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Maximal Operators along Surfaces of Revolution in Lebesgue Mixed Norm Spaces
Journal of Mathematical Analysis and Applications 258, 380᎐395 Ž2001. doi:10.1006rjmaa.2000.7211, available online at http:rrwww.idealibrary.com on
Maximal Operators along Surfaces of Revolution in Lebesgue Mixed Norm Spaces Hung Viet Le Mathematics Department, Oregon State Uni¨ ersity, Cor¨ allis, Oregon 97331-4605 E-mail: [email protected] Submitted by William F. Ames Received November 30, 1999
In this paper, we establish the boundedness of certain maximal operators along hyperspaces in Lebesgue mixed norm spaces. 䊚 2001 Academic Press
INTRODUCTION Maximal functions and singular integral operators have played an essential role in the study of harmonic analysis. This interesting topic has attracted several authors, and as a consequence, a great number of important results on this topic have been discovered. One may find a small portion of the results on maximal functions and singular integral operators in the references of this paper. Here we are interested in maximal operators along surfaces of revolution in the Lebesgue mixed norm spaces. We first begin by considering the maximal operator M defined by Mf Ž x, x n . s sup r)0
½
1 r
ny1
H< y
n
y ⌫ Ž y . . < dy
5
Ž x, y g R ny 1 ,
xn g R. ,
where ⌫ Ž y . ' ⌫ Ž< y <. is a real, measurable, and radial function defined on R ny 1. Here f is in L p Ž R n ., p ) 1. The analogue singular integral operator T is defined by Tf Ž x, x n . s p.¨ . h Ž y .
H
⍀Ž y. < y < ny1 380
0022-247Xr01 $35.00 Copyright 䊚 2001 by Academic Press All rights of reproduction in any form reserved.
f Ž x y y, x n y ⌫ Ž y . . dy,
381
LEBESGUE MIXED NORM SPACES
where h is an arbitrary bounded radial function, and ⍀ Ž y .r< y < ny 1 is the well-known Calderon᎐Zygmund kernel. The boundedness of these two ´ operators along surfaces was studied by several authors Žfor example, see w5, 7, 9x.. In w5x, without imposing any hypothesis on the surface Ž x, ⌫ Ž x .., LungKee Chen and Dashan Fan proved that the operator Tf Ž x, x n . is bounded on L p Ž R n . for 1 - p - ⬁, n G 3, provided that the one-dimensional maximal operator
M ⌫ g Ž x n . s sup r)0
½
1 r
H< t
n
y ⌫ Ž t . . < dt
5
is bounded on L p Ž R . for all p ) 1. Recently in w9x, it was shown that if ⌫ g C 2 w0, ⬁., ⌫ Ž0. s 0, and ⌫ is convex and increasing, then M is bounded on L p Ž R n . for all p ) 1, n G 3 Žsee w9x.. A more general result concerning M, including the case n s 2, can also be found in w10x. The purpose of this paper is to study this maximal operator M in a more general setting, i.e., in the Lebesgue mixed norm spaces. We denote x g R ny 1 , x n g R, and x s Ž x, x n . g R n. Let ¨ denote the unit vector in S ny 2 Ž n ) 2., and let d Ž ¨ . denote the Lebesgue measure on S ny 2 . Let ␦ a denote the dilation defined by ␦ a g Ž t . s g Ž at .. Also, let L p Ž Lq Ž S ny 2 ., R n . denote Lebesgue mixed norm spaces; i.e., if 5 h 5 L p Ž Lq . ' 5 5 h 5 Lq Ž S ny 2 . 5 L p Ž R n . 1rp
prq
s
ž
HR HS n
< hŽ ¨ , x . < d Ž ¨ . q
ny2
/
dx
- ⬁,
then we say that hŽ ¨ , x . g L p Ž Lq .. Define the maximal operator M¨ by M¨ f Ž x, x n . s sup r)0
½
1
r
H < f Ž x y t¨ , x r 0
n
y ⌫ Ž t . . < dt ,
5
where ⌫ Ž y . ' ⌫ Ž< y <. is again a real, measurable, and radial function on R ny 1. The Calderon᎐Zygmund method of rotation greatly reduces the study of ´ n-dimensional singular integral operators to the study of one-dimensional singular integral operators Žsee w3x.. Following this method of rotation, one
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HUNG VIET LE
writes Mf Ž x, x n . s sup r)0
F sup r)0
F '
HS HS
½ ½
ny2
ny2
1 r
ny1
1 r
HS H< t
HS H< t
sup r)0
½
1 r
n
H< t
n
y ⌫ Ž t . . < t ny 2 dt d Ž ¨ .
y ⌫ Ž t . . < dt d Ž ¨ .
n
y ⌫ Ž t . . < dt
5
5
5
d Ž ¨ .
M¨ f Ž x, x n . d Ž ¨ . .
This motivates us to investigate the boundedness of M¨ in L p Ž Lq Ž S ny 2 ., R n .. Lung-Kee Chen provided the best possible result on the boundedness of the maximal operator M t ¨ f Ž x . s sup r)0
½
1
r
H < f Ž x y ¨ . < dt r 0 t
5
in the Lebesgue mixed norm spaces Žsee w4x. Žhere, x g R n, ¨ g S ny1, t¨ s Ž t a1 ¨ 1 , . . . , t a n ¨ n ., and 0 - a1 - a2 - ⭈⭈⭈ - a n .. We would like to obtain a parallel result for our maximal operator M¨ , by following some ideas similar to those in the proof of w4x. THEOREM. Let ⌫ Ž y . ' ⌫ Ž< y <. be a real, measurable, and radial function defined on R ny 1, n G 2, which satisfies the following conditions: Either Ža. ⌫ Ž r . g L⬁Ž R ., Žb. ⌫ Ž r . is monotone on w0, ⬁., and Žc. < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁. or Žd. ⌫⬘Ž r . g L1 Ž R ., and Že. < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁.. Suppose that M ⌫ g is bounded on L p Ž R . for all p ) 1. Then 5 M¨ f 5 L p Ž Lq Ž S ny 2 ., R n . F C 5 f 5 L p Ž R n . , pro¨ ided that either 1-qF
2Ž n y 1 q  . ny1
and
qŽ n y 1 q 2 . n y 1 q 2q
- p F ⬁,
383
LEBESGUE MIXED NORM SPACES
or 2Ž n y 1 q  . ny1
-q
q Ž n y 1.
and
Here  s 1 if n s 2; otherwise  s
n y 1 q 2 1 3
- p F ⬁.
if n G 3.
COROLLARY. Let ⌫ be C 1 on its compact support. If ⌫ is strictly increasing on its compact support, and if ⌫⬘ is increasing on its support, then 5 M¨ f 5 L p Ž Lq Ž S ny 2 ., R n . F C 5 f 5 L p Ž R n . , where p and q satisfy exactly the same conditions as stated in the theorem. The proof of the our main theorem requires two essentials results from w7x and w10x. For convenience, we include those results below: THEOREM w10x Žto appear.. Let ⌫ be a real, measurable, and radial function defined on R ny 1. Then for n G 4, M is a bounded operator on L p Ž R n . for all p ) 1, pro¨ ided that the maximal operator M ⌫ g is bounded on L p Ž R . for all p ) 1. The conclusion also holds when n s 2 or 3, subject to the following additional conditions: Either Ža. ⌫ Ž r . g L⬁Ž R ., Žb. ⌫ Ž r . is monotone on w0, ⬁., and Žc. < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁. or Žd. Že.
⌫⬘Ž r . g L1 Ž R ., and < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁..
COROLLARY 1 w10x Žto appear.. Ža. Žb.
Suppose ⌫ g C 1w0, ⬁., and that
⌫ is strictly increasing on w0, ⬁., ⌫⬘ is increasing on Ž0, ⬁..
Then the maximal operator M is bounded on L p Ž R n . for all p ) 1, n G 4. Let ⌫ be C 1 on its compact support. If ⌫ is strictly increasing on its compact support, and if ⌫⬘ is increasing on its support, then the abo¨ e result also holds for n G 2. Remark. Many authors have obtained results somewhat similar to Corollary 1 above; for example, see w6, 9, 11, 16x. THEOREM C w7x. Let k , k g Z, be probability measures in R n. For 1 F m - n, define 0k ŽE. s k Ž E = R ny m . for e¨ ery Borel subset E of R m ; in terms of Fourier transforms, this means ˆ 0k Ž Ž0. . s ˆ k Ž Ž0., 0.. Let ak 4⬁y⬁
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HUNG VIET LE
be a lacunary sequence, i.e., a k ) 0, inf k g Z a kq1ra k 4 s a ) 1. Suppose that for some ␣ ) 0, Ž1. < ˆ k Ž Ž0., Ž1. . y ˆ k Ž Ž0., 0.< F C < akq1 Ž1. < ␣; Ž2. < ˆ k Ž Ž0., Ž1. .< F C < ak Ž1.
