Weighted Lp-Lq Mapping Properties of D-plan Transform on Euclidean Space, Real Hyperbolic Space and Sphere

7/31/2019 Weighted Lp-Lq Mapping Properties of D-plan Transform on Euclidean Space, Real Hyperbolic Space and Sphere

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Synopsis

Name of the Student : Ashisha Kumar

Roll Number : Y4108063

Degree for which submitted : Ph.D.

Department : Mathematics and Statistics

Thesis Title : Weighted Lp Lq Mapping Properties ofthe d-Plane Transform on Euclidean Space,Real Hyperbolic Space and

the Sphere

Thesis Supervisor : Swagato K. Ray

Month and year of submission : June, 2010

The problem of determining a function from the knowledge of its averages over

lower dimensional planes originated from the celebrated 1917 paper of J. Radon [R].

Given a nice function f on Rn, n 2 the Radon transform of f is defined by

Rf(, t) =

Rn1

f(t + y) dy Sn1, t R. (0.0.1)

One can analogously define the d-plane transform of a nice function as the integral

of the function over d-dimensional planes with respect to the d-dimensional Lebesgue

measure. Precisely, given 1 d n 1 and a d-dimensional linear subspace ofRn,

we define

Tdf(x, ) =

f(x y) dd(y), (0.0.2)

where d denotes the d-dimensional Lebesgue measure on .


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The problem of reconstructing f from Rf (or Tdf) has been studied by many

mathematicians and and it is still a very active area of research (see [M] and references

therein).

Our motivation in this thesis comes from a result of very different genre. In 1979

D. Solmon proved that if f Lp(Rn), 1 p < nd

, then the lower dimensional integral

involved in the definition of Tdf is well defined [S]. By a simple application of Fubinis

theorem one can see that the above observation is true for p = 1. But for p > 1 the

situation is more delicate. Roughly speaking, the above phenomena occurs because ofthe relation between the d-plane transform, the n-dimensional Fourier transform and

the Reisz potential. A natural question then is to ask: If f Lp(Rn), 1 p < nd ,

then what can one say about the size of Tdf? Of course the size should be measured

with respect to the natural measure on the set of d dimensional subspaces. The above

problem was completely settled for d = n 1 by D.M. Oberlin and E. M. Stein [OS].

Oberlin and Stein answered the question in the form of a mixed norm inequality for the

Radon transform. Their result is as follows: There exists a positive constant C such

that for all f Cc (Rn) the following inequality holds,Sn1

R

|Rf(, t)|qdtp/q

dn1()

1/p CfLp(Rn) , (0.0.3)

1 p < n/(n 1), 1/q = (n/p) n + 1, 1/p = 1 (1/p).

Here n1 denotes the normalized rotation invariant measure on the sphere.

We will follow the standard practice of using the letter C for a constant whose

value may change from one line to another. Here and everywhere else the constant C

which appears in the norm inequalities is independent of the functions.

Apart from the above result another intriguing observation related to the end-point


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estimate involving Lorentz norm was made in [OS] (we refer the reader to [SW3], p. 188

for relevant results regarding Lorentz spaces). They proved that the end-point estimate

of (0.0.3) holds for n 3, that is,Sn1

(suptR |Rf(, t)|)n dn1()1/n

CnfLn/(n1),1(Rn). (0.0.4)

But the above estimate fails for the case n = 2 (see the next paragraph).

A consequence of the above is that iff Ln/(n1),1(Rn) then its Radon transform

is well defined for almost every hyperplane. The failure of the end-point estimate for

n = 2 can be attributed to the existence of compact Kakeya sets in R2 of arbitrary

small Lebesgue measure ([OS], p. 642). Since the indicator function of a Kakeya set,

with radial symmetry, cannot have arbitrarily small L2,1 norm, one can still hope for

an end-point estimate for the radial functions in the case n = 2. Moreover, since the

behaviour of radial functions do not depend on the angular variable one would expect

the Radon transform of a radial function to be better behaved compared to general

functions. This viewpoint was adopted in [DNO] and the question regarding the end-

point estimate for the d-plane transform of radial functions was settled. It was proved

in [DNO] that if 1 d n 1 then there exists a constant C > 0 such that for all

radial functions f the following estimate holds,

sup(x,)RnGn,d|Tdf(x, )| CfLn/d,1(Rn). (0.0.5)

Here Gn,d stands for the set of d-dimensional linear subspaces ofRn.

There are two other instances in the literature where analogues of the above re-

sult appear for Radon transform on certain non Euclidean spaces. The first one is by

Cowling, Meda and Setti [CMS]. While working on Kunze-Stein phenomena on homo-

geneous trees they proved that the horospherical Radon transform of radial functions


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defines a continuous operator from L2,1 to L (see [CMS], Theorem 2.5). The second

result is by A. Ionescu [I]. Ionescu proved that for rank one Riemannian symmetric

spaces of non compact type the horospherical Radon transform of radial functions is

continuous from L2,1 to L (see [I], Proposition 2).

These results motivated us to consider the d-dimensional totally geodesic Radon

transform of radial functions on real hyperbolic space Hn and the sphere Sn. The

d-dimensional totally geodesic Radon transform of Lp functions on Hn has been dis-

cussed in several papers ([Str2], [BR1], [BR2], [Is]). A result in [BR1] says that: Iff Lp(Hn), 1 p < (n1)/(d1), then f is integrable over almost every d-dimensional

totally geodesic submanifold. In analogy with (0.0.5) it is only natural to enquire about

the validity of the end-point estimate for the d-dimensional totally geodesic Radon trans-

form of radial functions in this set up. We answer the question as follows (see Chapter

3, Theorem 4.3.7): The d-dimensional totally geodesic Radon transform restricted to

the class of radial functions defines a continuous linear map from L(n1)/(d1),1

(Hn

) to

L(R+) if n 3. We also show that the above result has no analogue for n = 2. As

a consequence it follows that if f is radial and f L(n1)/(d1),1, n 3, then f is

integrable over almost every d-dimensional totally geodesic submanifold. This is in the

same spirit as Rn, homogeneous trees and rank one symmetric spaces. However there are

some non Euclidean consequences of the above result [Corollary 4.3.9, Corollary 4.4.3].

One such is that if f L(n1)/(d1),1(Hn) and is radial then its Radon transform has anexponential decay at infinity. As a consequence it follows that the d-dimensional totally

geodesic Radon transform of radial functions is also continuous from L(n1)/(d1),1(Hn)

to Ln1,(R+). This is in sharp contrast with the Euclidean spaces.

We now consider the case of sphere. Here the situation is very different because of


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compactness. It is known from [Ru4] that in this case the d-dimensional totally geodesic

Radon transform is continuous from Lp(Sn) to Lp(SO(n + 1)/SO(n d) SO(d +1)),

1 p (see also [Str2]). Here the quotient space SO(n + 1)/SO(n d) SO(d + 1)

is viewed as the space of d-dimensional totally geodesic submanifolds of Sn. In this

case one can show that the exact analogue of (0.0.5), for functions which are invariant

under the action of SO(n), is not true (see Chapter 4, Example 4.4.1). It turns out

that one can prove a result analogous to (0.0.5) if the SO(n + 1) invariant measure on

SO(n + 1)/SO(n d) SO(d + 1) is considered along with a weight which is naturally

associated with the structure of the set of d-dimensional totally geodesic submanifolds

(Chapter 4, Theorem 4.4.2).

Coming back to Euclidean spaces we next consider a result of E. T. Quinto. In

[Q], Quinto proved the following weighted estimate for the Radon transform on Rn: For

all f Cc (Rn), n 3, there exists a positive constant C such that

Sn1R

|Rf(, t)|2 dt dn1() CRn

|f(x)|2xn1 dx. (0.0.6)

An obvious question related to this is to ask about the case n = 2. We prove that the

above inequality is equivalent to Pitts inequality for the Fourier transform and that

settles the case n = 2. The next question one asks is about the relevance of the power

(n 1) in the above inequality. In fact, we can ask the more general question: What

are all possible and such thatRSn1

|Rf(, t)|p|t| dt dn1() CRn

|f(x)|px dx. (0.0.7)

In Theorem 3.3.7, Chapter 3, we answer this question. Quintos proof of the L2 inequal-

ity (0.0.6) uses an analogue of Hecke-Bockner identity for Radon transform (which was

first proved in [L], Lemma 5.2). We found this technique useful in settling inequalities


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of the form (0.0.7) in general. The result of Quinto actually hints towards something

more general which we now explain. Since there is a relation between Radon transform

and Fourier transform by slice projection theorem ([M], Theorem 3.27) one would like

to know whether a weighted Fourier inequality (like Pitts inequality) would imply con-

tinuity of Radon transform on Lp spaces with power weight. In other words, we can

ask for an analogue of the result of [OS] on Lp spaces with power weight. We prove a

result in this direction in Theorem 3.3.10, Chapter 3. The results of chapter 3 has been

accepted for publication in Proc. Indian Acad. Sci. Math. Sci.

Our next aim is to consider the d-plane transform on weighted spaces. In this

context also we can ask for an inequality of the form (0.0.3) with weight. For functions

without any invariance, this seems to be a very difficult problem to tackle. So we

confine ourselves to the class of radial functions. Our main result in this direction is an

analogue of Theorem 1 of [DNO]. The Lp improving nature of the d-plane transform

plays a fundamental role here. It turns out that to have an Lp

Lq

boundedness of

the d-plane transform of radial functions on weighted spaces it is necessary to have

the condition p q. In Theorem 2.3.3 and Theorem 2.3.4 of Chapter 2, we prove the

best possible Lp Lq mapping property of the d-plane transform of radial functions on

Lp spaces with power weight. These results beg the question about how the d-plane

transform of radial functions behaves if p > q. This question is answered in Theorem

2.3.9. It turns out that in this case the operator in question can be continuous only

from Lp,s to Lq,s, 1 s , for certain values ofp and q which depend on the weights

we consider. In the case p > q the above result can be thought of as a substitute for an

Lp Lq inequality. Apart from the above results we also prove a mixed norm inequality

with power weight analogous to Theorem 1, (1.2) of [DNO]. The results of chapter 2

has been accepted for publication in Israel J. Math.


