Weighted weak type (1, 1) estimates for oscillatory singular integrals

Weighted weak type (1, 1) estimates for

oscillatory singular integrals

著者 Sato Shuichi journal or publication title Studia Mathematica volume 141 number 1 page range 1-24 year 2000-01-01 URL http://hdl.handle.net/2297/25018

FOR OSCILLATORY SINGULAR INTEGRALS

Shuichi Sato

Abstract. We consider theA

1-weights and prove the weighted weak type (1 1) inequali-ties for certain oscillatory singular integrals.

1. Introduction LetK2C1(

R

nnf0g) satisfy (1.1) jK(x)j
cjxj ;n jrK(x)j
cjxj ;n;1 (1.2) Z a<jxj<b K(x)dx= 0 for alla b(0< a < b). The smallest constant for which (1.1) holds will be denoted byC(K).

We consider an oscillatory singular integral operator:

T(f)(x) = p:v:Z R n eiP(x
y)_K(x ;y)f(y)dy= lim !0 Z jx;yj> eiP(x
y)_K(x ;y)f(y)dy

initially de ned forf 2S(

R

n) (the Schwartz space), whereP is a real-valued polynomial:

(1.3) P(x y) = X

jjM
jjN

axy:

The following results are known.

Theorem A.

(Ricci-Stein 9]) Let 1< p <1. Then,T is bounded onLp(

R

n) with the

operator norm bounded by a constant depending only on the total degree of P,C(K), p

and the dimension n.

1991Mathematics Subject Classi cation. Primary 42B20.

Key words and phrases. Oscillatory singular integrals, rough operators. 1

Theorem B.

(Chanillo-Christ 2]) The operatorT is bounded fromL1₍

_R

n_{) to the weak}

L1₍

_R

n_{) space:}

sup

>0jfx2

R

n:jT(f)(x)j> gj
ckfkL1

with a constantc depending only on the total degree ofP,C(K) and the dimension n. See also 1] and 3] for the weighted weak type (1 1) estimates for convolution operators with oscillating kernels.

Letwbe a locally integrable positive function on

R

n. We say thatw2A1 if there is

a constantcsuch that

(1.4) M(w)(x)
cw(x) a.e.

where M denotes the Hardy-Littlewood maximal operator. The smallest constant for which (1.4) holds will be denoted byC1(w).

In this note we shall prove thatT is bounded fromL1_{w toL}1
1

w (the weakL1w space)

forw2A1:

Theorem.

There exists a constant c depending only on the total degree of P, C(K),

C1(w) and the dimension nsuch that

sup >0w(fx2

R

n:jT(f)(x)j> g)
ckfkL 1 w wherew(E) =R Ew(x)dx andkfkL1 w= R jf(x)jw(x)dx.

The theorem will be proved by a double induction, as in 9] and 2]. Let P be a polynomial as in (1.3). We assume that there exists a multi-indexsuch thatjj=M

anda6= 0 for some. We write

(1.5) P(x y) = X

jjM

x_Q(y)

and de ne

L= maxfdeg(Q) :Q6= 0 jj=Mg:

Then 0
L
N. We assume thatL1 and

max

jj=M jj=L

jaj= 1:

Under this assumption on a polynomialP, we de ne

T1(f)(x) = Z

jx;yj>1

eiP(x
y)_K(x;y)f(y)dy:

Proposition.

Let >0. There exists a constant c depending only on the total degree of P, and the dimension nsuch that ifC(K), C1(w)
, then

sup >0w(fx2

R

n:jT 1(f)(x) j> g)
ckfkL 1 w:

Remark 1. By Theorem and the extrapolation theorem of Rubio de Francia, we get the

L_pw-boundedness of T for all p2 (1 1) and all w 2 Ap, where Lpw is the space of all

those measurable functionsf which satisfykfkL p w= (

R

jf(x)jpw(x)dx)1=p <1andAp

denotes the weight class of Muckenhoupt.

We shall give the outlines of the proofs of Theorem and Proposition in Sections 2 and 4, respectively. Our proof of Proposition is based on the techniques used in Christ 5] to prove the weak type (1 1) estimates for rough operators (see also Christ 6], Christ-Rubio 7] and Sato 10]). We also use the geometrical argument of Chanillo-Christ 2]. We have to prove a key estimate (Lemma 7 in x5) in the unweighted case in order to

apply the method of Vargas 11] involving an interpolation with change of measure. To prove Lemma 7, we need a geometrical result for polynomials (Lemma 5 inx5). We shall

prove Lemma 5 inx7 by using the results and the arguments appearing in the proof of

Chanillo-Christ 2,Lemma4.1].

Finally, we note that in this paper, the constants with the same notation are not necessarily the same at each occurrence.

2. Outline of proof of Theorem

To apply the induction argument of 9] we need some preparation. We may assume that M  1 and N  1 otherwise Theorem reduces to a well-known fact that the

operator:

A(f)(x) = p:v:Z

K(x;y)f(y)dy

is bounded fromL1_wtoL1
1

w (see, for example, 8]).