½
M¨ f Ž x, x n . s sup r)0
1
r
H f Ž x y t¨ , x r 0
n
y ⌫ Ž t . . dt
5
F 4 sup ⌬ k ) f Ž x, x n . , kgZ
where ⌬ k is defined on R by n
⌬ k ) f Ž x, x n . s
2
H1 f Ž x y 2
k
t¨ , x n y ␦ 2 k ⌫ Ž t . . dt.
It is therefore enough to show that sup k g Z < ⌬ k ) f < satisfies the conclusion of the theorem. For ␣ g C , we define a family of operators Tk,␣ ¨ f 4␣ by $
Tk␣, ¨ f Ž x, x n . s
1
Ž 1 q <2
k
x<
2 ␣ r2
.
' m k Ž ¨ , x . fˆŽ x .
2 i2 k tŽ ¨ ⭈ x . i x ␦ k ⌫Ž t . n 2
H1 e
e
Ž x s Ž x, x n . ;
dt fˆŽ x, x n .
x g R ny 1 , x n g R . .
Notice that Tk,0 ¨ f Ž x, x n . s ⌬ k ) f Ž x, x n .. To prove the theorem, we need the three lemmas below: LEMMA 1. sup < ⌬ k ) f < kgZ
L p Ž Lq .
F C5 f 5 p
for 1 F q F p F ⬁
Ž p ) 1. .
LEMMA 2. sup < Tk␣, ¨ f <
kgZ
F C5 f 52 L2 Ž L2 .
where  s 1 if n s 2, and  s
1 3
if n G 3.
for y - ᑬ ␣ - n,
385
LEBESGUE MIXED NORM SPACES
LEMMA 3.
If n ) ᑬ ␣ ) n y 1, then
sup < Tk␣, ¨ f <
kgZ
L p Ž Lq .
F C5 f 5 p
for 1 F q F ⬁, 1 - p F ⬁.
Applying the analytic interpolation theorem with mixed norms Žsee w1, 2, 17, 18x. to Lemmas 2 and 3, we get sup < ⌬ k ) f < kgZ
L p Ž Lq .
F C5 f 5 p
for 2Ž n y 1 q  . n y 1 q 2
- p,
q-
2Ž n y 1 q  . ny1
.
The theorem then follows by applying the real interpolation theorem to the above results and Lemma 1. Proof of Lemma 1. The case p s ⬁ is trivial. We shall consider the case 1 F q F p - ⬁ Ž p ) 1.. For x g R ny 1 , we write x s s¨ q x H , where ¨ g S ny 2 , s g R, x Hg R ny 2 , and s¨ is orthogonal to x H . Then ⌬ k ) f Ž x, x n . s
2
H1 f Ž x y 2 2
s
H1 f Ž x
'
H1 f
s
H
2
k
Ž xH , ¨.
H22
k
t¨ , x n y ␦ 2 k ⌫ Ž t . . dt
q Ž s y 2 k t . ¨ , x n y ␦ 2 k ⌫ Ž t . . dt
2
1
k
kq1
Ž s y 2 k t , x n y ␦ 2 ⌫ Ž t . . dt k
fŽ x H , ¨ . Ž s y t , x n y ⌫ Ž t . . dt
F 2 MŽ s, x n . fŽ x H , ¨ . Ž s, x n . for all k g Z. Here MŽ s, x n . fŽ x H , ¨ .Ž s, x n . is the maximal function along a hypersurface acting on the s and x n variables. Thus sup < ⌬ k ) f Ž x, x n . < s sup < ⌬ k ) fŽ x H , ¨ . Ž s, x n . < F 2 MŽ s, x n . fŽ x H , ¨ . Ž s, x n . . k
k
386
HUNG VIET LE
Now p
sup < ⌬ k ) f Ž x, x n . <
L pŽ R n.
k
p
ž sup < ⌬ ) f Ž x, x . < / dx dx s H H H ž sup < ⌬ ) f Ž s, x . < / s H H H ž sup < ⌬ ) f Ž s, x . < /
s
HRHR
k
ny1
n
R R ny2 R
R ny2 R R
k
Ž xH , ¨.
n
k
Ž xH , ¨.
n
k
k
HR HRHR< M
F2
Ž s, x n . fŽ x H , ¨ .
ny2
HR HRHR< f
FC
ny2
Ž xH , ¨.
HRHR HR< f Ž x
sC
H
ny2
sC
HRHR
ny1
n
k
p
p
ds dx H dx n ds dx n dx H
Ž s, x n . < p ds dx n dx H
Ž s, x n . < p ds dx n dx H
qs¨ , x n . < p ds dx H dx n
< f Ž x, x n . < p dx dx n s C 5 f 5 pp .
ŽThe second inequality follows from Theorem 1.. Thus sup < ⌬ k ) f < kgZ
L pŽ R n.
F C5 f 5 p.
Hence, for 1 F q F p - ⬁ Ž p ) 1., we have 5 sup < ⌬ k ) f < 5 Lq Ž S ny 2 . kgZ
L pŽ R n.
F 5 sup < ⌬ k ) f < 5 L p Ž R n . kgZ
Lq Ž S ny 2 .
F C 5 5 f 5 p 5 Lq Ž S ny 2 . F C5 f 5 p, where the first inequality follows from the Minkowski’s inequality for integrals. Lemma 1 is proved. Proof of Lemma 2. Let p be a positive Schwartz function with compact support in R ny 1 , p g C0⬁Ž R ny 1 ., such that HR ny 1 pŽ x . dx s 1. Let pk be defined on R ny 1 by pk Ž x . s Ž1rŽ2 k . ny 1 . pŽ ␦ 2yk x .. Then ˆ pk Ž x . s ˆ pŽ2 k x .. Let k be defined on R by
k ) g Ž x n . s
HR Ž y . g Ž x k
n
y y . dy s
2
H1 g Ž x
n
y ␦ 2 k ⌫ Ž t . . dt ,
387
LEBESGUE MIXED NORM SPACES
so that 1
ˆk Ž x n . s
2
k
H22
k
kq1
e i x n ⌫Ž t . dt.
Observe that sup < Tk␣, ¨ f < F
kgZ
ž
Ý < Tk␣, ¨ f y Ž pk m k . ) f < 2 k
1r2
/
H q M1H ⭈⭈⭈ Mny1 Mn⌫ f Ž x, x n . ,
where MiH is the Hardy᎐Littlewood maximal function acting on the ith variable, and Mn⌫ is the maximal function along a surface acting on the x n variable. To prove that sup k g Z < Tk,␣ ¨ f < is bounded on L2 Ž L2 ., it suffices, by the Plancherel theorem, to show that
ÝH
S ny2
k
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ .
is uniformly bounded for every x in R n. We have, on the one hand, < mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < s
s
F
1
Ž 1 q <2
k
x<
2 ␣ r2
2 i x ␦ k ⌫Ž t . n 2
H1
e
e
e itŽ ¨ ⭈2
½
e i tŽ ¨ ⭈2
2
H1
.
2 i x ␦ k ⌫Ž t . i tŽ ¨ ⭈2 k x . n 2
H1 e
k
x.
Ž 1 q <2 k x < 2 . k
dt y ˆ pk Ž x .
␣ r2
yˆ pk Ž x .
5
2 i x ␦ k ⌫Ž t . n 2
H1 e
dt
dt
x.
Ž 1 q <2 k x < 2 .
yˆ pk Ž x . dt.
␣ r2
The integrand immediately above is a smooth function of x and equals zero when x s 0. By the Mean Value Theorem, one has e itŽ ¨ ⭈2
k
x.
Ž 1 q <2 k x < 2 .
␣ r2
yˆ pk Ž x . F C <2 k x < ,
so that
HS
ny2
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ . F C <2 k x < 2 .
388
HUNG VIET LE
On the other hand, if n s 2, then x g R, and
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . s
2
1
Ž 1 q <2 k x < 2 .
2 i x ␦ k ⌫Ž t . iŽ2 k x .t n 2
H1 e
␣ r2
dt 2
1
q
e
Ž 1 q <2 k x < 2 .