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The thesis is organized as follows:

In Chapter 1, we briefly introduce the problems considered in the thesis.

In Chapter 2, we consider d-plane transform of radial functions and prove analogues

of the inequality (0.0.5) on Lp spaces with power weights.

In Chapter 3, we concentrate on Radon transform. The main results here are

analogues of (0.0.3) and (0.0.6) on Lp spaces with power weight.

In Chapter 4, we consider the totally geodesic d-dimensional Radon transform on

real hyperbolic space and sphere. The main results here deal with certain end-point

estimates of totally geodesic d-dimensional Radon transform of radial functions.

In Chapter 5, the thesis concludes with possible directions for further work. The

relevant references are appended at the end.


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Contents

1 Introduction 1

2 Weighted Estimates of the d-Plane Transform for Radial Functions

on Euclidean Space 12

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3 Norm Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.4 Mixed Norm Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 Mixed Norm Estimate for Radon Transform on weighted Lp Spaces 52

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.3 Main Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4 Estimates of the d-Plane Transform of Radial Functions on Real Hy-

perbolic Space and the Sphere 74

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.2 Notation and Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.2.1 Real Hyperbolic Space . . . . . . . . . . . . . . . . . . . . . . . . 75


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CONTENTS ix

4.2.2 The Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4.3 The d-Plane Transform on Real Hyperbolic Space . . . . . . . . . . . . . 87

4.4 The d-Plane Transform on the Sphere . . . . . . . . . . . . . . . . . . . . 100

5 Concluding Remarks 108

List of Papers Accepted for Publication 110

Bibliography 111


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List of Symbols

Rn = {(x1, . . . , xn) : xi R, }, n 1 and R the field of real numbers

Cn = {(z1, . . . , z n) : zi C, }, n 1 and C the field of complex numbers

x, y = x1y1 + + xnyn, xi, yi R, i = 1, . . . , n

x2 = |x1|2 + + |xn|2, where xi R

Re z : the real part ofz C

N : set of all natural numbers

Sn : the unit sphere in Rn+1, see page 84

Gl(n,R) : space of all invertible n n matrices over R

det A : determinant of the matrix A

diag (a1, . . . , an) : An n n matrix with diagonal entries a1, . . . , an and rest zero

O(n) : {A GL(n,R

) : det A = 1}SO(n) : {A GL(n,R) : det A = 1}

E : indicator function of the set E

Cc(X) : compactly supported continuous function on X

Cc (X) : infinitely differentiable functions in Cc(X)

G

H : G is diffeomorphic to H as homogeneous spaces

Rf(, t) : Radon transform, see page 1

H,t : hyperplane in Rn, see page 1

n1 : normalized rotation invariant measure on Sn1, see page 2

df() : distribution function of f, see page 3


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CONTENTS xi

Lp,q(M) : Lorentz space, see page 3

f(p,q) : Lorentz space norm of f, see page 3

Lp(M) : see page 4

f g, : see page 10

fp,,n, : see page 10

f(p,q),,n : see page 10

Tdf : d-plane transform off, see page 12

d : d-dimensional Lebesgue measure, see page 12

Gn,d : Grassmann manifold, see page 12

: element of Gn,d, see page 12

n,d : normalized O(n) invariant measure on Gn,d, see page 12

x : projection of x on , see page 14

Adf : d-dimensional Abel transform of a radial function f, see page 15

F : see page 17

H,d : see page 17

I1/2 : see page 21

f : see page 22

h,1 : see page 22

Et : see page 26

d(x, y) : Euclidean distance between x, y Rn, see page 32

: see page 35

I : Riesz potential of order , see page 43


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CONTENTS xii

gr,(q,) : see page 53

f : Fourier transform of f, see page 54

Gn21

l : see page 55

R(g) : dual Radon transform, see page 56

[x, y] : see page 75

J : see page 75

Hn : n-dimensional hyperbolic space, see page 75

O(n, 1) : see page 75

SO(n, 1) : see page 76

O0(n, 1) : see page 76

SO0(n, 1) : see page 76

d1(x, y) : hyperbolic distance between x, y Hn,see page 78

Rnd : see page 78, 83

Rd+1 : see page 78, 83

d : see page 80

Rd f : totally geodesic d-plane transform off on Hn, see page 82

R+d f : totally geodesic d-plane transform off on Sn, see page 84

d2(x, y) : spherical distance between x, y Sn, see page 85

f : see page 85, 88

A+d f : totally geodesic d-plane transform of a radial function f on Hn, see page 87

Ad f : totally geodesic d-plane transform of a radial function f on Sn, see page 91


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Chapter 1

Introduction

The central topic of this thesis is the study of continuity of Radon transform and

some of its generalizations on Lp spaces. Given a compactly supported smooth function

f on Rn we have the Radon transform of f which is the integral of the function over

(n 1)-dimensional planes of Rn with respect to the (n 1)-dimensional Lebesguemeasure of the plane. To make the concept precise one first parameterizes the (n

1)-dimensional planes ofRn

. It is done as follows. Any (n 1)-dimensional plane ofRn

can be written as

H,t = {x Rn : x, = t}, (1.0.1)

where Sn1, t R and x, denotes the Euclidean inner product. It followsfrom above that H,t = H,t . Every element y Rn can be uniquely written asy = t + y, where y belongs to the orthogonal complement of the one dimensional

subspace spanned by . We define

Rf(, t) =

H,t

f =

Rn1

f(t + y) dy . (1.0.2)

Thus f Rf defines a linear map from Cc (Rn) to a space of functions defined on(Sn1 R)/Z2. Fundamental questions related to Radon transform revolve around the


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following themes:

a) Reconstruction of function from its Radon transform.

b) Characterization of the image of different function spaces under Radon transform.

c) Description of the kernel of Radon transform.

We refer the reader to [He1], [M], [He3] and the references therein for research related to

questions of the above kind. However, a program of a very different nature was initiated

by D. Solmon in [S]. It is this direction which we are going to pursue in this thesis.We first note from (1.0.2) that the definition of Radon transform involves integration

over lower dimensional subsets and hence one cannot immediately talk about Radon

transform of Lp functions. But if f L1(Rn) then a simple application of Fubinistheorem shows that

supSn1 R

|Rf(, t)| dt R Rn1

|f(t + y)| dy dt = fL1(Rn) . (1.0.3)

Hence using the product measure on Sn1 R we can conclude from above that theRadon transform of a function f L1(Rn) makes sense for almost every plane. In [S],Solmon proved that the same phenomena is true if f Lp(Rn), 1 p < n/(n 1),but instead of (1.0.3) one gets a different size estimate in the form of a mixed norm

inequality. Solmons result is the following: If 1 p < n/(n1) then for all f Cc (Rn)

Sn1

R

|Rf(, t)|qdt2/q dn1() Cf2Lp(Rn), (1.0.4)n

p=

1

q+ n 1,

where n1 denotes the normalized rotation invariant measure on Sn1.

We will follow the standard practice of using the letter C for a constant whose value


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may change from one line to another. Occasionally the constant C will be suffixed to

show its dependency on important parameters. Here and everywhere else in this thesis

the constant C which appears in the norm inequalities is independent of the functions.

It turns out that the inequality (1.0.4) is not best possible. The best possible

estimate was obtained by D. M. Oberlin and E. M. Stein [OS]. They proved the following

mixed norm inequality for the Radon transform:

Sn1 R

|Rf(, t)|qdt

p/q

d

1/p CfLp(Rn) , (1.0.5)

where

1 p < nn 1 ,

1

q=

n

p n + 1, 1

p= 1 1

p.

Before we embark upon further studies of the behaviour of Radon transform we

need the following definitions and results for the Lorentz spaces (see [SW3], Chapter 5

or [G], Chapter 1 for details). Let (M, m) be a -finite measure space, f : M

C be

a measurable function and p [1, ), q [1, ]. We define

f(p,q) =

q

p

0

[f(t)t1/p]q dtt

1/qwhen q <

supt>0 t[df(t)]1/p when q =

(1.0.6)

(see [G], p. 48-50). Here df is the distribution function of f and f is the nonincreasing

rearrangement of f ([G], p. 45). That is, for > 0 and t > 0,

df() = m{x | |f(x)| > } and f(t) = inf{s | df(s) t}.

We take Lp,q(M) to be the set of all measurable functions f : M C such thatf(p,q) < . By L,(M) and (,) we mean respectively the space L(M) and


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the norm . For p, q [1, ) the following identity gives an alternative expressionof (p,q) which we will use (see [RS]).

q

p

0

(t1/pf(t))qdt

t= q

0

t[df(t)]

1/pq dt

t.

For p, q in the above range, Lp,p(M) = Lp(M) and if q1 q2 then f(p,q2) f(p,q1)and consequently Lp,q1(M) Lp,q2(M) (see [G], p. 49). The Lorentz space norm (p,q) is only a quasi-norm and this makes the space Lp,q(M) a quasi Banach space(see [G], p. 50). However for 1 < p , there is an equivalent norm p,q which makesit a Banach space (see [SW3], Theorems 3.21, 3.22, p. 204). Precisely, if f Lp,q(M),then

f(p,q) fp,q p/(p 1) f(p,q).

We now come back to the result of Oberlin and Stein. In the context of (1.0.5)

a remarkable observation was made in [OS]. It was shown that the end-point estimate

for the above theorem is true if n 3 but it is false if n = 2. The following theoremexplains the situation.

Theorem 1.0.1. If n 3, then the mixed norm estimate

Sn1

sup

t|Rf(, t)|

ndn1()

1/n CnfL nn1 ,1(Rn) (1.0.7)

is true for all functions in Ln

n1,1(Rn). Estimate (1.0.7) fails for the case n = 2.

