WEIGHTED Lp-Lq INEQUALITIES FOR THE FRACTIONAL INTEGRAL OPERATOR WHEN 1<q<p<∞

WEIGHTED L p

;L q

INEQUALITIES FOR

THE FRACTIONAL INTEGRAL OPERATOR

WHEN 1 < q < p < 1

Yves Rakotondratsimba

(Received March 23, 1998 Revised November 28, 1998)

Abstract. We nd necessary conditions and sucient conditions on weights u(:) andv(:) for which the fractional integral operatorI
is bounded from the

weighted Lebesgue spacesL p v into L q uwhenever 1 <q<p<1and 0<<n.

Actually such a boundedness is characterized for a large class of weights. AMS1991 Mathematics Subject Classi
cation. 42B25.

Key words and phrases. Weighted inequalities, Fractional integral operators.

x

1. INTRODUCTION

The fractional integral operator I

of order , 0<  <n, acts on locally integrable functions ofR n as (I
f)(x) = Z y2R n jx;yj
;n f(y)dy:

Our purpose in this paper is to derive conditions on weight functionsu(:)

and v(:) for which there is a constant C>0 such that

(1.1)
Z R n( I
f) q( x)u(x)dx  1 q C
Z R n f p( x)v(x)dx  1 p for allf(:)0

and for 1<q <p<1. The boundedness dened by (1:1) will be also denoted

by I
: L p v !L q u.

Since inequalities (1:1) have a fundamental role in Analysis (in deriving

weighted Poinca re and Sobolev inequalities, in estimating eigenvalues of some Schrodinger operators,....), they have been studied extensively by many au-thors for the range p  q. Recent papers on this topic can be found in

Sa-Wh-Zh] for the American school, in Ge-Go-Ko] for the Georgian school and in Ra2] for the author's contribution. Considering (1:1) for the range q<pwould enlarge for instance the available results (for pq) for weighted

Sobolev and Poincare inequalities.

A signicant attempt on a characterization for I
: L p v ! L q u with q < p,

based on a previous work of I. Verbitski Ve] and E. Sawyer Sa-Wh-Zh], was done by S. Zhao Zh]. In the present work we do not investigate on such a question since a necessary and sucient condition with general weight func-tions would be useless for practical computafunc-tions mainly when it is expressed in term of the operator I

itself and integrations over some set of cubes (see

for instance Theorem 1.2, p.98 in Zh]).

According to a work of I. Verbitski Ve], a necessary condition for the bound-ednessI
: L p v !L q u when q<p is (1.2) Z x2R n r( x)u(x)dx<1 withr = pq p;q and (x) = sup Q3x n jQj
n  1 jQj Z Q v 1;p 0 (y)dy  1 p 0  1 jQj Z Q u(y)dy  1 p o : Herep 0= p p;1 and

Qare arbitrary cubes with sides parallel to the coordinates

axes. Conversely in Ra1] (Theorem 2.1, p312), the boundednessI
: L p v !L q u

is seen to be held provided that for some t 1 t 2 >1 (1.3) Z x2R n r t 1 t 2( x)u(x)dx<1 where t 1 t 2( x) = sup Q3x n jQj
n  1 jQj Z Q v (1;p 0 )t1( y)dy  1 p 0 t 1  1 jQj Z Q u t2( y)dy  1 pt 2 o :

Obviously, by the Holder inequality, condition (1:3) is stronger than (1:2).

The interest on the implication (1:3) =)(1:1) is that the sucient condition

(1:3) is not expressed in term of I

. However the reader would be aware of

the diculty in checking (1:3). This problem is studied in the remainder of

results in Ra1].

