L1Estimates for Maximal Functions and Riesz Transform on Real Rank 1 Semisimple Lie Groups

journal of functional analysis 157, 327357 (1998) article no. FU983256

L 1 Estimates for Maximal Functions and Riesz Transform on Real Rank 1 Semisimple Lie Groups Takeshi Kawazoe Department of Mathematics, Keio University, 3-14-1, Hiyoshi, Kohokuku, Yokohama 223, Japan and Departement de Mathematiques, Universite de Nancy I, 54506 Vandoeuvre-les-Nancy, France E-mail: kawazoesfc.keio.ac.jp Received January 20, 1994; revised October 1, 1997; accepted February 17, 1998

Let G be a real rank one semisimple Lie group and K a maximal compact subgroup of G. Radial maximal operators for suitable dilations, the heat and Poisson maximal operators, and the Riesz transform, which act on K-bi-invariant functions on G, satisfy the L p-norm inequalities for p>1 and a weak type L 1 estimate. In this paper, through the Fourier theories on R and G we shall duplicate the Hardy space H 1(R) to a subspace H 1s (G) (s0) of L 1(G) and show that these operators are bounded from H 1s (G) to L 1(G).  1998 Academic Press

1. INTRODUCTION Let G be a real rank one connected semisimple Lie group with finite center, G=KAN an Iwasawa decomposition of G, and dg=dk da dn a corresponding decomposition of a Haar measure dg on G. Let L 1loc(GK) denote the space of locally integrable, K-bi-invariant functions on G and L p(GK) (0
|

f (a x n) dn

(x # R),

N

where A is parametrized as [a x ; x # R] (for the definition of \ see (3) below). We let F 1f (x)=e \xF f (x). Then the integral formula for the Iwasawa decomposition of G (cf. [6, p. 373]) yields that f # L 1(G) if and only if F 1|f | # L 1(R), and thus, 327 0022-123698 25.00 Copyright  1998 by Academic Press All rights of reproduction in any form reserved.

328

TAKESHI KAWAZOE

Proposition 1.1.

If f # L 1(G), then F 1f # L 1(R).

The reverse of this proposition is not true: Even if F 1f is well-defined for f # L 1loc(GK) and it belongs to L 1(R), f does not always belong to L 1(GK). The first problem we offer is the following, Problem A. For f # L 1loc(GK) find a condition on F 1f under which f belongs to L 1(GK). We next consider a problem concerning with a maximal operator on G. Let _: G  R + denote the K-bi-invariant function on G induced by the distance function on X=GK (cf. [13, p. 320]). For each r # R + we let B(r)=[ g # G; _(g)r] and / B(r) the characteristic function of B(r). The HardyLittlewood maximal operator M GHL on G is defined as follows: For f # L 1loc(GK) (M GHL f )( g)= sup

|B(r)| &1 (| f | V / B(r) )(g)

( g # G).

0
This definition makes sense for f # L p(X) ( p1). Clerc and Stein [3] and Stromberg [11] have shown the following, p Theorem 1.2. M G HL satisfies the L -norm inequalities for p>1, and a 1 weak type L estimate, that is, for all : # R + and all f # L 1(X)

c |[ g # G; (M G HL f )(g)>:]|  & f & L1(G) , : where c is independent of : and f. We here define a radial maximal operator on G as an analogue of the  one on R. We fix , # C  c (GK), the space of C , compactly supported K-bi-invariant functions on G, and suppose that , is sufficiently zero at the > origin of G. The radial maximal operator M G, ,, = (=0) on G is defined as 1 follows: For f # L loc(GK) > (1+r) &= |( f V , >r )( g)| (M G, ,, = f )( g)= sup

(g # G),

0
where the dilation , >r (r # R + ) of , is given by , >r (g)= |B(r)| &1 ,(a _( g)r )

( g # G).

This dilation is different from the one used in [7] and see Section 6 for > G G, > other dilations. Since M G, ,, = fcM HL f pointwisely, M ,, = also satisfies the property stated in Theorem 1.2. As Folland and Stein have shown in their book [5], when a Lie group G is of homogeneous type, a radial maximal

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

329

operator M G, on G is bounded from the atomic Hardy space H 1, 0(G) to L 1(G). More precisely, integrability of the maximal function M Gf = sup , M G, f where the supremum is taken over , in a suitable class, is equivalent to that f belongs to H 1, 0(G) (see [5, Chapter 3]). At present, we > have no definition of the (atomic) Hardy space on G on which M G, ,, = is bounded. The second problem, which was also treated in [7], is the following, Problem B. L (GK). 1

> Find a subspace of L 1(GK) on which M G, ,, = is bounded to

These two problems look unrelated each other. However, the answers obtained in this paper indicate that they are deeply related. Let C(*) (* # R) be the HarishChandra's C-function. Since C(*) has a meromorphic extension on C, the following definition of the Fourier multiplier C 1+ on R makes sense: (C 1+ F ) t (*)=C(&*&i\) &1 F t (*)

(* # R),

where F t denotes the Euclidean Fourier transform of F. Let H 1(R) be the H 1-Hardy space on R (cf. [10, Chap. 3]). Then our answer of Problem A can be stated as Answer A.

If C 1+ F 1f # H 1(R), then f # L 1(GK).

Let Qs (s0) denote the Fourier multiplier on R defined by (Qs F ) t (*)=

\

1+ |*| |*|

s

+F

t

(*)

(* # R).

We set H 1s (GK)=[ f # L 1loc(GK); Qs C 1+ F 1f # H 1(R)] and & f & H1s (G) =&Qs C 1+ F 1f & H1(R) . Answer A means that H 10(GK) is a subspace of L 1(GK) and moreover, the fact that (|*|1+ |*| ) s satisfies the Hormander condition on Fourier multiplier yields that H 1s (GK) (s>0) is a subspace of H 10(GK) (see Section 4). Our answer of Problem B is given as follows. > 1 Answer B. For $>0, M G, ,, 0 is a bounded operator of H 2+$(GK) to G, > 1 1 L (GK) and M ,, 1+$ is one of H 0(GK) to L (GK). 1

In order to understand a true character of C 1+ F 1f we need the Fourier analysis on G, so we refer to the Warner's book [14]. Let , *(g) ( g # G, * # R) be the zonal spherical function on G where the dual space of the Lie

330

TAKESHI KAWAZOE

algebra of A is identified with R. Then, for f # L 1(GK) the Fourier transform f (*) of f is defined by f (*)= G f (g) , *(g) dg. The inversion formula is of the following form: For f # C  c (GK) f (g)=

|

f (*) , *( g) |C(*)| &2 d*,

(1)

R

in which case, f (*) is a holomorphic function of exponential type. Hence, by regarding each K-bi-invariant function on G as en even function on R, by substituting the expansion of , * (see (5) and (6) below) into (1), and then, by shifting the integral line R to R+i\, we can rewrite (1) as follows [14, p. 356]: f (x)=e &\x

|

f (*) 8(*, x) C(&*) &1 e i*x d*

(x # R + )

R 

=e &2\x : e &2mx m=0

|

C(&*&i\) &1 1 m(*+i\) f (*+i\) e i*x d*.