2 i x ␦ k ⌫Ž t . yiŽ2 k x .t n 2
H1
␣r2
e
e
dt .
Using the same technique as in the proof of the Theorem in w10x for the case n s 2, we see that
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C
1 2 ᑬ␣
Ž 1 q <2 k x < .
<2 x < k
2
F
C <2 x < 2 ᑬ ␣q2 k
.
The last inequality follows if <2 k x < G 1, regardless of whether ᑬ ␣ ) 0 or ᑬ ␣ - 0. ŽIf ᑬ ␣ ) 0, it is clear that 1rŽ1 q <2 k x < 2 . ᑬ ␣ - 1r<2 k x < 2 ᑬ ␣ , while if ᑬ ␣ - 0, then we use the assumption that <2 k x < G 1, so that 1 q <2 k x < 2 F 2 <2 k x < 2 , which implies that Ž1 q <2 k x < 2 . < ᑬ ␣ < F Ž2 <2 k x < 2 . < ᑬ ␣ < , whence 1rŽ1 q <2 k x < 2 . ᑬ ␣ F Cr<2 k x < 2 ᑬ ␣.. For the case n G 3, we have
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . s
s
F
1
Ž 1 q <2 k x < 2 .
ᑬ␣
␥n
Ž 1 q <2
k
x<
2 ᑬ␣
.
␥n
Ž 1 q <2 k x < 2 .
ᑬ␣
2 i x ␦ k ⌫Ž t . Ž i2 k t .Ž ¨ ⭈ x . n 2
HS H1 ny2
e
e
2 i x ␦ k ⌫Ž t . i < 2 k x <Žcos .t n 2
2 i x ␦ k ⌫Ž t . i < 2 k x <Žcos .t n 2
H0 H1 H0 H1
e
e
e
e
2
d Ž ¨ .
dt 2
dt Ž sin . 2
dt
d .
Let ␦ g Ž0, 1.. This ␦ will be chosen later. We write
2 i x ␦ k ⌫Ž t . i < 2 k x <Žcos .t n 2
H0 H1
e
s
e
r2y ␦
H0
⭈⭈⭈ q
2
dt
r2q ␦
Hr2y␦
d ⭈⭈⭈ q
Hr2q␦ ⭈⭈⭈
ny3
d
389
LEBESGUE MIXED NORM SPACES
and denote the three integrals on the right-hand side of the above equation as I1 , I2 , and I3 , respectively. It is clear that I2 F 2 ␦ . Next, for I1 and I3 , we observe that 2 i x ␦ k ⌫Ž t . i2 k < x <Žcos .t n 2
H1 e
e
1
dt s
2
k
kq1
H22
k
e i x n ⌫Ž t . e i < x <Žcos .t dt
C
F
<2 x <
.
Here again, we integrate by parts and apply the same strategy as in w10x to obtain the above inequality. Therefore, I1 F F F
C
r2y ␦
H0
d
<2 x < 2 cos 2 k
C <2 k x <
H0
cos Ž r2 y ␦ . 2
C
F
<2 x < sin ␦ 2
k
1
r2y ␦
2
2
C <2 x < 2␦ 2 k
d
.
The last inequality follows because sinŽ ␦ . G 2 ␦r for 0 - ␦ - 1. By the same token, one has I3 F s
C
Hr2q␦ <2
k
C <2 k x < sin 2␦ 2
d F
x < cos 2
F
2
C <2 k x < 2␦ 2
C
1
<2 x < cos Ž r2 q ␦ . 2
k
2
.
Hence,
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C 2 ␣ r2
Ž 1 q <2 k x < .
½
1 <2 x < 2␦ 2 k
If <2 k x < ) 1, we choose ␦ s <2 k x
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C <2 k x <
2 ᑬ ␣ q2r3
<2 x <
2Ž ᑬ ␣ q  .
.
Therefore, if <2 k x < ) 1, we then have
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C k
,
5
q␦ .
390
HUNG VIET LE
where  s 1 if n s 2, and  s 13 if n G 3. Finally, ˆ p is a Schwartz function because p is. So < ˆ pŽ2 k x .< F Cr<2 k x < 2Ž ᑬ ␣q  . for <2 k x < large. Consequently,
HS
ny2
1
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ . F Cmin <2 k x < 2 ,
½
<2 x < k
2Ž ᑬ ␣ q  .
5
,
so that
ÝH k
S ny2
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ . F C
for every x g R n, if ᑬ ␣ ) y . Lemma 2 is proved. It remains to prove Lemma 3. The proof of this lemma requires the following lemma. LEMMA 4. Let be any real number. Let ⌫ satisfy the hypotheses in the theorem. Suppose that
½
M ⌫ g Ž x 2 . s sup r)0
1 r
H< t
2
y ⌫ Ž t . . < dt
5
is bounded on L p Ž R . for all p ) 1. Then M f Ž x 1 , x 2 . s sup r)0
½
1 r
H< t
1
y t , x 2 y ⌫ Ž t . . < dt
5
is bounded on L p Ž R 2 . for all p ) 1. Proof of Lemma 4. We use Theorem C in w7x to prove this lemma. There is nothing to prove if s 0. So we consider the case / 0. For k g Z, define k and 0k by
ˆk Ž 1 , 2 . s
1 2k
H< t
iŽ 1 tq 2 ⌫Ž t ..
k
ˆ 0k Ž 2 . s ˆ k Ž 0, 2 . s
1 2k
H< t
i 2 ⌫Ž t .
k
t g R, and
dt , dt ,
t g R.
Then k are finite Borel measures, k G 0. Also, we have
k ) f Ž x1 , x 2 . s
1 2k
H< t
1
y t , x 2 y ⌫ Ž t . . dt ;
and
0k ) g Ž x 2 . s
1 2k
H< t
2
y ⌫ Ž t . . dt.
x 1 , x 2 g R,
391
LEBESGUE MIXED NORM SPACES
It is obvious that sup < 0k ) g Ž x 2 . < F C sup k
r)0
½
1 r
H< t
2
y ⌫ Ž t . . < dt .
5
Thus, by hypothesis, sup k < 0k ) g Ž x 2 .< is bounded on L p Ž R . for all p ) 1. Observe that 1 < e i 2 ⌫Ž t . Ž e i 1 t y 1 . dt ˆk Ž 1 , 2 . y ˆ k Ž 0, 2 . < s k 2 < t
H
F F
1 2k 1 2k
H< t
i 1 t
k
y 1 < dt
H< t
k
F < 1 <
H< t
k
dt s 2 < 2 k 1 < .
Also, by the same technique as in the proof of the theorem in w10x, we obtain < ˆ k Ž 1 , 2 .< F C < 2 k 1
½
H
5
is bounded on L p Ž R 2 . for all p ) 1, and the bound does not depend on . Because f is in L p Ž R 2 . if and only if < f < is, we infer that 1 < f Ž x 1 y t , x 2 y ⌫ Ž t . . < dt sup r < t
½
H
5
is also bounded on L p Ž R 2 . for all p ) 1. Lemma 4 is proved. Proof of Lemma 3. For 0 - ᑬ ␣ - n, we let G ␣ denote the Bessel potentials defined on R ny 1 Žsee w14, p. 132x., the Fourier transform of ˆ ␣ Ž y . s Ž1 q < y < 2 .y ␣ r2 . Next, we define G k␣ Ž y . by Gˆk␣ Ž y . s Ž1 which is G 2 y␣ r2 k q <2 y < . . We then have Tk␣, ¨ f Ž x, x n . s
2
H1 HR
ny1
f Ž x y y y 2 k t¨ , x n y ␦ 2 k ⌫ Ž t . . G k␣ Ž y . dy dt ; x, y g R
s
2
H1 HR
ny1
f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . . G ␣ Ž y y t¨ . dy dt.
392
HUNG VIET LE
Because < G ␣ Ž y .< is controlled by Cr< y < ny ᑬ ␣ Ž n ) ᑬ ␣ ) 0. as < y < ª 0, and because < G ␣ Ž y .< is rapidly decreasing as < y < ª ⬁, we have C < y < F A
< G ␣ Ž y y t¨ . < F
< y y t¨ <
nyᑬ ␣
q
B < y < ) A < y
.
2rA n ŽHere A ) 2 and B ' 11 q 4 will do.. Therefore, y 2rA
f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . .
2
< Tk␣, ¨ f Ž x, x n . < F C
H1 HR
qC
< y y t¨ < nyᑬ ␣
ny1
f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . .
2
H1 HR
< y < F A dy dt
< y
ny1
< y < ) A dy dt.