Around the same time while working on the problem of Null Space of Radon trans-

form, E. T. Quinto proved the following weighted L2 estimate for the Radon transform

on Rn, n 3 (see [Q]):

Sn1R

|Rf(, t)|2 dt dn1() CRn

|f(x)|2xn1 dx. (1.0.8)


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Two natural questions associated with this result are as follows.

i) Is the above inequality true for n = 2?

ii) What are all possible power weights for which (1.0.8) is true? More generally, what

is the best possible Lp analogue of (1.0.8)?

It turns out that the answer to the first question is in the affirmative and follows

by a simple application of Pitts inequality [B] as well as a weighted version of Hardy-

Littlewood-Sobolev lemma (see [SW2], [Str1], [W]).

The result of Quinto actually hints towards something more general which we now

explain. Since there is a relation between Radon transform and Fourier transform by

slice projection theorem ([M], Theorem 3.27) one would like to know whether a weighted

Fourier inequality (like Pitts inequality) would imply continuity of Radon transform

on Lp spaces with power weight. In other words, we can ask for a mixed norm estimate

for the Radon transform analogous to (1.0.5) on Lp spaces with power weight. To be

precise, we will look for an estimate of the form

Sn1

R

|Rf(, t)|q|t| dtr/q

dn1()

1/r C

Rn

|f(x)|p|x| dx1/p

. (1.0.9)

These questions have been dealt with in Chapter 3, Theorem 3.3.7 and Theorem 3.3.10.

Unlike (1.0.5) our results are not best possible and it seems to us that for a complete

understanding of the situation, very different techniques may be required.

Our next object of study is certain generalization of Radon transform, namely the

d-plane transform. For 1 d n 1, the d-plane transform of a nice function isdefined as the integral of the function over d-dimensional planes ofRn with respect to

the d-dimensional Lebesgue measure. For a d-dimensional subspace ofRn and x Rn


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the d-plane transform Td of functions f defined on Rn is thus defined as

Tdf(x, ) = f(x y) dd(y) (1.0.10)where d denotes the d-dimensional Lebesgue measure on . If x denotes the pro-

jection ofx on then it is easy to see that Tdf(x, ) = Tdf(x, ). For d = n 1 itcan be easily seen that the operator Td and the Radon transform are related as follows,

Rf(, t) = Tn1(t, {}). (1.0.11)

Here {} denotes the orthogonal complement of the subspace spanR{}.

Regarding the absolute convergence of the lower dimensional integral involved in

the definition of d-plane transform, it was proved in [S] that if 1 p < nd

then d-plane

transform of an Lp function exists for almost every d-dimensional plane. A more general

version of the above result appears in [M] (see also [Ru1]).

Theorem 1.0.2. A nonnegative measurable function f defined onRn has the property

that Tdf exists almost everywhere and is locally integrable on the set of all d-planes if

and only if (1 + x)dnf(x) L1(Rn).

Research has been carried out to achieve an analogue of Theorem 1.0.5 for d-plane

transform ([C], [D1], [D2], [DO], [Gr], [Wo]) and several deep results have been obtained.

For instance, an end-point estimate analogous to (1.0.7) for Td, d > n2 , can be found in

[D1]. But the exact analogue of Theorem 1.0.1 is still not available. The failure of the

end-point estimate for n = 2 can be attributed to the existence of compact Kakeya sets

in R2 of arbitrary small Lebesgue measure ([OS], p. 642). Since the indicator function

of a Kakeya set, with radial symmetry, cannot have arbitrarily small L2,1 norm, one can

still hope for an end-point estimate for the radial functions in the case n = 2. Since the


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behaviour of radial functions does not depend on the angular variable one would expect

the Radon transform of a radial function to be better behaved compared to general

functions. This viewpoint was adopted in [DNO] and the question regarding the end-

point estimate for d-plane transform of radial functions was settled. For 1 d n 1,the following end-point estimate was proved in [DNO],

supx,

|Tdf(x, )| C fLnd ,1(Rn). (1.0.12)

Using the trivial L1L1 boundedness one can now prove the following Lp Lq mappingproperty of Td restricted to the class of radial functions:

Gn,d

|Tdf(x, )|q dnd(x)r/q

dn,d()

1/r C

Rn

|f(x)|p dx1/p

,

(1.0.13)

1 r , 1 p < nd

,n

p n d

q= d.

For definition of Gn,d we refer the reader to Chapter 2, page 12. We note that

(1.0.12) also implies the existence ofd-plane transform of radial functions f Lnd ,1(Rn).But it can be shown easily from Theorem 1.0.2 that the above is actually true for

all functions in Lnd

,1(Rn). We first show that the function (1 + x)dn belongs toL

nnd

,(Rn), which is the dual of Lnd

,1(Rn) (see [G], p. 52). In fact, it suffices to show

that the function = dn{xRn:x>1} Ln

nd,(Rn). A simple calculation shows

that the n-dimensional Lebesgue measure of the set

{x

Rn : (x) >

}is C

nnd if

0 < 1 and zero otherwise. Hence sup>0

df()ndn < . It now follows from (1.0.6)

that L nnd ,(Rn). An application of Holders inequality for Lorentz spaces (see [G],p. 72) then implies that f(1 + )dn L1(Rn) if f Lnd ,1(Rn).

Regarding the d-plane transform our goal is to prove analogues of (1.0.12) and

(1.0.13) for power weights. The Lp improving nature of the d-plane transform plays a


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8

fundamental role here. It turns out that to have an Lp Lq boundedness of the d-planetransform of radial functions on weighted spaces it is necessary to have the condition

p q. In Theorem 2.3.3 and Theorem 2.3.4 of chapter 2 we prove the best possibleLp Lq mapping property of the d-plane transform of radial functions on Lp spaceswith power weight. These results beg the question about how the d-plane transform

of radial functions behaves if p > q. This question is answered in Theorem 2.3.9. It

turns out that in this case the operator in question can be continuous only from Lp,s

to Lq,s, 1 s , for certain values of p and q which depend on the weights weconsider. In the case p > q the above result can be thought of as a substitute for an

Lp Lq inequality. In Chapter 2, we also prove a mixed norm inequality with powerweights, analogous to Theorem 1, (1.2) of [DNO]. Apart from the mapping property

these inequalities also prove the existence of Radon transform of radial functions on

certain weighted Lorenz spaces. This information cannot be derived from Theorem

1.0.2 as local integrability of Radon transform may fail because of weights.

There are two other instances in the literature where analogue of (1.0.12) appearfor Radon transform on certain non Euclidean spaces. The first one is by Cowling, Meda

and Setti [CMS]. While working on Kunze-Stein phenomena on homogeneous tree they

proved that the horospherical Radon transform of radial functions defines a continuous

operator from L2,1 to L. The second result is by A. Ionescu [I]. Ionescu proved that for

rank one, Riemannian symmetric spaces of non compact type the horospherical Radon

transform of radial functions is continuous from L2,1 to L.

These results motivated us to consider the d-dimensional totally geodesic Radon

transform of radial functions on real hyperbolic spaces Hn and on the sphere Sn. The

d-dimensional totally geodesic Radon transform of Lp functions on Hn has been dis-

cussed in several papers ([Str2], [BR1], [BR2], [Is]). However, we are not aware of any pa-

per (except [Str2]) which deals with the Lp Lq mapping property of the d-dimensional


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9

totally geodesic Radon transform on real hyperbolic spaces. A result in [BR1] says

that: If f Lp(Hn), 1 p < (n 1)/(d 1), then f is integrable over almost every

d-dimensional totally geodesic submanifold. In analogy with (1.0.12) it is only natural

to enquire about the validity of the end-point estimate for the d-dimensional totally

geodesic Radon transform of radial functions in this set up. We answer the question as

follows (Chapter 4, Theorem 4.3.7): The d-dimensional totally geodesic Radon trans-

form restricted to the class of radial functions defines a continuous linear map from

Ln1d1

,1(Hn) to L(R+) ifn 3. We also show that the above result has no analogue forn = 2. As a consequence it follows that if f is radial and f

L

n1d1

,1, n

3, then f is

integrable over almost every d-dimensional totally geodesic submanifold. This is in the

same spirit as Rn, homogeneous trees and rank one symmetric spaces. However there

are some non Euclidean consequences of the above result [Corollary 4.3.9, Corollary

4.4.3]. One such is that if f Ln1d1 ,1(Hn) and is radial then its Radon transform hasan exponential decay at infinity. As a consequence it follows that the d-dimensional

totally geodesic Radon transform of radial functions is continuous from Ln1d1

,1(Hn) to

Ln1,(R+). This is in sharp contrast with the Euclidean spaces.

We now consider the case of sphere. Here the situation is very different because of

compactness. It is known from [Ru4] that in this case the d-dimensional totally geodesic

Radon transform is continuous from Lp(Sn) to Lp(SO(n+1)/{SO(nd)SO(d+1)}),1 p (see also [Str2]). Here the quotient space SO(n+1)/{SO(nd)SO(d+1)}is viewed as the space ofd-dimensional totally geodesic submanifolds ofSn. In this case

one can show that the exact analogue of (1.0.12), for functions which are invariant

under the action of SO(n), is not true (see Chapter 4, Example 4.4.1). It turns out

that one can prove a result analogous to (1.0.12) if the SO(n + 1) invariant measure on

SO(n+1)/{SO(nd)SO(d+1)} is considered along with a weight which is naturallyassociated with the structure of the set of d-dimensional totally geodesic submanifolds

(Chapter 4, Theorem 4.4.2).