_

(2)

R

Since (F f ) t (*)= f (*) and 1 0 #1 [14, Proposition 9.2.2.3 and 9.1.5], the leading term corresponding to m=0 in the right side of (2) is nothing but e &2\x (C 1+ F 1f )(x). Roughly speaking, Answers A and B can be restated as follows: If the leading term of e 2\xf (x) (x # R + ) belongs to H 1(R), then f > 1 and M G, ,, 1+$ f belong to L (GK). We state the organization of this paper. In Section 3 we shall obtain a key lemma for the integrability of f satisfying C 1+ F 1f # H 1(R). Then the proof of Answer A and some basic properties of H 1s (GK) are given in Section 4. In Section 5 we shall obtain some criteria by which we can judge whether a radial maximal operator is bounded from H 1s (GK) to L 1(GK). > Actually, we apply a criterion to M G, ,, = and obtain the proof of Answer B in Section 6. Moreover, we consider the same problem for the heat and Poisson maximal operators M GH, = and M GP, = (=0) on G, which are defined as follows: For f # L 1loc(GK) (M GH, = f )(g)= sup (1+r) &= |( f V h r )( g)|

(g # G),

0
(M G (1+r) &= |( f V p r )( g)| P, = f )(g)= sup

(g # G),

0
2

2

2 12

where h r(*)=e &(* +\ ) r and p^ r(*)=e &(* +\ ) r. As shown by Stein [9] in great generality these operators also satisfy the property stated in Theorem 1.2. In Section 6 we shall show that M GH, 0 (resp. M GP, 0 ) is a bounded sublinear operator of H 11+$(GK) ($>0) (resp. H 112(GK))

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

331

to L 1(GK) and moreover, M GH, 1 and M GP, 2 are ones of H 10(GK) to L 1(GK). In Section 7 we shall treat the Riesz transform R G on G which is defined as follows: For f # L 1loc(GK) (R Gf )(g)=( |{| (&2) &12 )( f )( g)

( g # G),

where 2 is the Laplacian on G and |{| 2 (h)=2(h 2 )&22h } h for h # C (G). As studied by Anker [1] and Lohoue [8] R Gf is deeply related with G M GP, = f, especially, the L 1-norm of M G P, = f is controlled by the one of R f. 1 Therefore, we can expect that the (Hardy) space H s (GK) might be useful to obtain an L 1 estimate for R G. Indeed, we shall prove that R G is 1 (GK) to L 1(GK). bounded from H 12 2. NOTATIONS Let G be a real rank one connected semisimple Lie group with finite center and G=KAN an Iwasawa decomposition of G. Let a be the Lie algebra of A and F=a* the dual space of a. Let H be the unique element in a satisfying #(H)=1 where # is the positive simple root of (G, A) determined by N. We parametrize each element in A, a, and F as a x =exp(xH), xH, and x# (x # R) respectively. In what follows we often identify these spaces with R and also Fc , the complexification of F, with C without making mention of the identification. We put F(s)=[* # Fc ; |I(*)|
|

f ( g) dg=

G

|



f (x) D(x) dx, 0

where D(x)=(sinh x) m1 (sinh 2x) m2 (x # R + ), m 1 and m 2 are the multiplicities of # and 2# respectively. We put :=

m 1 +m 2 &1 2

and

\=

m 1 +2m 2 . 2

(3)

332

TAKESHI KAWAZOE

Then the order of D(x) is given by D(x)t

{

x 2:+1 e 2\x

(0
(4)

where the symbol ``t'' means that the ratio of the left side and the right side is bounded below and above by positive constants. Let L 1loc(GK) denote the space of locally integrable, K-bi-invariant functions on G. Let 1 L p(GK) (0
(x # R + )

(5)

and furthermore, 8(*, x) has the so-called Gangolli expansion: 

8(*, x)= : 1 m(*) e &2mx

(* # F, x # R + )

(6)

m=0

[14, 9.1.4 and 9.1.5]. Their explicit forms and some basic properties of C(*), 8(*, x), and 1 m(*), which will be used in the following arguments, are summarized in [4, Sections 2 and 3], and a sharp estimate for the derivatives of 1 m(*) will be obtained in the appendix of this paper (see Section 8). For f # L 1(GK) the Fourier transform f (*) (* # F) of f is defined by

f (*)=

|

f ( g) , *( g) dg

(* # F).

G

From the RiemannLebesgue's lemma on G [4, Lemma 11] it follows that f (*) is an even holomorphic function on F(\) satisfying | f (*)|  0 as |*|   in CL(F(\)) and hence, sup * # CL(F( \))

| f (*)| & f & L1(G) .

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

333

When f is in C  c (GK), the PaleyWiener theorem on G [14, 9.2.3] implies that f (*) is an even holomorphic function on Fc of exponential type, in which case, the Fourier inversion formula: f (g)=

|

f (*) , *( g) |C(*)| &2 d*

(g # G)

(7)

| f (*)| 2 |C(*)| &2 d*

(8)

R

and the Plancherel formula:

|



| f (x)| 2 D(x) dx=

0

|



0

hold. Thereby, the Fourier transform f [ f of C  c (GK) is uniquely extended to an isometry between L 2(GK) and L 2(R, |C(*)| &2 d*) [14, Theorem 9.2.2.13]. We now introduce some operators on G and R. In what follows most of operators on G and R are denoted by scripts: A, B, C, ..., excepting M and T. Especially, AB } } } Cf means that A(B( } } } (C( f )))) and A G an operator on G. For s # R we define the Abel transform F sf of f # L 1loc(GK) as F sf(x)=e s\xF f (x)=e (s+1) \x

|

f (a x n) dn

(x # R)

N

and the Fourier multipliers C s+ and 1 sm (m # N) on R as (C s+ F )(x)=

|

(1 sm F )(x)=

|

C(&*&is\) &1 F t (*) e i*x d*

(x # R),

1 m(*+is\) F t (*) e i*x d*

(x # R),

R

R

where F t denotes the Euclidean Fourier transform of F. Of course, these definitions make sense if the integrals of the right sides exist. Since f (*)=(F f ) t (*) (* # Fc ) for f # C  c (GK) [14, Proposition 9.2.2.3], the relation (F sf ) t (*)= f (*+is\)

(9)

holds if the both sides exist for * # Fc . We here suppose f # C  c (GK) and we observe that on F( \), f (*) is holomorphic and rapidly decreasing, C(&*) &1 is holomorphic and tempered, and 1 m(*) is holomorphic and uniformly dominated by a polynomial of m [14, Proposition 9.1.7.2 and p. 334]. Hence, by substituting the expansions in (5) and (6) into (7), by

334

TAKESHI KAWAZOE

changing the order of integration and summation, and by shifting the integral line from R to R+i\, we can deduce that f (x)=e &\x

|

f (*) 8(*, x) C(&*) &1 e i*x d*

(x # R + )

R 

=e &\x : e &2mx (1 0m C 0+ F 0f )(x) m=0 

=e &2\x : e &2mx (1 1m C 1+ F 1f )(x).

(10)

m=0

This manipulation is valid provided that f (*) is holomorphic on F( \) and satisfies sup 0!\  R+i! | f (*) C(&*) &1 | d*<. In this case, since |1 m(*)| &2mx 2: cm 2: (m # N) if I(*)\ (see Proposition 8.3 below) and   m m=0 e &(2:+1) tx as x  0 and t1 as x  , the right side of (10) is dominated by e &2\x (1+D(x))D(x) (see (4)). We then define

{

L(GK)= f # L 1loc(GK); f is holomorphic on F(\) and & f & L(G) = sup 0!\

|

=

| f (*) C(&*) &1 | d*< .

R+i!

3. A KEY LEMMA Lemma 3.1. Let T be a sublinear operator of R satisfying the following properties: there exists =>0 such that for each (1, 2, 0)-atom a with supp(a) /[x 0 &l, x 0 +l] (a)

&Ta& L2(R) &a& L2(R) ,

(b)

|(Ta)(x)| l = |x&x 0 | &(1+=)

if

|x&x 0 | 2l.

Then there exists a positive constant c=c = such that for any F # H 1(R)

"



e &2\x : e &2mx (T1 1m F) } D m=0

"

L1(R + )

c &F& H1(R) .

(11)

Proof. We abbreviate & } & L2(R) as & } & 2 . We first observe that &Ta& L1(R)   |x&x0 | 2l |(Ta)(x)| dx +  |x&x0 | >2l |(Ta)(x)| dx  (2l) 12 &a& 2 + l =  |x&x0 | >2l |x&x 0 | &(1+=) dxc, because a is a (1, 2, 0)-atom. Hence T is a bounded sublinear operator of H 1(R) to L 1(R). Let F # H 1(R). Since

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

335

1 0 #1 (cf. [14, 9.1.5]), the term corresponding to m=0 in the left side of (11) is dominated as &e &2\x (TF ) } D& L1(R + ) c &TF& L1(R) c &F& H1(R) .