Denote the two integrals inside the curly brackets on the right-hand side of the above inequality as I1 and I2 , respectively. Notice that I2 s F s F
½
ÝH ⬁
Ž 2 A.
⬁
Ý ⬁
Ý ls0
F
1 l
⬁
Ý ls0
n
< y
H< y <(2 A M f Ž x y 2 l
1
1 n
Ž 2 A. 2 l
1 l
1
2 Ž2 1 2l
kŽ ny1.
kq l
A.
ny1
5
1 < y
dy
dy
Mn f Ž x y 2 k y, x n .
l ls0 < y <(2 A
Ý
y, x n y ␦ 2 k ⌫ Ž t . . dt
< y
⬁
ls0
F
k
Mn f Ž x y 2 k y, x n .
H< y
ls0
s
2
H< y
n
H< y <(2
kql
H< y <(2
kql
A
A
dy k
y, x n . dy
Mn f Ž x y y, x n . dy
Mn f Ž x y y, x n . dy
H MŽ1, . . . , ny1. Mn f Ž x, x n .
H s 2 MŽ1, . . . , ny1. Mn f Ž x, x n . , H where MŽ1, . . . , ny1. is the Hardy᎐Littlewood maximal operator acting on the first n y 1 variables, and Mn is the maximal operator along a surface acting on the last variable.
393
LEBESGUE MIXED NORM SPACES
We now estimate the bound of I1. Since ¨ g S ny 2 , there is a ¨ i , say, ¨ 1 such that < ¨ 1 < 2 G 1rŽ n y 1.. Then f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . .
2
I1 F
H1 HR
< y 1 y t¨ 1 < nyᑬ ␣
ny1
< y < F A dy dt.
We let D s A q 2, ˜ x s Ž x 2 , . . . , x ny1 ., ˜ y s Ž y 2 , . . . , yny1 ., and, making a change of variable y 1 ¬ y 1 y t¨ 1 , we obtain I1 F s F F
2
H1 H< y
f Ž x 1 y 2 k y 1 y 2 k t¨ 1 , ˜ x y 2k˜ y, x n y ␦ 2 k ⌫ Ž t . . < y 1 < nyᑬ ␣ 2
½
H< y
1
y 2 k y 1 y 2 k t¨ 1 , ˜ x y 2k˜ y, x n y ␦ 2 k ⌫ Ž t . . dt
MŽ1, n. f Ž x y 2 k y, x n . < y 1 < ny ᑬ ␣ < y 1 < nyᑬ ␣
1
5
dy < y 1 < nyᑬ ␣
dy
MŽ1 , n. f Ž x y 2 k y, x n .
H< ˜y
dy dt
dy 1 dy. ˜
The next to the last inequality above follows due to Lemma 4, and MŽ1, n. f is the maximal function along a surface, acting on the first and last variables. But
H< y
MŽ1, n. f Ž x y 2 k y, x n . < y 1 < nyᑬ ␣
1
s F s
y⬁
MŽ1, n. f Ž x 1 y 2 k y 1 , ˜ x y 2k˜ y, x n .
ÝH
< y 1 < nyᑬ ␣
l ls0 < y 1 <(2 D
y⬁
Ý
1 l
nyᑬ ␣
ls0
Ž2 D.
y⬁
1
Ý ls0
s
dy 1
l
Ž2 D.
y⬁
Ý ls0
H< y <(2 D M 1
1
nyᑬ ␣
2k
H< y <(2 1
1
Ž 2 lD .
ny Ž ᑬ ␣ q1 .
Ž1 , n.
l
kq1
1 2
kql
D
D
dy 1
f Ž x1 y 2 k y1 , ˜ x y 2k˜ y, x n . dy 1
MŽ1, n. f Ž x 1 y y 1 , ˜ x y 2k˜ y, x n . dy 1
H< y <(2 1
kql
D
MŽ1 , n. f Ž x 1 y y 1 , ˜ x y 2k˜ y, x n . dy 1
394
HUNG VIET LE
F
y⬁
Ý ls0
1 l
Ž2 D.
M1H MŽ1 , n. f Ž x 1 , ˜ x y 2k˜ y, x n .
ny Ž ᑬ ␣ q1 .
y⬁
Ý w2 Ž ᑬ ␣q1.yn x
s D Ž ᑬ ␣q1.yn
l
M1H MŽ1 , n. f Ž x 1 , ˜ x y 2k˜ y, x n .
ls0
s
CM1H MŽ1 , n.
f Ž x1 , ˜ x y 2k˜ y, x n .
if ᑬ ␣ ) n y 1. Therefore, I1 F
H< ˜y
MŽ1, n. f Ž x y 2 k y, x n . < y 1 < nyᑬ ␣
1
FC
H< ˜y
H 1 MŽ1 , n.
f Ž x1 , ˜ x y 2k˜ y, x n . dy˜
y⬁
sC Ý
H 1 MŽ1 , n.
f Ž x1 , ˜ x y 2k˜ y, x n . dy˜
H< ˜y <(2
D
ls0
H< ˜y <(2 D M
y⬁
1
sC Ý
ls0
l
k
Ž2 .
ny2
y⬁
s C Ý D ny2 Ž 2 l . ls0 y⬁
dy 1 dy˜
kql
M1H MŽ1 , n. f Ž x 1 , ˜ x y˜ y, x n . dy˜ 1
ny 2
Ž2
kq l
D.
H< ˜y <(2
ny2
kql
D
M1H MŽ1 , n. f Ž x 1 , ˜ x y˜ y, x n . dy˜
l
H H F C Ý Ž 2 ny2 . MŽ2, x, x n . . . . , ny1. M1 MŽ1 , n. f Ž x 1 , ˜
ls0 H H s CMŽ2, . . . , ny1. M1 MŽ1 , n. f Ž x, x n . ,
provided that n y 1 - ᑬ ␣ - n. Consequently, < Tk␣, ¨ f Ž x, x n . < H H H F C MŽ2, . . . , ny1. M1 MŽ1 , n. f Ž x, x n . q MŽ1 , . . . , ny1. Mn f Ž x, x n . 4 .
This inequality holds for all k g ⺪. It is well known that the Hardy᎐ Littlewood maximal operators are bounded on L p Ž R n .. Thus, it follows from Theorem 1 that sup < Tk␣, ¨ f <
kgZ
L p Ž Lq .
F C5 f 5 p
for 1 F q F ⬁, 1 - p F ⬁. Lemma 3 is proved. Proof of the Corollary. One just needs to verify that M ⌫ g is bounded on L p Ž R . for all p ) 1 and that Mf is bounded on L p Ž R n . for all p ) 1,
LEBESGUE MIXED NORM SPACES
395
n G 2. The fact that M ⌫ g is bounded on L p Ž R . for all p ) 1 has been shown in the proof of Corollary 1 in w10x. Also, applying the result of Corollary 1 in w10x, we infer that Mf is bounded on L p Ž R n . for all p ) 1, n G 2. Therefore, the proof is finished.
REFERENCES 1. A. Benedek and R. Panzone, The spaces L p with mixed norm, Duke Math. J. 28 Ž1961., 301᎐324. 2. A. P. Calderon ´ and A. Zygmund, A note on the interpolation of sublinear operations, Amer. J. Math. 78 Ž1956., 282᎐288. 3. A. P. Calderon ´ and A. Zygmund, On singular integrals, Amer. J. Math. 78 Ž1956., 289᎐309. 4. L.-K. Chen, The maximal operators related to the Calderon-Zygmund method of rota´ tions, Illinois J. Math. 33, No. 2 Ž1989., 268᎐279. 5. L.-K. Chen and D. Fan, On singular integrals along surfaces related to block spaces, Integral Equations Operator Theory 29 Ž1997., 261᎐268. 6. M. Christ, J. Duoandikoetxea, and J. L. Rubio de Francia, Maximal operators related to the radon transform and the Calderon-Zygmund method of rotations, Duke Math. J. 53 ´ Ž1986., 189᎐209. 7. J. Duoandikoetxea and J. L. Rubio de Francia, Maximal and singular integral operators via Fourier transform estimates, In¨ ent. Math. 84 Ž1986., 541᎐561. 8. R. Fefferman, On an operator arising in the Calderon-Zygmund method of rotations and ´ the Bramble-Hilbert lemma, Proc. Nat. Acad. Sci. U.S. A. 80 Ž1983., 3877᎐3878. 9. W. Kim, S. Wainger, J. Wright, and S. Ziesler, Singular Integrals and maximal functions associated to surfaces of revolution, Bull. London Math. Soc. 28, No. 3 Ž1996., 291᎐296. 10. H. V. Le, On maximal operators along surfaces, Integral Equations Operator Theory, to appear. 11. A. Nagel and S. Waigner, Hilbert transform associated with plane curves, Trans. Amer. Math. Soc. 223 Ž1976., 235᎐252. 12. N. Riviere, ` Singular integrals and multiplier operators, Ark. Mat. 9 Ž1971., 243᎐278. 13. S. Sato, Maximal functions associated with curves and the Calderon-Zygmund method of ´ rotations, Trans. Amer. Math. Soc. 293 Ž1986., 799᎐806. 14. E. M. Stein, ‘‘Singular Integrals and Differentiability Properties of Functions,’’ Princeton Univ. Press, Princeton, NJ, 1970. 15. E. M. Stein, ‘‘Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals,’’ Princeton Univ. Press, Princeton, NJ, 1993. 16. E. M. Stein and S. Wainger, Problems in harmonic analysis related to curvature, Bull. Amer. Math. Soc. 84 Ž1978., 1239᎐1295. 17. E. M. Stein and G. Weiss, ‘‘Introduction to Fourier Analysis on Euclidean Spaces,’’ Princeton Univ. Press, Princeton, NJ, 1975. 18. J. L. Rubio de Francia, Vector valued inequalities for operators in L p spaces, Bull. London Math. Soc. 12 Ž1980., 211᎐215.
