FOURIER MULTIPLIERS FOR Lp ON CHE´BLI–TRIME`CHE … · classical Fourier multiplier theory to Lie groups and symmetric spaces (see [30, 2, 10, 23, 4]). In the consideration of

FOURIER MULTIPLIERS FOR Lp ONCHEÂ BLI±TRIMEÁ CHE HYPERGROUPS

WALTER R. BLOOM and ZENGFU XU

[Received 18 July 1997; revised 12 April 1999]

1. Introduction

Let m: Rn ! C be a bounded function on the euclidean space Rn and de®ne theoperator Tm associated with m by

�Tm f �b�l� � m�l�bf �l�:The multiplier theorem of HoÈrmander [18] gives a suf®cient condition on m for theoperator Tm to be bounded on Lp�Rn� whenever 1 < p < 1, namely that m satisfy

supl2R n

jljkjm�k��l�j < 1 for k � 0; 1; . . . ; � 12

n� � 1;

where � 12

n� is the integral part of 12

n.Over the past twenty years considerable effort has been made to extend the

classical Fourier multiplier theory to Lie groups and symmetric spaces (see [30, 2,10, 23, 4]). In the consideration of this problem a dichotomy has emerged, basedon the growth of the volume of balls centred at the identity as their radii becomelarge (polynomial or exponential growth). While in the case of polynomial growththe condition on a multiplier is similar to that on euclidean spaces (see [30, 2]),some analyticity of the multiplier is necessary for the operator to be bounded onLp when the volumes of balls grow exponentially (see [10, 23, 4]). In the settingof hypergroups a version of HoÈrmander's multiplier theorem was established in[25] on Bessel±Kingman hypergroups, a particular class of CheÂbli±TrimeÁchehypergroups with polynomial growth.

A hypergroup �K; �� is a locally compact space with a certain generalizedconvolution structure � on its measure space (see [5, Chapter 1] for thede®nition). Let «x be the point measure at x 2 K. Then the convolution «x � «y ofthe two point measures «x and « y is a probability measure on K with compactsupport. Unlike the case for groups this convolution is not necessarily a pointmeasure. The convolution between point measures extends naturally to allbounded measures on the hypergroup. In place of natural left translation of afunction f by x, available in the group case, the generalized (left) translation isintroduced on a hypergroup via

Tx f �y� :�Z

Kf �z��«x � «y� �dz�:

For every hypergroup admitting a Haar measure m the convolution of two

1991 Mathematics Subject Classi®cation: primary 43A62; secondary 43A15, 43A32.

Proc. London Math. Soc. (3) 80 (2000) 643±664. q London Mathematical Society 2000.

functions f and g is de®ned as

f � g�x� :�Z

Kf �y�Tx g�y�m�dy�:

The notion of an abstract algebraic hypergroup has its origins in the studies ofF. Marty and H. S. Wall in the 1930s, and harmonic analysis on hypergroupsdates back to J. Delsarte's and B. M. Levitan's work during the 1930s and 1940s,but the substantial development had to wait till the 1970s when Dunkl [14],Spector [22] and Jewett [19] put hypergroups in the right setting for harmonicanalysis. There have been many fruitful developments of the theory of hypergroupsand their applications in analysis, probability theory and approximation theory(see [5]).

Many examples of hypergroup structures on the half line R� � �0; 1� arise fromSturm±Liouville boundary value problems where the solutions coincide with thecharacters of the hypergroup in question. CheÂbli±TrimeÁche hypergroups (seeDe®nition (1.6) below) are a class of such òne-dimensional' hypergroups on R� withthe convolution structure related to the following second-order differential operator:

L � LA; x :� ÿ d 2

dx2ÿ A0�x�

A�x�d

dx�1:1�

where the function A is continuous on R�, twice continuously differentiable onR�� 0; 1�, and satis®es the following conditions (see [31]):

(1.2) A�0� � 0 and A�x� > 0 for x > 0;

(1.3) A is increasing and unbounded;

(1.4) A0�x�=A�x� � ��2a� 1�=x� � B�x� on a neighbourhood of 0 wherea > ÿ 1

2and B is an odd C 1-function on R;

(1.5) A0�x�=A�x� is a decreasing C 1-function on R��, and hencer :� 1

2lim x!�1�A0�x�=A�x��> 0 exists.

Such a function A is called a CheÂbli±TrimeÁche function.

(1.6) De®nition. A hypergroup �R�; �� is called a CheÂbli±TrimeÁchehypergroup if there exists a CheÂbli±TrimeÁche function A such that for any real-valued function f on R� that is the restriction of an even non-negative C 1-function on R, the generalized translation u�x; y� � Tx f �y� is the solution of thefollowing Cauchy problem:

�LA;x ÿ LA; y�u�x; y� � 0;

u�x; 0� � f �x�; uy�x; 0� � 0; x > 0:

�We denote by �R�; ��A�� the CheÂbli±TrimeÁche hypergroup associated with A.

Remark. In particular, if the function A is of the form A�x� :� x2a�1 witha > ÿ 1

2then �R�; ��A�� is a Bessel±Kingman hypergroup. In this case LA; x

de®ned by (1.1) is the radial part of the Laplace operator on euclidean space with

2a 2N. If A�x� :� �sinh x�2a�1�cosh x�2b�1 with a > b >ÿ 12

and a 6� ÿ 12

then�R�; ��A�� is a Jacobi hypergroup. In this case LA; x is the radial part of theLaplace±Beltrami operator on non-compact riemannian symmetric space of rank 1where 2a; 2b 2N.

644 walter r. bloom and zengfu xu

In this paper we consider Lp Fourier multipliers on general CheÂbli±TrimeÁchehypergroups. In § 2 we collect some basic facts on Fourier analysis on CheÂbli±TrimeÁche hypergroups and some useful estimates for their characters. Then in § 3we prove a version of HoÈrmander's multiplier theorem. In the ®nal section weinvestigate the Lp-boundedness of oscillating multipliers and the Lp ÿ Lq

boundedness of the Riesz potentials. Because of the possible exponential volumegrowth and the generalized convolution on the hypergroups, the standard methodsdo not apply. We employ the techniques for non-compact symmetric spaces toovercome the dif®culties caused by the exponential growth; and use properties ofthe Fourier transform and estimates for the characters to handle the problemsarising from the generalized translation.

2. Preliminaries on CheÂbli±TrimeÁche hypergroups

We begin by collecting some basic facts on harmonic analysis on CheÂbli±TrimeÁche hypergroups. For a general reference see [5].

The hypergroup �R�; ��A�� (see De®nition (1.6)) is non-compact and com-mutative with neutral element 0 and the identity mapping as the involution. Haarmeasure on �R�; ��A�� is given by q :� AlR� where lR� is the usual Lebesgue

measure on R�. For x0 2 R� and r > 0 we denote by B�x0; r� the open interval�maxf0; x0 ÿ rg; x0 � r�. The growth of the hypergroup is determined by thenumber r in (1.5). For the hypergroup to be of exponential growth it is necessaryand suf®cient that r > 0 (see [5, Proposition 3.5.55]), for then (1.5) implies

that A�x�> A�1�e2r�xÿ1� for x > 1. Otherwise we say that the hypergroup is ofsubexponential growth (which includes polynomial growth).