(12)

In estimating the rest terms, by substituting a (1, 2, 0)-atomic decomposition of F (cf. [5, Theorem 3.30]) it suffices to show that for any (1, 2, 0)atom a on R,

"



e &2\x : e &2mx (T1 1m a) } D m=1

"

c.

(13)

L1(R + )

We notice that 1 m(*+i\) (* # F) satisfies the Hormander condition with the constant cm 4: (see Corollary 8.4 below) and therefore, the multiplier theorem obtained by Taibleson and Weiss [12, Theorem 4.2] asserts that 1 1m a is a (1, 2, 0, 12)-molecule with &1 1m a& H1(R) cm 2:. Moreover, the proof of Theorem 2.9 in [12] yields that 1 1m a has a (1, 2, 0)-atomic decomposition such that  m 1 1m a= : d m k bk , k=0 &k2 2: where |d m m and, if we denote by x 0 the center of the support k | =c 2 m of a and we put _ m =&1 1m am 2:& &2 2 , b k is a (1, 2, 0)-atom on R supported m k m on the interval I k =[x; |x&x 0 | 2 _ m ]. Let l m k = |I k |. Since a and 1 m a are (1, 2, 0, 12)-molecules on R, it follows from the proof of Theorem 4.2 &2 and _ m =m &4: & |x&x 0 | 1 m a& 22 in [12] that _ m =m 4: &1 m a& &2 2 c&a& 2 &2 &a& 2 , that is,

_ m t&a& &2 2

and

k &2 lm k t2 &a& 2 .

(14)

We now split the region of integration in (13) into (0, 1] and (1, ). The integral over (1, ) is estimated as

|



1



: e &2mx |(T1 1m a)(x)| e &2\xD(x) dx m=1 

c : e &2m &T1 1m a& L1(R) m=1 

c : e &2mm 2: <. m=1

336

TAKESHI KAWAZOE

On the other hand, the estimate for (0, 1] is obtained as follows:

|

1



: e &2mx |(T1 1m a)(x)| e &2\xD(x) dx

0 m=1

c

|



1



m : e &2mx : |d m k | |(Tb k )(x)| D(x) dx

0 m=1 



c :

:

k=0 m=1

k=0

|

1

0

2:+1 e &2mx 2 &k2m 2: |(Tb m dx k )(x)| x

=I 1 +I 2 , where I 1 is the integral over the region D 1 : 0


I 1 c : 2 &k2 : 2 k= &a& &2= 2 k=0

m=1

|

e &2mxm 2:x 2:+1 |x&x 0 | &(1+=) dx

D1



c : 2 &k2 2 k= &a& &2= 2 k=0

|

&2

|x&x0 | >c 2k &a&2

|x&x 0 | &(1+=) dx



c : 2 &k2 <. k=0

In obtaining the estimate of I 2 we break the integral over D 2 as 

|

=:

D2

j=0

|

+ 2 &j&1
|

0x1 m m |x&x0 | 2lk , x0 >3lk

and we denote the corresponding terms by I 21 and I 22 respectively. As for k I 21 , if the integral does not vanish, we see that 2 & j&1 x 0 +2l m k 5l m  k &2 k k k &12 &k2 c2 &a& 2 by (14) and moreover, &Tb m & 2 &b m & 2 c(l m ) t2 &a& 2 by the property (a) of T. Therefore, taking the maximal values of e &2mx and x 2:+1 in the integrand and applying the Schwarz inequality to the integral, we can estimate I 21 as 

c : 2 &k2

: &2 2 &j c 2k &a&2

k=0 

c : 2 &k2 k=0

\

\

:

c : 2 &k2 <.

+

&j

m=1

&2 2 &j c 2k &a& 2



k=0



: e &m 2 m 2:2 & j(2:+1) 2 & j22 &k2 &a& 2

+

2 & j2 2 &k2 &a& 2

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

337

As for I 22 , if the integral does not vanish, we see that xx 0 +2l m k 5x 0 3 and xx 0 &2l m >x 3. Therefore, we can deduce that 0 k 

\

I 22 c : 2 &k2 k=0



+

: e &2mx03m 2: (5x 0 3) 2:+1 &Tb m k & L1(R) m=1



c : 2 &k2 <. k=0

This completes the proof of the lemma. K Remark 3.2. Let T be as in Lemma 3.1. Let T 8, D denote a pseudodifferential operator with the symbol _ 8, D (x, *) defined by e &2\x8(*+ i\, x) D(x) if x # R + and 0 otherwise. Then from (6) the conclusion of Lemma 3.1 can be restated as &T 8, D TF& L1(R) c &F& H1(R) for all F # H 1(R). Especially, taking the identity operator as T, we see that T 8, D is a bounded linear operator of H 1(R) to L 1(R). On the other hand, it follows from Proposition 8.3 below that d d*

}\ +

M

}

_ 8, D (x, *) c(1+ |*| ) &M

for M=0, 1. If the same estimate were also true for the derivative of x, or a modulus continuity of x, we would obtain the boundedness of T 8, D directly from Theoreme 9 in Coifman and Meyer [2].

4. L 1-CONDITION As stated in Theorem 1.1, F 1f is integrable on R if f is integrable on G. However, the reverse is not true. In this section we obtain a condition of F 1f under which f is integrable on G. Theorem 4.1. Let us suppose that C 1+ F 1f is well-defined for f # L (GK) and it belongs to H 1(R). Then f belongs to L 1(GK). In particular, 1 loc

& f & L1(G) c &C 1+ F 1f & H1(R) , where c is independent of f.

338

TAKESHI KAWAZOE

Proof. We first prove the inequality for f # L(GK) with &C 1+ F 1f & H1(R) <, in which case we see from (10) that 

f (x)=e &2\x : e &2mx (1 1m C 1+ F 1f )(x)

(x # R + ).

m=0

By taking the identity operator as T in Lemma 3.1 we can deduce that & f & L1(G) =& fD& L1(R + ) c &C 1+ F 1f & H1(R) .

(15)

For a general f # L 1loc(GK) with &C 1+ F 1f & H1(R) < we approximate f by functions in L(GK). Let  0 be an even C  function on R with  R  0(x) dx=1 and put  (=)(x)= 0(x) e =\x for = # R + . Since  0t (=*) is an even holomorphic function of exponential type, the PaleyWiener theorem (=) 7 ) (*)= on G yields that there exists , (=) # C  c (GK) such that (, t (=) &1 (=) &1 1  0 (=*). Let  $ (x)=$  ($ x) for $ # R + . Then F ,(=)(x)= (=) = (x) t because (F 1,(=) ) t (*)=(, (=) ) 7 (*+i\)= 0t (=(*+i\))=( (=) ) (*), and more= over, f V , (=) # L(GK) and C 1+ F 1f V ,(=) =(C 1+ F 1f ) V  (=) = . We here note the following, Lemma 4.2.

1 Let F be in H 1(R). Then F V  (=) = # H (R) for 0<=<1 and

&F&F V  (=) = & H1(R)  0

(=  0).

Proof. We refer to the notations and the results in Folland and Stein [5, Section 2]. Since & (=)& (1) c for all 0<=<1, it follows from Lemma 3.31 in [5] that there exists N1 such that sup 0<$< (M (N)(F V  (=) $ ))(x) c(M (1) F )(x). Especially, (M (N)(F V  (=) = ))(x)c(M (1) F)(x) and thus, 1 F V  (=) = # H (R) by Theorem 3.30 in [5]. The rest of the proof follows from the same argument as in the proof of Theorem 3.33 in [5]. K 1 1 1 1 1 Hence, C 1+ F 1f V ,(=) =(C 1+ F 1f ) V  (=) = # H (R) and &C + F f &C + F f V ,(=) & H1(R) (=)  0 as =  0. On the other hand, since f V , # L(GK), it follows from (15) that & f V , (=)& L1(G) c &C 1+ F 1f V ,(=) & H1(R) . Therefore, f V , (=) converges to a function h in L 1(GK). Clearly, h must be f, because (, (=) ) 7 (*)   0t (0)=1 as =  0. So, letting =  0, we obtain the desired inequality for f. This completes the proof of Theorem 4.1. K

We now define the Hardy space H 1s (GK) (s0) on G. Let Qs (s0) denote the Fourier multiplier of R defined by (Qs F ) t (*)=

\

1+ |*| |*|

s

+F

t

(*)

(* # R).