Maximal Operators along Surfaces of Revolution in Lebesgue Mixed Norm Spaces Hung Viet Le Mathematics Department, Oregon State Uni¨ ersity, Cor¨ allis, Oregon 97331-4605 E-mail: [email protected] Submitted by William F. Ames Received November 30, 1999
In this paper, we establish the boundedness of certain maximal operators along hyperspaces in Lebesgue mixed norm spaces. 䊚 2001 Academic Press
INTRODUCTION Maximal functions and singular integral operators have played an essential role in the study of harmonic analysis. This interesting topic has attracted several authors, and as a consequence, a great number of important results on this topic have been discovered. One may find a small portion of the results on maximal functions and singular integral operators in the references of this paper. Here we are interested in maximal operators along surfaces of revolution in the Lebesgue mixed norm spaces. We first begin by considering the maximal operator M defined by Mf Ž x, x n . s sup r)0
½
1 r
ny1
H< y
n
y ⌫ Ž y . . < dy
5
Ž x, y g R ny 1 ,
xn g R. ,
where ⌫ Ž y . ' ⌫ Ž< y <. is a real, measurable, and radial function defined on R ny 1. Here f is in L p Ž R n ., p ) 1. The analogue singular integral operator T is defined by Tf Ž x, x n . s p.¨ . h Ž y .
H
⍀Ž y. < y < ny1 380
0022-247Xr01 $35.00 Copyright 䊚 2001 by Academic Press All rights of reproduction in any form reserved.
f Ž x y y, x n y ⌫ Ž y . . dy,
381
LEBESGUE MIXED NORM SPACES
where h is an arbitrary bounded radial function, and ⍀ Ž y .r< y < ny 1 is the well-known Calderon᎐Zygmund kernel. The boundedness of these two ´ operators along surfaces was studied by several authors Žfor example, see w5, 7, 9x.. In w5x, without imposing any hypothesis on the surface Ž x, ⌫ Ž x .., LungKee Chen and Dashan Fan proved that the operator Tf Ž x, x n . is bounded on L p Ž R n . for 1 - p - ⬁, n G 3, provided that the one-dimensional maximal operator
M ⌫ g Ž x n . s sup r)0
½
1 r
H< t
n
y ⌫ Ž t . . < dt
5
is bounded on L p Ž R . for all p ) 1. Recently in w9x, it was shown that if ⌫ g C 2 w0, ⬁., ⌫ Ž0. s 0, and ⌫ is convex and increasing, then M is bounded on L p Ž R n . for all p ) 1, n G 3 Žsee w9x.. A more general result concerning M, including the case n s 2, can also be found in w10x. The purpose of this paper is to study this maximal operator M in a more general setting, i.e., in the Lebesgue mixed norm spaces. We denote x g R ny 1 , x n g R, and x s Ž x, x n . g R n. Let ¨ denote the unit vector in S ny 2 Ž n ) 2., and let d Ž ¨ . denote the Lebesgue measure on S ny 2 . Let ␦ a denote the dilation defined by ␦ a g Ž t . s g Ž at .. Also, let L p Ž Lq Ž S ny 2 ., R n . denote Lebesgue mixed norm spaces; i.e., if 5 h 5 L p Ž Lq . ' 5 5 h 5 Lq Ž S ny 2 . 5 L p Ž R n . 1rp
prq
s
ž
HR HS n
< hŽ ¨ , x . < d Ž ¨ . q
ny2
/
dx
- ⬁,
then we say that hŽ ¨ , x . g L p Ž Lq .. Define the maximal operator M¨ by M¨ f Ž x, x n . s sup r)0
½
1
r
H < f Ž x y t¨ , x r 0
n
y ⌫ Ž t . . < dt ,
5
where ⌫ Ž y . ' ⌫ Ž< y <. is again a real, measurable, and radial function on R ny 1. The Calderon᎐Zygmund method of rotation greatly reduces the study of ´ n-dimensional singular integral operators to the study of one-dimensional singular integral operators Žsee w3x.. Following this method of rotation, one
382
HUNG VIET LE
writes Mf Ž x, x n . s sup r)0
F sup r)0
F '
HS HS
½ ½
ny2
ny2
1 r
ny1
1 r
HS H< t
HS H< t
sup r)0
½
1 r
n
H< t
n
y ⌫ Ž t . . < t ny 2 dt d Ž ¨ .
y ⌫ Ž t . . < dt d Ž ¨ .
n
y ⌫ Ž t . . < dt
5
5
5
d Ž ¨ .
M¨ f Ž x, x n . d Ž ¨ . .
This motivates us to investigate the boundedness of M¨ in L p Ž Lq Ž S ny 2 ., R n .. Lung-Kee Chen provided the best possible result on the boundedness of the maximal operator M t ¨ f Ž x . s sup r)0
½
1
r
H < f Ž x y ¨ . < dt r 0 t
5
in the Lebesgue mixed norm spaces Žsee w4x. Žhere, x g R n, ¨ g S ny1, t¨ s Ž t a1 ¨ 1 , . . . , t a n ¨ n ., and 0 - a1 - a2 - ⭈⭈⭈ - a n .. We would like to obtain a parallel result for our maximal operator M¨ , by following some ideas similar to those in the proof of w4x. THEOREM. Let ⌫ Ž y . ' ⌫ Ž< y <. be a real, measurable, and radial function defined on R ny 1, n G 2, which satisfies the following conditions: Either Ža. ⌫ Ž r . g L⬁Ž R ., Žb. ⌫ Ž r . is monotone on w0, ⬁., and Žc. < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁. or Žd. ⌫⬘Ž r . g L1 Ž R ., and Že. < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁.. Suppose that M ⌫ g is bounded on L p Ž R . for all p ) 1. Then 5 M¨ f 5 L p Ž Lq Ž S ny 2 ., R n . F C 5 f 5 L p Ž R n . , pro¨ ided that either 1-qF
2Ž n y 1 q  . ny1
and
qŽ n y 1 q 2 . n y 1 q 2q
- p F ⬁,
383
LEBESGUE MIXED NORM SPACES
or 2Ž n y 1 q  . ny1
-q
q Ž n y 1.
and
Here  s 1 if n s 2; otherwise  s
n y 1 q 2 1 3
- p F ⬁.
if n G 3.
COROLLARY. Let ⌫ be C 1 on its compact support. If ⌫ is strictly increasing on its compact support, and if ⌫⬘ is increasing on its support, then 5 M¨ f 5 L p Ž Lq Ž S ny 2 ., R n . F C 5 f 5 L p Ž R n . , where p and q satisfy exactly the same conditions as stated in the theorem. The proof of the our main theorem requires two essentials results from w7x and w10x. For convenience, we include those results below: THEOREM w10x Žto appear.. Let ⌫ be a real, measurable, and radial function defined on R ny 1. Then for n G 4, M is a bounded operator on L p Ž R n . for all p ) 1, pro¨ ided that the maximal operator M ⌫ g is bounded on L p Ž R . for all p ) 1. The conclusion also holds when n s 2 or 3, subject to the following additional conditions: Either Ža. ⌫ Ž r . g L⬁Ž R ., Žb. ⌫ Ž r . is monotone on w0, ⬁., and Žc. < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁. or Žd. Že.