One of the motivations of our present work is the observation that condition (1:3) is not always applicable due to the high integrability required for the

weights u(:) and v 1;p 0 (:). Indeed taking v 1;p 0 (x) = jxj ;nln ;p 0 (jxj ;1) for jxj < 1 2 then R jxj<R v (1;p 0 )t(

x)dx = 1, for all t > 1 and R < 1 2, though R jxj<R v 1;p 0

(x)dx < 1. However for such a weight v(:) (see Corollary 2.4)

the boundedness (1:1) can be held. Our second motivation is that a simple

characterization for the two-weight inequality (1:1) with 1<q <p <1 can

be derived for a large class of weight functions including those of radial and monotone ones.

Necessary conditions forI
: L p v !L q uwith

q<pwill be stated in Theorem

2.1. These conditions are of two types: the Hardy conditions and the Muck-enhoupt condition. In general they are not together sucient to derive the above boundedness. However in Theorem 2.2, we will see that with a slight strong version of the Muckenhoupt condition then inequality (1:1) can be

de-rived. Consequently, a characterization for (1:1) for many usual weights will

be found in Proposition 2.3. Concrete and explicit examples, which cannot be decided from results in Ra1] and Zh], will be given in Corollary 2.4.

Our results, stated inx2, are based on the "principle of three parts proof"

already used by the author in Ra2] to tackle the boundedness problem for the case p  q. Two useful basic lemmas are given in x3. And the proofs of all

results are performed in the last section.

x

2. RESULTS

Throughout this paper it is always assumed that 0<<n 1<q <p<1 p 0= p p;1  q 0= q q;1  r= pq p;q or 1 r = 1 q ; 1 p  and u(:) v 1;p 0

(:) are weight functions:

We rst give some natural necessary conditions for the boundedness (1:1)

to be satised.

Theorem 2.1.

Assume the boundednessI
: L p v !L q

u does hold. Then

(2.1)Z x2R n   Z jxj<jyj jyj (
;n)q u(y)dy  1 q  Z jzj<jxj v 1;p 0 (z)dz  1 q 0  r v 1;p 0 (x)dx<1 (2:1 ) Z x2R n   Z jxj<jzj jzj (
;n)p 0 v 1;p 0 (z)dz  1 p 0  Z jyj<jxj u(y)dy  1 p  r u(x)dx<1

and for each integer N 1

(2.2) 1 X k=;1 ; A N( k)  r <1 where (2.3) A N( k) = 2 k(
;n)  Z k ;N k +N u(y)dy  1 q  Z k ;N k +N v 1;p 0 (z)dz  1 p 0 :

Conditions (2:1) and (2:2) will be often referred as Hardy and

Mucken-houpt conditions respectively, and (2:1

) is named as the dual condition of

(2:1). Both the Hardy conditions and the Muckenhoupt condition would not

be sucient in general to imply the boundedness (1:1) as it is the case for pq.

Our next main result states that this boundedness can be obtained just by using a (sligthly) stronger condition than (2:2). Precisely

Theorem 2.2.

The boundedness I
: L p v ! L q

u does hold provided that the

Hardy conditions (2:1) and (2:1

) are satised and

(2.4) 1 X k=;1 ; e A(k)  r <1 where (2.5) e A(k) = 2 kn
n + 1 q ; 1 p ]  sup 2 k ;1 <jxj<2 k +1 u(x)  1 q  sup 2 k ;1 <jzj<2 k +2 v 1;p 0 (z)  1 p 0 :

Observe that condition (2:4) is stronger than (2:2) with N = 1, since for

some xed constant c>0, which only depends onn,,pand q: A(k) =A

1( k)c

e

A(k) for all integersk:

But (2:4) is not too far from the necessary condition (2:2) since for a large class

of weights it turns out that e

A(k) c 1

A

N(

k), for some constant c 1

> 0 and

integer N 1 which only depends on these weights. Precisely, an additional

property required for each weight to realize this last inequality is the condition

H. That isw(:)2Hwhenever (2.6) sup 4 ;1 R<jyj<4R w(y) C R n Z 2 ;N R<jyj<2 N R w(z)dz for all R>0.

Here the integer N  1 and the constant C > 0 depend only on w(:). For

a radial and monotone weight w(:), property (2:6) is fullled with N = 3

and C > 0 only depending on n but not on w(:). There exists also

non-necessarily monotone weight for which (2:6) is satised, as the case ofw(x) = w 1( x)1I jxj<1( x)+w 2( x)1I jxj>1( x) withw 1( :) andw 2(

:) are radial and monotone

(a proof is given in Ra1]).