We write a polynomial in (1.3) as follows:

P(x y) =XM j=0 X jj=j x_{Q(y) =}XM j=0Pj(x y)

say. We further decomposePj as follows:

Pj(x y) =XN t=0 X jj=j jj=t axy=XN t=0Pjt(x y)

say. Forj= 1 2 ::: M andk= 0 1 ::: N, de ne (2.1) Rjk(x y) =j ;1 X s=0Ps(x y) + k X t=0Pjt(x y): Note thatRjN =Pj s=0Ps (j= 1 2 ::: M).

Proposition

A(j k)

.

Let >0. There exists a constant c depending only on , j,N

and the dimension n such that if C(K), C1(w)
 and if Rjk is a polynomial as in

(2:1), then sup >0w(fx2

R

n:jTjk(f)(x)j> g)
ckfkL 1 w where Tjk(f)(x) = p:v: Z R n eiRjk(x
y)K(x ;y)f(y)dy:

Then, Theorem follows from PropositionA(M N). We shall prove it by double induc-tion. We rst note thatA(1 0) follows from theL1_w;L1

1

w boundedness of the operator

A.

Next, we observe that ifM2 and ifA(j N) (1
j
M;1) is true, so isA(j+1 0)

since

Rj+1
0(x y) =RjN(x y) + X jj=j+1

a0x

and hence jTj+1
0(f)(x)j=jTjN(f)(x)j. Thus, to complete the induction starting from

A(1 0) and arriving at A(M N), it suces to prove A(j k+ 1) by assuming A(j k) (0
k < N, 1
j
M). To achieve this, putR =Rj
k+1, R0=Rjk, Tj
k+1 =S. We

note that R(x y) =R0(x y) + X jj=j jj=k+1 axy:

We may assume Cjk = maxjj=j
jj=k+1

jaj6= 0. Then, by a suitable dilation we may

assumeCjk= 1. This can be seen as follows. We rst note that, fora >0,

S(f)(ax) = p:v:Z

eiR(ax
ay)_Ka(x

;y)f(ay)dy

where Ka(x) = anK(ax). Assume the boundedness of S for the case Cjk = 1. Then,

choosingato satisfyaj+k+1_{Cjk= 1, and using the dilation invariance of both the class}

A1 and the class of the kernels satisfying (1.1) and (1.2), we get

w(fx2

R

n:jS(f)(x)j> g) =wa(fx2

R

n:jS(f)(ax)j> g)
c ;1 Z jf(ax)janw(ax)dx =c;1 kfkL1 w:

We split the kernel K as K = K0 +K1, where K0(x) = K(x) if

jxj
1 and

K1(x) =K(x) if

jxj>1. Assuming Cjk = 1, we consider the corresponding splitting

S=S0+S1: S0(f)(x) = p:v: Z eiR(x
y)_K0(x;y)f(y)dy S1(f)(x) = Z eiR(x
y)_K 1(x ;y)f(y)dy:

In the next section, we shall prove

(2.2) sup_>0w(fx2

R

n:jS0(f)(x)j> g)
ckfkL1 w

while by Proposition we have

(2.3) sup_>0w(fx2

R

n:jS

1(f)(x)

j> g)
ckfkL 1 w:

Combining (2.2) and (2.3), we shall complete the proof ofA(j k+ 1), which will nish the proof of Theorem.

3. Estimate forS0

In this section, we shall prove, under the assumption made inx2, that ifC(K),C1(w)

 ( >0), thenS0 satis es (2.2) with a constantc depending only onj,N, andn.

First, we shall prove

(3.1) w(fx2B(0 1) :jS0(f)(x)j> g)
c ;1

Z

jyj<2

jf(y)jw(y)dy

whereB(x r) denotes the closed ball with centerxand radiusr >0.

Lemma 1.

Letw2A1. LetT be an operator of the form:

T(f)(x) = p:v:Z R n K(x y)f(y)dy= lim_ !0 Z jx;yj> K(x y)f(y)dy a.e. for f 2L1w, where the kernelK satisesjK(x y)j
c0jx;yj

;n. For >0, put T(f)(x) = p:v: Z jx;yj< K(x y)f(y)dy: Suppose sup >0w(fx2

R

n :jT(f)(x)j> g)
cwkfkL1 w:

Then, there exists a constantc depending only on the dimension nsuch that sup

>0w(fx2

R

n:jT(f)(x)j> g)
c(cw+c0C1(w))kfkL1 w:

Proof. The proof is similar to that ofLemmain 9, p. 187]. We shall prove (3.2) w(fx2B(h =4) :jT(f)(x)j> g)
(2cw+cc0C1(w)) ;1 Z jy;hj<5=4 jf(y)jw(y)dy

uniformly in h2

R

n. Integrating both sides of the inequality in (3.2) with respect toh,

Splitf into 3 pieces: f =f1+f2+f3, wheref1(y) =f(y) if jy;hj< =2,f1(y) = 0

otherwise f2(y) = f(y) if =2
jy;hj < 5=4, f2(y) = 0 otherwise f3(y) = f(y) if jy;hj5=4,f3(y) = 0 otherwise. Note that ifjx;hj
=4, thenT(f1)(x) =T(f1)(x)

sincejy;hj< =2 andjx;hj
=4 implyjx;yj< . So by the assumption onT, we

have w(fx2B(h =4) :jT(f1)(x)j> g) =w(fx2B(h =4) :jT(f1)(x)j> g)
w(fx:jT(f1)(x)j> g)
cw ;1 kf1kL 1 w
cw ;1 Z jy;hj<5=4 jf(y)jw(y)dy:

Next, ifjx;hj
=4 and=2
jy;hj<5=4, then=4
jx;yj<3=2, and so jT(f2)(x)j
cc0

;n Z

jy;hj<5=4

jf2(y)jdy:

Hence, by Chebyshev's inequality,

w(fx2B(h =4) :jT(f2)(x)j> g)
cc0 ;1w(B(h =4));n Z jy;hj<5=4 jf2(y)jdy
cc0C1(w) ;1 Z jy;hj<5=4 jf(y)jw(y)dy:

Finally, if jx;hj
=4 and jy;hj5=4, then jx;yj, and so T(f3)(x) = 0.

