Lerner Pointwise Inequalities

8/12/2019 Lerner Pointwise Inequalities

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J. Math. Anal. Appl. 289 (2004) 248259www.elsevier.com/locate/jmaa

On some pointwise inequalities

A.K. Lerner

Department of Mathematics, Bar-Ilan University, 52900 Ramat Gan, Israel

Received 7 February 2003

Submitted by R.H. Torres

Abstract

In a recent paper (Studia Math. 138 (2000) 285291) we proved pointwise estimates relating someclassical maximal and singular integral operators. Here we show that, in a sense, there are moreexible inequalities which not only imply the previously known results but also give something new.In particular, they hold for the multilinear CaldernZygmund operators. This result gives a newapproach to a recent work by Grafakos and Torres, and unies some classical inequalities by Cotlar

and Coifman and Fefferman. 2003 Elsevier Inc. All rights reserved.

Keywords: Multilinear CaldernZygmund operators; Pointwise inequalities

1. Introduction

Let be a non-negative, locally integrable function on R n . Given a measurable set E ,let (E) = E (x) dx . The non-increasing rearrangement of a measurable function f onR n with respect to is dened by

f (t) = sup(E) = t

inf x E

f(x) (0 < t < ).

If 1 we use the notation f (t) .Let us consider the maximal function

m f(x) = supQ x

(f Q ) |Q | (0 < < 1),

E-mail address: [email protected]: http://www.math.biu.ac.il/~lernera.

0022-247X/$ see front matter 2003 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2003.09.051


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A.K. Lerner / J. Math. Anal. Appl. 289 (2004) 248259 249

where the supremum is taken over all cubes Q containing x , Q denotes the indicatorfunction of Q and |Q | denotes the Lebesgue measure of Q .

In [10,11], we have established several pointwise estimates involved m and some clas-sical operators in harmonic analysis. The following inequalities hold for any appropriate f and all x R n (see [10]):

m (Mf )(x) c,n f #(x) + Mf ( x) ( 0 < < 1), (1)

m (T f)(x) c,n Mf (x) + T f (x) ( 0 < < 1), (2)

where Mf and f # are the HardyLittlewood and FeffermanStein maximal operators,respectively, T f is the maximal CaldernZygmund singular integral operator. Such in-equalities easily imply rearrangement, L p (for satisfying A Muckenhouptscondition)and BLO-norm estimates, i.e., many classical results concerning the above-mentioned op-erators.

The aim of this paper is to show that there are more exible pointwise estimates which,in particular, imply (1) and (2). First of all we mention the following simple

Proposition 1. For any f L 1loc(Rn ) and all x R n ,

m f(x)1

f #(x) + Mf (x ) ( 0 < < 1) (3)

and

Mf (x)2

1 f #(x) + m f (x) ( 0 < < 1). (4)

It turns out that (3) and (4) yield more information than (1). The following main lemmaclaries the sense of these inequalities.

Lemma 2. Let f, g and h be non-negative functions on R n . Suppose that for any 0 < , < 1 , there exist constants A , B > 0 so that

m f(x) A g(x) + h(x) (5)

and

h(x) B g(x) + m f(x) (6)

for all x R n . Assume also that A . Then

(i) if f (+ ) = 0 and h (+ ) = 0 , thenmax f L p , h L p cp, g L p (0 < p < );

(ii) if g L , then

max m f BLO , h BLO c g (0 < < 1).

For example, Proposition 1 combined with this lemma immediately gives the followingnew result which can be viewed as an analogue of the Bennett theorem [1] saying thatMf : BMO BLO.


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250 A.K. Lerner / J. Math. Anal. Appl. 289 (2004) 248259

Corollary 3. If f BMO and m f < , then m f BLO and

m f BLO c,n f (0 < < 1).

Our main theorem states that inequalities exactly of the same type as (5) and (6) holdfor the CaldernZygmund operators. This gives a unied approach to the classical resultsdue to Cotlar [8, p. 56] and Coifman and Fefferman [3] which say about certain rela-tions between maximal and singular integral operators. Recently these results have beengeneralized by Grafakos and Torres [6] to the case of multilinear singular integrals. Esti-mates (5) and (6) work in this case as well, and we will state our theorem in this context.Let f = (f 1 , . . . , f m ) . Denote by T ( f ) and T ( f ) the m -linear CaldernZygmund and

maximal truncated operators, respectively (see [6] or Section 3 below for precise deni-tions).

Theorem 4. Let T be an m-linear CaldernZygmund operator. Then for all f in any product of L qj (R n ) spaces, with 1 qj < , and for all x R n ,

m T ( f ) (x) c1m

j = 1

Mf j (x) + T ( f )(x ) ( 0 < < 1) (7)

and

T ( f )(x) c2m

j = 1

Mf j (x) + m T ( f ) (x) ( 0 < < 1) (8)

with constants ci independent of f and x .