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10

We now introduce some notation which will be used throughout the thesis. If f is

a complex valued measurable function on Rn and R then

fp,,n =

Rn

|f(x)|px dx1/p

and

Lp(Rn) = {f : Rn C : f is measurable and fp,,n < },

for 1 p < . Since the function x is a nonnegative function we have that

L

(Rn

) = L

(Rn

). When = 0 then L

p

(Rn

) = L

p

(Rn

) and fp,0,n is the usualLp norm. By Lp,q (R

n) we will denote the Lorentz space with respect to the measure

d(x) = xdx. That is,

f(p,q),,n =

q0

t[df(t)]

1p

qdtt

1/q, q < ,

supt>0 t[df(t)]1p , q =

,

(1.0.14)

where df(t) = {x Rn : |f(x)| > t}. Then Lp,q (Rn) is defined as the set of allmeasurable functions f : Rn C such that f(p,q),,n < .

For nonnegative functions f and g the symbol f(x) g(x) will mean that thereexist positive constants C1 and C2 such that C1g(x) f(x) C2g(x) in the appropriatedomain of the functions f and g.

Finally, given p [1, ), p will denote the index conjugate to p, that is, 1/p +1/p = 1.

The thesis is organized as follows:

In Chapter 2, we consider the d-plane transform of radial functions and prove


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11

analogues of Theorem 1.0.12 on Lp spaces with power weights.

In Chapter 3, we concentrate on Radon transform. The main results here are

analogues of Theorem 1.0.5 and Theorem 1.0.8 on Lp spaces with power weight.

In Chapter 4, we consider the totally geodesic d-dimensional Radon transform on

real hyperbolic space and on the sphere. The main results here deal with certain end-

point estimates of totally geodesic d-dimensional Radon transform of radial functions.

In Chapter 5, the thesis concludes with possible directions for further work. The

relevant references are appended at the end.


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Chapter 2

Weighted Estimates of the d-PlaneTransform for Radial Functions on

Euclidean Space

2.1 Introduction

We start with the notion of the d-plane transform on Rn

. For d 1 and n 2 let Gn,ddenotes the Grassmannian manifold ofd-dimensional linear subspaces in Rn. Since Gn,d

can be viewed as the compact homogeneous space O(n)/{O(d) O(n d)} (see [M],p. 141) there exists an O(n) invariant nonnegative Borel measure on Gn,d which is

unique up to a constant. Let n,d denotes the unique O(n) invariant, nonnegative Borel

measure on Gn,d such that n,d(Gn,d) = 1. Iff is a complex valued measurable function

on Rn and

Gn,d then for every x

Rn, the d-plane transform Tdf off is defined as

Tdf(x, ) =

f(x y) dd(y) (2.1.1)

where d denotes the d-dimensional Lebesgue measure on .

The following theorem, proved in [DNO], is our main concern in this chapter.


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2.1 Introduction 13

Theorem 2.1.1. For radial functions the inequality

Gn,d

|Tdf(x, )|q dnd(x)r/q dn,d()1/r

CRn

|f(x)|p dx1/p(2.1.2)

holds if and only if

1 r , 1 p < nd

,n

p n d

q= d,

and the inequality

Rn

Gn,d

|Tdf(x, )|r dn,d()q/r

dx

1/q CRn

|f(x)|p dx1/p

(2.1.3)

holds if and only if

1 < p n

p d. (2.1.4)

Here nd denotes the (n d)-dimensional Lebesgue measure on .

The proofs of the above inequalities were achieved by proving all the required

end-point estimates. This seems to be the main advantage of dealing with the radial

functions. Our aim in this chapter is to prove analogues of Theorem 2.1.1 on Lp spaces

with power weight. We will prove inequalities of the form

Gn,d

|Tdf(x, )|qxdnd(x)r/q

dn,d()

1/r Cfp,,n,(2.1.5)

Rn

Gn,d


x dx1/q Cfp,,n,(2.1.6)

for appropriate values of ,,p,q and r. These inequalities trivially specialize to


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2.2 Preliminaries 14

Theorem 2.1.1 for = = 0. The results we obtain regarding (2.1.5) are best possible,

in fact, we prove all the required end-point estimates also (see Theorem 2.3.7 and

Theorem 2.3.9). But the results regarding (2.1.6) are far from being complete. We

could not prove the required end-point estimates neither we could prove the expected

result for all possible admissible weights. One of the reasons for this is our lack of

understanding of the end-point estimate of the Riesz potential for radial functions on

weighted Lp spaces (see Theorem 2.4.4 and Theorem 2.4.5). However, our results for

the cases d = 1 and d = n 1 are significantly better compared to the other values ofd (see Theorem 2.4.2).

This chapter is organized as follows. In the next section we explain the notation

and gather relevant results on d-plane transform. In section 3 and section 4 we will

prove our main results.

2.2 Preliminaries

Most of our notation related to d-plane transform is standard and can be found in [M].

Here we shall recall a few of them. For x Rn by x (respectively x), we denotethe projection of x on (respectively on ) and we have x = x + x. Using the

translation invariance of Lebesgue measure on Rd it now follows from (2.1.1) that

Tdf(x, ) =

f(x y) dd(y). (2.2.1)

We now specialize to the case of radial functions. Iff is a radial function defined on

Rn then we will sometime consider f as a function defined on (0, ). For 1 d n 1we define the d-dimensional generalization of the Abel transform (see [He1]) of radial


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2.3 Norm Inequality 15

functions defined on Rn as,

Adf(x) = Rd fx2 + y2 dd(y) = cd

x f(s)(s2

x2

)

d2

2 s ds (2.2.2)

where x Rnd and x, y stand for norms on Rnd and Rd respectively. Thus Adis a map from a space of radial functions on Rn to a space of radial functions on Rnd.

It thus follows from (2.1.1) and (2.2.2) that for nice radial functions f on Rn one has

Tdf(x, ) = Tdf(x, ) = Adf(x) = cd

x

f(s)(s2 x2)d22 s ds. (2.2.3)

Hence for each fixed Gn,d and nonnegative, measurable and radial functions wehave

|Tdf(y, )|ry dnd(y) =Rnd

|Adf(x)|rx dnd(x), (2.2.4)

for 1 r (for r = , (2.2.4) has to be interpreted accordingly).

2.3 Norm Inequality for the d-Plane Transform of

Radial Functions

In this section we will prove an inequality of the form

supGn,d

|Tdf(x, )|qx dnd(x)1/q

Cfp,,n, (2.3.1)

for appropriate values ofp,q, and . Since Gn,d is a finite measure space, an inequality

of the form (2.1.5) then will follow from (2.3.1). To prove (2.3.1), in view of (2.2.4), it


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is enough to prove an inequality of the form

Adfq,,nd Cfp,,n. (2.3.2)

To prove the above inequality we first search for a necessary relation on p,q,,using

the dilates of a radial function. For > 0 we define f(x) = f(x), x Rn. Using(2.2.2) it is easy to see that Adf(y) =

dAdf(y), y Rnd. So

fp,,n = +np fp,,n, and Adfq,,nd =

+ndq

dAdfq,,nd. (2.3.3)

Thus for the validity of (2.3.2) we must have the following relation between ,,p,q,d,

+ n

p= d +

+ n dq

. (2.3.4)

The advantage of dealing with the class of radial functions is the fact that inequal-

ities of the form (2.3.2) can be reduced to the study of Lp Lq mapping propertiesof certain convolution operators on the multiplicative group (0, ) equipped with theHaar measure dt/t. We believe that this fact is known to the experts but since we could

not locate it in the literature we elaborate on it for the sake of completeness. By using

the relation (2.3.4) and the definition (2.2.2) we write

Adf

q,,nd

=

0

t

f(r)(r2 t2) d22 r drq t+nd1 dt1/q=

0

t

f(r)r+np

1 t

2

r2

d22

(t/r)+np

d dr

r

q

dt

t

1/q

=

0

0

f(r)r+np

1 t

2

r2

d22

(t/r)+np

d (0,1) (t/r)dr

r

q

dt

t

1/q= F H,dLq((0,),dt/t), (2.3.5)


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where

F(t) = f(t)t

+n

p , (2.3.6)

H,d(t) = t+np

d(1 t2)d22 (0,1)(t) (2.3.7)

and the convolution is on the multiplicative group (0, ). Since

FLp((0,),dt/t) = fp,,n,

to prove (2.3.2) it suffices to prove inequalities of the form

0

|F H,d(t)|q dtt

1/q Cp,q

0

|F(t)|p dtt

1/p, (2.3.8)

for appropriate values of, , p and q. Based on (2.3.8) we now make two observations.

First, for (2.3.2) to hold it is necessary that p q as convolution operators are Lp

improving (see [H]). But (2.3.4) then implies that ( + n)/p

d + (+ n

d)/p, that

is,

p 1 + d

. (2.3.9)

This is a nontrivial condition on p only in the case > . Our second observation is

that the nature of the kernel H,d is more involved for d = 1 compared to the case d 2because of the appearance of the factor (d 2)/2 in the expression of H,d. As a resultwe will see some differences in the mapping properties of the operator Td for d = 1 and

d 2.

Next we need to determine necessary conditions on and under which (2.3.2) can

hold for p, q greater than or equal to one. By modifying a standard example available

in [OS] we will show below that for p > ( + n)/d and p = ( + n)/d > 1 the integral

in (2.2.2) is infinite for large values of x. Hence a necessary condition for (2.3.2) to


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be true is

1 p < + nd

or 1 = p = + n

d. (2.3.10)

Example 2.3.1. Given > 0 we define

f(x) =1

xd(log x) {x:x>e}(x), x Rn.

We will first show that f Lp(Rn, xdx) ifp > ( + n)/d and f L+nd (Rn, xdx)if > d/( + n). Using the change of variable u = log x it follows that

fpp,,n =1

eu(+ndp)up du.

The above integral is certainly finite if p > ( + n)/d. If p = ( + n)/d then we need

the condition p > 1 for finiteness of the above integral. Now we will show that Adf

takes the value infinity on a set of positive measure. Since Adf is a radial function in

the following we are going to treat it as a function on (0, ). Now suppose that d = 1.