⌫⬘Ž r . g L1 Ž R ., and < ⌫⬘Ž r .< is increasing on supp ⌫⬘ l w0, ⬁..
COROLLARY 1 w10x Žto appear.. Ža. Žb.
Suppose ⌫ g C 1w0, ⬁., and that
⌫ is strictly increasing on w0, ⬁., ⌫⬘ is increasing on Ž0, ⬁..
Then the maximal operator M is bounded on L p Ž R n . for all p ) 1, n G 4. Let ⌫ be C 1 on its compact support. If ⌫ is strictly increasing on its compact support, and if ⌫⬘ is increasing on its support, then the abo¨ e result also holds for n G 2. Remark. Many authors have obtained results somewhat similar to Corollary 1 above; for example, see w6, 9, 11, 16x. THEOREM C w7x. Let k , k g Z, be probability measures in R n. For 1 F m - n, define 0k ŽE. s k Ž E = R ny m . for e¨ ery Borel subset E of R m ; in terms of Fourier transforms, this means ˆ 0k Ž Ž0. . s ˆ k Ž Ž0., 0.. Let ak 4⬁y⬁
384
HUNG VIET LE
be a lacunary sequence, i.e., a k ) 0, inf k g Z a kq1ra k 4 s a ) 1. Suppose that for some ␣ ) 0, Ž1. < ˆ k Ž Ž0., Ž1. . y ˆ k Ž Ž0., 0.< F C < akq1 Ž1. < ␣; Ž2. < ˆ k Ž Ž0., Ž1. .< F C < ak Ž1.
½
M¨ f Ž x, x n . s sup r)0
1
r
H f Ž x y t¨ , x r 0
n
y ⌫ Ž t . . dt
5
F 4 sup ⌬ k ) f Ž x, x n . , kgZ
where ⌬ k is defined on R by n
⌬ k ) f Ž x, x n . s
2
H1 f Ž x y 2
k
t¨ , x n y ␦ 2 k ⌫ Ž t . . dt.
It is therefore enough to show that sup k g Z < ⌬ k ) f < satisfies the conclusion of the theorem. For ␣ g C , we define a family of operators Tk,␣ ¨ f 4␣ by $
Tk␣, ¨ f Ž x, x n . s
1
Ž 1 q <2
k
x<
2 ␣ r2
.
' m k Ž ¨ , x . fˆŽ x .
2 i2 k tŽ ¨ ⭈ x . i x ␦ k ⌫Ž t . n 2
H1 e
e
Ž x s Ž x, x n . ;
dt fˆŽ x, x n .
x g R ny 1 , x n g R . .
Notice that Tk,0 ¨ f Ž x, x n . s ⌬ k ) f Ž x, x n .. To prove the theorem, we need the three lemmas below: LEMMA 1. sup < ⌬ k ) f < kgZ
L p Ž Lq .
F C5 f 5 p
for 1 F q F p F ⬁
Ž p ) 1. .
LEMMA 2. sup < Tk␣, ¨ f <
kgZ
F C5 f 52 L2 Ž L2 .
where  s 1 if n s 2, and  s
1 3
if n G 3.
for y - ᑬ ␣ - n,
385
LEBESGUE MIXED NORM SPACES
LEMMA 3.
If n ) ᑬ ␣ ) n y 1, then
sup < Tk␣, ¨ f <
kgZ
L p Ž Lq .
F C5 f 5 p
for 1 F q F ⬁, 1 - p F ⬁.
Applying the analytic interpolation theorem with mixed norms Žsee w1, 2, 17, 18x. to Lemmas 2 and 3, we get sup < ⌬ k ) f < kgZ
L p Ž Lq .
F C5 f 5 p
for 2Ž n y 1 q  . n y 1 q 2
- p,
q-
2Ž n y 1 q  . ny1
.
The theorem then follows by applying the real interpolation theorem to the above results and Lemma 1. Proof of Lemma 1. The case p s ⬁ is trivial. We shall consider the case 1 F q F p - ⬁ Ž p ) 1.. For x g R ny 1 , we write x s s¨ q x H , where ¨ g S ny 2 , s g R, x Hg R ny 2 , and s¨ is orthogonal to x H . Then ⌬ k ) f Ž x, x n . s
2
H1 f Ž x y 2 2
s
H1 f Ž x
'
H1 f
s
H
2
k
Ž xH , ¨.
H22
k
t¨ , x n y ␦ 2 k ⌫ Ž t . . dt
q Ž s y 2 k t . ¨ , x n y ␦ 2 k ⌫ Ž t . . dt
2
1
k
kq1
Ž s y 2 k t , x n y ␦ 2 ⌫ Ž t . . dt k
fŽ x H , ¨ . Ž s y t , x n y ⌫ Ž t . . dt
F 2 MŽ s, x n . fŽ x H , ¨ . Ž s, x n . for all k g Z. Here MŽ s, x n . fŽ x H , ¨ .Ž s, x n . is the maximal function along a hypersurface acting on the s and x n variables. Thus sup < ⌬ k ) f Ž x, x n . < s sup < ⌬ k ) fŽ x H , ¨ . Ž s, x n . < F 2 MŽ s, x n . fŽ x H , ¨ . Ž s, x n . . k
k
386
HUNG VIET LE
Now p
sup < ⌬ k ) f Ž x, x n . <
L pŽ R n.
k
p
ž sup < ⌬ ) f Ž x, x . < / dx dx s H H H ž sup < ⌬ ) f Ž s, x . < / s H H H ž sup < ⌬ ) f Ž s, x . < /
s
HRHR
k
ny1
n
R R ny2 R
R ny2 R R
k
Ž xH , ¨.
n
k
Ž xH , ¨.
n
k
k
HR HRHR< M
F2
Ž s, x n . fŽ x H , ¨ .
ny2
HR HRHR< f
FC
ny2
Ž xH , ¨.
HRHR HR< f Ž x
sC
H
ny2
sC
HRHR
ny1
n
k
p
p
ds dx H dx n ds dx n dx H
Ž s, x n . < p ds dx n dx H
Ž s, x n . < p ds dx n dx H
qs¨ , x n . < p ds dx H dx n
< f Ž x, x n . < p dx dx n s C 5 f 5 pp .
ŽThe second inequality follows from Theorem 1.. Thus sup < ⌬ k ) f < kgZ
L pŽ R n.
F C5 f 5 p.
Hence, for 1 F q F p - ⬁ Ž p ) 1., we have 5 sup < ⌬ k ) f < 5 Lq Ž S ny 2 . kgZ
L pŽ R n.
F 5 sup < ⌬ k ) f < 5 L p Ž R n . kgZ
Lq Ž S ny 2 .
F C 5 5 f 5 p 5 Lq Ž S ny 2 . F C5 f 5 p, where the first inequality follows from the Minkowski’s inequality for integrals. Lemma 1 is proved. Proof of Lemma 2. Let p be a positive Schwartz function with compact support in R ny 1 , p g C0⬁Ž R ny 1 ., such that HR ny 1 pŽ x . dx s 1. Let pk be defined on R ny 1 by pk Ž x . s Ž1rŽ2 k . ny 1 . pŽ ␦ 2yk x .. Then ˆ pk Ž x . s ˆ pŽ2 k x .. Let k be defined on R by
k ) g Ž x n . s
HR Ž y . g Ž x k
n
y y . dy s
2
H1 g Ž x
n
y ␦ 2 k ⌫ Ž t . . dt ,
387
LEBESGUE MIXED NORM SPACES
so that 1
ˆk Ž x n . s
2
k
H22
k
kq1
e i x n ⌫Ž t . dt.
Observe that sup < Tk␣, ¨ f < F
kgZ
ž
Ý < Tk␣, ¨ f y Ž pk m k . ) f < 2 k
1r2
/
H q M1H ⭈⭈⭈ Mny1 Mn⌫ f Ž x, x n . ,
where MiH is the Hardy᎐Littlewood maximal function acting on the ith variable, and Mn⌫ is the maximal function along a surface acting on the x n variable. To prove that sup k g Z < Tk,␣ ¨ f < is bounded on L2 Ž L2 ., it suffices, by the Plancherel theorem, to show that
ÝH
S ny2
k
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ .
is uniformly bounded for every x in R n. We have, on the one hand, < mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < s
s
F
1
Ž 1 q <2
k
x<
2 ␣ r2
2 i x ␦ k ⌫Ž t . n 2
H1
e
e
e itŽ ¨ ⭈2
½
e i tŽ ¨ ⭈2
2
H1
.