It is clear, by Chebyshevs inequality, that

m f(x) = m |f |1/ 1

M | f | (x)

1/

( > 0),

and therefore (8) easily implies a multilinear version of Cotlars inequality proved in [6]:

T ( f )(x) cm,n, m

j = 1

Mf j (x) + M T ( f ) (x)

1/ ( > 0).

On the other hand, (8) combined with (7) gives, by part (i) of Lemma 2, a multilinearversion of the CoifmanFefferman theorem [6]. Moreover, our approach gives L p boundsfor T ( f ) and T ( f ) at once. Also we get the following BLO-estimate.

Corollary 5. For all f in any product of L qj L (R n ) spaces, with 1 qj < , T ( f )belongs to the space BLO , and

T ( f ) BLO cm

j = 1

f j .


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The paper is organized as follows. In Section 2 we establish several properties of thefunction m f and prove Proposition 1 and Lemma 2. In Section 3 we list main factsconcerning the multilinear CaldernZygmund operators and prove Theorem 4. Section 4contains some concluding remarks.

2. Proofs of Proposition 1 and Lemma 2

Recall that the HardyLittlewood and FeffermanStein maximal functions are denedby

Mf (x) = supQ x

1|Q | Q f(y) dyand

f #(x) = supQ x

1|Q |

Q

f(y) f Q dy ,

respectively, where f Q = (1/ |Q | ) Q f and the supremum is taken over all cubes Q con-taining x .Proof of Proposition 1. For any Q containing x we get

(f Q ) |Q | (f f Q ) Q |Q | + | f Q |

1 |Q | Q |f f Q | + | f Q | 1 f

#(x) + Mf (x),

which gives (3). To prove (4), we can assume that f 0, and then use the fact that| f |#(x) 2f #(x) . For any Q x ,

f Q inf y Q

f(y) f Q + f(y)

(f f Q ) Q (1 ) |Q | + (f Q ) |Q |1

1 f #(x) + m f(x),

and this implies (4).

We recall also several well-known denitions used below.The space BLO [4] consists of all functions f L 1loc(R

n ) such that

f BLO = supQ

f Q inf Q

f < .

A weight satises Muckenhoupts condition A if there exist c, > 0 such that forany Q and E Q ,

(E) c |E | / |Q | (Q).


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We will need some properties of the operator m . The following two lemmas say thatthe function m f is a pointwise majorant of | f | ; nevertheless, m f and f have equivalentweighted rearrangements with respect to A .

Lemma 6. For any measurable function f and for a.e. x R n ,

f(x) m f (x) ( 0 < < 1).

Proof. Let x be the point of the approximately continuity of f (see [12, p. 132]). The setof such points have full measure. For any > 0 there exists a cube Q centered at x and aset E Q such that |E | > |Q | and | f(x) | | f(y) | + for all y E . It follows from thisthat

f(x) inf y E

f(y) + (f Q ) |Q | + m f(x) + .

Letting 0 completes the proof.

Lemma 7. Let A . For any measurable function f and all t > 0 ,

(m f ) (t) f (ct ) ( 0 < < 1), (9)

where c = c(,) .

This lemma was proved in [11] (cf. inequality (3.1) therein).

Remark 8. In the case of the Lebesgue measure we can take in (9) c = / 3n , that is,

(m f ) (t) f (t/ 3n ). (10)

Next, the operator m satises the following subadditivity and submultiplicativity prop-erties.

Lemma 9. For any measurable functions f, g and for all x R n ,

m (f + g)(x) m/ 2f(x) + m/ 2g(x),

m (fg)(x) m/ 2f(x)m / 2g(x).

Proof. This lemma follows immediately fromthe simple propertiesof rearrangements(see

[2, p. 67])(f + g) (t ) f (t / 2) + g (t / 2)

and

(fg) (t ) f (t / 2)g (t / 2).

In the next two lemmas we consider the composition of m with itself and with maximaloperators. We will use the fact that if cubes Q 1 and Q 2 intersect, then either Q 1 3Q 2 orQ 2 3Q 1 .


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Lemma 10. For any measurable function f and all x R n ,

m (m f )(x) m/ 9n f (x) ( 0 < , < 1).

Proof. Let Q be any cube containing x . For all y Q we get

m f(y) = max supQ y, Q 3Q

(f Q ) |Q | , supQ y, Q 3Q

(f Q ) |Q |

max m (f 3Q )(y),m / 3n f(x) .

Hence, from this and (10) we obtain

(m f ) Q |Q | max m (f 3Q ) |Q | , m / 3n f(x)

max (f 3Q ) |Q | / 3n , m / 3n f(x)

max m/ 9n f(x),m / 3n f(x) m/ 9n f(x).

Taking the upper bound over all Q x yields the lemma.