Combining these, we get (3.2). This completes the proof of Lemma 1. Now we turn to the proof of (3.1). Ifjxj
1 andjyj
2, then

exp(iR(x y));exp 0 B B @i 0 B B @R0(x y) + X jj=j jj=k+1 ay+ 1 C C A 1 C C A
cjx;yj

wherec depends only onk j andn. Hence, ifjxj
1, jS0(f)(x)j
U 0 B B @exp 0 B B @i X jj=j jj=k+1 ay+ 1 C C Af(y) 1 C C A(x) +cI(f)(x) where U(f)(x) = p:v:Z eiR0(x
y)K 0(x;y)f(y)dy I(f)(x) = Z jx;yj<1 jx;yj ;n+1 jf(y)jdy:

Note that U(f)(x) = U(fB(0
2))(x), I(f)(x) = I(fB(0
2))(x) if jxj < 1. By the

induction hypothesisA(j k) and Lemma 1, we see thatU is bounded fromL1_wtoL1
1

w .

On the other hand, since

Z jx;yj<1 jx;yj ;n+1w(x)dx= X j0 Z 2j;1 jx;yj2 j jx;yj ;n+1w(x)dx
c X j0 2j2;jn Z jx;yj2 j w(x)dx
cM(w)(y)

by Chebyshev's inequality we have

w(fx2B(0 1) :I(f)(x)> g)
 ;1 Z jyj<2  Z jx;yj<1 jx;yj ;n+1w(x)dx ! jf(y)jdy
cC1(w) ;1 Z jyj<2 jf(y)jw(y)dy:

Combining these results, we get (3.1). Similarly we can prove

(3.3) w(fx2B(h 1) :jS0(f)(x)j> g)
c ;1

Z

jy;hj<2

jf(y)jw(y)dy

wherec is independent ofh2

R

n. To see this, we rst note that

S0(f)(x+h) = p:v: Z eiR(x+h
y+h)_K0(x;y)f(y+h)dy and R(x+h y+h) =R1(x y h) + X jj=j jj=k+1 axy:

We can apply the induction hypothesisA(j k) to the operator p:v:Z

eiR1(x
y
h)K(x

;y)f(y)dy

to get its boundedness fromL1_wtoL1
1

w . Thus, by the same argument that leads to (3.1)

we get w(fx2B(h 1) :jS0(f)(x)j> g) =hw(fx2B(0 1) :jS0(f)(x+h)j> g)
c ;1 Z jyj<2 jf(y+h)jw(y+h)dy
c ;1 Z jy;hj<2 jf(y)jw(y)dy

where hw(x) =w(x+h), and we have used the translation invariance of the classA1.

4. Outline of proof of Proposition

We may assume f 2S(

R

n). By Calderon-Zygmund decomposition at height  >0

we have a collectionfQgof non-overlapping closed dyadic cubes and functions g bsuch

that f =g+b (4.1) 
jQj ;1 Z Qjfj
c (4.2) v(Q)
cvkfkL1 v= for allv 2A1 (4.3) kgk 1
c (4.4) kgkL1 v
cvkfkL1 v for allv 2A1 (4.5) b=X Q bQ (4.6) supp(bQ)Q (4.7) Z bQ= 0 (4.8) kbQkL1
cjQj: (4.9)

Remark 2. In this note we do not use (4.8).

Let a polynomialP be as in Proposition. We assume, as we may, that M 1 as in

the outline of the proof of Theorem in x2. We write P as in (1.5). Then, let q(y) = P

jjLcy

 _{be the coecient ofx}M_{1 . By a rotation of coordinates and a normalization,}

and by discarding a negligible dierence, we see that to prove Proposition we may study

T1assuming maxjj=L

jcj= 1 in this case the condition max

jj=M
jj=L

jaj= 1 may

not hold (see 2, p. 151] and Sublemma 2 inx7).

We pick a non-negative'2C 1 0 (

R

n) such that supp(')f1=2
jxj
2g 1 X j=0'(2 ;jx) = 1 if jxj1: PutKj(x y) ='(2;j(x ;y))K 1(x y), whereK1(x y) =e iP(x
y)_K 1(x ;y) (K 1(x) is as inx2) and decomposeK 1(x y) asK1(x y) = P 1 j=0Kj(x y). De ne Vj(f)(x) = Z Kj(x y)f(y)dy for j 0 and put V(f)(x) = 1 X j=1Vj(f)(x):

Then T1 = V0+V. In the following, we study V only, since we easily see that V0 is

We set (see 5, 6, 7]) Bi= X jQj=2 in bQ (i1) B0= X jQj1 bQ:

PutU =Q~, where ~Qdenotes the cube with the same center asQand with sidelength

100 times that of Q. Here and in the sequel all cubes we consider have sides parallel to the coordinate axes.