The multiplicative functions on �R�; ��A�� coincide with the solutionsJl �l 2 C� of the differential equation

LJl�x� � �l2 � r2�Jl�x�; Jl�0� � 1; J0l�0� � 0;

and the dual space cR� can be identi®ed with the parameter set R� È i �0; r�.For 0 < p < 1 the Lebesgue space Lp�R�; A dx� is de®ned as usual, and we

denote by k f kp; A the Lp-norm of f 2 Lp�R�; A dx�. For f 2 L1�R�; A dx� theFourier transform of f is given by

bf �l� :�ZR�

f �x�Jl�x�A�x� dx:

(2.1) Theorem (Levitan±Plancherel; see [5, Theorems 2.2.13 and 3.5.54(ii)]).There exists a unique non-negative measure p on cR� with support �0; 1� such

that the Fourier transform induces an isometric isomorphism from L2�R�; A dx�onto L2�cR�; p�, and for any f 2 L1 Ç L2�R�; A dx�,Z

R�j f �x�j2A�x� dx �

ZcR� j bf �l�j2 p�dl�:

To determine the Plancherel measure p we must place a further (growth)restriction on A. A function f is said to satisfy Condition (H) if for some a > 0, fcan be expressed as

f �x� � a2 ÿ 14

x2� z�x�

645fourier multipliers on cheÂbli±trimeÁche hypergroups

for all large x, where Z 1

x 0

xg�a�jz�x�j dx < 1

for some x0 > 0 and z�x� is bounded for x > x0; here g�a� � a� 12

if a > 12

andg�a� � 1 otherwise. For x > 0 we put

G�x� :� 1

4

�A0�x�A�x�

�2

� 1

2

�A0�x�A�x�

�0ÿ r2:

(2.2) Theorem (see [6, Proposition 3.17]). Suppose that G satis®es Condition(H) together with one of the following conditions:

(i) a > 12;

(ii) a 6� ja j;(iii) a � a < 1

2andZ 1

0x1=2ÿaz�x�J0�x�A�x�1=2 dx 6� ÿ2a

��MA

porR 1

0 xa�1=2z�x�J0�x�A�x�1=2 dx � 0 where MA :� lim x! 0� xÿ2aÿ1A�x�and z�x� � G�x� � � 1

4ÿ a2�=x2.

Then the Plancherel measure p is absolutely continuous with respect to Lebesguemeasure and has density jc�l�jÿ2 where the function c satis®es the following: thereexist positive constants C1; C2; K such that, for any l 2 C with Im�l�< 0,

C1jlja�1=2 < jc�l�jÿ1 < C2jlja�1=2 for jlj< K; a > 0;

C1jlja�1=2 < jc�l�jÿ1 < C2jlja�1=2 for jlj > K:

Remark. Using the proof of [6, Proposition 3.17], K. TrimeÁche has obtainedthe following estimates (see [28; 29, Proposition 6.I.12 and Corollary 6.I.5]). Ifthe function A0�x�=A�x� is such that

there exists d > 0 such that for all x 2 �x0;�1� (for some x0 > 0),

A0�x�A�x� �

2r� eÿdxD�x� if r > 0;

2a� 1

x� eÿdxD�x� if r � 0

8<:where D is a C 1-function bounded together with its derivatives,

then there exist positive constants C1, C2, K such that

(i) if r > 0 and a > ÿ 12

then

C1jlj2a�1 < jc�l�jÿ2 < C2jlj2a�1 for l 2 C; jlj > K;

(ii) if r > 0 and a > ÿ 12

then

C1jlj2 < jc�l�jÿ2 < C2jlj2 for l 2 C; jlj< K;

(iii) if r � 0 and a > 0 then

C1jlj2a�1 < jc�l�jÿ2 < C2jlj2a�1 for l 2 C; jlj< K:

This result is a particular case of Theorem (2.2).


In the sequel we assume that A satis®es the conditions in Theorem (2.2) witha > 0, and for each k 2N, �A0�x�=A�x��k� is bounded for large x 2 R�. In

addition we assume that in the case r � 0, A satis®es A�x� � O�x2a�1� �x! 1�.The following result can be found in [6, Lemmas 2.6 and 3.28].

(2.3) Lemma. We have

A�x�, x2a�1 �x! 0��;and if r > 0 then

A�x�, e2r x �x! �1�:

Let «x be the unit point mass at x 2 R�. For any x; y 2 R� the probabilitymeasure «x � « y is q-absolutely continuous with

supp�«x � «y� Ì � jxÿ y j; x� y�:�2:4�We denote by Tx f the generalized translation of a function f by x 2 R� de®ned by

Tx f �y� :�ZR�

f �z��«x � «y� �dz�:�2:5�

We now list some useful properties and estimates for characters.

(2.6) Lemma (see [9, 1, 27]). (i) For each l 2 C, Jl is an even C 1-functionand l 7! Jl�x� is analytic.

(ii) We have jJl�x�j< 1 for x 2 R�, l 2 C and jIm�l�j< r.

(iii) For each l 2 C; Jl has a Laplace representation

Jl�x� �Z x

ÿ xe�ilÿr� tnx �dt� for x 2 R�;

where nx is a probability measure on R supported in �ÿx; x�.

(2.7) Lemma. Let l � y� ih 2 C. Then

(i) jJl�x�j< ejhjxJ0�x�,(ii) eÿr x < J0�x�< C�1� x�eÿr x.

Proof. The lemma follows from the Laplace representation of Jl in Lemma(2.6) and the following estimate given in [1]:

jJl�x�j< CA�1� x�eÿr x for x; l 2 R�:

(2.8) Lemma (see [7, Lemma 2.4]). For each k 2N0 we have

jJ�k�l �x�j<CA�1� l�k if lx < 1; x < 1;

CA xA�x�ÿ1=2 if lx < 1; x > 1;

CA A�x�ÿ1= 2jc�l�j�1� l�k if lx > 1:

8><>:We also have the following alternative estimate:

jJ�k�l �x�j< CA A�x�ÿ1=2�lx�1=2ÿajc�l�j�1� l�k for lx < 1; x > 1:


For an q-measurable subset E we denote by jE j its Haar measure and xE itscharacteristic function. In the sequel we shall use C to denote a positive constantthe value of which may vary from line to line. Dependence of such constantsupon parameters of interest will be indicated through the use of subscripts.

3. Fourier multipliers

We begin by introducing Fourier multipliers for the Lp-functions on �R�; ��A��.For a bounded function m on �R�; ��A�� consider the operator Tm de®ned by

�Tm f �b�l� :� m�l�bf �l�:�3:1�By Theorem (2.1), Tm is bounded on L2�R�; A dx�. The function m is said to be aFourier multiplier on Lp �1 < p < 1� if the operator Tm maps Lp continuously intoitself. We denote by mp�R�; ��A�� the Banach space of all convolution operatorsthat are bounded from Lp�R�; A dx� to Lp�R�; A dx� with the usual operatornorm, and by Mp the set of all Fourier multipliers of Lp with the norm inheritedfrom mp�R�; ��A��. It is not dif®cult to see that each Fourier multipliercommutes with the translation operator Tx . Note also that, by duality, Mp �Mp 0

when 1=p� 1=p 0 � 1.The following result shows that for p 6� 2 some analyticity of the function m is

necessary for Tm to be bounded on Lp�R�; A dx� when r > 0. This newphenomenon, different from the euclidean case, arises from the exponentialgrowth of the hypergroups.