Therefore a (simple) characterization forI
: L p v ! L q

u for weights having

Proposition 2.3.

Let u(:), v 1;p 0 (:) 2 H. The boundedness I
: L p v ! L q u

does hold if and only if both the Hardy conditions (2:1), (2:1

) and the

Muck-enhoupt condition (2:2) are satised.

Note that in this result, the integerN 1 involved in condition (2:2) would

depend on properties Hbut not directly on the weights.

We will end with explicit examples showing the gain over results in Ra1] and Zh].

Corollary 2.4.

Dene the weight functions

u(x) =jxj ;n1I jxj< 1 2( x) +jxj ;n1I jxj> 1 2( x) v(x) =jxj (p;1)nlnp( jxj ;1)1I jxj< 1 2( x) +jxj ;n1I jxj> 1 2( x):

Suppose 0<, (n;)q< and <np. ThenI
: L p v !L q u if and only if

i)p< ii) <(n;)q and iii)+  q <  p : Also set u  (x) =jxj ;n ln;q jxj ;1 1Ijxj< 1 2( x) +jxj (1;q)(;n) 1Ijxj> 1 2( x) v  (x) =jxj (1;p)(;n) 1Ijxj< 1 2( x) +jxj (1;p)(;n) 1Ijxj> 1 2( x): Suppose 0<, (n;)p 0 < and<nq 0. Then I
: L p v  !L q u  if and only if iv)q 0 < v)  <(n;)p 0 and vi) +  p 0 <  q 0 :

As mentioned in the introduction, for these examples the boundednessI
: L p v !L q

u is not obtainable from criterion (1

:3) since R jxj<R v (1;p 0 )t( x)dx=1 and R jxj<R u  t

(x)dx=1, for allt>1 and R< 1

2. Also criteria given in Zh]

seem to be dicult to apply for these concrete and explicit examples.

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3. BASIC LEMMAS

First we state a basic Lemma needed for the proofs of Theorem 2.1 and Corollary 2.4.

Lemma 3.1.

Let 0a<band 0 <. Then there is a constant c>0 such

that for allh(:) 0:

(3.1)  Z h(x)dx  1+ =c Z h Z h(y)dy i  h(x)dx:

Identity (3:1) can be obtained from the corresponding one-dimensional

re-sult after using polar coordinates.

The next result is about then-dimensional weighted Hardy inequality
Z x2R n h Z jyj<jxj f(y)dy i q w(x)dx  1 q
cA
Z x2R n f p( x)v(x)dx  1 p (3.2) for allf(:)0.

Lemma 3.2.

Suppose that for some constantA>0

(3.3) Z x2R n 
Z jxj<jyj w(y)dy  1 q
Z jzj<jxj v 1;p 0 (z)dz  1 q 0  r v 1;p 0 (x)dx
A r :

Then inequality (3:2) is satis
ed for a constant c>0 which only depends on n,p and q. Conversely the Hardy condition (3:3) is a necessary condition for

inequality (3:2) to hold.

A proof of this result was given by P. Dr avel, H. Heinig and A. Kufner Dr-He-Ku] (see Theorem 2.2, p7-8).

x

4. PROOFS OF RESULTS

Proof of Theorem 2.1.