Whenx2

R

nnU, we observe that

(4.10) V(b)(x) =V 0 @ X i0 Bi 1 A(x) =X i0 X j1 Z Kj(x y)Bi(y)dy=X i0 X ji+1 Z Kj(x y)Bi(y)dy =X s1 X js Z Kj(x y)Bj;s(y)dy= X s1 X js Vj(Bj;s)(x):

Inx5 we shall prove the following.

Lemma 2.

Suppose w2A1. There exists an  >0 such that, for any positive integer s,       X js Vj(Bj;s)       2 L2 w
c2 ;s kfkL1 w:

Inx6 we shall prove the following.

Lemma 3.

Suppose w2A1. Letkk2
w denote the operator norm onL2w. Then, there

exist constantsc, >0 such that

kVjk2
w
c2

;j for all j 1:

Assuming Lemmas 2 and 3, we now prove Proposition. From Lemma 3 we easily see thatV is bounded onL2_{w. By this boundedness, (4.1), (4.4), (4.5), (4.10) and Lemma 2}

we have (4.11) w(fx2

R

nnU :jV(f)(x)j> g)
w(fx2

R

nnU:jV(g)(x)j> =2g) +w(fx2

R

nnU :jV(b)(x)j> =2g)
c ;2 kgk 2 L2 w+c ;2       X s1 X js Vj(Bj;s)       2 L2 w
c ;1 kfkL 1 w+c ;2 0 @ X s1 1=2₂;s=2 kfk 1=2 L1 w 1 A 2
c ;1 kfkL 1 w:

On the other hand, by (4.3) we see that

(4.12) w(U)
cw

;1 kfkL

1 w:

Combining (4.11) and (4.12), we get the boundedness of V from L1_{w to L}1
1 w . This

5. Proof of Lemma 2

In this section we shall prove Lemma 2 inx4. Fork m1, put

(5.1) Hkm(x y) = Z Kk(z x)Km(z y)dz = Z e;iP(z
x)+iP(z
y)K(z ;x)K(z;y)'k(z;x)'m(z;y)dz: ThenV kVm(f)(x) =R Hkm(x y)f(y)dy, whereV

k denotes the adjoint ofVk.

Lemma 4.

Letkm1. Then,Hkm(x y) = 0 unlessjx;yj
42k and

(1) jHkm(x y)j
c2 ;kn, (2) jHkm(x y)j
c2 ;kn2;m jq(x);q(y)j ;1=M:

Proof. We prove the estimate (2) only since the other assertions immediately follow from the de nition ofHkm in (5.1). We rst note that

(@=@z1)M(P(z x);P(z y)) =M!(q(x);q(y)):

Hence, from van der Corput's lemma it follows that

Z b

a ei(P(z
x)

;P(z
y))dz

1
cjq(x);q(y)j ;1=M

for anyaandb(see 2, p.152]).

Therefore by integration by parts in variablez1in the formula of (5.1), and by using

the estimates in (1.1), we easily get the conclusion.

For the rest of this noteP(x) will denote a real-valued polynomial on

R

n.

Denition 1.

For a polynomialP(x) =P

jjNax

 _{of degreeN, de ne} kPk= max

jj=N jaj:

Denition 2.

For a polynomialP and >0, let

R(P ) =fx2

R

n:jP(x)j
g:

Letd(E F) denote the distance between setsE and F. We now state a geometrical lemma for polynomials, which will be proved inx7.

Lemma 5.

Let k, m be integers such that k  m. Suppose N  1. Then, for any

polynomial P of degree N satisfying kPk = 1 and any  > 0, there exists a positive

constant Cn
N
depending only onn,N and such that  x2B(a 2k) :d ; x R(P 2Nm)
2m
Cn
N
2(n ;1)k2m uniformly in a2

R

n.

Let >0 and letfBjgj

0 be a family of measurable functions such that

(5.2)

Z

QjBjj
jQj

for all cubesQin

R

n with sidelength`(Q) = 2j. Then we have the following.

Lemma 6.

Let the kernelsHji be as in Lemma4. Then, we can nd a constantcsuch that _j X i=sxsup2R n Z Bi ;s(y)Hji(x y)dy
c2 ;s

for all integers j ands such that0< s
j.

Denition 3.

For m 2

Z

(the set of all integers), letDm be the family of all closed

dyadic cubesQwith sidelength`(Q) = 2m. Proof of Lemma 6. Fixx2

R

n. Let

F=  Q2Di ;s:Q \B(x 2j +2₎ 6 = (0< s
i
j): Then clearlyP Q2F jQj
c2jn. DecomposeF =F0F1, where F0= n Q2F :Q\R(q();q(x) 2L (i;s)) 6 = o

andF1=FnF0. Then by Lemma 5 we have

(5.3) X

Q2F 0

jQj
c2

(n;1)j2i;s:

By Lemma 4 (1), (5.2) and (5.3), we see that

X Q2F 0 Z QjBi ;s(y)Hji(x y) j dy
c2 ;jn X Q2F 0 Z QjBi ;s(y) jdy (5.4)
c2 ;jn X Q2F 0 jQj
c2 ;jn2(n;1)j2i;s =c2i;j;s:

Next, by Lemma 4 (2), (5.2) and the estimateP Q2F 1 jQj
c2jn, we have X Q2F1 Z QjBi ;s(y)Hji(x y) jdy
c2 ;jn2;i2;L(i;s)=M X Q2F1 Z QjBi ;s(y) jdy (5.5)
c2 ;jn2;i2;L(i;s)=M X Q2F1 jQj
c2 ;i2;L(i;s)=M:

From (5.4) and (5.5) it follows that

Z jBi ;s(y)Hji(x y) jdy= X Q2F Z QjBi ;s(y)Hji(x y) jdy =X1 =0 X Q2F Z QjBi ;s(y)Hji(x y) jdy
c  2i;j;s+ 2;i2;L(i;s)=M  :

Thus we see that j X i=sxsup2R n Z jBi ;s(y)Hji(x y) jdy
c j X i=s  2i;j;s+ 2;i2;L(i;s)=M 
c2 ;s:

This completes the proof of Lemma 6. By Lemma 6 we readily get the following.

Lemma 7.

LetfBjgj

0 be as in Lemma 6. Suppose P

j0

kBjkL1<1. Then, for any

positive integers, we have

      X js Vj(Bj ;s)       2 L2
c2 ;s X j0 kBjkL1:

Proof. Leth idenote the inner product inL2. Using Lemma 6, we see that       X js Vj(Bj ;s)       2 L2
2 X js j X i=s jhVj(Bj ;s) Vi( Bi ;s) ij
2 X js j X i=s hBj ;s V  jVi(Bi ;s) i
2 X js j X i=s kBj ;s kL 1 kV  jVi(Bi ;s) kL 1
c2 ;s X js kBj ;s kL 1:

This completes the proof of Lemma 7.

Denition 4.

For eachj0, letGj be a family of non-overlapping closed dyadic cubes

Qsuch that`(Q)
2j. We suppose that ifQ2Gj,R2Gk andj6=k, thenQandRare

non-overlapping and thatP j0 P Q2G j jQj<1. PutG=j 0 Gj.

Let >0. To eachQ2G we associatefQ 2L1such that Z jfQj
jQj supp(fQ)Q: We de ne Ai= X Q2Gi fQ:

Lemma 8.

Letv be a locally integrable positive function and letsbe a positive integer. Then       X js Vj(Aj ;s)       2 L2 v
c 2X Q2G jQjinf Q MM(v)

whereinfQf = infx2Qf(x).

Proof. The proof we give here is essentially the same as that in 11]. We include the proof for the sake of completeness. We may assume P

Q2G

jQjinfQMM(v)<1. Let h iv denote the inner product inL2v. Then, ifs
i
j, we see that

hVj(Aj ;s) Vi( Ai ;s) iv = Z Z Kj(x y)Aj ;s(y)dy Z Ki(x z)Ai ;s(z)dz  v(x)dx = Z Aj ;s(y) Z Ai ;s(z) Z Kj(x y)Ki(x z)v(x)dx  dz dy: Put v(y zi j) = 2;in2;jn Z B(y
2j+2) \B(z
2 i+2) v(x)dx:

LetcQ denote the center of a cubeQ. IfQ2Gj ;s,R

2Gi

;sand ifB(y 2

j+2_{) intersects}

B(z 2i+2_{) for somey}2Qand somez2R, thenRB(cQ n1=22j+10). Thus we have j X i=s jhVj(Aj ;s) Vi( Ai ;s) ivj
c j X i=s Z jAj ;s(y) j Z jAi ;s(z) jv(y zi j)dz dy
c X Q2Gj;s Z jfQ(y)j j X i=s X R2Gi;s Z jfR(z)jv(y zi j)dz dy
c X Q2Gj;s 2;jn Z jfQ(y)jdy j X i=s X R2Gi;s RB(c Q
n 1=22j+10) inf_R M(v) Z jfR(z)jdz =I say. Since inf_R M(v) Z jfR(z)jdz
jRjinf R M(v)
 Z RM(v)(z)dz

and cubes inGare non-overlapping, I
c X Q2G j;s 2;jn Z B(cQ
n 1=22j+10) M(v)(z)dzZ jfQ(y)jdy
c X Q2Gj;s inf_Q MM(v)Z jfQ(y)jdy
c 2 X Q2Gj;s jQjinf Q MM(v):

Therefore, we get the conclusion by summing overjs, since       X js Vj(Aj ;s)       2 L2 v
2 X js j X i=s jhVj(Aj ;s) Vi( Ai ;s) ivj:

We prove Lemma 2 by the interpolation argument of 11] between the estimates of Lemma 7 and Lemma 8.

Lemma 9.