Lemma 11. For any f L 1loc(Rn ) and all x R n ,

m (Mf )(x) cn, Mf (x ) ( 0 < < 1), (11)

m (f #)(x) cn, f #(x) ( 0 < < 1). (12)

Proof. We prove only the rst inequality, since the proof of the second one is essentiallythe same. Let Q be any cube containing x . For all y Q we get

Mf (y) = max supQ y, Q 3Q

1|Q |

Q

| f | , supQ y, Q 3Q

1|Q |

Q

| f |

max M(f 3Q )(y),3n

|3Q | 3Q

| f | M(f 3Q )(y) + 3n Mf (x).

Hence, by the weak type (1, 1) property of M ,

(Mf ) Q |Q | M(f 3Q ) |Q | + 3n Mf (x)

3n

|Q |

3Q

| f | + 3n Mf (x) (9n / + 3n )Mf (x).

Taking the upper bound over all Q x yields (11).

Remark 12. It is interesting to note that inequality (12) may be deduced from (11), usingthe fact that f #(x) M M # f(x) , where M

# f(x) is the so-called local sharp maximal

function (cf. [7,9]).

Now we state the main lemma from [10], which is a key ingredient in the proving of Lemma 2.


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Lemma 13. Let f and g be non-negative functions on R n . Suppose that for any , 0 < 1/ 2 , there exists a constant c > 0 so that

m f(x) c g(x) + f(x)

for all x R n , and let A . Then

(i) there exists a constant c > 0 so that

f (t) c g (2t ) + f (2t ) (t > 0);

(iii) if f (+ ) = 0 , then

f L p cp, g L p (0 < p < );(iii) if g L , then

f BLO c g .

Proof of Lemma 2. Using (5), (6) and Lemmas 6, 9 and 10, we obtain

m (m f )(x) m/ 9n f(x) A/ 9n g(x) + h(x)

(A / 9n + B )g(x) + m f(x) (13)

and

m h(x) B m/ 2g(x) + m/ 29n f(x) (B + A/ 29n )m / 2g(x) + h(x). (14)

From this and (iii) of Lemma 13 we immediately obtain BLO-estimates. Further, by (9),f (+ ) = 0 implies (m f ) (+ ) = 0. Hence, (13), Lemma 6 and (ii) of Lemma 13 give

f L p m f L p cp, g L p .

To get L p -bound for h, we proceed exactly as in the proving of Lemma 13, that is, wechoose depending on such that

h L p cp, m h h L p ,

whenever h (+ ) = 0. Then, using (14) and (9), we obtain

h L p c m / 2g L p c g L p .

The lemma is proved.

3. Proof of Theorem 4

Let K(x,y 1, . . . , y m ) be a locally integrable function dened off the diagonal x = y1 = = ym in (R n )m+ 1 , which satises the size estimate

K(y 0, y 1, . . . , y m )A

mk,l = 0 |yk yl |

mn (15)


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and, for some > 0, the regularity condition

K(y 0, . . . , y j , . . . , y m ) K(y 0, . . . , y j , . . . , y m )A |yj yj |

mk,l = 0 |yk y l |

mn + , (16)

whenever 0 j m and |yj yj | 12 max0 k m |yj yk | .

Consider multilinear operators T initially dened on the m -fold product of Schwartzspaces and taking values into the space of tempered distributions,

T : S( R n ) S( R n ) S (R n ).

We say that T is an m-linear CaldernZygmund operator if, for some 1 qj < , itextends to a bounded multilinear operator from L q1 L qm to L q , where 1 /q = 1/q 1 + + 1/q m , and there is a kernel K satisfying (15) and (16) such that

T (f 1 , . . . , f m )(x) = (R n )m

K(x,y 1, . . . , y m )f 1(y 1) . . . f m (y m ) dy 1 . . . dy m , (17)

whenever f 1, . . . , f m are C functions with compact support and x / mj = 1 supp f j .The multilinear CaldernZygmund theory has been recently developed by Grafakos

and Torres (see [5,6]). In particular, it was shown in [5] that any m-linear CaldernZygmund operator T is bounded from L r1 L rm to L1/( 1/r 1++ 1/r m ) for all1 < r j < , and the representation (17) still holds for L r j functions as long as x /

mj = 1 suppf j . Moreover, T maps L

1 L 1 into L 1/m, , that is,

T (f 1 , . . . , f m ) (t )

c

t m

m

j = 1 f j 1 . (18)

We will occasionally write y = (y 1 , . . . , y m ) , f = (f 1 , . . . , f m ) . Given a point x R n , set S (x) = {y : max1 j m |x y j | }, U (x) = {y S (x) : |x y1 |2 + +|x ym |2 > 2}.