The implication (1:1) =)(2:1). Observe that jxj ;n Z jyj<jxj f(y)dy
2 n; Z jx;yj<2jxj jx;yj ;n f(y)dy
2 n; (I  f)(x)

for allf(:)0. So the boundednessI  : L p v !L q

uimplies the Hardy inequality

(3:2) withw(x) =jxj q(;n)

u(x). Consequently, condition (3:3) arises because

of the second part of Lemma 3.2. The Hardy condition (2:1) is nothing else

than (3:3) due to this choice ofw(:).

The implication (1:1) =) (2:1

). By a duality argument, inequality (1 :1) is equivalent to I  : L q 0 u 1;q 0 ! L p 0 v 1;p

0. By analogue arguments as used for the

implication (1:1) =) (2:1), this last boundedness implies condition (2:1 ),

sincep 0

<q 0.

The implication (1:1) =) (2:2). Let us x nonnegative integers N, M  1

and dene the function

g NM( x) = N X 2k(;n) r p
Z 2 k ;N 0 <jzj<2 k +N 0 u(z)dz  r pq
Z 2 k ;N 0<jzj<jxj v 1;p 0 (z)dz  r pq 0

v 1;p 0 (x)1I 2 k ;N 0 <jxj<2 k +N 0( x) Here N 0

1 is the integer (N) involved in condition (2:2). Obviously B(NM) = N X k=;M  2k(;n)
Z 2 k ;N 0<jzj<2 k +N 0 u(z)dz  1 q  (4.1)
Z 2 k ;N 0<jzj<2 k +N 0 v 1;p 0 (z)dz  1 p 0  r <1

and it can be assumed thatB(NM)>0. The points keys for obtaining (2:2)

are (4.2) Z x2R n g p NM( x)v(x)dx
c 0 B(NM) and (4.3) Z x2R n( I  g NM) q( x)u(x)dxcB(NM)

for some constants c 0,

c > 0 which do not depend on the integers N and M. Indeed with (4:3) and (4:2), the boundedness I

 : L p v ! L q u yields
B(NM) 1 q
c 1
B(NM)  1 p

. This last inequality, point (4:1) and 1 r = 1 q ; 1 p >0 lead to B(NM)
c r 1 :

The Muckenhoupt condition (2:2) arises from this last estimate by letting NM !1. At this point, the proof of (1:1) =)(2:2) is now reduced to that

of (4:2) and (4:3).

Inequality (4:2) follows after using the denition of g NM( :), the identity p(1 ;p 0) + 1 = (1 ;p 0), the identity (3 :1) (with h(:) = v 1;p 0 (:) and  = r q

0) and the fact that 1I 2 k ;N 0<j:j<2 k +N 0( :) = N0;1 X l=;N 0 1I2 k +l <j:j<2 k +l+1(:) almost everywhere. Indeed n Z R n g p NM( x)v(x)dx
c(N 0) N 0 ;1 X l=;N 0 N X k=;M 2k(;n)r
Z 2 k ;N 0 <jzj<2 k +N 0 u(z)dz r q  Z k +l k +l+1 h Z k ;N 0<jzj<jxj v 1;p 0 (z)dz i r q 0 v 1;p 0 (x)dx

c(N 0) N X k=;M 2k(;n)r
Z 2 k ;N 0<jzj<2 k +N 0 u(z)dz  r q  Z 2 k ;N 0 <jxj<2 k +N 0 h Z 2 k ;N 0<jzj<jxj v 1;p 0 (z)dz i r q 0 v 1;p 0 (x)dx =c 1( N 0) N X k=;M 2k(;n)r
Z 2 k ;N 0<jzj<2 k +N 0 u(z)dz  r q 
Z 2 k ;N 0 <jzj<2 k +N 0 v 1;p 0 (z)dz  r p 0 : Estimate (4:3) is based on (4.4) Z 2 k ;N 0 <jyj<2 k +N 0 g NM( y)dyc2 k(;n) r p 
Z 2 k ;N 0 <jzj<2 k +N 0 u(z)dz  r pq
Z 2 k ;N 0 <jzj<2 k +N 0 v 1;p 0 (z)dz  r pq 0 +1 :