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

Definition 4.3.

339

For s0

H 1s (GK)=[ f # L 1loc(GK); Qs C 1+ F 1f is well-defined and is in H 1(R)] and & f & H1s (G) =&Qs C 1+ F 1f & H1(R) . Theorem 4.4. If s$s0, then H 1s$(GK)/H 1s (GK)/L 1(GK) and for all f # H 1s$(GK) & f & L1(G) c & f & H1s (G) c$ & f & H1s$(G) , where c and c$ are independent of f. Proof. Theorem 4.1 asserts that & f & L1(G) c & f & H10(G) . Let Rs denote the Fourier multiplier on R defined by (Rs F) t (*)=F t(*)(|*|+1|*|) s (* # R). We notice that (|*|1+ |*| ) s (s0) satisfies the Hormander condition and thus, Rs is a bounded linear operator on H 1(R) (see [12, Theorem 4.2]). Therefore, for all f # H 1s$(GK), & f & H1s (G) =&Rs$&s Qs$ C 1+ F 1f & H1(R) c &Qs$ C 1+ F 1f & H1(R) = & f & Hs$1 (G) . K Theorem 4.5. For s0, H 1s (GK) & L(GK) is dense in H 1s (GK). Proof. We retain the notation used in the proof of Theorem 4.1 and suppose that f # H 1s (GK). Since Qs C 1+ F 1f # H 1(R) and Qs C 1+ F 1f V ,(=) = 1 1 (=) (cf. [4, Theorem 5]), it follows from Qs C 1+(F 1f V  (=) = )=(Qs C + F f ) V  = (=) 1 Lemma 4.2 that f V , # H s (GK) & L(GK) and & f& f V , (=)& H1s (G)  0 (=  0). K Remark 4.6. H 10(GK) contains all f # C  c (GK) with  G f (g) dg=0. Indeed, we suppose that supp( f )/B(r), and we observe that f (*) is holomorphic on Fc of exponential type r and C(&*) &1 is holomorphic and tempered on the upper half plane (cf. [14, 9.2.3] and [4, Lemma 8]). We here recall the technique used in the proof of the PaleyWiener theorem (cf. [14, 9.2.3]). Hence, shifting the integral line R of the integral  R C(&*&i\) &1 f (*+i\) e i*x d* defining C 1+ F 1f to R+i' ('  +), we can deduce that supp(C 1+ F 1f )/(&, r]. Thereby, &C 1+ F 1f & L2(R) = &e \x (C 0+ F 0f )& L2(R) e \r &C 0+ F 0f & L2(R) =e \r & f & L2(G) by the Plancherel formulas on R and G, and similarly, &x(C 1+ F 1f )& L2(R) c(1+re \r ) & f & L2(G) . On the other hand, since , i\ #1 and C(&i\)=1 (cf. [4, Section 3 and Lemma 8]), we have  R (C 1+ F 1f )(x) dx=C(&i\) &1 f (i\)= G f (g) dg=0. These follow immediately that C 1+ F 1f is a (1, 2, 0, 12)-molecule on R and hence, in H 1(R) (see [12, Theorem 2.9]).

340

TAKESHI KAWAZOE

5. CRITERIA FOR BOUNDEDNESS We fix 0
(g # G).

r1
In this section we obtain some conditions on , under which T G,, r1, r2 is a bounded sublinear operator of H 1s (GK) (s0) to L 1(GK). Let B r1, r2 be the set of all functions ;(r, *) on (r 1 , r 2 )_R for which there exists a continuous function 3=3 ; on R such that (a 1 ) 3(*) # L 1(R), (a 2 ) (b)

lim 3(*)=0,

(16)

|*|  

d d*

}\ +

M

}

;(r, *) r M3(r*)

(M=0, 1, 2),

and B + r1, r2 the subset of B r1, r2 defined by replacing (16) (a 1 ), (a 2 ) with (a 1 ) (a 2 )

*3(*) # L 1(R),

(17)

lim *3(*)=0.

|*|  

Lemma 5.1. If ,(r, *+i\) # B r1, r2 , then the maximal operator defined by sup r1 1, and a weak type L1(R) estimate. Proof. Since (F 1,(r, } ) ) t (*)=,(r, *+i\) (see (9)), it follows from (16) of ,(r, *+i\) that |x MF 1,(r, })(x)|| R (dd*) M ,(r, *+i\)} e i*x d*|&3& L1(R) r M&1 (M=0, 1, 2) and thus, |F 1,(r, } )(x)| cr &1(1+r &1 |x| ) &2 (x # R). This inequality easily yields that sup r1
|(F 1,(r, } ) V a)(x)| cl |x&x 0 | &2

r1
where c is independent of a.

if

|x&x 0 | 2l,

341

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

Proof. The moment condition  R a(x) dx=0 and the mean value theorem yield that for x # R + (F 1,(r, } ) V a)(x)=

|

F 1,(r, } )(x& y) a( y) dy

=

|

(F 1,(r, } )(x& y)&F 1,(r, } )(x&x 0 )) a( y) dy

=

R

|

R l

&l

\

d 1 (x& y 0 ) (x 0 & y) a( y) dy, F dx ,(r, } )

+

where y 0 is on the line segment from y to x 0 . Since (dF 1,(r, } ) dx) t (*)= i*,(r, *+i\) (see (9)), it follows from (16)(b) and (17) of ,(r, *+i\) that

}

(x& y 0 ) 2

d 1 (x& y 0 ) = F dx ,(r, } )

d d*

2

} } | \ + (*,(r, *+i\)) e R

2r

|

|3(r*)| d*+r 2 R

|

i*(x& y0 )

d*

}

|*3(r*)| d* R

=2 &3& L1(R) +&*3& L1(R) . Let |x&x 0 | 2l. Since |x 0 & y| l and thus, |x&x 0 | |x& y 0 | &1 1+ |x 0 & y 0 | |x& y 0 | &1 2. Therefore, if |x&x 0 | 2l, sup r1
Proof. From Theorem 4.5 it suffices to obtain the boundedness for f # H 10(GK)&L(GK). Since ( f V ,(r, } )) 7 (*)=f (*) ,(r, *)=f (*)(F 0,(r, } ) ) t (*) (cf. [4, Theorem 5]), the same manipulation as in (10) yields that for x # R+ 

( f V ,(r, } ))(x)=e &2\x : e &2mx (F 1,(r, } ) V (1 1m C 1+ F 1f ))(x) m=0

and thereby, 

(T G,, r1, r2 f )(x)e &2\x : e &2mx sup m=0

|(F 1,(r, } ) V (1 1m C 1+ F 1f ))(x)|.

(18)

r1
Since ,(r, *+i\) # B + r1, r2 , Lemmas 5.1 and 5.2 imply that the maximal operator sup r1
342

TAKESHI KAWAZOE

(a) and (b) in Lemma 3.1. Therefore, the result follows from Lemma 3.1 and Definition 4.3. K Let B $r1, r2 ($ # R) be the set of all functions ;(r, *) on (r 1 , r 2 )_R for which there exists a continuous function 3=3 ; on R such that (a 1 ) (a 2 )

* $+13(*) # L 1(R), lim * $+13(*)=0, d d*

}\ +

(b)

(19)

|*|  

M

}

;(r, *) r M+$3(r*)

(M=0, 1, 2).