Following [6], dene the truncated and modied truncated operators by

T (f 1 , . . . , f m )(x)

= |x y1 |2++| x ym |2> 2

K(x,y 1, . . . , y m )f 1(y 1) . . . f m (y m ) dy 1 . . . dy m

and

T (f 1 , . . . , f m )(x) = y / S (x) K(x,y 1, . . . , y m )f 1(y 1) . . . f m (y m ) d y,respectively, and the associated maximal operators by

T ( f )(x) = sup> 0

T ( f )(x)

and

T ( f )(x) = sup> 0

T ( f )(x) .


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It is easy to see, by (15), that for any > 0,

U (x)

K(x,y 1, . . . , y m )f 1(y 1) . . . f m (y m ) d y cm

j = 1

Mf j (x),

and since

T ( f)(x) = U (x)

K(x,y 1, . . . , y m )f 1(y 1) . . . f m (y m ) d y + T ( f )(x),

we have

T ( f )(x) cm

j = 1

Mf j (x) + T ( f)(x) (19)

and

T ( f )(x) cm

j = 1

Mf j (x) + T ( f)(x). (20)

Let B = B(x,r) be an arbitrary ball centered at x with radius r . As in [6], we denotef 0j = f j B , f

j = f j f

0j . Then

f 1(y 1) . . . f m (y m ) =m

j = 1

f 0j (y j ) + f j (y j ) =

1 ,..., m { 0,}f 11 (y1) . . . f

mm (ym ).

Lemma 14. Let z, B(x,r/ 2) .

(i) Suppose that j 1 = = j l = 0 for some {j 1, . . . , j l } { 1, . . . , m } , where 1 l < m .Then

T f 11 , . . . , f mm (z) c

m

j = 1

Mf j (x) ; (21)

(ii) If 1 = = m = , then

T f 1 , . . . , f

m (z) T f 1 , . . . , f

m ( ) c

m

j = 1

Mf j (x). (22)

This lemma is contained implicitly in [6].

Proof of Theorem 4. Let Q be an arbitrary cube containing the point x , and let B =B(x, 2) , where is the diameter of Q . Set f 0j = f j B , f

j = f j f

0j . Note that

T f 1 , . . . , f m (x) = T 2(f 1 , . . . , f m )(x)

1 ,..., m { 0,}T f 11 , . . . , f

mm (x),

(23)


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where the sum extends over all i taking values {0, } except for the cases 1 = =m = 0 and 1 = = m = . Hence, applying (21) and (20), we get

T f 1 , . . . , f m (x) c

m

j = 1

Mf j (x) + T ( f )(x) cm

j = 1

Mf j (x) + T ( f)(x).

From this and (22), for any z Q we have

T f 1 , . . . , f m (z) c

m

j = 1

Mf j (x) + T f 1 , . . . , f

m (x)

c

m

j = 1Mf j (x) + T ( f )(x),

and therefore, using again (21), we obtain

T ( f)(z) =1 ,..., m { 0,}

T f 11 , . . . , f m

m (z)

cm

j = 1

Mf j (x) + T f 01 , . . . , f 0

m (z) + T f 1 , . . . , f

m (z)

cm

j = 1

Mf j (x) + T f 01 , . . . , f 0

m (z) + T ( f )(x).

From this and the weak-type estimate (18) we get

T ( f ) Q |Q | T f 01 , . . . , f 0

m |Q | + cm

j = 1

Mf j (x) + T ( f )(x)

c( |Q | )m

m

j = 1 B(x, 2) f (y j ) dy j + cm

j = 1

Mf j (x) + T ( f)(x)

cm

j = 1

Mf j (x) + T ( f )(x).

Taking the upper bound over all Q x completes the proof of (7).Let now B = B(x,) and Q be the cube centered at x with diameter . Set f 0j = f j B ,

f j = f j f 0j . By (21)(23), for any z Q we have

T ( f)(x) cm

j = 1

Mf j (x) + T f 1 , . . . , f

m (x)

cm

j = 1

Mf j (x) + T f 1 , . . . , f

m (z)


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cm

j = 1

Mf j (x) + T f 01 , . . . , f 0m (z) + T ( f )(z) .

Taking the supremum over all > 0 and then inmum over all z Q , and using (18), weobtain

T ( f )(x) cm

j = 1

Mf j (x) + inf z Q

T f 01 , . . . , f 0

m (z) + T ( f )(z)

cm

j = 1

Mf j (x) + T f 01 , . . . , f 0m (1 ) |Q | + T ( f ) Q |Q |

cm

j = 1

Mf j (x) + c(( 1 ) |Q | )m

m

j = 1 B(x,) f (y j ) dy j + m T ( f ) (x)c

m

j = 1

Mf j (x) + m T ( f ) (x).

From this and (19) we have (8). The theorem is proved.

4. Concluding remarks

In [1], it was proved that a non-negative function f belongs to BLO iff there are func-

tions h L

and g BMO with Mg

Lerner Pointwise Inequalities

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