial of degree less than
Since
+
yi
+tkk
e )d tl' ° *dtk
tkkf(y+t e,
was arbitrary
I
al
Pa(Y)
in the domain
yi
in
is a polynomial in each of
its variables separately and it follows that Pa(y) is a polynomial
in
in the domain
y
supp f C B(0, r)
Tr
0
+ V.
0
on
[Tro +
It follows that the polynomials
on
+ V. Since
Tr0
+ V.
f(y+t le 1+
t1'
.
r>0
such that
then
f
Tr0
If we choose
V
(V \ B(0, r))] C Tro
Pa(Y)
+
V.
vanish on an open subset of
is connected the polynomials vanish identically
But then (5.9) shows that for all y E V,
. +tkek)
tk. Since
is orthogonal to all polynomials in the variables
f has compact support,
f
vanishes on
The convolution of two integrable functions f
given by
the function
and
h
Tr
0
is
+ V.
36
f *h(x) =
f(y)h(x-y)dy
and it is well known that
(f*g)A = (27)
n /2A /\
f g.
Thus it follows from Lemma (1.5) that
(5.10)
LTr(f *g) = LTrf
(L7g).
Theorem (3. 6) shows that when k < n/2
and
formula
f E Dk,
(5.10) holds almost everywhere on Tri for almost every
Corollar
V,
70
5.11). Let
be as in Theorem (5. 7
ciently large integers
(5.12)
E
.
L1(Rn)
have com act su ort and
Assume in addition that for suffi-
n>0
Vn = {x" E V d(x", rroi \V) > 1/n)
(where
d
denotes distance), is an o en, unbounded, connected sub-
set of V. Then if Lf vanishes in a neighborhood of
vanishes almost everywhere on
Proof. Let
El) E Cx0(Rn)
unit ball, and satisfy
Tr0
7
0
El V,
+ V.
be nonnegative, have support in the
f
37
4)(x)dx = 1.
If
(x)
then
4)
is in C,0
oo
*f
n>0
has support in B(0, p) + supp
f
f
For
p >0
-11(x /p),
p
in
L1 (R
sufficiently large,
n
as
)
p
and
0.
in (5.12) is an open, unbounded,
Vn
connected subset of V. Suppose that f has support in
B(0, r).
Since
Lf
Vn n B(0, r+1)
is compact and contained in
vanishes in an E -neighborhood of
ciently small
E ( n ) > 0.
L(
(5. 13)
1T
If
Ix
for suffi-
El (Vn n B(0, r+1))
Tr0
From (5.10) we have
*f)(x") =
and
> r +1
y" E B(0, p)
p
V,
L
0 < p < 1,
Tr
p
(y")L f(x"-y")dy"
.
Tr
then
Ix" - y"1 > r
for all
and Corollary (5.4) implies
L(4 p*f)(x") =0.
Tr
If
Ix "1 < r+1
(Tro,x")
and
X" E
TT-1-
0
vanishes in an E -neighborhood of
Lf
and (5.13) shows that L(
1T
p
*f)(x")
vanishes if
38
0<p<E<1
and
is sufficiently close to
Tr
(5.7) implies that
vanishes on
10p *f
Tro
Tro.
Now Theorem
+ V. Letting
n
go to
infinity we get the desired result.
Remark (5.14). The assumption of the unboundedness of
is necessary. Indeed consider the function with support in the
rectangle
I x.
<n defined by
1<
,
0 <x1 <1
1,
f(x) =
-1 < x < 0
-1,
If
Tr =[e
,
ek],
=
0 < r < 1,
E Tr-Hx.1
and
r, j = (n-k+1),
<V
then LTrf
vanishes in a neighborhood of
vanish on
Tr + V.
Corollary (5.13). If
fE
L1 (Rnhas compact support and for
each (n-1)-dimensional subspace
does not intersect
side
K,
ii,
LTrf(x") = 0
whenever x" + IT
then f vanishes almost everywhere out-
K.
Proof. Set k = (n-1)
V
Tr 0 V while f does not
is a half line.
Thus
Tr0
in Corollary (5.11). In this case the set
+V
is a halfspace. Corollary (5.11)
39
implies that f vanishes almost everywhere on each halfspace not
containing
K.
Since
K
is convex we are done.
40
THE RANGE OF L
6.