Proposition 5.4. If ,(r, *+i\) # B $r1, r2 ($2), then T G ,, r1, r2 is a bounded sublinear operator of H 1$(GK) to L 1(GK). Proof. As we have deduced (18), we see that for f # H 1$(GK) & L(GK) 

(T G,, r1, r2 f )(x)e &2\x : e &2mx |(T1 1m Q$ C 1+ F 1f )(x)|

(x # R + ),

m=0

where T is a maximal operator on L 1loc(R) defined by (TF )(x)= sup r1
|

F t (*) R

\

|*| 1+ |*|

$

+ ,(r, *+i\) e

i*x

d*

(x # R).

Then, as in the proof of Proposition 5.3 it suffices to show that (|*|1 + |*|) $ ,(r, * + i\) # B + r1, r2 . Indeed, it follows from (19)(b) of ,(r, *+i\) that for M=0, 1, 2 d d*

M

}\ + \\

|*| 1+ |*|

M

c : m=0

$

+ ,(r, *+i\)+} d |*| d r } r } \d*+ \1+ |*| + } } \d*+ M&m

$

$

&$

m

,(r, *+i\)

}

M

c : $($&1) } } } ($&M+m+1) |r*| $&M+m r M&m } r m3(r*) m=0

cr M9(r*) 3(r*), where 9(*)=($($&1) |*| $&2 +$ |*| $&1 + |*| $ ). Therefore, if we define 3 in (16) and (17) by 93, the above calculation and (19)(a 1 ), (a 2 ) imply that K (|*|1+ |*| ) $ ,(r, *+i\) # B + r1, r2 .

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

343

+ Let B $, r1, r2 ($ # R) be the set of all functions ;(r, *) on (r 1 , r 2 )_R for which there exists a continuous function 3=3 ; on R such that

* $+13(*) # L 1(R),

(a 1 )

lim * $+13(*)=0,

(a 1 )

|*|  

(b 1 )

| ;(r, *)| 3(r*),

(b 2 )

}\d*+ ;(r, *)} r 3(r*), d }\d*+ ;(r, *)} r 3(r*).

d

(20) $

2

(b 3 )

2+$

+ G Proposition 5.5. Let r 1 1 and $2. If ,(r, *+i\) # B $, r1, r2 , then T ,, r1, r2 1 1 is a bounded sublinear operator of H $&1+=(GK) to L (GK) for =>0.

Proof. We first notice that under the assumption that r 1 1 and $2, + ;(r, *) in B $, r1, r2 also satisfies (b 4 ) (b 5 )

| ;(r, *)| r $&13(r*), d d*

}\ +

M

}

;(r, *) r M+$3(r*)

(M=0, 1, 2).

In what follows we modify the proofs of Lemmas 5.1 and 5.2. As in the proof of Theorem 4.4, Rs (s # R) denotes the Fourier multiplier on R defined by (Rs F) t (*)=F t (*)(|*|1+ |*|) s (* # R). Since (R$&1 F 1,(r, } ) ) t (*) =,(r, *+i\) ( |*|1+ |*| ) $&1, we see from (a), (b 2 ), and (b 4 ) of ,(r, *+i\) that |(R$&1 F 1,(r, } ) )(x)| cx &1(($&1) &* $&23& L1(R) +&* $&13& L1(R) ) (21) and similarly from (a) and (b 5 ) that |(R$ F 1,(r, } ) )(x)| crx &2($($&1) &* $&23& L1(R) +$ &* $&13& L1(R) +&* $3& L1(R) ). We here fix r and x, and we observe that (Rz F 1,(r, } ) )(x) makes sense for z # C with $&1R(z)$ and, as a function of z, it is holomorphic on $&1
344

TAKESHI KAWAZOE

Since |(R$&1+= F 1,(r, } ) )(x)| r &1 &3& L1(R) , we have |(R $&1+= F 1,(r, } ) )(x)|  cr &1(1+r &1 |x| ) &(1+=) (0=1). In particular, if 0<=1, the maximal function defined by sup r1
}|

R

(R$&1 F 1,(r, } ) )(x& y) a( y) dy

}

c |x&x 0 | &1 (($&1) &* $&23& L1(R) +&* $&13& L1(R) ) if |x&x 0 | 2l. Moreover, |((Rz F 1,(r, } ) ) V a)(x)| &Rz F 1,(r, } ) &  &a& L1(R)  r &1 &3& L1(R) for $&1R(z)$. Then, applying Three Lines Lemma again, we see that for 0=1, |((R$&1+= F 1,(r, } ) ) V a)(x)| cl = |x&x 0 | &(1+=) if |x&x 0 | 2l. We have therefore proved that if 0<=1, the maximal operator defined by sup r1
e &2\x : e &2mx sup m=0

|((R$&1+= F 1,(r, } ) ) V (1 1m Q$&1+= C 1+ F 1f ))(x)|,

r1
the desired result follows from Lemma 3.1, Definition 4.3, and Theorem 4.4. K

6. MAXIMAL OPERATORS We apply the criteria obtained in the previous section to some radial maximal operators. Actually, as ,(r, g) in Section 5, we shall take ,
MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

345

, >r (g), , r ( g), h r( g), and p r( g), respectively (see AE below). For simplicity, we denote henceforth the set of bounded sublinear operators of H 1s (GK) to L 1(GK) by (H 1s , L 1 ) and that on L p(GK) by (L p, L p ). We also say that an operator T of L 1(GK) is of weak type (1, 1) provided there exists a constant c such that c |[ g # G; |(Tf )( g)| >:]|  & f & L1(G) : for all f # L 1(GK) and for every :>0. A. Let A <$ (G) ($ # R) be the set of all functions , # L 1(GK) satisfying |(dd*) M ,(*)| (1+ |*| ) &$ (* # F(\)) for M=0, 1, 2. Definition 6.1.

For f # L 1loc(GK)

< (M G, $, 0 f )( g)= sup

< , # A$ (G)

|( f V ,
( g # G),

0
where , 2, then M $, 0 # (H 0 , L ).

Proof. We easily see that if $>2, ,(r*) belongs to B + 0,  and hence, by Proposition 5.3 we have the desired result. K B. Let A N (G) (N # N) be the set of all functions , # C N (GK) whose restriction on A satisfies supp(,)/[&1, 1], &(ddx) n ,&  1 (0nN), and ,(x)=O(x N ). Definition 6.3.

For =0 and f # L 1loc(GK)

> (M G, N, = f )( g)= sup

(1+r) &= |( f V , >r )(g)|

(g # G),

, # AN(G) 0
where , >r ( g)=(1|B(r)|) ,(_(g)r). Theorem 6.4. (1)

> p p M G, N, = is in (L , L ) for p>1 and of weak type (1, 1),

(2)

> 1 1 If N6, then M G, N, 0 # (H $ , L ) for $>2,

(3)

> 1 1 If N4 and =>1, then M G, N, = # (H 0 , L ).

> Proof. (1) is obvious from Theorem 1.2 because M G, N, = f is pointwisely dominated by M GHL f. To prove (2) and (3) we first obtain the following,

346

TAKESHI KAWAZOE

Lemma 6.5. d d*

}\ +

Let , # A 2N (G) (N # N). Then, for M=0, 1, 2,

M

}

(, >r ) 7 (*+i\) cr M(1+r) s (1+ |r*| ) &2s

(0sN).