Indeed from this last inequality it follows that

Z R n( I  g NM) q (x)u(x)dx c(N 0) 1 X k=;1 Z 2 k ;N 0<jxj<2 k +N 0 (I  g NM) q( x)u(x)dx c(N 0) N X k=;M Z 2 k ;N 0 <jxj<2 k +N 0 h Z 2 k ;N 0 <jyj<2 k +N 0 jx;yj ;n g NM( y)dy i q u(x)dx c 1 N X k=;M 2k(;n)q
Z 2 k ;N 0<jyj<2 k +N 0 g NM( y)dy  q
Z 2 k ;N 0<jzj<2 k +N 0 u(z)dz  c 2 N X k=;M 2k(;n)q(1+ r p )
Z 2 k ;N 0<jzj<2 k +N 0 u(z)dz  r p +1 
Z 2 k ;N 0 <jzj<2 k +N 0 v 1;p 0 (z)dz  q( r pq 0 +1) =c 2 N X k=;M  2k(;n)
Z 2 k ;N 0 <jzj<2 k +N 0 u(z)dz  1 q
Z 2 k ;N 0 <jzj<2 k +N 0 v 1;p 0 (z)dz  1 p 0  r =c 2 B(NM):

To derive (4:4), the point is the identity (3:1). Indeed, for each integer k 2 f;M::: Ng,

Z 2 k ;N 0<jyj<2 k +N 0 g NM( y)dy2 k(;n) r p
Z 2 k ;N 0 <jzj<2 k +N 0 u(z)dz  r pq  Z 2 k ;N 0<jyj<2 k +N 0 h Z 2 k ;N 0<jzj<jyj v 1;p 0 (z)dz i r pq 0 v 1;p 0 (y)dy c2 k(;n) r p
Z 2 k ;N 0<jzj<2 k +N 0 u(z)dz  r pq 
Z 2 k ;N 0<jzj<2 k +N 0 v 1;p 0 (z)dz  r pq 0 +1 :

Proof of Theorem 2.2.

Since (I  f)(x) =A 1( x) + ( e I  f)(x) +A 3( x) for all f(:)0 with A 1( x) = Z jyj 1 2 jxj jx;yj ;n f(y)dy (e I  f)(x) =A 2( x) = Z 1 2 jxj<jyj<2jxj jx;yj ;n f(y)dy A 3( x) = Z 2jxjjyj jx;yj ;n f(y)dy

then to get the boundedness I  : L p v ! L q

u it is sucient to estimate each of Z R n A q i( x)u(x)dx,i2f123g, byC
Z R n f p (x)v(x)dx  q p , withC a

nonnega-tive and xed constant. Observe that A 1( x)
cjxj ;n R jyj<jxj f(y)dy, since 1 2 jxj < jx;yj when-ever jyj
1 2

jxj. By the Hardy condition (2:1) which is (3:3) with w(x) = jxj

(;n)q

u(x)] and by Lemma 3.2, the conclusion arises since Z R n A q 1( x)u(x)dx
c q Z R n h Z jyj<jxj f(y)dy i q jxj (;n)q u(x)dx
C
Z n f p( x)v(x)dx  q p :

Note that A 3( x)
c R jxj<jyj jyj ;n f(y)dy, since 1 2 jyj < jx;yj whenever

2jxj<jyj. At this stage the conclusion also follows since Z R n A q 3( x)u(x)dx
c q Z R n h Z jxj<jyj jyj ;n f(y)dy i q u(x)dx
C
Z R n f p( x)v(x)dx  q p :

This last Hardy inequality can be also obtained from Lemma 3.2 by using a duality argument. Indeed the problem is reduced to inequality (3:2) with q, p,w(:) andv(:) respectively replaced by p

0, q 0, j:j (;n)p 0 v 1;p 0 (:) andu 1;q 0 (:). Therefore (2:1

) yields the corresponding condition (3 :3).