By writing s = ez and x = ey it follows from (2.2.2) that for y > 1

A1f(ey) =

y

(1 e2(yz))12z dz

y

z dz,

which is infinite if 1. For d 2 and t = x > e, we use the description of thefunction f and (2.2.2) to get that

Adf(t) 2t

1(log s)

1 t2s2 d22 ds

s (3/4)d22

2t

1s(log s)

ds,

which is infinite if 1. So, if p > ( + n)/d or p = ( + n)/d > 1 we can alwayschoose 0 < < 1 to construct a radial function f Lp(Rn) whose d-plane transformdoes not exist.


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Next we will give an example to show that for any p in (0, 1) and for any in R

there exist a radial function g in Lp(Rn, xdx) such that Tdg(x, ) is infinite on a set

of positive measure.

Example 2.3.2. Let g(x) = (log x)1(1, e)(x) then g Lp(Rn, xdx) for allp (0, 1) but for all x such that x < 1/2

Tdg(x, ) = C

x

(log s)1(1, e)(s)(s2 x2)

d22 s ds

= Ce

1

(log s)1(s2 x2)d22 s ds = .

This proves the necessity of (2.3.10).

From (2.3.4) and (2.3.10) it now follow that the admissible values of and are

given by

d n, d n. (2.3.11)

In the following we will prove a precise version of (2.3.1). It turns out that the case

d = 1 differs considerably from the cases d 2. For the readers benefit we put them asseparate theorems.

Theorem 2.3.3. Letp,q,d satisfy the relations d 2 and (2.3.4).

a) If > > d n and1 +

d p < + n

d

then there exist a positive constant C such that for all radial functions f Cc(Rn)

supGn,d


Cp,qfp,,n. (2.3.12)


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b) If > d n, and1 p < + n

d

then (2.3.12) holds for all radial function f Cc (Rn).

c) If = d n, then (2.3.12) holds only for p = 1 and q = .

Proof. We first note that as d 2 and (+n)/dp > 0, we have H,d Ls((0, ),dt/t)for s [1, ]. Hence a) and b) follow simply by applying Youngs inequality in (2.3.5).

Next we will prove c). For = d n we have from (2.3.4) that

dp

=+ n d

q.

If > = d n then according to (2.3.4) the only possible choice for p, q is p =1, q = . Similarly, if = = d n then the possible choices are p = 1, 1 q

. In any case, we have that H(dn),d is a bounded function and hence the required

inequality in c) follows again by Youngs inequality. Now we show that if = =

d n then (2.3.12) does not hold for p = 1 and 1 q < . We consider the radialfunction f(x) = {xRn:1xe}(x)xd which is in L1(Rn, xdndx). Viewing this asa function on (0, ) we get Fdn(t) = (1,e)(t) and Hdn,n(t) = (1 t2) d22 (0,1)(t) (see(2.3.7)). To prove the claim, in view of (2.3.5), it suffices to show that Fdn Hdn,n /Lq((0, ),dt/t). It follows that for t < 1,

Fdn Hdn,n(t) =t

t/e

(1 r2) d22 drr

. (2.3.13)

Since (1 t2)(d2)/2 (1 r2)(d2)/2 (1 t2e2 )(d2)/2 it follows that

limt0

(Fdn Hdn,n)(t) = limt0

tt/e

(1 r2) d22 drr

= 1.


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This proves the claim.

In the next theorem we present the case d = 1. Using the change of variable s2

= u

and x2 = v it follows from (2.2.2) that T1f(x, ) = A1f(x) = I1/2 g(v) whereg(v) = f(

v) and

I1/2 g(v) =

v

g(u)(u v)1/2 du,

(see [Ru3], p.23). The Lp Lq mapping property of the operator I12 on weighted spaces

is available in [Ru3]. Though the following theorem is essentially contained in [Ru3],

Chapter 2, we include the proof for the sake of completeness.

Theorem 2.3.4. (1) Let d = 1 and p, q satisfy (2.3.4) together with

1

2+

1

q>

1

p. (2.3.14)

a) If > > 1 n and 1 + ( ) p < ( + n) then there exists a constant

C > 0 such that for all radial functions f Cc(Rn

)

supGn,1

|T1f(x, )|qx dn1(x)1/q

Cp,qfp,,n. (2.3.15)

b) If > 1 n, and 1 p < + n then (2.3.15) holds for all radialfunctions f Cc (Rn).

(2) If d = 1 and p, q satisfy (2.3.4) together with

1

2+

1

q=

1

p(2.3.16)

then a) and b) are valid for p > 1.

Proof. As in Theorem 2.3.3 we first observe that H,1 Ls((0, ),dt/t) for s < 2. An


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application of Youngs inequality then finishes the proof of (1) as 1/2 + 1/q > 1/p.

To prove (2) we realize the functions F and H,1 on R instead of the multiplicative

group (0, ). Let f(x) = F(ex) and

h,1(x) = H,1(ex) = (1 e2x) 12e+npp x(,0)(x),

for x R. We note that fp,,n = fLp(R). We write h,1(x) = h1(x) + h2(x) whereh1(x) = h,1(x)(1,0)(x) and h2(x) = h,1(x)(,1)(x). Since h1(x) (x)1/2 it

follows thatf h,1Lq(R) C(I1/2fLq(R) + f h2Lq(R))

where I1/2 stands for the one dimensional Riesz potential of order half see (2.4.31). By

Hardy-Littlewood-Sobolev lemma we have

I1/2fLq(R) CfLp(R).

On the other hand, since h2 Ls(R), s [1, ] (as ( + n) > p) it follows from Youngsinequality that

f h2Lq(R)) CfLp(R).

Remark 2.3.5. The following remarks are in order.

i) We first note that the condition (2.3.14) in Theorem 2.3.4 is necessary. To see this

we consider (x) = (x) where is a nonnegative measurable function defined

on R and > 0. It follows by a standard computation that for > 1,

h1Lq(R) 12 1q h1Lq(R).


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We also observe that if the operator F F h,1 is continuous from Lp to Lq

then so is the operator F Fh1. Hence for all 1 we will have the inequality

12 1q h1Lq(R) C

1pLp(R).

Thus it is necessary for p, q to satisfy the relation

1

2+

1

q 1

p. (2.3.17)

ii) In Theorem 2.3.4 we did not talk about the case = 1 n. It can be shownexactly as in Theorem 2.3.3 that one cannot have any Lp Lq boundedness inthis case if p = 1 and 1 q < . However the case p = 1, q = differs fromTheorem 2.3.4. We will show by an example that A1f may not exist for some

f L1(Rn, x1ndx). For x Rn let f(x) = [1,2](x)(x 1)1/2 so thatf1,1n,n < . From 2.2.2 we have

A1f(t) =0

(r 1)1/2(r2 t2)1/2[1,2](t,)(r) r dr.

It is clear that the above integral diverges if t 1.

iii) If = = 0 then (2.3.4) implies that p q. But this is not true in general, forexample, = n = p = 2, = 0 and d = q = 1 satisfies (2.3.4).

Next we will prove the end-point results for Theorem 2.3.3 and Theorem 2.3.4. We

need the following elementary observation to start with.

Lemma 2.3.6. Letn N and 1. If x1 x2 xn 0 are real numbers thenthe following inequality holds,


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n

i=1(1)i1xi

n

i=1(1)i1(xi) (2.3.18)

Proof. It is enough to prove (2.3.18) for > 1 and n = 2m + 1, m 1 because if nis even, we can add one more term which is zero. We will prove it by induction on m.

For a fixed positive real number a we consider the function f(x) = (x + a) x wherex R and > 1. Since the function is clearly increasing it follows that for x < y we

have (x + a) x (y + a) y.

If we choose x = x3, y = x2 and a = x1 x2 it then follows that

(x1 x2 + x3) x1 x2 + x3 , (2.3.19)

which corresponds to the case m = 1. Let us now assume that the result is true

for m, that is,

2m+1

i=1

(1)i1xi

2m+1

i=1

(1)i1(xi). (2.3.20)

We now need to prove the result for m+1. As the numbers {xi}2m+2i=1 are decreasingit follows easily that

2m+1i=1

(1)i1xi x2m+2 x2m+3. (2.3.21)


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Thus using (2.3.21) it follows from (2.3.19) and (2.3.20) that

2m+1i=1

(1)i1

xi x2m+2 + x2m+3

2m+1i=1

(1)i1xi

x2m+2 + x2m+3

2m+1

i=1

(1)i1(xi) x2m+2 + x2m+3.

This completes the proof.

Theorem 2.3.7. If d 2 and , satisfy (2.3.11) then there exists a positive constantC such that for all radial functions f the following estimate holds

supGn,d


Cf(+nd ,1),,n, (2.3.22)

if and only if q = . For d = 1 the above statement holds if + n 2.

Proof. In view of (2.2.3) it is enough for us to prove that for all nonnegative radial

functions f

Adfq,,nd Cf(+nd ,1),,n, (2.3.23)

if and only if q = .

We note that if > d n and p = ( + n)/d then only q = satisfies (2.3.4).

But if = d n then q [1, ] satisfies (2.3.4). We will first show that for q = the result is true. Since L(Rnd, xdx) = L(Rnd, dx), it is enough for us to provethat

Adf Cf(+nd ,1),,n. (2.3.24)

It is enough to prove the theorem for functions of the form li=1Ei where Ei = {x


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Rn : ai x < bi}, bi < ai+1, a1 0 and i = 1, . . . , l 1 (see [SW3], p. 195, [DNO],p. 269 and Lemma 4.3.5, Chapter 4). An explicit calculation using (2.2.2) shows that

Ad(Ei)(t) = C

(b2i t2)d/2 (a2i t2)d/2, t < ai

(b2i t2)d/2, ai t bi

0, bi < t

(2.3.25)

where t = x, x Rnd. If we denote E = li=1Ei then by using linearity of Adand (2.3.25) we get that

Ad(E)(t) = C

li=1

(b2i t2)d/2 (a2i t2)d/2

, if t < a1.

l

i=j+1

(b2i t2)d/2 (a2i t2)d/2 + (b2j t2)d/2,if aj t < bj, 1 j l.

li=j+1

(b2i t2)d/2 (a2i t2)d/2

,

if bj t < aj+1, 1 j l 1.