(3.2) Lemma. Assume that r > 0 and 1 < p < 1. Then every m 2Mp extends toan even function analytic and bounded in the interior of the stripF« � fz 2 C: jIm�z�j< «rg where « � j2=pÿ 1j. If p � 1 then m extends to aneven function analytic inside F1 and continuous and bounded on the closed strip F1.

Proof. We follow the proof of [10, Theorem 1] in the case of non-compactsymmetric spaces. By duality we need only consider 1 < p < 2. For 1 < p < 2,Lemma (2.7) gives

kJlkp 0; A ��Z 1

0jJl�x�j p

0A�x� dx

�1=p 0

�3:3�

<

�Z 1

0�1� x� p 0e p 0x � j Im�l�jÿr�A�x� dx

�1=p 0

< Cp 0

for each l 2 C with Im�l� < «r, where 1=p� 1=p 0 � 1. Using the fact thatTx Jl�y� � Jl�x�Jl�y� we obtain Tm Jl � m�l�Jl. Since Tm is also Lp 0-bounded,

we have Tm Jl 2 Lp 0 �R�; A dx� and kTm Jlkp 0; A < CA; p 0 kJlkp 0; A. Thus m isbounded inside the strip F«. The analyticity come from the analyticity of the mapl 7! Jl from the interior of F« to Lp 0 �R�; A dx� using Lemma (2.6).

For p � 1 we can similarly extend m to an analytic function in the interior ofF1, and also a bounded and continuous function on F1 since by Lemma (2.6)(ii)we have kJlk1 < 1 for all l 2F1.


(3.4) De®nition. Let r > 0. For a positive integer N we say that a boundedfunction m satis®es a HoÈrmander condition of order N (and denote this bym 2M�2; N �) if m extends to an even analytic function inside F«, and thederivatives m�i � extend continuously to the whole of F« and satisfy

supl2F«

�1� jlj�i jm�i ��l�j < 1 for i � 0; 1; . . . ;N:

For m 2M�2; N � set

kmkM�2;N � :� max0 < i < N

supl2F«

�1� jlj�i jm�i ��l�j:

We now give a version of the HoÈrmander±Mihlin multiplier theorem on thehypergroup �R�; ��A��.

(3.5) Theorem. Let 1 ggj< CAkmkM�2;N �gÿ1k f k1; A where g > 0;

and it is bounded on Lp�R�; A dx� for 1 < p < 1, that is,

kTm f kp; A < CA; pkmkM�2;N �k f kp; A:

We shall postpone the proof and give more general characterizations of Mp

which include the above theorem (see Corollary (3.40)). For this purpose weintroduce some function spaces on �R�; ��A�� as in [4].

(3.6) De®nition. For 1 < q < 1, 1 < r < 1 and ÿ1 < j; t < 1, the weightedSobolev space H j; t

q; r �R� is the space of all tempered distributions h on R that canbe written as

h�l� �X1k�0

hk�2ÿ k l��3:7�

where hk 2 H jq �R�, supp�h0� Ì �ÿ4; 4�, supp�hk� Ì fl 2 R: 1

4< jlj< 4g (with

k > 1), andP1

k�0�2�t�1=q�kkhkkH jq�r < 1. Here

H jq �R� �

�f :

ZRj�1� D�j=2 f �x�jq dx < �1

�is the usual Sobolev space on R, where D is the Laplacian on R. Set

khkH j; tq; r� inf

�X1k�0

�2�t�1=q� kkhkkH jq�r�1=r

where the in®mum is taken over all the representations (3.7) of h. For r � 1 wede®ne H j; t

q;1�R� by the usual modi®cation. Also we denote by H j; tq; r �R�� the space

of all even h in H j;tq; r �R� restricted to R�. In the sequel we shall abbreviate

H j; tq; r �R�� by H j; t

q; r , and H j; tq;q by H j; t

q .

(3.8) De®nition. Let 1 < q < 1, 1 < r < 1, 1=q < j < 1, ÿ1 < t < 1 and0 < n < 1. We denote by Hj; t; n

q; r the space of all functions m: Fn ! C satisfyingthe following conditions:


(i) m is even and continuous;

(ii) m is analytic inside Fn;

(iii) m has at most polynomial growth at in®nity; and

(iv) if h�l� � m�l� inr� for l 2 R, then h 2 H j; tq; r .

We abbreviate Hj; t;2=qq;q by Hj; t;2=q

q .

Setting

kmkH j; t; nq; r

:� km�´� inr�kH j; tq; r

we see that Hj; t; nq; r is a Banach space (see [4, Lemma 9]).

(3.9) Lemma. The following embedding relationships hold:

Hj;ÿ1=2; n2;1 Ì Hj; t;n

2;2 for t < ÿ 12;

Hj;ÿ1=2; n2;1 Ì Hj; t;n

2;1 for t < ÿ 12;

Hj=q;ÿ1=q;2=qq;1 Ì Hj=q; t= q;2=q

q for t < ÿ1 �1 < q < 1�;Hj=2;ÿ1=2; n

2;1 Ì H j=2;ÿ1=22;1 ;

Hj=q;ÿ1=q; nq;1 Ì H j=q;ÿ1=q

q;1 �1 < q < 1�;H j=2;ÿ1=2

2;1 Ì H j=2;a�1=2ÿj=2;12 ;

H j=2;ÿ1=22;1 Ì H j=2;a�1=2ÿj =2;1

2 ;

H j=q;ÿ1=qq;1 Ì H j=q; t=q;2 =q

q �1 < q < 1; t < ÿ1�:

Proof. The lemma follows from a straightforward calculation using De®nitions(3.6) and (3.8).

For m 2Hj;t; nq; r or H j; t

q; r let K denote the kernel obtained as the Fouriertransform of m in the distributional sense, so that Tm f � K � f . Choose an evenC 1-function w on R such that w�x� � 1 for jxj< 1

2and w�x� � 0 for jxj> 1, and

®x once and for all a kernel decomposition K � K 0 � K 1 where K 0 � Kw andK 1 � K�1ÿ w�. Consequently we have the operator decomposition Tm � T 0

m � T 1m

where T 0m f � K 0 � f and T 1

m f � K 1 � f . It turns out that when r > 0 the Lp-boundedness of T 1

m requires some analyticity of m, while T 0m behaves like its

euclidean analogue. Throughout the remainder of the paper we shall alwaysassume that m is in fact rapidly decreasing (that is, m is in the usual Schwartzspace) though none of our estimates will depend upon the actual rate of decrease.It suf®ces to ¯atten m or, equivalently, to regularize K in the standard way.

(3.10) Lemma. Suppose that m 2Hj=2;a��1ÿj�=2;12 for some j > 2 when

r > 0, or m 2 Hj=2;a��1ÿj�=22 for some j > 2a� 2 when r � 0. Then

K 1 2 L1�R�; A dx� and

kK 1k1; A <

�CAkmkH

j = 2; a� �1ÿ j� = 2; 1

2

if r > 0;

CAkmkH

j = 2; a� �1ÿ j� = 2

2

if r � 0:


Proof. The lemma can be proved similarly to that of [4, Theorem 11] usingLemma (2.3) and Theorems (2.1) and (2.2) together with interpolation.