The real task is now to prove that

Z R n( e I  f) q( x)u(x)dx
C
Z R n f p( x)v(x)dx  q p

for all f(:) 0. For this purpose, it is convenient to introduce the following

notations: E k= fz2R n  2k <jxj
2 k+1 g F k = fy2R n  2k;1 <jyj
2 k+2 g C(x) =fz2R n  12jxj <jzj<2jxjg U k= sup x2E k u(x) W k= sup z2F k v 1;p 0 (z): Then v 1;p 0 (z)
W k whenever z2C(x) and x2E k.

By the Holder inequality and this observation, for each x2E k (e I  f)(x)

Z z2C(x) jx;zj ;n v 1;p 0 (z)dz  1 p 0
Z y2C(x) jx;yj ;n f p( y)v(y)dy  1 p
c 1
jxj  W k  1 p 0
Z y2C(x) jx;yj ;n f p( y)v(y)dy  1 p
c 2
2k W k  1 p 0
Z y2C(x) jx;yj ;n f p (y)v(y)dy  1 p :

Using this last inequality, the Holder inequality, the Fubini theorem, conditions (2:4) and (2:5) and nally another Holder inequality, the conclusion arises as

follows Z R n( e I  f) q u(x)dx= 1 X Z x2E k (e I  f) q u(x)dx

c 3 1 X k=;1
2k W k  q p 0 U k Z x2Ek h Z y2C(x) jx;yj ;n f p( y)v(y)dy i q p dx
c 3 1 X k=;1 2k q p 0 +kn(1; q p ) U k W q p 0 k  Z x2E k
Z y2C(x) jx;yj ;n f p( y)v(y)dy  dx  q p
c 3 1 X k=;1 2k q p 0 +kn(1; q p ) U k W q p 0 k  Z y2Fk f p( y)v(y)
Z x2Ek jx;yj ;n dx  dy  q p
c 4 1 X k=;1 2k q p 0 +kn(1; q p )+k q p U k W q p 0 k Z y2F k f p (y)v(y)dy  q p =c 4 1 X k=;1  2kn
n + 1 q ; 1 p ] U 1 q k W 1 p 0 k  q Z y2Fk f p( y)v(y)dy  q p
c 4 1 X k=;1 e A q( k) Z y2F k f p( y)v(y)dy  q p
c 4  1 X j=;1 e A qp p;q( j)  1; q p  1 X k=;1 Z y2F k f p (y)v(y)dy  q p
c 4 A r(1; q p )  1 X k=;1 Z 2 k ;1 <jyj<2 k +Z 2 k <jyj<2 k +1 + Z 2 k +1 <jyj<2 k +2 f p( y)v(y)dy  q p =c 5 A q Z R n f p( y)v(y)dy  q p :

Proof of Proposition 2.3.

For the necessary part, by Theorem 2.1, the Hardy conditions (2:1) and

(2:1

) and also the Muckenhoupt condition (2

:2) are implied by the

bound-edness I  : L p v ! L q

u. Here, the integer

N  1 involved in condition (2:2)

and (2:3) can be chosen as a common constant resulting from the assumptions u(:), v

1;p 0

(:)2H (see denition and (2:6)).

For the sucient part, by Theorem 2.2, the boundednessI  : L p v !L q uwill

hold whenever the Hardy conditions (2:1), (2:1

Muckenhoupt condition (2:4) are satised. Usingu(:), v 1;p 0 (:)2H then e A(k)
cA N(

k) for all integers k:

Here c > 0 just depends on n, , q, p and on a common constant involved

in assumptionsu(:), v 1;p

0

(:)2H. This last inequality means that the

Muck-enhoupt condition (2:2) implies the stronger one (2:4), and consequently the

conclusion arises immediately.

Proof of Corollary 2.4.