0, bl < t.

(2.3.26)

For a fixed t > 0, let Et denotes the set {x E : x > t} so that Et E andhence the corresponding distribution functions satisfy the relation dEt (s) dE(s) forall s > 0. Thus, to prove the theorem it suffices to prove that for all t > 0

Ad(E)(t) CEt(+nd ,1),,n, (2.3.27)


39/127


where C is independent of t. The distribution function dEt (s) with respect to the

measure xdx is constant on the set (0, 1) and is zero otherwise. From (1.0.14) we

see that

Et(+nd ,1),,n = C10

dEt (s)

d+nds, (2.3.28)

where C is independent of t. Using polar coordinates on Rn it now follows that

Et(+nd ,1),,n = C

l

i=1

[bi+n ai+n]

d+n

, if t < a1.

l

i=j+1

[bi+n

ai+n

] + bj +n t+nd

+n

,

if aj < t < bj, 1 j l.l

i=j+1

[bi+n ai+n]

d+n

,

if bj < t < aj+1, 1 j l 1.

0, if bl < t.

(2.3.29)

Now we will consider t (aj, bj ) for a fixed j and prove (2.3.27). Other cases can bedealt with exactly the same way. As bdl > a

dl > > adj+1 > bdj > td and ( + n)/d 1

we can use Lemma 2.3.6 to get that

Et(+nd ,1),,n = C

li=j+1

[(bdi )+nd (adi )

+nd ] + (bdj )

+nd (td)+nd

d+n

Cl

i=j+1

(bid aid) + bj d td. (2.3.30)

For a > 0, 1 and x y 0 we have that (x + a) x (y + a) y (seeLemma 2.3.6). Thus for each fixed i and d 2 we can choose = d/2, a = b2i a2i ,


40/127


x = a2i , y = a2i t2 to get the inequality

bd

i ad

i (b2

i t2

)d/2

(a2

i t2

)d/2

.

Now by summing over i, j + 1 i l, (2.3.27) follows from (2.3.30) and (2.3.26)(note that bdj td (b2j t2)d/2 again by Lemma 2.3.6). This takes care of the cased 2. For d = 1 we choose = ( + n)/2 (which is greater than 1 by hypothesis) andx, y, a exactly as before to get that

li=j+1

[bi+n ai+n] + bj +n t+n

1+n

(2.3.31)

li=j+1

(b2i t2)

+n2 (a2i t2)

+n2

+ (b2j t2)

+n2

1+n

,

for all t (aj , bj). As

(b2

l t2)

+n2 > (a2

l t2)

+n2 >

> (b2

j t2)

+n2

we can apply Lemma 2.3.6 for = + n in (2.3.31) to get (2.3.27). This completes the

proof for the case q = .

To see that (2.3.23) does not hold for q < we consider the function {xRn:a

for all t [0, a]. This simply means that the xdndx measure of the set {x Rnd :AdE(x) > s} is infinite for 0 < s < . Hence AdE / Lq,(Rnd, xdndx) for1 q < but clearly E L+nd ,1(Rn, xdx).

Remark 2.3.8. The occurrence of the condition + n 2 in the last theorem canbe explained as follows. If p = + n, q = and d = 1 then (2.3.17) and (2.3.4)


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together imply that + n 2. In the case = 0 (see Theorem 2.1.1) the condition isautomatically satisfied as n 2.

We have seen that for LpLq boundedness of the operator Ad the condition p qisnecessary simply because the problem is equivalent to the boundedness of a convolution

operator. However, the same cannot be said about the mapping property of the operator

Ad on Lorentz spaces. In fact, our next result shows that the condition p q is notnecessary for boundedness of Ad on Lorentz spaces. Recall that the conditions p qand (2.3.4) forced us to restrict the range ofp in the interval [1 + (

)/d, ( + n)/d)

in Theorem 2.3.3 and Theorem 2.3.4. So it is natural to ask about the mapping property

of Ad when > > d n and 1 p < 1 + ( )/d. In this case it is easy to see ,invoking (2.3.4), that the range of p is given by ( + n)/(+ n) p < 1 + ( )/dand if p = ( + n)/(+ n) then q = 1. We now prove the following result.

Theorem 2.3.9. If d 1 and > > d n then for all radial functions the followinginequality holds,

supGn,d

|Tdf(x, )| x dnd(x) Cf(+n+n ,1),,n. (2.3.32)

Consequently

supGn,d

Tdf(, )Lq,s(Rnd,xdx) Cf(p,s),,n, (2.3.33)

where 1 s ,

+ n

+ n p < 1 +

dand

+ n

p=

+ n dq

+ d.


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Proof. As before, using (2.2.3), it suffices to prove

Adf1,,nd Cf(+n+n ,1),,n. (2.3.34)

We begin by considering the set E as in Theorem 2.3.7. Note that the expression

of Ad(E)(t) is given by (2.3.26). Using this we get

Ad(E)1,,nd

= Ca10 l

i=1(b2i t2) d2 (a2i t2) d2 t+nd1dt

+l

j=1

bjaj

l

i=j+1

(b2i t2)

d2 (a2i t2)

d2

+ (b2j t2)

d2

t+nd1dt

+l1

j=1

aj+1bj

l

i=j+1

(b2i t2)

d2 (a2i t2)

d2

t+nd1dt

= Cl

i=1

bi0

(b2i t2)d2 t+nd1dt

ai0

(a2i t2)d2 t+nd1dt

. (2.3.35)

Using the change of variables t = bi sin (respectively t = ai sin ,) we get that the

above integrals are constant multiples of bi+n (respectively ai

+n), i = 1, . . . , l . Here

the constant is B

d+22

, +nd2

, where B denotes the Beta function (note that it is well

defined as > d n). Hence we rewrite (2.3.35) as follows

Ad(E)1,,nd = C

l

i=1 b+ni a+ni

. (2.3.36)

For p = ( + n)/(+ n) (note that p > 1) we apply Lemma 2.3.6 in (2.3.36) to get

Ad(E)p1,,nd C

li=1

b(+n)pi a(+n)pi

= C

l

i=1

b+ni a+ni

. (2.3.37)


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2.4 Mixed Norm Inequality 31

Since E(p,1),,n =

li=1

bi+n ai+n

1/pthe inequality (2.3.34) follows from (2.3.37)

and Theorem 3.13, Chapter 5 of [SW3]. For (2.3.33) we observe from (2.3.4) that if

p = 1+()/d then p = q. Since we have from Theorem 2.3.3 and Theorem 2.3.4 thatthe operator Ad, d 1 is strong type (1 + ( )/d, 1 + ( )/d) we can interpolate([SW3], P.197) with (2.3.32) to get (2.3.33).

Remark 2.3.10. We make the following observations.

i) If = = 0 then (2.3.4) reduces to n/p n/q = d(1 1/q). Thus p q impliesp = q = 1 and hence (2.3.34) reduces to the trivial estimate AdfL1(Rnd) fL1(Rn).

ii) The main difference between the weighted case and the case = = 0 appears when

> . In this case the range of p is 1 < ( + n)/(+ n) p < ( + n)/d. Unlike thelater case one cannot get Lp Lq boundedness of the operator Ad for all values of p inthe above range, instead one lands up with Lorentz space estimates. These estimates

change to strong (p,q) estimates only when p

q, that is p

1 + (

)/d.

2.4 Mixed Norm Inequality for the d-Plane

Transform of Radial Functions

In this section our aim is to prove the following inequality

Rn

Gn,d


x dx1/q C

Rn

|f(x)|p x dx1/p

.

(2.4.1)

We first look at necessary conditions on , , p, q and r under which (2.4.1) can hold.

We recall from section 2 that (2.3.10) gives the necessary condition for the existence of


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d-plane transform of the function f Lp(Rn, xdx).

Given a measurable function f we consider f, > 0 as before and get

Rn

G(n,d)


x dx1/q

= +nqd

Rn

G(n,d)


x dx1/q

. (2.4.2)

Thus from (2.3.3) and (2.4.2) it follows that the necessary condition on ,,p,q for

(2.4.1) to hold is given by + n

p=

+ n

q+ d. (2.4.3)

Using (2.3.10), above relation also suggests following restriction on the range of ,

> d n, > n. (2.4.4)

We will now show that for the existence of left hand side integral of the inequality

(2.4.1) we need the following condition on ,q,d,r

+ n

q > 0 with < 1 we have

n,d{ : d(x, ) } = n1(A), (2.4.6)

where A =

y Sn1 : ni=d+1 y2i 1/2 /x .We first need to estimate the measure of the set A. We will do so by using bispher-


45/127


ical coordinates which we explain in the following (see [M], p. 134). If y = (y1, . . . , yn)

then for 1 d < n let y = (y1, . . . , yd, 0, . . . , 0) and y = (0, . . . , yd+1, . . . , yn) so that

y = y + y. Let Sd1 = Sn1 Rd and Snd1 = Sn1 Rnd be the copies of theunit spheres ofRd and Rnd in Rn. Bispherical coordinates are defined by the map :

Sd1 Snd1 [0, /2] Sn1 given by (a,b,) = a cos + b sin . Given y Sn1

by writing cos =d

i=1 y2i

1/2it follows that y = (y/ cos )cos + (y/ sin )sin ,

that is, is surjective. It is known that is injective almost everywhere. One now has

the following formula regarding integrals of measurable functions on Sn1,

Sn1

f() dn1()

=

Sd1

Snd1

/20

f(a cos + b sin )sinnd1 cosd1 da db d, (2.4.7)

where da, db denote the normalized rotation invariant measures on Sd1 and Snd1

respectively. We now observe that using bispherical coordinates we can write A =

{(a,b,) : a Sd1, b Snd1, sin /x}. So from (2.4.7) it follows that

n1(A) = C

/20

{:sin /x}()sinnd1 cosd1 d.