Proof. We recall that |(dd*) m , *+i\(x)| x m, i\(x)x m (* # F, x # R + ), , * is an eigenfunction of the LaplaceBeltrami operator 0 on G with eigenvalue & p(*)=&* 2 &\ 2, and the radial component of 0 on R + is of the form D &1 } (ddx(D } ddx)). For these facts we refer to [4], Lemma 14, Proposition 3, and Section 2 respectively. We now apply them to the integral defining the Fourier transform (, >r ) 7 (*+i\) and thereby, we can deduce that for each m, n # N and 0nN d d*

}\ +

m



( p(*+i\) n } (, >r ) 7 (*+i\))

|

 0

}

d d*

c |B(r)| &1 r m

m

\ +, | }\ D(x)

0 n, >r (x) }

r

0

}

} d x d D(x) , \ r +} D(x) dx. dx \ dx ++

*+i\

(x) D(x) dx

n

&1

(22)

For simplicity, we put D=D &1(dDdx) and 8(x)=,(xr) (r # R + ), and we observe that (ddx) p 8(x)tr &2Nx 2N& p r &2nx 2n& p if 00. Hence we have d dx

2n

"\ + 8 " d d 8} "\dx+ \dx+ D } D " d "\dx+ 8 } D " p

cr &2n, 

q

s

cr &2n

if p+q+s<2n and q>0,

cr &2n(1+r) s

if p+s=2n.



p

s



Then, by using these estimates to handle the derivatives of ,(xr) in (22), we can deduce that |(dd*) m ( p(*+i\) n } (, >r ) 7 (*+i\))| cr m&2n(1+r) n. Specifically, letting m=0, we obtain the inequality for M=0 and s=n. Since | p(*+i\) n } (dd*) m (, >r ) 7 (*+i\)| is dominated by |(dd*) m ( p(*+i\) n } i n m&i (, >r ) 7 (*+i\))|+c  m (, >r ) 7 (*+i\)|, the i=1 |(dd*) ( p(*+i\) ) } (dd*) rest of the inequalities follow by induction and interpolation. K As for (2) let , # A N (G) (N6). By Lemma 6.5, (, >r ) 7 (*+i\) # B + 0, 1 if 0 1 1 5.3, 5.4, and Theorem 4.4 it follows that M G, N, 0 # (H $ , L ) for $>2. As for (3) let , # A N (G) (N4). Since (1+r) &= (, >r ) 7 (*+i\) # B + 0,  (=>1) by

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

347

> 1 1 Lemma 6.5, it follows from Proposition 5.3 that M G, N, = # (H 0 , L ) for =>1. K

C. Let A N (G) be the same as in B. We here introduce a dilation which preserves the L 1-norm on G (see [7]). Definition 6.6.

For =0 and f # L 1loc(GK)

 (M G, N, = f )( g)= sup

(1+r) &= |( f V , r )( g)|

(g # G),

, # AN(G) 0
where , r ( g)=(1r)(1D(_(g))) D(_( g)r) ,(_(g)r). Theorem 6.7. (1) (2)

 p p M G, N, = is in (L , L ) for p>1 and is of weak type (1, 1), G,  If N6, then M N, 0 # (H 10 , L 1 ) for $>1,

(3)

 1 1 If N4 and =>1, then M G, N, = # (H 0 , L ).

Proof. It is easy to verify that &, r & L1(G) =&, 1 & L1(G) and moreover, , (x)c, >r (x) if 01 and of weak type (1, 1). We now suppose that , # A 2N (G) and we observe that for each 0n2N and 0xr, (ddx) n ,(xr)tcr &2Nx 2N&n and  r

n

d dx

x r+x cr &n r x

n

x

x

\ + \ + \ + D \ r + cx D \ r + , d 1+x \dx+ D(x) c \ x + D(x) . D

n

&n

n

&1

&1

Then the same argument as in the proof of Lemma 6.5 yields that, for M=0, 1, 2, d d*

}\ +

M

}

(, r ) 7 (*+i\) cr M(1+r) s (1+ |r*| ) &2s

(0sN),

and therefore, the desired result follows. K D. The heat maximal operator M G H, = (=0) on G is given as follows. Definition 6.8.

For =0 and f # L 1loc(GK)

(M GH, = f )(g)= sup (1+r) &= |( f V h r )( g)| 0
where h r( g)= R e &(*

2 +\2 ) r

, *( g) |C(*)| &2 d*.

(g # G),

348

TAKESHI KAWAZOE

Theorem 6.9. p p (1) M G H, = is in (L , L ) for p>1 and of weak type (1, 1), G 1 1 (2) M H, 0 # (H $ , L ) for $>1, 1 1 (3) M G H, 1 # (H 0 , L ). Proof. (1) is well-known (cf. [9, p. 73] and [1, Corollary 3.2]). Since 2 2, + h r(*+i\)=e &* re &2i\*r, we see that h r2(*+i\) # B + 0, 1 if 0
For =0 and f # L 1loc(GK)

(M GP, = f )( g)= sup (1+r) &= |( f V p r )( g)|

(g # G),

0
2 12 r

where p r( g)= R e &(* +\ )

, *(g) |C(*)| &2 d*.

Theorem 6.11. p p (1) M G P, = is in (L , L ) for p>1 and of weak type (1, 1), 1 1 (2) M G P, 0 # (H 12 , L ), 1 1 (3) M G P, 2 # (H 0 , L ). Proof. As shown by Stein [9, p. 48], it is well-known that M GP, = is in (L p, L p ) for p>1 and of weak type (1, 1). Let us recall the calculation in [9, p. 49] and modify it slightly: The subordination formula gives an integral representation of f V p r such as ( f V p r )( g)=

1 r2

|



( f V h y )( g) ,

0

y

\r + dy 2

( g # G),

where ,( y)=1(2? 12 ) e &14 yy &32, and the integral by part yields the estimate of (1+r) &2 ( f V p r )(g) such as 1 (1+r) 2 r 2



} | \| 0

1 0< y< y

 sup _

y

0

|

y

0

1 (1+r) 2 r 2

1 d y ( f V h s )(g) ds (1+ y) , 2 1+s dy r

+ \

1 |( f V h s )(g)| ds 1+s

|



0

C(M G H, 1 f )( g),

y

d

y

} dy \ (1+ y) , \r ++} dy 2

\ ++ dy }

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

349

where C=&y,& L1(R) +&(1+ y) y } d,dy& L1(R) . Therefore, it follows from 0 G, 1 Theorem 6.9 that M GP, 2 # (H 10 , L 1 ). Let M G, P, 0 (resp. M P, 0 ) denote the maximal operator defined by replacing sup 0
7. OTHER OPERATORS A. We shall define the Fourier multiplier on G corresponding to an even bounded function m(*) on F. Definition 7.1.

For f # L 1loc(GK)

(T Gm f )(g)=

|

f (*) m(*) , *( g) |C(*)| &2 d*

(g # G).

F

We easily see that T Gm # (L 2, L 2 ) by the Plancherel formula on G. We now suppose that m(*) has a holomorphic extension on F( \) and there exists s0 such that sup &m !, s &<,

where

0!\

m !, s(*)=m(*+i!)

\

s |*+i(!&\)| . 1+ |*+i(!&\)|

+

Then by the same manipulation as in (10) we see that for f # L(GK), T Gm f has the expansion: 

(T Gm f )(x)=e &2\x : e &2nxT m\, 0 1 1c C 1+ F 1f (x)

(x # R + )

n=0

=(T 8, D T m\, s )(Qs C 1+ F 1f )(x) D(x) &1, where T w is the Fourier multiplier on R defined by (T w F ) t (*)= w(*) F t (*) (* # R) and T 8, D is the pseudo-differential operator given in Remark 3.2. Since T 8, D is a bounded operator of H 1(R) to L 1(R), we can obtain from Definition 4.3 and Theorem 4.5 that

350

TAKESHI KAWAZOE

Theorem 7.2. Let m(*) be an even holomorphic function on F(\). If there exists s0 such that sup 0!\ &m !, s &< and T m\, s is bounded on H 1(R), then T Gm # (H 1s , L 1 ). B. We last treat the Riesz transform on G, and we henceforth use the standard notation as in [9]. Especially, we denote the Laplacian on G by 2 and the covariant differentiation on G by {. We put |{| 2 (h)= 2(h 2 )&22h } h for h # C (G). Then the Riesz transform R G on G is given as follows: Definition 7.3.

For f # L 1loc(GK)

(R Gf )( g)=( |{| (&2) &12 )( f )( g)

(g # G).