LetF denote the family of dyadic cubes arising from the Calderon-Zygmund

decomposition inx4. Then, for all t >0, we have Z X js Vj(Bj;s)(x) 2 min(v(x) t)dx
c 2 X Q2F jQjmin t2;s inf Q MM(v) 

wheres is a positive integer andv is a locally integrable positive function. Proof. We de ne Ft=  Q2F : inf Q MM(v)< t2;s  andF  t =FnFt. Forj1, put B0 j= X jQj=2 jn Q2F t bQ B00 j = X jQj=2 jn Q2F  t bQ and B0 0= X jQj1 Q2Ft bQ B00 0 = X jQj1 Q2F  t bQ:

ThenBj =B0 j+B0 0 j forj0. Hence Z X js Vj(Bj;s)(x) 2 min(v(x) t)dx
2 Z X js Vj(B0 j;s)(x) 2 min(v(x) t)dx+ 2 Z X js Vj(B00 j;s)(x) 2 min(v(x) t)dx
2 Z X js Vj(B0 j;s)(x) 2 v(x)dx+ 2 Z X js Vj(B00 j;s)(x) 2 tdx =I+II say.

Applying Lemma 8 with Aj =c1B 0

j (see (4.7) and (4.9)), we get

I
c 2 X Q2Ft jQjinf Q MM(v) =c2 X Q2Ft jQjmin t2;s inf Q MM(v)  : By Lemma 7 withBj =c2B 0 0

j (see (4.7) and (4.9)), we have

II
ct2 ;s X j0 kB 00 jkL1
c2t2 ;s X Q2F  t jQj =c2 X Q2F  t jQjmin t2;s inf Q MM(v)  :

(Here c1 and c2 are normalizing constants.) Combining the estimates for I and II, we

get the conclusion.

Now we nish the proof of Lemma 2. Multiplying both sides of the inequality in Lemma 9 byt; (

2(0 1)), then integrating them on (0 1) with respect to the measuredt=t

and using Z 1 0 min(A t)t ; dt t =cA1; (A >0) for some c >0 we get Z X js Vj(Bj;s)(x) 2 v(x)1;dx
c 2 X Q2F jQj2 ;sinf Q MM(v)1; (5.6)
c2 ;s X Q2F Z Qjf(x)jdxinf Q MM(v)1;
c2 ;s Z jf(x)jMM(v)(x)1 ;dx

where the second inequality follows from (4.2).

If w2 A1, then w1+ 2 A1 for some > 0 so substituting w1+ forv and putting

= =(1 + ) in (5.6), we get Lemma 2.

6. Proof of Lemma 3 In this section we shall prove Lemma 3 inx4.

Lemma 10.

Letkk2 denote the operator norm on L2. Then, for j1, kVjk2
 CM
L2;j=2;min(L=M
M=L)j=2 (M 6 =L) CMj1=22;j (M=L):

Estimates of this kind have been obtained in Ricci-Stein 9]. Here we give an alterna-tive proof.

Proof of Lemma 10. Fixx. Let

E=R(q();q(x) 2 ;jM) \B(x 2j +2_{) and F =B(x 2}j+2₎ nE: De ne E =  Q2D ;jM=L]:Q \E6=

where a] denotes the greatest integer not exceedinga. Then, by Lemma 4 (1) and Lemma 5 we have

Z EjHjj(x y)jdy
X Q2E Z QjHjj(x y)jdy (6.1)
c2 ;jn X Q2E jQj
c2 ;j2;jM=L: For= 0 1 2 :::, let F =B(x 2j+2)\ ; R(q();q(x) 2 ;jM++1) nR(q();q(x) 2 ;jM+) : ThenF = 1 =0F. For 0

j(M+L);1, let F =  Q2D ;(jM;;1)=L]:Q \F 6= :

Then by Lemma 5 we have

jFj
X Q2F

jQj
c2j(n

;1)2;(jM;;1)=L]:

So by Lemma 4 (2) we see that

Z F jHjj(x y)jdy
c2 ;jn2;j2j;=M jFj
c2 ;j2;=M2;(jM;;1)=L]
c2 ;j2;jM=L2;(1=M;1=L):

Thus, ifM6=L, j(M+L);1 X =0 Z F jHjj(x y)jdy (6.2)
c2 ;j2;jM=L(2;(1=M;1=L) ;1) ;1(2;j(M+L)(1=M;1=L) ;1) =c2;j(2;(1=M;1=L) ;1) ;1(2;jL=M ;2 ;jM=L) and ifM=L, (6.3) j(M+L) ;1 X =0 Z F jHjj(x y)jdy
c2 ;2jj(M+L):

Finally, by Lemma 4 (2) we have

Z j(M+L)F jHjj(x y)j dy
c2 ;jn2;j2;jL=M  j(M+L) F (6.4)
c2 ;j2;jL=M:

By (6.1), (6.2), (6.3) and (6.4) we see that sup x2R n Z jHjj(x y)jdy
 CM
L2;j;min(L=M
M=L)j (M 6 =L) CMj2;2j (M=L):

We have the same estimate for sup_yR

jHjj(x y)j dx. From these results we get the

conclusion since kV  jVj(f)kL2
sup_x Z jHjj(x y)jdy 1=2 sup_y Z jHjj(x y)jdx 1=2 kfkL2 andkV  jVjk2=kVjk22.

Now we prove Lemma 3. It is easy to see that kVjk2
w
cw. By interpolation with

change of measure between this estimate and that of Lemma 10, we get (6.5) kVjk2
w
cw
2

;(1;)j=2

for all  2(0 1). We havew1+ 2 A1 for some  >0  so substituting w1+ forw and

putting= 1=(1 +) in (6.5), we get the desired estimate. 7. Proof of Lemma 5

Our proof is an application of the methods appearing in the proof of 2,Lemma4.1]. We use some tools and results given in 2].