Using the change of variables t = sin it follows from above that

n1(A) =

10

{t: t/x}(t)tnd1(1 t2)d22 dt

/x/2x

tnd11 cd 2x2d22

dt,

where cd = 1 if d 2 and cd = 1/4 if d = 1. It now follows from the above expressionthat

n1(A) C

xnd

1 cd 2

x2 d2

2

. (2.4.8)


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Using the direct sum decomposition x = x + x it follows that the condition

d(x, ) implies that x . So, if d(x, ) then for f = B(0,1) it follows

from (2.2.3) that for all x Rn with x >

Tdf(x, ) = cd

1x

(s2 x2)d22 s ds

cd1

(s2 2)d22 s ds

=cd2

(1 2)d/2. (2.4.9)

So, for x > it follows from (2.4.8) and (2.4.9) that

Rn

Gn,d

Tdf(x, )r dn,d()

q/rx dx

Rn

{Gn,d: d(x,)}

Tdf(x, )r dn,d()

q/rx dx

Rnc

x

(nd)q

r

1 cd

2

x

2

(d2)q2r

x dx,

where c is given by (2.4.9). It is now clear that the integrand in the right hand

side integral behaves like x q(nd)r at infinity. Necessity of (2.4.5) now follows bydemanding finiteness of the integral on right hand side.

To proceed further we need the following facts.

a) Iff L1(Rn) or f is a nonnegative measurable function on Rn then

Gn,d

yndf(y) dd(y)

dn,d() = Cn,d

Rn

f(x) dx, (2.4.10)


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(see [M], p. 147). Hence we deduce

Gn,d Tdf(x, ) dn,d() = Cn,d Rn f(x y)ydn dy = Idf(x), (2.4.11)where y denotes the norm of y Rn.

b) IfF is a nonnegative measurable function on Gn,nd then

Gn,d

F() dn,d() =

Gn,nd

F() dn,nd(), (2.4.12)

(see [M], p. 144). We note that for d = n 1 we have Gn,nd = Gn,1 = { : Sn1} where = spanR{}. If F is a nonnegative measurable function onGn,n1 then from (2.4.12) it follows that

Gn,n1

F( ) dn,n1( ) =

Gn,1

F( ) dn,1()

= C

Sn1F( ) dn1(), (2.4.13)

where denotes the orthogonal complement of .

We will require the following lemma which follows easily by repeated applications

of Youngs inequality.

Lemma 2.4.1. If p, p1, p2, r 1, q rp2 and

1 +1

r+

1

q=

1

rp2+

1

p+

1

p1(2.4.14)

then for all measurable functions f , g , h defined on a locally compact abelian group G

the following inequality holds

(f g)r hLq/r(G) frLp(G)grLp1 (G)hLp2(G). (2.4.15)


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Proof. Since q rp2 and p2 1 we can choose s 1 such that

1 +r

q=

1

s+

1

p2. (2.4.16)

As q r 1 then we have by Youngs inequality

(f g)r hLq/r(G) (f g)rLs(G) hLp2(G)

= (f g)rLrs(G) hLp2(G) (2.4.17)

Since (2.4.14) and (2.4.17) implies 1 + 1rs

= 1p

+ 1p1

it follows by another application of

Youngs inequality that

(f g)Lrs(G) fLp(G) gLp1(G) (2.4.18)

(2.4.1) now follows from (2.4.17) and (2.4.18).

We will now prove (2.4.1) for d = 1 and d = n 1.

Theorem 2.4.2. (1) Let n > 2, and d = n 1. If , , p, q, r satisfy (2.3.10),(2.4.3), (2.4.4), (2.4.5) and

p q, r q, 1 r < (2.4.19)

then there exists C > 0 such that for all f Cc (Rn),

Rn

Gn,n1

|Tn1f(x, )|r dn,n1( )q/r

x dx1/q Cp,q,rfp,,n.

(2.4.20)


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(2) Let n 2 and d = 1. If , , p, q, r satisfy (2.3.10), (2.4.3), (2.4.4), (2.4.5),(2.4.19) and

1

p 0, (+n)rq 1 < 0 and n 3. These

conditions hold by (2.4.5) and (2.4.4). Now we can use lemma 2.4.1 to dominate I by

FpGp1H1/rp2 provided there exist p1, p2 which satisfy the conditions of the lemmaand all the above quantities are finite. It follows from (2.4.14) that 1 p2 < isequivalent to

1 +1

q 1

p 1

p1< 1 +

1

r+

1

q 1

p

and q rp2 is equivalent to

1p1 1 + 1r 1p .

To show the existence of such a p1 [1, ) we just need to make sure that 1r > 0,1 + 1

r 1

p> 0, 1 + 1

q 1

p 1 and 1 + 1

r 1

p 1 + 1

q 1

p. The first and second follows

as r < . The third and fourth inequality follow from p q and r q respectively.Since FLp(0,) = Cfp,,n we are through with the case d = n 1.

Next we consider the case d = 1. The proof of this case is similar to the previous

one. As before, if x = x x then it follows from (2.2.3) that

I =

Rn

Gn,1

|T1f(x, )|rdn,1()q/r

x dx1/q

= C

Rn

Gn,1

0

f(s)

s2 x2 1

2

[0,1]x/ss dsr dn,1()q/r x dx1/q

= C

Rn

Sn1

0

f(s)

s2 x21

2

[0,1]x/ss dsr dn1()q/r x dx1/q .

Let {x/x = u1, u2, . . . , un} be an orthonormal basis ofRn. If an unit vector makesan angle with u1 and i with ui, i = 2, 3, . . . , n then x = x, =

x cos so


52/127


that x = x sin . Hence

I = CRn

0

0 2

0

0 f(s)s2 x2 sin2 n32 [0,1]x sin /ss dsr

sinn2 sinn3 2 . . . sin n2 d d2 . . . d n1q/r x dx1/q

C

Rn

0

0

f(s)

s2 x2 sin2 n32 [0,1]x sin /ss dsr sinn2 dq/r x dx1/q

= C

0

0

0

f(s)

s2 t2 sin2 n32 [0,1]t sin /s s ds

r

sinn2

dq/r t+n1 dt1/q= C

0

/20

0

f(s)

s2 t2 sin2 n32 [0,1]t sin /s s dsr sinn2 dq/r t+n1 dt1/q .

Using the change of variable t sin = u we get

I C0t

0

0|f(s)|(s2 u2)

12[0,1](u/s) s dsr

t/u2n(t2 u2) 12 duq/r t+n1 dt1/q=

0

0

0

|f(s)|s1++nq u/s+nq 1 u2s2

12

[0,1]

u/sds

s

r

t/u2n+ (+n)rq t2u2

11

2

[1,)

t/udu

u

q/rdt

t

1/q

= (F G)r H1/rLq/r((0,),dt/t),

where

F(t) = |f(t)|t1++nq = |f(t)|t+np , G(t) = t+nq (1 t2)12 [0,1](t),

H(t) = (t2 1)12 t2n+ (+n)rq [1,)(t). (2.4.24)


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It is easy to observe that both G(t), H(t) Lp(0, ),dt/t only for 1 p < 12 provided+ n > 0, (+n)rq + 1 n < 0. These conditions hold by (2.4.5) and (2.4.4). As before

we will now use lemma 2.4.1. It follows from (2.4.14) that 1 p2 < 2 is equivalent to

1 +1

q 1

p 1

p1< 1 +

1

2r+

1

q 1

p, (2.4.25)

and q rp2 is equivalent to1

p1 1 + 1

r 1

p. (2.4.26)

To show the existence of such a p1 [1, 2), in view of (2.4.25) and (2.4.26), we need toshow that p, q and r satisfy the following conditions

r < ,

1 +1

r 1

p>

1

2,

1 +1

2r+

1

q 1

p>

1

2,

1 +

1

q 1

p 1,1 +

1

r 1

p 1 + 1

q 1

p.

First two conditions are there in the hypothesis, third and fourth conditions follow from

the relations p q and r q respectively.

Remark 2.4.3. We have following observations to make.

i) From (2.1.4) of Theorem 2.1.1 it follows easily that r < , r q, p q. Thus ap-pearance of these conditions in Theorem 2.4.2 is not entirely unexpected. Though

we could not prove that the condition r q is necessary in the following we showthe necessity of the condition p q. Since Gn,d is compact we can put r = 1 in


54/127


(2.4.20) and (2.4.22) which by an application of (2.4.11) gives the boundedness

of Riesz potential on weighted Lp spaces. The necessity of the condition p q

follows from Theorem 2.4.4.

ii) Now we show that the condition 1p 1 we deduce from (2.4.29).

(F G2)r

H21/r

Lq/r(R,dx)

=

R

R

R

F(x y)G2(y) dyr

H2(z x) dxq/r

dz

1/q

=

R

R

R

F(x y)G2(y/) dy

rH2

z x

dx

q/rdz

1/q.

As G2(y/) G2(y)1/2 for y (1, 0) and H2(y/) H2(y)1/2 for y (0, 1)it follows that

(F G2)r H21/rLq/r(R,dx) 12 12r1q (F G2)r H21/rLq/r(R,dx). (2.4.30)

Now by allowing tending to infinity it follows that 1p 12 12r 1q < 0.

To proceed further we will need to talk about the Lp Lq mapping property of

Riesz potential on L

p

spaces with power weight. We define the Riesz potential as follows

If(x) =

Rn

f(x y) yn dz, 0 < < n. (2.4.31)

The following theorem was proved in [SW2] (See also [Str1], [Co], [B]).