2 i x ␦ k ⌫Ž t . i tŽ ¨ ⭈2 k x . n 2
H1 e
k
x.
Ž 1 q <2 k x < 2 . k
dt y ˆ pk Ž x .
␣ r2
yˆ pk Ž x .
5
2 i x ␦ k ⌫Ž t . n 2
H1 e
dt
dt
x.
Ž 1 q <2 k x < 2 .
yˆ pk Ž x . dt.
␣ r2
The integrand immediately above is a smooth function of x and equals zero when x s 0. By the Mean Value Theorem, one has e itŽ ¨ ⭈2
k
x.
Ž 1 q <2 k x < 2 .
␣ r2
yˆ pk Ž x . F C <2 k x < ,
so that
HS
ny2
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ . F C <2 k x < 2 .
388
HUNG VIET LE
On the other hand, if n s 2, then x g R, and
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . s
2
1
Ž 1 q <2 k x < 2 .
2 i x ␦ k ⌫Ž t . iŽ2 k x .t n 2
H1 e
␣ r2
dt 2
1
q
e
Ž 1 q <2 k x < 2 .
2 i x ␦ k ⌫Ž t . yiŽ2 k x .t n 2
H1
␣r2
e
e
dt .
Using the same technique as in the proof of the Theorem in w10x for the case n s 2, we see that
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C
1 2 ᑬ␣
Ž 1 q <2 k x < .
<2 x < k
2
F
C <2 x < 2 ᑬ ␣q2 k
.
The last inequality follows if <2 k x < G 1, regardless of whether ᑬ ␣ ) 0 or ᑬ ␣ - 0. ŽIf ᑬ ␣ ) 0, it is clear that 1rŽ1 q <2 k x < 2 . ᑬ ␣ - 1r<2 k x < 2 ᑬ ␣ , while if ᑬ ␣ - 0, then we use the assumption that <2 k x < G 1, so that 1 q <2 k x < 2 F 2 <2 k x < 2 , which implies that Ž1 q <2 k x < 2 . < ᑬ ␣ < F Ž2 <2 k x < 2 . < ᑬ ␣ < , whence 1rŽ1 q <2 k x < 2 . ᑬ ␣ F Cr<2 k x < 2 ᑬ ␣.. For the case n G 3, we have
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . s
s
F
1
Ž 1 q <2 k x < 2 .
ᑬ␣
␥n
Ž 1 q <2
k
x<
2 ᑬ␣
.
␥n
Ž 1 q <2 k x < 2 .
ᑬ␣
2 i x ␦ k ⌫Ž t . Ž i2 k t .Ž ¨ ⭈ x . n 2
HS H1 ny2
e
e
2 i x ␦ k ⌫Ž t . i < 2 k x <Žcos .t n 2
2 i x ␦ k ⌫Ž t . i < 2 k x <Žcos .t n 2
H0 H1 H0 H1
e
e
e
e
2
d Ž ¨ .
dt 2
dt Ž sin . 2
dt
d .
Let ␦ g Ž0, 1.. This ␦ will be chosen later. We write
2 i x ␦ k ⌫Ž t . i < 2 k x <Žcos .t n 2
H0 H1
e
s
e
r2y ␦
H0
⭈⭈⭈ q
2
dt
r2q ␦
Hr2y␦
d ⭈⭈⭈ q
Hr2q␦ ⭈⭈⭈
ny3
d
389
LEBESGUE MIXED NORM SPACES
and denote the three integrals on the right-hand side of the above equation as I1 , I2 , and I3 , respectively. It is clear that I2 F 2 ␦ . Next, for I1 and I3 , we observe that 2 i x ␦ k ⌫Ž t . i2 k < x <Žcos .t n 2
H1 e
e
1
dt s
2
k
kq1
H22
k
e i x n ⌫Ž t . e i < x <Žcos .t dt
C
F
<2 x <
.
Here again, we integrate by parts and apply the same strategy as in w10x to obtain the above inequality. Therefore, I1 F F F
C
r2y ␦
H0
d
<2 x < 2 cos 2 k
C <2 k x <
H0
cos Ž r2 y ␦ . 2
C
F
<2 x < sin ␦ 2
k
1
r2y ␦
2
2
C <2 x < 2␦ 2 k
d
.
The last inequality follows because sinŽ ␦ . G 2 ␦r for 0 - ␦ - 1. By the same token, one has I3 F s
C
Hr2q␦ <2
k
C <2 k x < sin 2␦ 2
d F
x < cos 2
F
2
C <2 k x < 2␦ 2
C
1
<2 x < cos Ž r2 q ␦ . 2
k
2
.
Hence,
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C 2 ␣ r2
Ž 1 q <2 k x < .
½
1 <2 x < 2␦ 2 k
If <2 k x < ) 1, we choose ␦ s <2 k x
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C <2 k x <
2 ᑬ ␣ q2r3
<2 x <
2Ž ᑬ ␣ q  .
.
Therefore, if <2 k x < ) 1, we then have
HS
ny2
< mk Ž ¨ , x . < 2 d Ž ¨ . F
C k
,
5
q␦ .
390
HUNG VIET LE
where  s 1 if n s 2, and  s 13 if n G 3. Finally, ˆ p is a Schwartz function because p is. So < ˆ pŽ2 k x .< F Cr<2 k x < 2Ž ᑬ ␣q  . for <2 k x < large. Consequently,
HS
ny2
1
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ . F Cmin <2 k x < 2 ,
½
<2 x < k
2Ž ᑬ ␣ q  .
5
,
so that
ÝH k
S ny2
< mk Ž ¨ , x . y ˆ pk Ž x . ˆk Ž x n . < 2 d Ž ¨ . F C
for every x g R n, if ᑬ ␣ ) y . Lemma 2 is proved. It remains to prove Lemma 3. The proof of this lemma requires the following lemma. LEMMA 4. Let be any real number. Let ⌫ satisfy the hypotheses in the theorem. Suppose that
½
M ⌫ g Ž x 2 . s sup r)0
1 r
H< t
2
y ⌫ Ž t . . < dt
5
is bounded on L p Ž R . for all p ) 1. Then M f Ž x 1 , x 2 . s sup r)0
½
1 r
H< t
1
y t , x 2 y ⌫ Ž t . . < dt
5
is bounded on L p Ž R 2 . for all p ) 1. Proof of Lemma 4. We use Theorem C in w7x to prove this lemma. There is nothing to prove if s 0. So we consider the case / 0. For k g Z, define k and 0k by
ˆk Ž 1 , 2 . s
1 2k
H< t
iŽ 1 tq 2 ⌫Ž t ..
k
ˆ 0k Ž 2 . s ˆ k Ž 0, 2 . s
1 2k
H< t
i 2 ⌫Ž t .
k
t g R, and
dt , dt ,
t g R.
Then k are finite Borel measures, k G 0. Also, we have
k ) f Ž x1 , x 2 . s
1 2k
H< t
1
y t , x 2 y ⌫ Ž t . . dt ;
and
0k ) g Ž x 2 . s
1 2k
H< t
2
y ⌫ Ž t . . dt.
x 1 , x 2 g R,
391
LEBESGUE MIXED NORM SPACES
It is obvious that sup < 0k ) g Ž x 2 . < F C sup k
r)0
½
1 r
H< t
2
y ⌫ Ž t . . < dt .
5
Thus, by hypothesis, sup k < 0k ) g Ž x 2 .< is bounded on L p Ž R . for all p ) 1. Observe that 1 < e i 2 ⌫Ž t . Ž e i 1 t y 1 . dt ˆk Ž 1 , 2 . y ˆ k Ž 0, 2 . < s k 2 < t
H
F F
1 2k 1 2k
H< t
i 1 t
k
y 1 < dt
H< t
k
F < 1 <
H< t
k
dt s 2 < 2 k 1 < .