Denition 5.

Supposen2. Let Sm=fQm+ (0 0 ::: 0 j) :j2

Z

g where m= (m1 m2 ::: mn;1) 2

Z

n ;1 andQ m = 0 1]n+ (m1 m2 ::: mn;1 0). We callSma strip.

Denition 6.

Supposen2. Form2

Z

n

;1, we de ne

Im=fQm+ (0 0 ::: 0 j) :j1< j < j2g

wherej1,j22

Z

f;1 1gandQm is as in De nition 5. We callIman interval.

Denition 7.

For a setE

R

n, we put

N(E) =fx2

R

n :d(x E)
1g:

LetP be a polynomial of degreeN as in Lemma 5. We consider R(P ) for >0

(see De nition 2).

Lemma 11.

Suppose that n  2 and N  1. There exists a positive integer Cn
N

depending only onn andN such that for i= 1 2 ::: Cn
N we can ndUi2O(n) (the

orthogonal group) and families of cubesJm
iSm(m2

Z

n

;1) so that (1) N(R(P )) SC nN i=1 Ui(Li) where Li= 8 < : Q:Q2  m2Z n;1 Jm
i 9 =  

(2) card(Jm
i)
cfor some constant c depending only onn,N and.

Remark 3. If Lemma 11 holds, then we have, for any >0,

fx:d(x R(P ))
g CnN

 i=1 Ui(

Li)

for some positive integerCn
N
depending only onn,N and, whereUi andLi are as

in Lemma 11. This can be proved by considering a nite number of polynomials which are de ned by translating P and by applying Lemma 11 to each of them. (See 2, p. 149].)

To prove Lemma 11, we need the follwing results given in 2].

Sublemma 1.

Supposen2. For any positive integerN, there exists a positive integer

Cn
N depending only on nandN such that for any stripS, any polynomial P of degree

N and any  >0

fQ2S:Q\R(P )6=g

Sublemma 2.

Supposen2. For any positive integerN, there exist positive constants

An
N andBn
N depending only onnandN such that

An
NkPk
kPk
Bn
NkPk

for every polynomial P of degreeN and every2O(n), whereP(x) =P(x).

Sublemma 3.

Suppose n2. For any positive integer N, there exists a positive

con-stant Cn
N depending only on nand N such that for any polynomial P of degree N we

can nd2O(n) so that

min

1jn

kDj(P)kCn
NkPk

whereDj=@=@xj.

Now we prove Lemma 11. We use induction on the polynomial degreeN. LetA(N) be the assertion of Lemma 11 for polynomials of degreeN.

Proof ofA(1). LetP(x) =Pn

i=1aixi+b. First, we consider the casejanj= 1. Now we

show that ifI is an interval such that each cube ofI intersectsR(P ), then card(I)
c

for somec depending only onnand. Lety2Q2I satisfyjP(y)j
. We note that

P(y+den);P(y) =dan for d2

R

whereej is the element of

R

nwhose jth coordinate is 1 and whose other coordinates are

all 0. Therefore, ify+den 2Q 0 2I, we see that inf z2Q 0 jP(z)jjP(y+den)j; n X i=1 jaijjdanj;; n X i=1 jaijjdj;;n:

This easily implies that card(I)
c.

By this and Sublemma 1, there exists a constantc depending only onn and such that

card(fQ2S:Q\R(P )6=g)
c

for all stripsS. Therefore, if we put

Jm=fQ2Sm:d(Q R(P ))
1g

then card(Jm)
cfor somec depending only onnand  andN(R(P ))L, where L= ( Q:Q2  m2Z n;1 Jm ) :

Next, we consider any polynomialP of degree 1 such that kPk= 1. Then ifP1(x) =

P(Ux) for a suitableU 2 O(n), we have DnP1 = 1. Hence, by what we have already

proved we get N(R(P1 ))  L. It follows that N(R(P ))  U(L) since N(R(P 

U )) =U;1

Now we assumeA(N;1) (N 2) and proveA(N). For a polynomialP of degreeN

such thatkPk= 1, we choose 2O(n) as in Sublemma 3. Put

E0=R(P )\ 0 @ n  j=1 R(Dj(P) ) 1 A

and for= (1 2 ::: n)2f;1 1gn put

E=fx2R(P ) :jDj(P)(x)>  for j= 1 2 ::: ng: Then R(P ) =E0 0 @  2f;1
1g n E 1 A and so (7.1) N(R(P )) =N(E0) 0 @  2f;1
1g n N(E) 1 A:

We separately treat the 2n+ 1 sets on the right hand side. First, clearly (7.2) N(E0) n  j=1 N(R(Dj(P) )):

Since Cj =kDj(P)k1 ( this means that c ;1

kDj(P )k
c for somec >1

depending only on nand N) andR(Dj(P ) ) =R(C ;1

j Dj(P) C ;1

j ) we can

apply the induction hypothesisA(N;1) to the right hand side of (7.2).