(3.11) Corollary. Suppose that m 2Hj=2;a��1ÿj�=2;1

2 for some j > 2 when

r > 0 or m 2 Hj=2;a��1ÿj�= 22 for some j > 2a� 2 when r � 0. Then

T 1m 2mp�R�; ��A�� for 1 0;

CA; pkmkH

j = 2; a� �1ÿ j� = 2

2

if r � 0:

(

(3.12) Lemma. Let 2 < q < 1 and consider j; t 2 R with j � t > 2a� 1.Suppose that m 2Hj=q; t=q;2=q

q with j > 2 when r > 0, or m 2 H j= q; t=qq with

j > 2a� 2 when r � 0. Then T 1m 2mp�R�; ��A�� for j1=pÿ 1

2j< 1=q and

kT 1m km p

<CA; p;qkmk

Hj = q; t = q; 2 = q

qif r > 0;

CA; p;qkmkH

j = q; t = qq

if r � 0:

(

Proof. The lemma can be proved by interpolation in the same way as in theproof of [4, Theorem 13].

For the Lp-boundedness of T 0m we need some estimates for K 0 .

(3.13) Lemma. Let j > 2a� 2.

(i) If m 2Hj= 2;ÿ1= 2;12;1 when r > 0, or m 2 H j=2;ÿ1=2

2;1 when r � 0, then

K 0 2 L1�R�; A dx� and

kK 0k1; A <CAkmk

H j = 2;ÿ1 = 2; 12; 1

if r > 0;

CAkmkH j = 2;ÿ1 = 2

2; 1

if r � 0:

(

(ii) If m 2Hj=2;ÿ1=2;12;1 when r > 0, or m 2 H j=2;ÿ1=2

2;1 when r � 0, then K 0

satis®es HoÈrmander's condition that, for y; y0 2 R�,Zjxÿ y 0j> 2jyÿ y0j

jTy K 0�x� ÿ Ty 0K 0�x�jA�x� dx <

CAkmkH j = 2;ÿ1 = 2; 1

2;1if r > 0;


2;1if r � 0:

(

Proof. We follow the idea of [4, Theorem 14] but use properties of the Fouriertransform and estimates for the characters to handle the generalized translation.We only give the proof of (ii). Assertion (i) can be proved in a similar way.

Recall that supp�w� Ì �ÿ1; 1� and

Ty K 0�x� �Z x� y

j xÿ y jK�u�w�u��«x � «y� �du�:

Now jxÿ y0j> 2jyÿ y0 j implies jxÿ yj> jyÿ y0j. Therefore Ty K 0�x� � 0 andTy 0

K 0�x� � 0 if jxÿ y0j> 2jyÿ y0j and jyÿ y0j > 1. Thus we need only considerthe case when jyÿ y0j< 1. Also observe that if jxÿ y0 j > 2 then jxÿ y j>jxÿ y0j ÿ jyÿ y0j > 1 and hence Ty K 0�x� � Ty 0

K 0�x� � 0. Similarly if jxÿ yj > 2


then Ty K 0�x� � Ty 0K 0�x� � 0. ThereforeZjxÿ y0j> 2j yÿ y0j

jTy K 0�x� ÿ Ty0K 0�x�jA�x� dx

�Z

E y; y 0

jTy K 0�x� ÿ Ty 0K 0�x�jA�x� dx

<

ZE y; y 0

jTy K�x� ÿ Ty 0K�x�jA�x� dx

�Z

E y; y 0

jTy K 1�x� ÿ Ty 0K 1�x�jA�x� dx

:� I1�y; y0� � I2�y; y0�;where Ey; y 0

:� fx 2 R�: 2 j yÿ y0j< jxÿ y0j< 2; jxÿ yj< 2g. By Lemmas (2.10)and (2.9) we have

I2�y; y0�< 2kK 1k1;A <CAkmk

H j = 2;ÿ1 = 2; 12;1

if r > 0;


2;1if r � 0:

(It suf®ces to prove that

I1�y; y0�<CAkmk

H j = 2;ÿ1 = 22; 1

if r > 0;


2;1if r � 0:

(�3:14�

Let l denote the Abel transform A of K (see [27] or [8] for the de®nition ofthe Abel transform on a CheÂbli±TrimeÁche hypergroup). Then m � bK � F0�l� (see[27; 8, Theorem 4.33]) where F0 is the classical Fourier transform. Fix a dyadicdecomposition m�l� �P1

k�0 mk�2k l� and corresponding K�x� �P1k�0 Kk�x� and

l�u� �P1k�0 lk�u� where cKk�l� � mk�2ÿ k l�. Then cKk�l� � mk�2ÿ k l� � F0 lk�l�.

We can assume that mk 2S�R��, the usual Schwartz space, otherwise we can usethe standard regularization. Thus lk 2S�R�� and Kk 2S�R�; ��A��, the general-ized Schwartz space (see [8]). To establish (3.14) we introduce smooth cut-offfunctions. Let y be an even C 1-function on R such that y�x� � 1 for jx j< 1

4and

y�x� � 0 for jx j> 12, and set y j�x� � y�2 jx� for each j 2N0. Then y j 2 C 1

satis®es y j�x� � 1 for jx j< 2ÿ jÿ2 , y j�x� � 0 for jx j> 2ÿ jÿ1 andj �d i=dxi�y j�x�j< Ci 2

i j for i � 0; 1; 2; . . . : Put lk j � �1ÿ y j� lk and let

Kk j �Aÿ1�lk j� and mk j � F0�lk j�. Observe that lk ÿ lk j is supported in

fu: juj< 2ÿ jÿ1g. Using properties of the Abel transform given in [27, Theorem6.4] we have

Kk�x� � Kk j�x� if x > 2ÿ jÿ1:�3:15�Let k0 � k0�y; y0� be the integer satisfying 1 < 2 k 0 jyÿ y0j < 2. Then

I1�y; y0�<Xk 0ÿ1

k�0

ZE y; y 0

jTy Kk�x� ÿ Ty 0Kk�x�jA�x� dx�3:16�

�X1k� k 0

ZE y; y 0

jTy Kk�x� ÿ Ty 0Kk�x�jA�x� dx


<Xk 0ÿ1

k�0

ZE 1; k

y; y 0


�Xk 0ÿ1

k�0

ZE 2; k

y; y 0


�X1k� k 0

ZE y; y 0


:�Xk 0ÿ1

k�0

I�1�k �y� �

Xk 0ÿ1

k�0

I�2�k �y� �

X1k� k 0

I�3�k �y�;

where

E 1; ky; y 0

:� fx 2 Ey; y 0: jxÿ y0j > 2ÿ k�1g

and

E 2; ky;y 0

:� fx 2 Ey; y 0: jxÿ x0j< 2ÿ k�1g:

Using Lemma (2.3) we obtain, for any R 2 �0; 2�,�Z 2

Rjxÿ y0jÿjA�x� dx

�1=2

<CA Rÿj= 2�a�1 if y0 < 2R;

CA Rÿj= 2�1=2A�y0�1=2 otherwise,

(�3:17�

where CA is independent of R. Put

Ej :� fx 2 Ey; y0: 2ÿ j < jxÿ y0j< 2ÿj�1; jxÿ yj< 2g for j � 0; 1; . . . ; k ÿ 1:

Then for x 2 Ej we have jxÿ y0j > 2ÿ j and jxÿ y0j> jxÿ y0j ÿ jyÿ y0j >12jxÿ y0j > 2ÿ jÿ1. Hence by (2.4), (2.5) and (3.15) we have