Ludwig [101, and later Lax and Phillips [ 9 ] , derived necessary
and sufficient conditions that a measurable function 0,o, 0 sn-1,
t
E
R
be the Radon transform of a square integrable function with
support in an n-dimensional, compact, convex set. The same is done
here for the operator L. Although our proof fails in the case of
hyperplanes, Ludwig's theorem establishes the result. Moreover
when
k <- n-2 Corollary (5.11) gives a characterization of the sup-
port of f which, in some cases, is sharper than the convex hull
f.
Lemma (6.1). If f
yE R
(6.2)
E
n
L1 (R)
has compact support and
then
P
(y) =<y, x">m L7 f(x")dx",
11
J-
is- inde endent of the choice of the k-s ace.
Tr
C
varies in Gn, k the integrals
y E Tr,
Moreover as
Tr
a homogsneotzspolL-
nornial of degree m on Rn
Proof. Taking p(t)
Pm(y)
tm
<
Tr -I-
,
in Lemma (1.3) gives
x">m L f(x")dx"
=
<y, x>mf(x)dx.
41
The integral on the right is independent of
and obviously defines
7
a homogeneous polynomial of degree m on Rn
Let
f E L1(Rn)
have compact support. We investigate the
polynomials
(6. 3)
)
Pm(
= Pm(f,
<x,
=
f(x)dx,
E
Rn.
Rn
These can also be defined for
E
simply by replacing
Cn
by
in (6.3). Lemmas (6.4) and (6. 6) were established by K. T. Smith in
an unpublished paper where he gives a new proof of Ludwig's theorem.
Lemma (6.4 . If
Fourier transform of
has compact support, then the
f E L1 (Rn )
f
extends to a complex analytic function on
with the expansion
oo
(6.5)
=
-mPm( )
(27)-n/2
m!
m=-0
Proof. The extension of the Fourier transform is given by the
Laplace transform
AfR)
(27)-n/2
e
-i<x,
f(x)dx .
42
En
The integral converges absolutely for every
even after
differentiating under the integral sign, and does determine an entire
function on
Tn.
On the other hand if the support of f is contained in B(0, r),
then from (6.3)
1Pm()1 <WI
I
This shows that the series on the right of (6.5) converges absolutely
En
for every
and also determines an entire function. Thus to
prove (6. 5) it is enough to show it for
gE
as two distinct
Rn,
entire functions cannot agree on Rn
If
0
is a direction in Rn and
D
(
a
a
agi
agn
)
then
Taylor's formula gives
oo
<0, D>rn ?1(0)Till
f(TO) =
m!
m---0
while
Pm(0)
(2T)n/2(<x, 0>m
f(xi
))n
I
(27)11(<0,D>mlf\)(0)
-
Comparison of the two gives formula (6.5) for
=T
0.
The following theorem of Smith gives a characterization of the
43
possible sequence of polynomials P
that can arise from a square
integrable function f with support in
B(0, r).
Lemma (6.6). Let
{13
be a sequence of polynomials such
}
that
P m is homogeneous of degree
Pm() <
ill
m-
for all m and all
E
-m
P IY1 (
m!
c) The sum
)
(which converges by b) is square
integrable on Rn.
Then there is a square integrable
such that
Pm() = Pm(f,
f
with support in the ball B(0, r)
for all
Proof. Define the entire function f by the formula (6.5).
According to c),
A
f
is square integrable on the real space, so it is
the Fourier transform of a square integrable f,
to show that
f
and the problem is
has support in B(0, r). By virtue of b)
141 <(Z7)
rl
c-11/2
e
11+'
+1.1
< cle
By the Paley-Wiener theorem [3] f must vanish outside the cube
Q
fx
1
<- r, i = 1,
By a rotation of coordinates
becomes any cube that circumscribes
outside the ball.
B(0, r),
so
f
Q
must vanish
44
The Sobolev space Hs, s > 0,
consists of those functions
11
on the fiber bundle
T(Gn, k
satisfying
g(Tr,x)
sA
2
ilgis=g
)
IgTr(V)12dVd1J- <
Gn, k
These spaces will be denoted
Theorem (6.7
K
Let
HS(T).
or
be a measurable function on
Tr, x
There exists a square integrable function
T(Gn, k).
in
.