Next, we x  and consider N(E). Pick O 2 O(n) such that O(en) = n ;1=2. De ne D  0=D0n 8 < : Q2D0: 0 @ n  j=1 R((Dj(P))O ) 1 A \Q6= 9 =  :

Since k(Dj(P))Ok1 by Sublemmas 2 and 3, we can apply the hypothesis

A(N;1) along with Remark 3 to

G= 8 < : Q2D0: 0 @ n  j=1 R((Dj(P))O ) 1 A \Q6= 9 =  to get (7.3) N(G)iU 0 i(L 0 i)

for someU0 i2O(n) and someL 0 i such that L 0 i = 8 < : Q:Q2  m2Z n;1 J0 m
i 9 =  for someJ0

m
i (Sm) satisfying card(J 0

m
i)
c.

We have to study O;1

 (E)\(D 

0). First, we note that if O;1

 (E) intersects Q, Q2D  0, then (7.4) ₁min jn jDj(P)(Oy)>  for all y 2Q:

This can be seen as follows. Suppose that there arej0andy02Qsuch thatj 0Dj0(P



)(Oy0)
. Then, since we have j 0Dj0(P

)(Ox)>  for some x 2 Q, by the

intermediate value theorem we can nd z 2Q such thatjDj 0(P

)(Oz)j
. This

contradicts the fact that Q2D  0. By (7.4) we have (7.5) O;1  (E)\(D  0)  Q2D0: min 1jn jDj(P )(Oy)>  for all y2Q and R(P O )\Q6=  :

For a stripS, put

E=  Q2S : min 1jn jDj(P)(Oy)>  for all y2Q and R(P O )\Q6=  :

We shall show card(E)
Cn
N.

We rst see thatE is a union of at mostCn
N intervals. Put E 0 =  Q2S: min 1jn jDj(P)(Oy)j>  for all y2Q and R(P O )\Q6=  : Then E 0 = 0 @ n \ j=1(S nfQ2S:R((Dj(P))O )\Q6=g) 1 A \fQ2S:R(PO )\Q6=g:

We observe that the complement of a nite union of intervals in a stripS is also a nite union of intervals, and the intersection of nite unions of intervals is also a nite union

of intervals. Hence, by Sublemma 1 we see thatE 0 is a union of at mostC n
N intervals : E 0 = iJi.

Consider anyJi. Then by the intermediate value theorem we have either

min 1jn jDj(P)(Oy)>  for all y2fQ:Q2Jig or min 1jn jDj(P)(Oy)<; for all y2fQ:Q2Jig:

ThusE is a union of a subfamilyfIigoffJig: E =iIi.

LetI be any interval infIig. We need the following (see 2, p. 151]).

Sublemma 4.

There exists a constantcn depending only onnsuch that ifx y2I and

yn;xn cn, then y;x= n X i=1iO ;1  ei

for somei2

R

such that ii3.

Proof. We see that

O(y;x) = n X i=1(yi ;xi)Oei= n;1 X i=1(yi ;xi)Oei+ (yn;xn)n ;1=2 =Xn i=1  n;1=2(y n;xn)i+bi  ei

for somebi2

R

such thatjbij
c, which is feasible sincejyi;xij
1 fori= 1 2 ::: n;1.

This readily implies the conclusion.

Put Y = P O. Then rY(x) = O ;1  (r(P )(Ox)) so, if x y 2 I and yn;xn cn, by Sublemma 4 we have Y(y);Y(x) = Z 1 0 hy;x (rY)(x+t(y;x))idt = Z 1 0 n X i=1i  O;1  ei O;1  (r(P)(O(x+t(y;x))))  dt =Z 1 0 n X i=1iDi(P )(O(x+t(y;x)))dt  n X i=1ii 3n >3

whereh idenotes the inner product in

R

n. SinceR(Y )\Q6=for allQ2I, we can

conclude that card(I)
cn+ 3.

Combining the above results, we have card(E)
Cn
Nas claimed. From this and (7.5)

we easily see that

(7.6) N ; O;1  (E)\(D  0) L

whereL= Q:Q2 m 2Z

n;1Jm for some JmSm with card(Jm)
Cn
N.

By (7.3) and (7.6) we have N ; O;1  (E) N(G)N ; O;1  (E)\(D  0) (iU 0 i(L 0 i))L

and so, observingN ; O;1  (E) =O;1  N(E), (7.7) N(E)(iOU 0 i(L 0 i))O(L): SinceN(R(P )) =  ;1 N(R(P )), by (7.1), (7.2) withA(N;1) and (7.7) we

getA(N). This completes the proof of Lemma 11.

Proof of Lemma 5. We see thatR(P 2Nm) = 2mR( ~P 1), where

~

P(x) = 2;NmP(2mx):

Note thatkP~k= 1. (See 2, p. 151].) This observation enables us to assumem= 0 to

prove Lemma 5. Clearly, we may also assume= 1. Thus it suces to show, fork0,

(7.8)  x2B(a 2k) :d(x R(P 1))
1
Cn
N2 (n;1)k uniformly in a2

R

n.

If n = 1, (7.8) easily follows from Chanillo-Christ 2, Lemma 3.2] (see also 4]). Supposen2. Then, (7.8) follows from Lemma 11 with = 1 and the obvious estimate:

B(a 2k)\Ui(Li)
c2(n ;1)k

whereUi(Li) is as in Lemma 11. This completes the proof of Lemma 5. References

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Dept. of Mathematics, Faculty of Education, Kanazawa University, Kanazawa 920-1192, Japan