Ty Kk�x� � Ty Kk j�x�:�3:18�To estimate I

�1�k we ®rst consider the case when y0 < 2ÿ k�1. Then by (3.17)

(with R � 2ÿ k�, (3.18), Theorems (2.1) and (2.2), the Schwarz inequality and the

fact that �Ty f �b�l� � Jl�y�bf �l�, we obtain

I�1�k < CA 2 k�j=2ÿaÿ1�

�ZE 1; k

y; y 0

j�xÿ y0�j=2�Ty Kk�x� ÿ Ty 0Kk�x��j2A�x� dx

�1=2

< CA 2k�j= 2ÿaÿ1��Xkÿ1

j�0

2ÿjj

ZE j

j�Ty Kk j�x� ÿ Ty 0Kk j�x��j2A�x� dx

�1=2

< CA 2k�j= 2ÿaÿ1��Xkÿ1

j�0

2ÿjj

Z 1

0jmk j�l��Jl�y� ÿ Jl�y0��j2 jc�l�jÿ2 dl

�1=2

< CAjyÿ y0j2k�j= 2ÿaÿ1��Xkÿ1

j�0

2ÿ jj

Z 1

0

Z 1

0jmk j�l�J0l�y�j2jc�l�jÿ2 dl du

�1=2

;


where y � y0 � u�yÿ y0�. We now use Lemma (2.8) and Theorem (2.2) to obtain

I�1�k < CAjyÿ y0j2 k�j=2ÿaÿ1�

�Xkÿ1

j�0

2ÿjj

Z 1

0jmk j�l��1� l�a�3=2j2 dl

�1=2

< CAjyÿ y0j2 k�j=2ÿaÿ1��Xkÿ1

j�0

2ÿ jjk lk jk2

H a� 3 = 22

�1=2

:

Arguing as in [4, Lemma 15] we have, for j1; j2 > 0,�Xkÿ1

j�0

�2ÿj2 jk lk jkHj1

2�2�1=2

< C 2k�1=2�j1ÿj2�kmkkHj 2

2:�3:19�

Thus using (3.19) (with j1 � a� 32

and j2 � 12j) we obtain, for y0 < 2ÿ k�1 ,

I�1�k < CA 2 kjyÿ y0j kmkkH j = 2

2

:�3:20�Similarly, we can prove (3.20) for y0 > 2ÿ k�1 . Consequently,Xk 0ÿ1

k�0

I�1�k < CAkmk

H j = 2;ÿ1 = 22;1

:�3:21�

A similar argument gives Xk0ÿ1

k�0

I�2�k < CAkmk

H j = 2;ÿ1 = 22;1 :

�3:22�

For I�3�k we apply (3.17) (with R � jyÿ y0j), (3.18) and Theorems (2.1) and

(2.2) to obtain

I�3�k <

CAjyÿ y0jÿj=2�a�1Ik if y0 < 2jyÿ y0j;CAjyÿ y0jÿj=2�a�1A�y0�1=2Ik if y0 > 2jyÿ y0j;

(where

Ik ��Xk 0

j�0

2ÿ jj

Z 1

0jmk j�l��Jl�y� ÿ Jl�y0��j2jc�l�jÿ2 dl

�1=2

:

Then using Lemma (2.8), Theorem (2.2) and (3.19) yields

I�3�k <

CA�2 kjyÿ y0j�ÿj=2�a�1kmkkH j = 22

if y0 < 2jyÿ y0j;CA�2kjyÿ y0j�ÿj=2�1=2kmkkH j = 2

2

if y0 > 2jyÿ y0j;

8<:and hence Xk0ÿ1

k�0

I�3�k < CAkmk

H j = 2;ÿ 1 = 22;1 :

�3:23�

Thus (3.14) follows from (3.16), (3.21)±(3.23) and Lemma (3.9). This completesthe proof of the lemma.

(3.24) Lemma. For any x0 2 R� let I denote the interval �x0 ÿ 12; x0 � 1

2�Ç R�.

Suppose that k�x; y� 2 L2�I ´ I; q q�, where q is the Haar measure on


�R�; ��A�� (restricted to I ) and Kf , de®ned by

K f �x� :�Z

If �y�k�x; y�q�dy� for x 2 I; f 2 L2�I; q�;

satisfy the following two conditions:

(a) there exists a constant C1 > 0 independent of x0 such that

kK f kL 2�I;q� < C1k f kL 2�I;q�;

(b) there exists a constant C2 > 0 independent of x0 such that, for any y; y0 2 I,Zj xÿ y 0j > 2 j yÿ y 0j

jk�x; y� ÿ k�x; y0�jq�dx�< C2:

Then for any p 2 �1; 2� there exist constants Ap depending only on p and C1, C2

such that, for any f 2 L2 Ç Lp�I; q�,kK f kL p�I;q� < Apk f kL p�I;q� if 1 ggj< A1gÿ1k f kL 1�I;q� if p � 1; g > 0:

Proof. By Lemma (2.3) it is straightforward to verify that �I; q; d � is ahomogeneous space where d�x; y� � jxÿ yj for x; y 2 I (see [11, Chapter III] fordetails of homogeneous spaces). Thus the lemma can be proved as in [11, pp. 66±75].

(3.25) Corollary. Let j > 2a� 2.

(i) If m 2Hj= 2;ÿ1= 2;12;1 when r > 0, or m 2 H j=2;ÿ1=2

2;1 when r � 0, then

T 0m 2mp�R�; ��A�� for 1 0;

CA; pkmkH j = 2;ÿ 1 = 2

2; 1

if r � 0:

((ii) If m 2Hj=2;ÿ1=2;1

2;1 when r > 0, or m 2 H j=2;ÿ1=22;1 when r � 0, then

T 0m 2mp�R�; ��A�� for 1 0;

CA; pkmkH j = 2;ÿ1 = 2

2;1if r � 0:

(Also T 0

m is of weak-type �1; 1�:

jfx 2 R�: jT 0m f �x�j > ggj<

CA; pkmkH j = 2;ÿ1 = 2; 1

2;1gÿ1k f k1;A if r > 0;

CA; pkmkH j = 2;ÿ1 = 2

2;1gÿ1k f k1;A if r � 0

8<:for any g > 0 and f 2 L1�R�; A dx�.

Proof. Assertion (i) follows immediately from Lemma (3.13)(i). To prove (ii)we ®x f 2 L1�R�; A dx�. For each j 2N0 let Ij � � j; j� 1� and fj � x I j

f . Thenf �P1

j�0 fj and

T 0m f �

X1j�0

T 0m fj:�3:26�


Let Jj :� � jÿ 1; j� 2� and ®x a dyadic decomposition of m:

m�l� �X1k�0

mk�2ÿ k l��3:27�

where mk�2ÿ k l� � fk�l�m�l� and fk�l� � f�2ÿ k l� for k > 1, f0�l� �P0k�ÿ1 f�2ÿ k l� and f is an even C 1-function on R satisfying supp�f�Ì

fl 2 R: 12

< jlj< 2g andP1

k�ÿ1 f�2ÿ k l� � 1 �l 6� 0�. Let K �Pk Kk be the

corresponding decomposition with cKk�l� � mk�2ÿ k l� and K 0k � Kk w as before.