Theorem 2.4.4. If 0 < < n, 1 < p q < and + n > 0 then there exists aconstant C such that for all f

Cc (R

n) the inequality

Ifq,,n Cfp,,n

holds, if

+ n

n< p,

p

q 0 and + n

q+ =

+ n

p.


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In [B] a different proof of Theorem 2.4.4 was given which uses only Youngs in-

equality. The end-point estimates for the Theorem 2.4.4 was proved in [W]. In [W] it

was also shown that the conditions on , , p, q in Theorem 2.4.4 are necessary. But

it turns out that if we are concerned only about radial functions then the condition

p

q

0 in the above theorem is not necessary. In the following we prove an analogueof Theorem 2.4.4 for radial functions with a restriction on the range of. This result is

not new. In fact, a more general version of the result is available in [Ru2] (see Theorem

3, P. 754). But Theorem 3 of [Ru2] deals with more complicated weight functions than

ours and hence it becomes little difficult to link it with the result we need.

To proceed we will need the following well known formula ( see [G], D.3 ).

Sn1

f(x.) dn1() = |Sn2|11

f(tx)(1 t2)n32 dt, x Rn, (2.4.32)

for all measurable functions f defined on R.

Theorem 2.4.5. If n 2, 1 < < n, 1 p q < and + n > 0 then thereexists a constant C such that for all compactly supported, smooth radial functions f the

inequality

Ifq,,n Cfp,,n

holds, if

+ n

n< p and

+ n

p=

+ n

q+ .

Proof. It is enough to prove the result for nonnegative measurable functions. We note

that If is a radial function as f is radial. Using polar coordinates and (2.4.32), we


57/127


write

If(x) = 0 f(r)Sn1 x2 + r2 2xrx/x, n2 dn1() rn1 dr= C

0

f(r)r1

11

(1 t2)n32 x2

r2+ 1 2x

rt

n2

dt

dr

In the last line we have used (2.4.32). Since f is nonnegative so is If hence we have

Ifq,,n =

Rn

|If(x)|q x dx1/q

= Rn

0f(r)r11

1(1 t2)n32

x2

r2+ 1 2x

rt

n2

dt

dr

qx dx

1/q

= C

0

0

f(r)r11

1

(1 t2)n32

u2

r2+ 1 2u

rt

n2

dt

dr

qu+n1 du

1/q

= C0

0

f(r)r+n+q u/rn+q 11

(1 t2)n32

u2

r2+ 1 2u

rt

n2

dt

dr

r

qdu

u

1/q= CF G(Lq(0,),du/u) (2.4.33)

where

F(u) = f(u)u++nq = f(u)u

+np ,

G(u) = u+nq

11

(1 t2)n32 (u2 + 1 2ut)n2 dt

.

Note that F(Lp(0,),du/u) = fp,,n.


58/127


Now by applying Youngs Inequality in (2.4.33) we have

IfLq(Rn,xdx) CF(Lp(0,),du/u)G(Ls(0,),du/u)= fp,,nG(Ls(0,),du/u) (2.4.34)

provided there exists s [1, ] such that

1

s+

1

p= 1 +

1

q(2.4.35)

and G Ls(0, ),du/u. Hence the result will follow from (2.4.34) once we show thatG Ls((0, ), du/u), 1 s .

We first consider the function

H(u) =

11

(1 t2)n32 (u2 + 1 2ut)n2 dt, u 0 (2.4.36)

and show that it is continuous. This will imply that G is continuous on [0, ). Since(1 + u2 2ut) (u 1)2 for all t [1, 1] it follows that for u [0, 1/2]

(1 t2)n32 (1 + u2 2tu)n2 (1 t2)n32 |u 1|n, as < n.

Since the right hand side is integrable with respect to t it follows by dominated con-

vergence theorem that H is a continuous in [0, 1/2] and H(0) > 0. For u > 1/2 using

(1 + u2 2ut) 2u(1 t) 0 it follows that

(1 t2)n32 (1 + u2 2tu)n2 (1 t)32 (1 + t)n32 (2u)n2 .

Using the fact that 1 < < n it follows that the right hand side is an integrable

function oft and hence an application of dominated convergence theorem shows that H


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is continuous in (1/2, ). Since H(u)/u(n) = H(1/u) it follows that H(u)/u(n) H(0) > 0 as u . Since G(u) = u+nq H(u) we conclude that G(u) u+nq

around zero and G(u) u+n

q n+ around infinity. Thus G L(0, ), du/u L1

(0, ), du/u as + n > 0 and +nq

n + = +np

n < 0.

We will now prove the last result of this chapter. The next result is analogous to

Theorem 2.4.2 but with a restriction on the range of and . The method of proof is

very similar to that of [DNO].

Theorem 2.4.6. Letn > 2, 1 < d < n

1, and + d

0. If , , p, q , r also

satisfy (2.4.3), (2.4.5) and

a) 1 < p < +nd

b) 1 r <

then there exists a C > 0 such that for all f Cc (Rn), f radial

Rn

Gn,d

|Tdf(x, )|rdn,d()q/r

xdx1/q Cp,q,rfp,,n. (2.4.37)

Proof. Using (2.4.5), (2.4.3) and a) we note that for a fixed r in b) the best possible

range of p is given as follows

( + n)r

n d + dr < p < + n

d. (2.4.38)

Note that (+n)rnd+dr

+nn

1.

We will first choose an r0 [1, ) and prove (2.4.37) for all values of p satisfying(2.4.38). Since r0 is arbitrary this will prove the result. We will first prove the upper


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end-point estimate for p = ( + n)/d. Next for each p given by (2.4.38) we will prove

(2.4.37) for indicator functions with r replaced by r0. Viewing the later estimate as a

restricted weak type estimate we will use an interpolation argument.

We define the sublinear operator S given by

Sf(x) =

Gn,d

|Tdf(x, )|r0dn,d()1/r0

, x Rn. (2.4.39)

For f = E with E a rotationally symmetric, measurable subset ofRn it follows by

using (2.2.3) and Theorem 2.3.7 that,

supxRn,Gn,d

TdE(x, ) = supx

RndAdE(x)

E(+nd ,1),,n. (2.4.40)

Hence using compactness of Gn,d we obtain

SE CE(+nd ,1),,n. (2.4.41)

Next we choose p0 and q0 such that p0 =(+n)r0

nd+dr0and +np0 =

+nq0

+ d. We further choose

a p1 in (p0, ( + n)/d) and q1 such that+np1

= +nq1

+ d. Note that q1 (q0, ). We willnow prove that

SEq1,,n CE(p1,1),,n, (2.4.42)

with E as before. Assuming (2.4.42) for the moment, we get by interpolation ([G],

Theorem 1.4.19) that

Sfq,,n Cfp,,n (2.4.43)

where p1 < p q0r0

+nnd

1 we can apply Theorem 2.4.5 to get

SE

q1,,n

C

E

(r01)/r0

(+nd ,1),,n

E1/r0

p2,,n

, (2.4.44)

if p2 satisfies the conditions of Theorem 2.4.5, that is,

p2 1, + nn

< p2 q1r0

<


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and

+ n

p2=

(+ n)r0q1

+ d. (2.4.45)

We need to make sure that such a p2 exists. Now, p2 >+n

n is equivalent to+n

q1< ndr0

which follows from the relation q1 > q0 and (2.4.5). Alsoq1r0

p2 is equivalent top2 d which holds as d 1. Using (2.4.44) we get

SEq1,,n CE(r01)/r0(+nd ,1),,nE1/r0

p2,,n

= CE(r01)/r0(+nd ,1),,nE1/r0(p2,1),,n

= C

E

xdx (r01)dr0(+n)+ 1r0p2= C

E

xdx1/p1

= CE(p1,1),,n.

Note that in the last but one step we have used (2.4.45) and (2.4.3) to get

(r0 1)dr0( + n)

+1

r0p2=

1p1

.

This completes the proof.

Remark 2.4.7. We would like to make the following comments.

i) Using (2.4.3) we observe that the necessary condition p q is equivalent to

p d which is redundant as we have assumed that + d and 1 < p.

ii) The above method of proof can be applied for other admissible values of and

but that produces severe restriction on range of r. For instance, it can be

shown that if d n < 0 and + d 0 then the above result holds for( + n d)/r > (+ n)/d. The situation gets more complicated if one considersthe case > + d 0. It seems to us that a different method of proof has to be


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adopted to tackle the case of all possible admissible values of and .

iii) Clearly Theorem 2.4.2 is more complete than Theorem 2.4.6. The main reason for

this is the following fact. Suppose that d = n 1 or d = 1. For each fixed x Rn

let G = {A SO(n) : A(x) = x}. Then G is conjugate to SO(n 1) in SO(n). Itis then easy to see that for radial functions f one has Tdf(x, ) = Tdf(x, A()) for

A G. That is, Tdf(x, ) is invariant under the action ofSO(d)SO(nd).But this phenomena fails if 1 < d < n 1.


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Chapter 3

Mixed Norm Estimate for Radon

Transform on weighted Lp Spaces

3.1 Introduction

In this chapter we will deal with the Radon transform of functions defined on Rn, n 2,which are not necessarily radial. We start by recalling the following results regarding

continuity of Radon transform. The first one is from [OS].

Theorem 3.1.1. There exists a positive constant C such that for all f Cc (Rn) thefollowing inequality holds,

Sn1

R

|Rf(, t)|q dtp/q

dn1()

1/p CfLp(Rn) ,

where 1 p < nn1

and 1q

= np

n + 1.

The next one was proved in [Q].

Theorem 3.1.2. For all f Cc (Rn), n 3, there exists a positive constant C suchthat

RSn1|Rf(, t)|2 dt dn1() C

Rn

|f(x)|2 xn1 dx.


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3.1 Introduction

Documents

Weighted Lp-Lq Mapping Properties of D-plan Transform on Euclidean Space, Real Hyperbolic Space and Sphere