Hs(T(Gn,k))
such that for every
L7f
g
Tr
with support
Tr,
a. e.
on
Tr
-I-
if and only if
g E Hk
(T)
a_<y,x>
k-space
gTr(x)dx,
y E Tr-L,
is independent of the
varies in Gn,k these integrals define a homogeneous polynomial of degree m
in
Tr C yL_
and as
7
Rn.
g(Tr,x) = 0
whenever
x + Tr
there exists a constant c > 0
<c
g
L (Tr-L)
for all
does not meet K.
such that
1T
G
n,k
45
Proof. When k = n-1,
this is the theorem of Ludwig [10], and
our proof fails. We prove the theorem when k < n-2. The necessity
of the conditions has already been shown. We proceed to establish
the sufficiency.
For each
(6.8)
P
and
Tr
Tr,m (
)
Rn
E
Tri,
g (x)dx .
<x,
=
define
Tr
Tr
Condition (ii) shows that
(
) = P 772/M (
if
)
E TrtcmTrt.
Moreover (iii), (iv), and Lemma (6.4) give that
A7T , M
oo
(6.9)
i
g
(
)
Tr
=
_rn,
(2Tr)(k-n)/2
P
m!
(
)
for each
Tr.
m=0
Now condition (ii) implies that
(6.10)
Pm( )
= (2Tr)
-n/ ZP,Tr
is a homogeneous polynomial of degree
the polynomials
P
111
on R. We claim that
satisfy the conditions of Lemma (6.6). We
have already shown that (a) holds. Moreover (6.10), (6.9), (i), and
46
Lemma 2.2) show that
00
(6.11)
f(
i-mP
) =
(
m!
is square integrable on Rn. Thus (c) also holds. To establish (b)
in Lemma (6.6), note that each polynomial P
has a unique
ril
extension to
Tri-='RE
given by writing
E
in place of
'IT
,
E
1T
j
in (6.8). If K C B(0, r), then
(iii) and (iv) imply that
pTr mg) < crm
I
Moreover since
P
=
E
I
(27)-n/2P
G
Tr,
on
TT
unique extension of P 111
Thus if
n
E
I
But
agrees with PTr
(En
/71
on
is perpendicular to a real k-space, then
Prn(
{x e Rn: <x,
to
it follows that the
I
< I Pit,
= 0}
=I.(
I
<
Crnil OM
is a subspace of dimension >
.
Thus
(b) of Lemma (6.6) is satisfied. Now, if f is the inverse Fourier
transform of Af,
Lemma (6.6) implies that
f
has support in the
47
ball
B(0, r). Consequently, Lemma (1.5), (6.11), (6.10), and (6.9)
give that for each
(L f)
7
(2Tr)
1c/2^
f =A
g
Tr
Thus
Lf = g. The fact that
Theorem (5.5).
on
f has support in
K follows from
48
7.
THE NULL SPACE OF
Let K be a fixed compact subset of Rn. We denote the
square integrable functions on Rn with support in
Lemma (7.1). For each k-space,
is continuous from L2(K)
Proof. Let f
function of
into
L2(K)
E
the transformation
E
inequality in
and we have
ILf II2?
L
it
denote the characteristic
and let x
K. Lemma ( 1 . ) shows that for each
L2(IT)
L2(K).
L2 (Tr .)
for almost every x"
TT
Tr,
K by
Tr,
f(x', x") L2(n)
E
Thus we may use the Cauchy Schwarz
=S IL fl 2dX"
L
1-11
Tr
X(xl,xll)f(x',xu)dx1I2 dx"
=
71-1-
IT
Oci,x")dx1)1
<
IT
/Z
IT
If(x',x")I 2dxf)1
(
/
2dx"
Tr
< C(k,K)IIf 1122
where
C(k, K) =
sup
E TT
x(x
x u)dx
it
Theorem(7.Z). The null space
is a closed subspace of
49
2
L (K) whose orthogonal complement consists of the functions in
2
that are constant in the direction
L (K)
or more precisel on_
Tr,
the intersection of k-planes in the direction
Proof. Lemma (7.1) shows that
with
ir
is Continuous
Lir
is closed. Without loss of generality assume
It = [e1,. ,
function h is constant on k-planes in the direction
if h
f
is a function of x" = (0...0, x
Xh
where
function of
X
,x n )
,
It
SO
.
,
ek]. A
It
if and only
alone. Suppose that
is the characteristic function of K and h is a
alone. If
x"
k+1
K.
then
gE
Tr
<g,f> =
S.