For each N 2N0 set K N �PNk�0 Kk, K 0;N �PN

k�0 K 0k and T 0;N

m f � K 0;N � f .

Then from (3.27) we see that jPNk�0 mk�2ÿ k l�j< jm�l�j. Also by [4,

Proposition 23(iii)] we have m 2 L1�R�� and

kmk1 < CkmkH j = 2;ÿ 1 = 2

2;1:�3:28�

Hence in view of [15, Chapter 3] and (3.28) we have, for f 2 L2�R�; A dx�,kT 0;N

m f k2;A < CkK N � f k2;A�3:29�

� C

X1k�0

mk�2ÿ k ´�bf 2;j

< Ckmk1kbf k2;j

< CkmkH j = 2;ÿ 1 = 2

2;1k f k2;A

and

kT 0m f ÿ T 0;N

m f k2;A � kK 0 � f ÿ K 0;N � f k2;A�3:30�

< C

�K ÿXN

k�0

Kk

�� f

2;A

� C

�mÿXN

k�0

mk�2ÿ k ´��bf

2;j

! 0 �N ! 1�:Recall that supp�w� Ì �0; 1� and supp� fj� Ì � j; j� 1�. Thus supp�T 0;N

m fj� Ì Jj and,by (3.29),

kT 0;Nm fjkL 2�Jj;q� < Ckmk

H j = 2;ÿ 1 = 22;1

k fjkL 2�Jj;q��3:31�where C is independent of m, fj and j. By the proof of Lemma 3.13 we haveZ

j xÿ y 0j> 2 j yÿ y 0jjTy K 0;N�x� ÿ Ty 0

K 0;N�x�jq�dx�< CkmkH j = 2;ÿ 1 = 2

2;1�3:32�

for any y; y0 2 R�, where C is an absolute constant. Now takek�x; y� � kN�x; y� � Tx K 0;N�y�. Then kN 2 L2�Jj ´ Jj; q q� and by (3.31),(3.32), Lemma (3.24) and the proof of [11, Theorem III(2.4)],

jfx 2 Jj: jT 0;Nm fj�x�j > ggj< Ckmk

H j = 2;ÿ 1 = 22;1

gÿ1k fjkL 1�Jj;q��3:33�


for any g > 0, where C is independent of j. Using (3.30) we can ®nd asubsequence such that T 0;Ns

m fj�x� ! T 0m fj�x� �s! 1� q-a.e. x 2 R�. Hence (3.33)

gives, for any g > 0,

jfx 2 Jj: jT 0m fj�x�j > ggj< Ckmk

H j = 2;ÿ 1 = 22;1

gÿ1k fjkL 1�Jj;q�:�3:34�Thus T 0

m fj is ®nite for q-almost every x 2 R�. We now observe that for anyJ 2D��R�� with supp�J� Ì R�nJj, w�x� fj � J�x� � 0 for all x 2 R�. HenceZ 1

0T 0

m fj�x�J�x�q�dx� � 0:

This implies that supp�T 0m fj� Ì Jj and for each x 2 R� the series in (3.26) has at

most three terms: T 0m fjÿ2, T 0

m fjÿ1 and T 0m fj if x 2 Ij. Therefore for any g > 0,

Eg Ì[1j�2

�Ejÿ2;g=3 È Ejÿ1;g=3 È Ej;g= 3�È E0;g È E0;g=2 È E1;g=2�3:35�

where

Eg � fx 2 R�: jT 0m f �x�j > gg

and

Ej;l � fx 2 Jj: jT 0m fj�x�j > lg �l > 0�:

It readily follows by (3.34) and (3.35) that, for any g > 0,

jEgj< CkmkH j = 2;ÿ 1 = 2

2;1gÿ1

X1j�0

k fjkL 1�Jj;q��3:36�

� CAkmkH j = 2;ÿ 1 = 2

2;1gÿ1k f k1;A:

Using (3.28) and Theorem 2.1 we have, for f 2 L2�R�; A dx�,kT 0

m f k2;A < CkmkH j = 2;ÿ 1= 2

2;1 :�3:37�Therefore applying (3.36) and (3.37) together with Marcinkiewicz's interpolationtheorem and duality, we deduce that for f 2 Lp�R�; A dx� �1 < p < 1� ,T 0

m f 2 Lp�R�; A dx� and kT 0m f kp;A < Ckmk

H j = 2;ÿ 1 = 22;1

. Now (ii) follows from

Lemma (3.9) and this completes the proof of the corollary.

(3.38) Corollary. Suppose that m 2 H j=q;ÿ1=qq;1 �R�� for some 2 < q < 1 and

j > 2a� 2. Then T 0m 2mp�R�; ��A�� for j1=pÿ 1

2j< 1=q and

kT 0mkm p

< CkmkH

j = q;ÿ 1 = qq;1

:

Proof. In view of the decomposition (3.26) we see that the corollary followsusing a similar argument to that in the proof of [4, Corollary 19].

The following result can be veri®ed straightfowardly using De®nitions (3.6) and (3.8).

(3.39) Theorem. (i) If r > 0 and j > 2a� 2 then

(a) Hj=2;ÿ1=2;12;1 Ì Mp for 1 ggj< CAgÿ1kmkH j = 2;ÿ 1 = 2; 1

2;1k f k1;A with g > 0;

and

(c) for any 2 < q < 1, Hj=q;ÿ1=q;2=qq;1 Ì Mp for j1=pÿ 1

2j< 1=q.

(ii) If r � 0 and j > 2a� 2 then

(a) H j=2;ÿ1=22;1 �R�� Ì Mp for 1 ggj< CA gÿ1kmkH j = 2;ÿ 1= 2

2;1k f k1;A with g > 0;

and

(c) for any 2 < q < 1, H j=q;ÿ1=qq;1 �R�� Ì Mp for j1=pÿ 1

2j< 1=q.

Proof. The theorem follows immediately from Corollary (3.11), Lemma(3.12), Corollaries (3.25) and (3.38) and Lemma (3.9).

(3.40) Corollary. (i) Assume r > 0 and let 0 < n < 1, N � �n�a� 1�� 1,0 < g < 1 and ÿ1 < d < �gÿ 1�N. Suppose that m is an even bounded functionsuch that m extends analytically inside Fn and the m�i � �i � 0; 1; . . . ;N � extendcontinuously to Fn and satisfy

jm�i ��l�j< B�jlj � 1�dÿg i for l 2Fn; i � 0; 1; . . . ;N:�3:41�(1) If 0 < n < 1 then m 2Mp for j2=pÿ 1j< n and kmkMp

< CAB.

(2) If n � 1 then

(a) m 2Mp for 1 lgj< CA Blÿ1k f k1;A with l > 0;

if g � 1 and d � 0, and

(c) m 2M1 with kmkM1< CA B in all other cases.

(ii) Assume r � 0 and let n, N, g and d be as in (i). If m is an even bounded functionon R satisfying (3.41) for l 2 R� then assertions (1) and (2) in (i) are valid.

Proof. The proof is in fact a straightforward calculation. We only show (i)(1)and simply note that the others can be handled similarly.