R Gf is a K-bi-invariant function on G. Indeed, (&2) &12 =T G1- p where p(*)=* 2 +\ 2 and we notice that for h # C (G), |{h| 2 (g)=  nm=1 |X i h| 2 ( g) (g # G), here [X i ; 1in] is denoted as an orthonormal basis of the Lie algebra g of G and each X i is regarded as a left (or right) invariant differential operator on G, and furthermore, if h is K-bi-invariant on G, |{h| 2 ( g) is simply expressed as c |(ddx) h(a x )| 2 provided _( g)=x (cf. [4, Section 2]). Theorem 7.4. (2)

(1)

R G is in (L p, L p ) for p>1 and of weak type (1, 1),

R G # (H 112 , L 1 ).

Proof. (1) is well-known (see [1, Corollary 5.2]). As for (2) we may assume that f # H 112(GK) & L(GK) and then, we have 

((&2) &12f )(x)=e &2\x : e &2mx (T 1- p\, 0 1 1m C 1+ F 1f )(x)

(x # R + ).

m=0

As remarked after Definition 7.3, in order to obtain the estimate of R Gf it suffices to calculate the action of ddx on the right side of the above equation. Actually, ddx acts on e &2\x, e &2mx, and (T 1- p\, 0 1 1m C 1+ F 1f )(x), so we denote the results by (I 1 f )(x), (I 2 f )(x), and (I 3 f )(x), respectively. Since P(*+i\) &12 ( |*| +1+ |*| ) 12 satisfies the Hormander condition, it follows from Theorem 7.2 that &I 1 f } D& L1(R+ ) c & f & H11 (G) . Similarly, since (ddx) 2 (T 1- p\, 0 1 1m C 1+ F 1f )(x)=(T 1- p\, 0 } (i*) 1 1m C 1+ F 1f )(x) and p(*+i\) &12 (i*) satisfies the Hormander condition, we have &I 3 f } D& L1(R+ ) c & f & H10(G) . In obtaining the estimate for I 2 f, we rewrite it as follows: 

(I 2 f )(x)=e &2\x : (&2m) e &2mx (# 1m Q12 C 1+ F 1f )(x) m=1

(x # R + ),

351

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

where # 1m is the Fourier multiplier on R defined by (# 1m F) t (*)=1 m(*+i\) p(*+i\) &12

\

|*| 1+ |*|

+

12

F t (*)

(* # R).

We observe from (25), (26) and Lemma 8.1 below that for M=0, 1, d d*

}\ +

M

}

1 m(*+i\) cm 2:+1(1+ |*|) &M

\ |2\&i*| m |m+\&i*|

and hence, d d*

M

}\ + \

+\

1 m(*+i\) p(*+i\) &12 }

|*| 1+ |*|

12

+ } cm

2:&1

|*| &M.

Therefore, 1 m(*+i\) p(*+i\) &12 } ( |*|1+ |*| ) 12 satisfies the Hormander condition with the constant cm 4:&2. We now repeat the proof of Lemma 3.1 when T is the identity operator, in which case we replace 1 1m and F by # 1m and Q12 C 1+ F 1f , respectively. Fortunately, the extra order of m that appears in the differentiation of e &2mx is cancelled, because the operator norm cm 2: of 1 1m is changed to cm 2:&1 of # 1m . So the exactly same proof of Lemma 3.1 is valid in this case and it follows that &I 2 f } 2& L1(R+ )  c &Q12 C 1+ F 1f & H1(R) =c & f & H 112(G) . This completes the proof of (2). K Remark 7.5. Theorem 6.11(2) also follows from Theorem 7.4(2) and [1, Corollary 6.3]. Remark 7.6. Let R=T i*|*| be the Riesz transform on R and Q the Fourier multiplier on R defined by (QF ) t (*)=

\

* 2 +4\ 2 *2 &1

+

14

} ei 2

&1 tan &1(2\*)

}

* } F t (*) |*|

(* # R).

&1

We observe that e &i 2 tan (2\*) } *|*| } (* 2 +4\ 2 ) &14 (1+ |*| ) 12 satisfies the Hormander condition and hence, if QF # H 1(R), then Q12 F # H 1(R). In fact, Definition 4.3 and Theorem 7.4(2) yield that if f # H 112(GK), then &R Gf & L2(G) +&QC 1+ F 1f & L1(R) c &QC 1+ F 1f & H1(R) .

(23)

We now suppose that f # L 1loc(GK) satisfies &R Gf & L1(G) +&QF 1f & L1(R) <. If we take X 1 in a in the remark after Definition 7.3, we see that |{h| (a x g) |(ddx) h(a x g)| (g # G) for h # C (G), and thereby,

352

TAKESHI KAWAZOE

&R Gf & L1(G) =

| | A



| | A



|

N

}

d ((&2) &12 f )(a x n) e 2\x dn dx dx

}

d

A

=

|(R Gf )(a x n)| e 2\x dn dx N

} dx |

d

}

((&2) &12 f )(a x n) dn e 2\x dx

N

" dx F

&1 (&2) &12 f

} e 2\x

&1

"

L1(R)

=&RQF 1f & L1(R) ,

&1

because (*+2i\)(* 2 +2i\*) &12 =e i 2 tan (2\*) } (* 2 +4\ 2*2 ) 14. Therefore, the assumption on f yields that QF 1f # H 1(R) and &QF 1f & H1(R) c(&R Gf & L1(G) +&QF 1f & L1(R) ).

(24)

If we let \  0 formally, then F 1f goes to an even function f on R, R G to R on R, and C 1+ , Q to the identity operators on R, so from (23) and (24) we can recover the classical result: & f & H1(R) t&Rf & L1(R) +& f & L1(R) (cf. [10, p. 123]). We last restrict our attention to functions f # L(GK) such that QC 1+ F 1f # L 1(R) (see Section 2 and Remark 7.6), and we define L 1(GK)=[ f # L(GK); & f & L1(G) =& f & L(G) +&QC 1+ F 1f & L1(R) <] and H 1(GK)=[ f # L(GK); & f & H1(G) =& f & L(G) +&QC 1+ F 1f & H1(R) <]. Obviously, H 1(GK)/L 1(GK) and &R Gf & L1(G) +& f & L1(G) c & f & H1(G) (see (23)). We now suppose that f # L 1(GK) and we recall the proof of Theorem 7.4(2). Especially, |(R Gf )(x)| = | 3k=1 (I k f )(x)| (x # R + ) where (I 1 f )(x)+(I 3 f )(x) 

=e &2\x : e &2mx (T i(*+2i\)(*2 +2i\*)12 1 1m C 1+ F 1f )(x) m=0 

=e &2\x : e &2mx (1 1m RQC 1+ F 1f )(x), m=0 

I 2 f (x)=e &2\x : (&2m) e &2mx (# 1m Q12 C 1+ F 1f )(x). m=1

(x # R + )

353

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

Therefore, we can deduce that for x # R + 

|(RQC 1+ F 1f )(x)| e 2\x |(R Gf )(x)| + : e &2mx |(1 1m RQC 1+ F 1f )(x)| m=1 

+ : 2me &2mx |(# 1m Q12 C 1+ F 1f )(x)|. m=1

Since &(QC 1+ F 1f ) t &  &QC 1+ F 1f & L1(R) & f & L1(G) and &(C 1+ F 1f ) t & L1(R)  & f & L(G) & f & L1(G) , it follows that |(1 1m RQC 1+ F 1f )(x)| cm 2:& f & L1(G) and similarly, |(# 1m Q12 C 1+ F 1f )(x)| = |(# 1m Q12 Q &1 )(QC 1+ F 1f )(x)| cm 2:&1 & f & L1(G) (see Section 2 and the proof of Theorem 7.4(2)). Hence we have,

|

 1

|(RQC 1+ F 1f )(x)| dx

|

c







|

1

i(*+2i\) f (*+i\) C(&*&i\) &1 e i*x d* (* 2 +2i\*) 12

}

|(R Gf )(x)| D(x) dx+c & f & L1(G) : m 2:

1

m=1

e &2mx dx

c(&R Gf & L1(G) +& f & L1(G) ). On the other hand, since |(RQC 1+ F 1f )(x)| =

}|

R

=e \x

}|

R

i(*+i\) f (*) C(&*) &1 e i*x d* (* 2 +\ 2 ) 12

(x # R)

}

c & f & L(G) e \x, it follows that  1& |(RQC 1+ F 1f )(x)| dxc & f & L1(G) . Then, we can deduce that &RQC 1+ F 1f & L1(R) c(&R Gf & L1(G) +& f & L1(G) ). In particular, & f & H1(G) = & f & L(G) + &QC 1+ F 1f & H1(R)  c(& f & L(G) +&QC 1+ F 1f & L1(R) +&RQC 1+ F 1f & L1(R) ) c(&R Gf & L1(G) +& f & L1(G) ). We have therefore proved the following, Theorem 7.7. f # H 1(GK) if and only if f # L 1(GK) and R G # L (GK). Especially, 1

& f & H1(G) t&R Gf & L1(G) +& f & L1(G) .