Also, by the same technique as in the proof of the theorem in w10x, we obtain < ˆ k Ž 1 , 2 .< F C < 2 k 1
½
H
5
is bounded on L p Ž R 2 . for all p ) 1, and the bound does not depend on . Because f is in L p Ž R 2 . if and only if < f < is, we infer that 1 < f Ž x 1 y t , x 2 y ⌫ Ž t . . < dt sup r < t
½
H
5
is also bounded on L p Ž R 2 . for all p ) 1. Lemma 4 is proved. Proof of Lemma 3. For 0 - ᑬ ␣ - n, we let G ␣ denote the Bessel potentials defined on R ny 1 Žsee w14, p. 132x., the Fourier transform of ˆ ␣ Ž y . s Ž1 q < y < 2 .y ␣ r2 . Next, we define G k␣ Ž y . by Gˆk␣ Ž y . s Ž1 which is G 2 y␣ r2 k q <2 y < . . We then have Tk␣, ¨ f Ž x, x n . s
2
H1 HR
ny1
f Ž x y y y 2 k t¨ , x n y ␦ 2 k ⌫ Ž t . . G k␣ Ž y . dy dt ; x, y g R
s
2
H1 HR
ny1
f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . . G ␣ Ž y y t¨ . dy dt.
392
HUNG VIET LE
Because < G ␣ Ž y .< is controlled by Cr< y < ny ᑬ ␣ Ž n ) ᑬ ␣ ) 0. as < y < ª 0, and because < G ␣ Ž y .< is rapidly decreasing as < y < ª ⬁, we have C < y < F A
< G ␣ Ž y y t¨ . < F
< y y t¨ <
nyᑬ ␣
q
B < y < ) A < y
.
2rA n ŽHere A ) 2 and B ' 11 q 4 will do.. Therefore, y 2rA
f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . .
2
< Tk␣, ¨ f Ž x, x n . < F C
H1 HR
qC
< y y t¨ < nyᑬ ␣
ny1
f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . .
2
H1 HR
< y < F A dy dt
< y
ny1
< y < ) A dy dt.
Denote the two integrals inside the curly brackets on the right-hand side of the above inequality as I1 and I2 , respectively. Notice that I2 s F s F
½
ÝH ⬁
Ž 2 A.
⬁
Ý ⬁
Ý ls0
F
1 l
⬁
Ý ls0
n
< y
H< y <(2 A M f Ž x y 2 l
1
1 n
Ž 2 A. 2 l
1 l
1
2 Ž2 1 2l
kŽ ny1.
kq l
A.
ny1
5
1 < y
dy
dy
Mn f Ž x y 2 k y, x n .
l ls0 < y <(2 A
Ý
y, x n y ␦ 2 k ⌫ Ž t . . dt
< y
⬁
ls0
F
k
Mn f Ž x y 2 k y, x n .
H< y
ls0
s
2
H< y
n
H< y <(2
kql
H< y <(2
kql
A
A
dy k
y, x n . dy
Mn f Ž x y y, x n . dy
Mn f Ž x y y, x n . dy
H MŽ1, . . . , ny1. Mn f Ž x, x n .
H s 2 MŽ1, . . . , ny1. Mn f Ž x, x n . , H where MŽ1, . . . , ny1. is the Hardy᎐Littlewood maximal operator acting on the first n y 1 variables, and Mn is the maximal operator along a surface acting on the last variable.
393
LEBESGUE MIXED NORM SPACES
We now estimate the bound of I1. Since ¨ g S ny 2 , there is a ¨ i , say, ¨ 1 such that < ¨ 1 < 2 G 1rŽ n y 1.. Then f Ž x y 2 k y, x n y ␦ 2 k ⌫ Ž t . .
2
I1 F
H1 HR
< y 1 y t¨ 1 < nyᑬ ␣
ny1
< y < F A dy dt.
We let D s A q 2, ˜ x s Ž x 2 , . . . , x ny1 ., ˜ y s Ž y 2 , . . . , yny1 ., and, making a change of variable y 1 ¬ y 1 y t¨ 1 , we obtain I1 F s F F
2
H1 H< y
f Ž x 1 y 2 k y 1 y 2 k t¨ 1 , ˜ x y 2k˜ y, x n y ␦ 2 k ⌫ Ž t . . < y 1 < nyᑬ ␣ 2
½
H< y
1
y 2 k y 1 y 2 k t¨ 1 , ˜ x y 2k˜ y, x n y ␦ 2 k ⌫ Ž t . . dt
MŽ1, n. f Ž x y 2 k y, x n . < y 1 < ny ᑬ ␣ < y 1 < nyᑬ ␣
1
5
dy < y 1 < nyᑬ ␣
dy
MŽ1 , n. f Ž x y 2 k y, x n .
H< ˜y
dy dt
dy 1 dy. ˜
The next to the last inequality above follows due to Lemma 4, and MŽ1, n. f is the maximal function along a surface, acting on the first and last variables. But
H< y
MŽ1, n. f Ž x y 2 k y, x n . < y 1 < nyᑬ ␣
1
s F s
y⬁
MŽ1, n. f Ž x 1 y 2 k y 1 , ˜ x y 2k˜ y, x n .
ÝH
< y 1 < nyᑬ ␣
l ls0 < y 1 <(2 D
y⬁
Ý
1 l
nyᑬ ␣
ls0
Ž2 D.
y⬁
1
Ý ls0
s
dy 1
l
Ž2 D.
y⬁
Ý ls0
H< y <(2 D M 1
1
nyᑬ ␣
2k
H< y <(2 1
1
Ž 2 lD .
ny Ž ᑬ ␣ q1 .
Ž1 , n.
l
kq1
1 2
kql
D
D
dy 1
f Ž x1 y 2 k y1 , ˜ x y 2k˜ y, x n . dy 1
MŽ1, n. f Ž x 1 y y 1 , ˜ x y 2k˜ y, x n . dy 1
H< y <(2 1
kql
D
MŽ1 , n. f Ž x 1 y y 1 , ˜ x y 2k˜ y, x n . dy 1
394
HUNG VIET LE
F
y⬁
Ý ls0
1 l
Ž2 D.
M1H MŽ1 , n. f Ž x 1 , ˜ x y 2k˜ y, x n .
ny Ž ᑬ ␣ q1 .
y⬁
Ý w2 Ž ᑬ ␣q1.yn x
s D Ž ᑬ ␣q1.yn
l
M1H MŽ1 , n. f Ž x 1 , ˜ x y 2k˜ y, x n .
ls0
s
CM1H MŽ1 , n.
f Ž x1 , ˜ x y 2k˜ y, x n .
if ᑬ ␣ ) n y 1. Therefore, I1 F
H< ˜y
MŽ1, n. f Ž x y 2 k y, x n . < y 1 < nyᑬ ␣
1
FC
H< ˜y
H 1 MŽ1 , n.
f Ž x1 , ˜ x y 2k˜ y, x n . dy˜
y⬁
sC Ý
H 1 MŽ1 , n.
f Ž x1 , ˜ x y 2k˜ y, x n . dy˜
H< ˜y <(2
D
ls0
H< ˜y <(2 D M
y⬁
1
sC Ý
ls0
l
k
Ž2 .
ny2
y⬁
s C Ý D ny2 Ž 2 l . ls0 y⬁
dy 1 dy˜
kql
M1H MŽ1 , n. f Ž x 1 , ˜ x y˜ y, x n . dy˜ 1
ny 2
Ž2
kq l
D.
H< ˜y <(2
ny2
kql
D
M1H MŽ1 , n. f Ž x 1 , ˜ x y˜ y, x n . dy˜
l
H H F C Ý Ž 2 ny2 . MŽ2, x, x n . . . . , ny1. M1 MŽ1 , n. f Ž x 1 , ˜
ls0 H H s CMŽ2, . . . , ny1. M1 MŽ1 , n. f Ž x, x n . ,
provided that n y 1 - ᑬ ␣ - n. Consequently, < Tk␣, ¨ f Ž x, x n . < H H H F C MŽ2, . . . , ny1. M1 MŽ1 , n. f Ž x, x n . q MŽ1 , . . . , ny1. Mn f Ž x, x n . 4 .
This inequality holds for all k g ⺪. It is well known that the Hardy᎐ Littlewood maximal operators are bounded on L p Ž R n .. Thus, it follows from Theorem 1 that sup < Tk␣, ¨ f <
kgZ
L p Ž Lq .
F C5 f 5 p
for 1 F q F ⬁, 1 - p F ⬁. Lemma 3 is proved. Proof of the Corollary. One just needs to verify that M ⌫ g is bounded on L p Ž R . for all p ) 1 and that Mf is bounded on L p Ž R n . for all p ) 1,
LEBESGUE MIXED NORM SPACES
395
n G 2. The fact that M ⌫ g is bounded on L p Ž R . for all p ) 1 has been shown in the proof of Corollary 1 in w10x. Also, applying the result of Corollary 1 in w10x, we infer that Mf is bounded on L p Ž R n . for all p ) 1, n G 2. Therefore, the proof is finished.
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