By Theorem (3.39) it suf®ces to show that m 2HN;ÿ n=2; n2= n;1 . Let h�l� � m�l� inr�

for l 2 R, and ®x a decomposition of h as before: h�l� �P1k�0 hk�2ÿ kl� where

hk�2ÿ k l� � h�l�fk�l�. Thus

kmkHN;ÿ n = 2; n

2 = n;1< sup

k

khkkH N2 = n

< supk

XN

i�0

�Z 1

0jh�i �k �l�j2= n dl

�n=2


< supk

XN

i�0

�Z 4

1=4jh�i ��2 k l�2k i j2= n dl

�n= 2

� supk

XN

i�0

�Z 2 k� 2

2 kÿ 2jh�i ��u�ui j2 = n 2ÿ k du

�n=2

< supk

XN

i�0

�Z 2 k� 2

2 kÿ 2jm�i ��u� inr��u� 1�i j2= n 2ÿ k du

�n=2

< CA B supk

XN

i�0

�2 k�n�dÿg i� i �=2

< CA B;

and we have ®nished.

Remark. Theorem (3.5) is included in this corollary.

4. Oscillating multipliers

In this section we apply the results in § 3 to investigate Lp�R�; ��A��-boundedness for oscillating multipliers mb; d, which are multipliers of type

mb; d�l� :� �l2 � r2�ÿb=2ei �l2�r2�d = 2

if r > 0;

w�l�lÿbei jl jd if r � 0;

8<:where Re�b�> 0, d > 0 and w is an even C 1-function on R vanishing in aneighbourhood of the origin and is identically 1 in a neighbourhood of in®nity.The euclidean analogue of mb;d is the radial multiplier de®ned by

emb; d�z� :� w�z�jz jÿbei j z j d for Re�b�> 0; d > 0;

where w is a C 1 radial function on Rn that vanishes for jz j< 1 and equals 1 forjz j> 2.

Oscillating multipliers have already been studied extensively in the setting ofeuclidean spaces (see, for example, [13, 21]), and some of these results have beengeneralized to some Lie groups of polynomial growth and non-compact symmetricspaces (see [20, 3, 17, 12]). Oscillating multipliers are interesting because oftheir intimate connection with the Cauchy problem for the SchroÈdinger and thewave equations.

Let Tb;d denote the convolution operator associated to mb;d de®ned by (2.1).

(4.1) Theorem. Suppose that 1 0 and a > 1 then Tb; d is bounded on Lp�R�; A dx� if and only if p � 2.

(ii) If r > 0 and a < 1 then Tb; d is bounded on Lp�R�; A dx� if Re�b� >�2a� 2�dj1=pÿ 1

2j.

(iii) If r� 0 then Tb;d is bounded on Lp�R�; A dx� if Re�b� > �2a� 2�dj1=pÿ 12j.


Proof. By Theorem (2.1) we have, for f 2 L2�R�; A dx� ,kTb; d f k2; A < Cb; dk f k2; A:

Hence Tb; d is always bounded on L2�R�; A dx�. A straightforward calculationshows that if r > 0 then mb; d is unbounded on each strip F« where « � j2=pÿ 1jand p 6� 2. Thus (i) follows from Lemma (2.2).

To prove (ii) we can restrict ourselves to the case 1 �2a� 2�d j1=pÿ 12j where « � 2=pÿ 1. By induction, a direct calculation

yields, for k 2N0 ,

jm�k�b; d�l�j< CA;b; d�1� jlj�ÿRe�b�� k�dÿ1�ec jlj dÿ 1

for l 2F«;�4:2�where c is some positive constant depending only on d and A. Let h�y� �mb; d�y� i«r� and ®x a dyadic decomposition h�y� �P1

k�0 hk�2ÿ ky� where

hk�2ÿ ky� � h�y�fk�y� and fk is de®ned as in the proof of Corollary (2.25). By(4.2) we have, for every s 2N0,

khkkH s2 = «

<Xs

j�0

�Z 1

0jh� j �k �y�j2=« dy

�«=2

�4:3�

<Xs

j�0

�Z 2 k� 2

2 kÿ 2jm� j �b; d�y� i«r�y jj2= « dy

�«= 2

< CA;b; d 2ÿ k�Re�b�ÿ d s�:

By interpolation (4.3) holds for any non-negative number s. Hence usingDe®nitions (2.4) and (2.6) together with (4.3) (with s � Re�b�=d) we obtain

kmb; dkHRe�b� = d;ÿ « = 2; «

2 = «;1< CA;b; d:

Thus (ii) is proved. Similarly we can prove (iii) using Theorem (2.39)(ii)(c).

We now proceed to establish some results concerning the Riesz potentials. Forb > 0 put mb�l� :� �l2 � r2�ÿb=2 and let Kb be the inverse Fourier transform ofmb in the distributional sense. As before we write Kb � K 0

b � K 1b where K 0

b � Kbw

and K 1b � Kb�1ÿ w�. For r > 0 each mb is a particular type of oscillating

multiplier operator.

(4.4) Lemma. Suppose that r > 0.

(i) If b > 2a� 2 then K 0b 2 Lp�R�; A dx� for 1 < p < 1.

(ii) If b � 2a� 2 then K 0b �x�, log x (as x! 0�) and K 0

b 2 Lp�R�; A dx� for1 < p < 1.

(iii) If b < 2a� 2 then K 0b �x�, xbÿ�2a�2� (as x! 0) and K 0

b 2 Lp�R�; A dx�for 1 2a� 2 then clearly mb 2 L1�R�; jc�l�jÿ2 dl�.Hence K 0

b is bounded and compactly supported and (i) follows immediately.


To prove the rest of the lemma we observe that

Kb�x� �1

G�b�Z 1

0tb=2ÿ1ht�x� dt�4:5�

since

rÿb=2 � 1

G� 12b�Z 1

0t b= 2ÿ1eÿr t dt:

Here ht is the heat kernel on �R�; ��A�� de®ned by

ht�x� �Z 1

0eÿt �l2�r2�Jl�x�jc�l�jÿ2 dl:

In view of [1, TheÂoreÁme II.2], Lemma (2.6)(ii), Theorem (2.2) and Lemma (2.3),we have for t > 1,

0 < ht�x�< CA eÿr2t:�4:6�By [16, Theorem 3.1], there exist m1 and m2 2 R such that, for x; t > 0,

CA;1 tÿaÿ1em 1teÿ x 2=4 t xa�1=2��A�x�p < ht�x�< CA;2tÿaÿ1em 2teÿx 2= 4 t xa�1=2��

A�x�p :�4:7�

Therefore using (4.6) and (4.7) we obtainZ 1

1tb=2ÿ1ht �x� dt < CA�4:8�

and Z 1

0t b=2ÿ1ht �x� dt <

xa�1=2��A�x�p Z 1

0t b=2ÿ2ÿaeÿx 2=4 t dt < CA

xbÿaÿ3=2��A�x�p eÿx 2=8�4:9�

if x > 1, and Z 1

0t b= 2ÿ1ht �x� dt ,

Z 1

0t b=2ÿ2ÿaeÿx 2=4 t dt�4:10�

,x bÿ2aÿ2 if b < 2a� 2;

j log xj if b � 2a� 2

(if 0 < x < 1. Thus (ii) and (iii) follow from (4.5), (4.8) and (4.10).