Rn
Tr
since the inner integral is L7g(x")
Suppose
hçgdxd
gxhd
h1
O,
TT
.
It is sufficient to find a function h of x"
f EV/ft
alone such that
L f = L Xh
TT
7T
Indeed if such an h
is found, then
L (f-xh) = 0,
SO
f
Xh
But by the above
cflL
7
Thus
f = xh,
xh
ILL
lv
So (f-xh) E IL.
L
50
To find
note that
h
Lf
L xh
Tr
Tr
if and only if
(7.3)
f(x x")dx =
X (x', x")h(x")dx1 = 6(x ")h(x")
Tr
where
Define
xh
E
6(x")
LX
II
h(x) by the above formula. We need only show that
L2 (K). Now
(7.4)
1 Xh 1dx
=
S,5
5(x )1h(x )12dx
SiTi.II
But since
xf = f,
XIII2dxidx!'
Tr
1T
II
.
7.3) and the Cauchy-Schwarz inequality give
(6( "))2 1 h(x") 1
2
Xf(xl x")dx? 1
=
Tr
It follows from (7.4) that
s1 xh 1 2dx
Rn
.
L (R n )
_11f1122
2
< 6(x") slf, 2dx1
Tr
.
51
and we are done.
We now describe an iterative scheme due to Kacmarz [2] that is
being used to detect brain tumors.
In the brain tumor work the
method is used in two dimensions to reconstruct cross sections. The
method is general and is presented here for functions in Rn and
k-planes.
Suppose that
and that the x-rays
is a compact subset of Rn,
K
L11
f,
i
f+
Tr
f
0
E
L2(K)
f E L2(K),
are known. Let P. be
1,
the projection on the closed plane
that
in
L2(K). Let
be an arbitrary initial guess and define
i mod(J).
fl = Pifl -1
The characterization of the null space in Theorem (7.2) shows that
L f (x") =
Tr.
and
f
If
Tr.
f(x"),
i
i mod(J),
- f1
1- is constant on planes with direction
P = P3.. P1,
then a theorem of Amemiya and Ando [1],
shows that
Pnf 0
where
P
Tr.
f
GM 0
strongly as n
00,
52
crn.
=
i=1
and
(f+cn r. )
Pam_ is the projection on cr)1..
Thus the iterative scheme of projections converges strongly
to the projection of the initial guess,
fo,
onto the plane cM.
53
BIBLIOGRAPHY
Amemiya, I. and Ando, T. Convergence of random products of
contractions in Hilbert space. Szeged (26) (1965), p. 239-244.
,
,
Durand, E., Solutions Numeriques des Equations Algebriques.
Tome II, p. 120, Masson, (1960), (Paris).
Gelfand, I. M. , Graev, M.I. , and Vilenkin, N., Generalized
Functions. Vol. 5, (1962), (Moscow).
Gordon, R. , and Herman, G. T, Three dimensional reconstruction from projections: a review of algorithms. International
Review of Cytology, ed. Bourne, G.H. and Danielli, J.R. , (1973),
(Academic Press, New York).
Guenther, R. and Smith, K. T. Reconstruction of objects from
x-rays and the location of brain tumors. (to appear)
,
,
b. Helgason, S Differential operators on homogeneous spaces.
Acta. Math. (1965), p. 239-299.
The Radon transform on Euclidean spaces,
compact two-point homogeneous spaces and Grossmann manifolds.
Acta Math. (1965), p. 153-180.
,
John, F. , Bestimmung einer Function aus ihren Integralen aber
gewisse Mannigfaligkeiten. Math. Ann. 100 (1934), p. 488-520.
Lax, P.D. and Phillips, 8. , The Paley-Wiener theorem for
the Radon transform. Comm. on Pure and Applied Math. , (1970),
,
Vol. XXIII, p. 409-424.
Ludwig, D., The Radon transform on Euclidean space. Comm.
on Pure and Applied Math., (1966), Vol. XIX, p. 49-81.
Nachbin, L., The Haar Integral, Van Nostrand, Princeton,
N. J. , (1965).
Radon, J., Uber die Bestimmung von Functionen durch ihre
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Sachs, Akad. Wiss. Leipzig, Math. -Nat. Kl. 69, (1917),
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Smith, KT and Solmon, D.C. , Lower dimensional integrability
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