Proof for the rst example. Necessity of conditions i), ii), iii). Suppose that

I  : L p v !L q

u. Then by Theorem 2.1, the Hardy and Muckenhoupt conditions

(2:1), (2:1

) and (2

:2) are satised.

If
p, then the Hardy condition (2:1

) does not hold since: Z jzj>100 jzj (;n)p 0 v 1;p 0 (z)dz = Z jzj>100 jzj (;  p )p 0 jzj ;n dz=1:

Similarly the Hardy condition (2:1) is not satised if (n;)q
. And the

fact that +  q ;  p

< 0 is an immediate consequence of the Muckenhoupt

condition (2:2) since 1> 1 X k=;1 (A 2( k)) r >c 1 X k=100 2k+  q ;  p ]r

for some xed constant c>0.

Suciency of conditions i), ii) and iii). In view of Theorem 2.2, to get the boundedness I  : L p v ! L q

u, the task is to check conditions (2

:1), (2:1 ) and

(2:4).

To deal with the Hardy condition (2:1) the idea is to divide the integral

with respect to x into ones on the regions jxj< 1

2 and jxj>

1

2. This division

is required because of the nature of the weights u and v, and actually each

region is associated to two integrals. For example, corresponding to jxj < 1 2 we have to evaluate I 11= Z jxj< 1 2 
Z jxj<jyj< 1 2 jyj (;n)q u 1( y)dy  1 q
Z jzj<jxj  1( z)dz  1 q 0  r  1( x)dx and I 12= Z jxj< 1 2 Z jzj<jxj  1( z)dz  r q 0  1( x)dx where u 1( z) = u(z) = jzj ;n and  1( z) = v 1;p 0 (z) = jzj ;nln ;p 0 (jzj ;1) for jzj< 1 2. The integralI

11 can be bounded just by using (

;n)q+ >0 since I 11
c 1 Z jxj< 1 ln; p 0 r pq 0( jxj ;1)  1( z)dz
c 2 Z jxj< 1  1( z)dz=c 3 :

AndI 12 is carried by using (3 :1) in Lemma 3.1 as I 12
Z jxj< 1 2  1( z)dz  r q 0 +1 =c 4 :

The estimates for the two integrals corresponding to the regionjxj> 1

2 require

the use ofii), iii) and<np.

The dual Hardy condition (2:1

) can be checked by using the same

argu-ments as for (2:1). Here the estimates for the two integrals corresponding to

the regionjxj< 1

2 require the use of (3

:1) in Lemma 3.1 and 0<(n;)q < .

And for the integrals corresponding to the regionjxj> 1

2, assumptions i), iii)

and  >0 are needed.

To check the condition (2:4) observe that by the denition of the weights

then for some xed constantc>0: e A(k)
2 ;k n p 0 = c2 k+  q ;n] for allk
;4 and e A(k)
c2 k+  q ;  p ]

for all k  1. Therefore condition (2:4) follows from

assumptionsii) and iii).

Finally we will now end with the

Proof for the second example. By duality the boundedness I  : L p v  ! L q u 

holds if and only if I  : L p v ! L q u with p = q 0, q = p 0, v = (u )1;q 0 and u = (v )1;p 0 . Observe that u(y) =  jyj ;n for jyj< 1 2 jyj ;n for jyj> 1 2 and v(z) =  jzj n(p;1)ln ;p (jzj ;1) for jzj< 1 2 jzj ;n for jzj> 1 2.

So using the rst example, when  > 0,

(n;)p 0 = ( n;)q < ,<nq 0 = np thenI  : L p v  !L q u  (or equivalently I  : L p v ! L q u) if and only if q 0 = p < ,  < (n;)p 0 = ( n;)q and +  p 0 = +  q <  p =  q 0.

Acknowledgement

The author is indebted to the referee for valuable comments regarding the rst version of this paper.

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13 bd de l'Hautil 95 092 Cergy Pontoise France E-mail: [email protected]