It remains to show (iv). From Lemma (2.3), (4.5), (4.8) and (4.9) we see thatK 1

b 2 L1�R�; A dx�. Note that mb satis®es the conditions in Corollary (2.40)(i)(1)(with « � j2=pÿ 1j, g � 1, d � b and any positive integer N ) and Lemma (2.2).Arguing as in the proof of the corollary we see that mb 2Hk;ÿ1=2;«

2;1 for anypositive integer k and 0 < « < 1. Now we can follow the proof of [4, Proposition5] to show that K 1

b 2 Lp�R�; A dx� for 1 < p < 2; in fact the main problem is toestimate Ij �

R j�1j jK 1

b �x�j pA�x� dx for 1 < p < 2. Now HoÈlder's inequality gives

Ij <

�Z j�1

jA�x� dx

�1ÿp=2�Z j�1

jjK 1

b �x�j2 A�x� dx

�p=2

< CA e2r�1ÿp=2� j�Z j�1

jjKb�x�j2 A�x� dx

�p =2

:


Also by Lemma (2.9) we have mb 2Hk;ÿ1=2;«2;1 Ì Hk;a�1=2ÿ k;«

2;2 for any 0 < « < 1and positive integer k > a� 1, and hence mimicking the proof of [4, Proposition5] we obtain

Ij < CA jÿ kkmbkH k; a� 1 = 2ÿ k; 2 = pÿ 12; 2

:

Finally, for 1 a� 1,�Z 1

0jK 1

b �x�j p q�dx��1=p

< CA; pkmbkH k; a� 1 = 2ÿ k; 2 = pÿ 12; 2

< CA; pkmbkH k;ÿ 1 = 2; 2 = pÿ 12;1

:

Therefore by duality we have K 1b 2 Lp�R�; A dx� for 1 < p < 1. On the other

hand, (ii) and (iii) show that K 0b 2 L1�R�; A dx�. So if K 1

b 2 L1�R�; A dx� thenmb should be continuous on F1, which is in fact not the case. ThereforeK 1

b 62 L1�R�; A dx� and (iv) is proved.

(4.11) Theorem. Suppose that r > 0, b > 0 and 1 < p; q < 1.

(i) We have kKb � f k p; A < Cp; bk f k p; A if and only if 1 2a� 2,

(b) b � 2a� 2 and 1 �2a� 2�=b or p � �2a� 2�=b with q < 1,

(2) 1 < p < �2a� 2�=b and 1=pÿ b=�2a� 2�< 1=q,

(3) p � 1 and 1ÿ b=�2a� 2� < 1=q < 1.

Proof. Assertion (i) is an immediate consequence of Corollary (2.40)(i)(1)(with « � j2=pÿ 1j, g � 1 and d � b) and Lemma (2.2). Now we assume that p < q.

If b > 2a� 2 then by Lemma (4.4)(i) and (iv), Kb 2 Lp�R�; A dx� (with1 < p < 1). Hence, for any 1 < p0 < 1 and 1 < q0 < 1,

kKb � f k p 0;A< kKbk p 0;A

k f k1;A if f 2 L1�R�; A dx�;kKb � f k1 < kKbkq 0

0;Ak f kq 0; A if f 2 Lq 0�R�; A dx�

and by Riesz±Thorin interpolation

kKb � f kq; A < Cp;q;bk f k p; A for b > 2a� 2; 1 < p < q < 1:�4:12�If b � 2a� 2 then by Lemma (4.4)(ii) and (iv), Kb 2 Lp�R�; A dx� for

1 < p < 1, and similarly use Riesz±Thorin interpolation to obtain (4.12) for1 < p < q < 1 or 1 < p < q < 1. In this case kKb � f k1 < Ck f k1; A does nothold. Otherwise we would have kKb � htk1 < Ckhtk1; A � C and hence by Lemma

(4.4)(iv), kK 0b � htk1 < C. However ht � K 0

b �x� ! K 0b �x� (as t! 0�) for q-almost

every x 2 R� then shows that K 0b 2 L1�R�; A dx� while, by Lemma (4.4)(ii),

K 0b �x�, log x (as x! 0�). This is a contradiction.If b < 2a� 2 then by Lemma (4.4)(iv), K 1

b 2 Lp�R�; A dx� for 1 < p < 1 and


hence f 7! K 1b � f is bounded from Lp�R�; A dx� to Lq�R�; A dx� as above for

1 < p < q < 1. Thus f 7! Kb � f is bounded from Lp�R�; A dx� to Lq�R�; A dx� ifand only if f 7! K 0

b � f is. By Lemma (4.4)(iii), we recall that K 0b 2 Lp�R�; A dx�

for 1 �2a� 2�=b(the conjugate index of �2a� 2�=�2a� 2ÿ b�),

kK 0b � f kp 0

0; A < kK 0

b kp 00; Ak f k1; A;

kK 0b � f k1 < kK 0

b k p 00; Ak f k p0; A:

Again by Riesz±Thorin interpolation we have for �2a� 2�=b < p < q < 1 or�2a� 2�=b < p < q < 1,

kK 0b � f kq; A < Cp;q;bk f kp; A:

Arguing as in the euclidean case (see [24, Chapter 5]) we see that kK 0b � f kq; A <

Ck f kp; A does not hold for q � 1 and p � �2a� 2�=b. Also as in the euclideancase f 7! Kb � f is of weak-type �1; n=�nÿ b��. Therefore by interpolation (see

[26, § 1.18.9, Theorem 2]), f 7! K 0b � f is of type � p; q0� for 1 < p < �2a� 2�=b

and 1=q0 � 1=pÿ b=�2a� 2�. Recall that f 7! K 0b � f is of type � p; p� for

1 1=pÿ b=�2a� 2�. As in the caseb � 2a� 2 we see that f 7! K 0

b � f is not bounded from L1�R�; A dx� toL1�R�; A dx�. For p 6� 1 or q < 1 we recall by Lemma (4.4) thatK 0

b 2 Lp�R�; A dx� if and only if p < �2a� 2�=�2a� 2ÿ b�. Thus if f 7! K 0b � f

is bounded from Lp�R�; A dx� to Lq�R�; A dx� for some pair � p; q� with q > pthen kKb�gkq; A � kKb � Kgkq; A < CkKgkp; A for p < �2a� 2�=�2a� 2ÿ b�since K 1

b 2 Lp�R�; A dx� for 1 < p < 1. Consequently, K 0b�g 2 Lq�R�; A dx� and

q < �2a� 2�=�2a� 2ÿ bÿ g�. Therefore f 7! K 0b � f is not bounded from

Lp�R�; A dx� to Lq�R�; A dx� if 1=q < 1=pÿ b=�2a� 2� and the proof is complete.

Arguing similarly to the euclidean case (see [24, Chapter 5]) we obtain thefollowing result.

(4.13) Theorem. Suppose that r � 0, 0 1 then

kKb � f kq; A < Cp;q;bk f kp; A for f 2 Lp�R�; A dx�:(b) If p � 1 then, for g > 0,

jfx 2 R�: j f � Kb�x�j > ggj< Cq;b�gÿ1k f k1; A�q for f 2 L1�R�; A dx�:

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Walter R. Bloom and Zengfu XuDivision of Science and EngineeringMurdoch UniversityPerthWA 6150Australia

[email protected]@starwon.com.au

Documents

FOURIER MULTIPLIERS FOR Lp ON CHE´BLI–TRIME`CHE … · classical Fourier multiplier theory to Lie groups and symmetric spaces (see [30, 2, 10, 23, 4]). In the consideration of