354

TAKESHI KAWAZOE

8. APPENDIX We shall obtain an estimate for the derivatives of 1 m(*) (see (6)), which yields the Hormander condition of 1 m . In what follows we denote 1 m by 1 2m and refer to the notation and the proof of Lemma 7 in FlenstedJensen [4]. Actually, 1 m (m # N) is recurrently defined by 1 0 =1, 1 2n+1 =0 (n # N) and m&1

4m(m&i*) 1 2m(*)= : (2k&i*+\) 4((:&;)+$ m k (2;+1)) 1 2k(*), k=0 m where $ m k =0 for k#m+1 (mod 2), $ k =1 for k#n (mod 2). For each m # N we put

C k(*)=4k(k&i*)

R k(*)=4%(2k&i*+\),

and

where %=:&; if k#m+1 and %=\ if k#m. Then it follows that m&1

1 2m(*)=C m(*) &1 : R k(*) 1 2k(*) k=0 m&1

=

R k(*) R 0(*) . ` 1+ C m(*) k=1 C k(*)

\

+

Therefore, if we put c k(*)= |C k(*)|, r k(*)= |R k (*)| and m&1

b 0(*)=1

and

b m(*)=c m(*) &1 ` b k(*) c k(*)

for m1,

k=0

we easily see that |1 2m(*)| b m(*)=

\ | \&i*| m&1 r k(*) . ` 1+ m |m&i*| k=1 c k(*)

\

+

(25)

Lemma 8.1. Let *=!+i' and suppose that '\. Then there exists a positive constant c such that for all m # N m&1

` k=1

Proof.

r k(*)

\1+c (*)+ cm

2:+1

.

k

Since '\, we have |2k+\+'&i!||k+'&i!| 2 and thereby,

\

k

r k(*) c k(*)

2

+ (2%) . 2

355

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

Then it follows that m&1

` k=1

\

1+

r k(*) exp c k(*)

+

m&1

\ exp \ \

r k(*) c k(*)

:

k=1

+ 2k &1 +(:&;)

: k#m 1km&1

2k &1

: k#m+1 1km&1

+

c exp((\+:&;) log m) K

cm 2:+1

Lemma 8.2. Let I(*)0 and fix M # N. Then there exists a positive constant c such that for all m # N M

d d*

R 0(*)

r 0(*)

1

}\ + C (*) } c c (*) (1+ |*| ) , d 1 R (*) r (*) }\d*+ \1+ C (*)+} c \1+c (*)+ (k+ |*| )

(1)

M

m

m

M

(2) Proof.

k

k

k

(M1).

M+1

(1) d d*

}\ +

(2)

k

M

R 0(*) m&\ =c C m(*) C m(*)(m&i*) M

} }

}

c

m&\ r 0(*) c m(*) ( \&i*)(m&i*) M

c

r 0(*) 1 . c m(*) (1+ |*| ) M

}

}

For M1 d d*

M

R k(*) C k(*)

}\ + \ +} k&\ =c } C (*)(k&i*) } k&\ r (*) c 1+ \ c (*)+} ((k +%(2k+\))&(k+%) i*)(k&i*) } r (*) 1 c 1+ \ c (*)+ (k+ |*| ) . K 1+

M

k

k

M

2

k

k

M+1

k

356

TAKESHI KAWAZOE

We now suppose that '=I(*)0 and observe by Lemma 8.2 that d d R 0(*) m&1 R k(*) 1 2m(*) = ` 1+ d* d* C m(*) k=1 c k(*)

}\ +

} }\ + +

\

m&1

R 0(*) d C m(*) d*

+

R k(*)

\ + ` \1+ C (*)+} k

k=1

m&2

\

cb m(*) (1+ |*| ) &1 + : (1+ j+ |*| ) &2 j=0

cb m(*)(1+ |*| )

&1

.

+ (26)

Hence, by Lemma 8.1 and the above estimate we can deduce the following, Proposition 8.3. Let M=0, 1. If I(*)\, then there exists a positive constant c such that for all m # N d d*

}\ +

M

}

1 2m(*) cm 2:(1+ |*| ) &M.

Corollary 8.4. Suppose that '\. Then there exists a positive constant c such that for all m # N R 2M&1

|

R< |!| 2R

d d*

}\ +

M

}

2

1 2m(!+i') d!cm 4:

for M=0, 1 and R>0.

ACKNOWLEDGMENTS This work was accomplished in 1993, while visiting Universite de Nancy I. I am grateful to Departement de Mathematiques for their hospitality, and to Professor J.-Ph. Anker for stimulating and helpful discussions. I also thank Professor N. Lohoue for many valuable comments.

REFERENCES 1. J.-Ph. Anker, Sharp estimates for some functions of the Laplacian on noncompact symmetric spaces, Duke Math. J. 65 (1992), 257297. 2. R. R. Coifman and Y. Meyer, Au-dela des operateurs pseudo-differentiels, Asterisque 57 (1978). 3. J. L. Clerc and E. M. Stein, L p-multipliers for non-compact symmetric spaces, Proc. Nat. Acad. Sci. USA 71 (1974), 39113912. 4. M. Flensted-Jensen, PaleyWiener type theorems for a differential operator connected with symmetric spaces, Ark. Mat. 10 (1972), 143162.

MAXIMAL FUNCTIONS AND RIESZ TRANSFORM

357

5. G. B. Folland and E. M. Stein, ``Hardy Spaces on Homogeneous Groups,'' Math. Notes, Vol. 28, Princeton Univ. Press, Princeton, NJ, 1982. 6. S. Helgason, ``Differential Geometry and Symmetric Spaces,'' Academic Press, San Diego, 1962. 7. T. Kawazoe, Atomic Hardy spaces on semisimple Lie groups, Japan J. Math. 11 (1985), 293343. 8. N. Lohoue, Transformes de Riesz et fonctions sommables, Amer. J. Math. 114 (1992), 875922. 9. E. M. Stein, ``Topics in Harmonic Analysis. Related to the LittlewoodPaley Theory,'' Ann. Math. Stud., Vol. 63, Princeton Univ. Press, Princeton, NJ, 1970. 10. E. M. Stein, ``Harmonic Analysis. Real-Variable Methods, Orthogonality, and Oscillatory Integrals,'' Princeton Math. Ser., Vol. 43, Princeton Univ. Press, Princeton, NJ, 1993. 11. J. Stromberg, Weak type estimates for maximal functions on non-compact symmetric spaces, Ann. of Math. 114 (1981), 115126. 12. M. H. Taibleson and G. Weiss, The molecule characterization of certain Hardy spaces, Asterisque 77 (1980), 68149. 13. V. S. Varadarajan, ``Harmonic Analysis on Real Reductive Groups,'' Lecture Notes in Math., Vol. 576, Springer-Verlag, BerlinNew York, 1977. 14. G. Warner, ``Harmonic Analysis on Semi-Simple Lie Groups II,'' Springer-Verlag, Berlin New York, 1972.