A NEW PROOF OF THE BOUNDEDNESS OF MAXIMAL …diening/archive/newmax6.pdf · a new proof of the boundedness of maximal operators on variable lebesgue spaces d. cruz ... are simpler

A NEW PROOF OF THE BOUNDEDNESS OF MAXIMALOPERATORS ON VARIABLE LEBESGUE SPACES

D. CRUZ-URIBE, SFO, L. DIENING, AND A. FIORENZA

Abstract. We give a new proof using the classic Calderon-Zygmund decompo-sition that the Hardy-Littlewood maximal operator is bounded on the variableLebesgue space Lp() whenever the exponent function p() satisfies log-Holder con-tinuity conditions. We include the case where p() assumes the value infinity. Thesame proof also shows that the fractional maximal operator M, 0 < < n, mapsLp() into Lq(), where 1/p() 1/q() = /n.

1. Introduction

Given a measurable function p() : Rn [1,], let ,p() = {x Rn : p(x) =}.We define the variable Lebesgue space Lp() to be the set of functions such that forsome > 0,

p()(f/) =

Rn\,p()

(|f(x)|

)p(x)dx+ 1fL(,p()) 0 : p()(f/) 1

}.

These spaces are a special case of the Musielak-Orlicz spaces (cf. Musielak [18]) andgeneralize the classical Lebesgue spaces: if p(x) = p0, then L

p() = Lp0 .

Variable Lebesgue spaces have been known since the 1930s, but have become thefocus of intense investigation in the past fifteen years. (See [8, 10, 22] for furtherhistory and applications.) A central problem has been to extend the techniques ofharmonic analysis to these spaces, which in turn leads naturally to the study of the

Date: March 11, 2008.1991 Mathematics Subject Classification. 42B25,42B35.Key words and phrases. variable Lebesgue space, maximal operators, Calderon-Zygmund

decomposition.The first author is partially supported by the Stewart-Dorwart faculty development fund of the

Department of Mathematics of Trinity College. The first two authors would like to thank CarloSbordone and also the Dipartimento di Costruzioni e Metodi Matematici in Architettura, Universitadi Napoli, for its generous hospitality.

1

2 D. CRUZ-URIBE, SFO, L. DIENING, AND A. FIORENZA

Hardy-Littlewood maximal operator and the closely related fractional maximal oper-ator. The purpose of this paper is to give a new and simpler proof of the boundednessof these operators on variable Lebesgue spaces.

Before stating our main result, we first make a few key definitions. Given ,0 < n, we define

Mf(x) = supQ3x|Q|/n

Q

|f(y)| dy,

where the supremum is taken over all cubes Q Rn that contain x. (Equivalently,cubes may be replaced by balls containing x.) When = 0 this is the Hardy-Littlewood maximal operator and we write Mf instead of M0f . For > 0 this isthe fractional maximal operator introduced by Muckenhoupt and Wheeden [17].

Given a function r() : R [0,), we say that r() is locally log-Holder continuous,and write r() LH0, if there exists a constant C0 such that

|r(x) r(y)| C0 log |x y|

, x, y Rn, |x y| < 1/2.

Similarly, we say that r() is log-Holder continuous at infinity, and write r() LH,if there exists constants C and r() such that

|r(x) r()| Clog(e+ |x|)

, x Rn.

We say r() is (globally) log-Holder continuous if r() LH0 LH and we writer() LH.Remark 1.1. The LH condition is equivalent to the uniform continuity condition

|r(x) r(y)| Clog(e+ |x|)

, x, y Rn, |y| |x|.

The LH condition was originally defined in this form in [4].

Finally, given a set E Rn, letp(E) = ess inf

xEp(x), p+(E) = ess sup

xEp(x);

If E = Rn, then we simply write p and p+.Theorem 1.2. Given , 0 < n, let p() : Rn [1,] be such that 1/p() LHand 1 < p p+ n/. Define the exponent function q() by

1

p(x) 1q(x)

=

n, x Rn,

where we let 1/ = 0 and 1/0 =. ThenMfq() Cfp().

BOUNDEDNESS OF MAXIMAL OPERATORS 3

Remark 1.3. The constant in the conclusion of Theorem 1.2 depends on the dimensionn, the log-Holder constants of 1/p(), p, and p() (if this value is finite).

Remark 1.4. The assumption that 1/p() LH implies that 1/q() LH as well.Further, if p+ < , then the assumption 1/p() LH is equivalent to assumingp() LH, since 1p(x) 1p(y)

= p(x) p(y)p(x)p(y) p(x) p(y)(p+)2

.Theorem 1.2 combines a number of results that have been proved by the authors

and others. We first consider the case = 0, that is, norm inequalities for theHardy-Littlewood maximal operator. In [6] Theorem 1.2 was proved with the strongerassumption that p+ 0t{Mf(x)>t}q() Cfp().

Our proofs of Theorems 1.2 and 1.6 have several features that we want to highlight.First, each proof uses the machinery of Calderon-Zygmund cubes, which are of greatimportance in harmonic analysis on the classical Lebesgue spaces. This machinerywas not used in proving earlier versions of Theorem 1.2 (though some of it was used


in [7].) We believe that these techniques will be applicable to other problems invariable Lebesgue spaces.

Second, our proofs, especially in the case p+ < n/, are simpler than previousproofs. The proof in [2] for the Hardy-Littlewood maximal operator depends on thefollowing estimate: if fp() 1, then there exists a function S() L1 and C > 0such that for every ball B and x B,

(1.2)

(B

|f(y)| dy)p(x)

C(B

|f(y)|p(y)/p dy)p

+ S(x).

The proof of this inequality required considering separately the averages of f{|f |1}and f{|f |1}, and then subdivided the argument further by considering the distanceof the ball from the origin in comparison to its radius. Our proof still requires that wedivide f into its large and small parts, but the Calderon-Zygmund decomposition pro-vides the natural family of cubes on which to consider the averages. Furthermore,the structure of the proof makes clear the role played by the log-Holder continuityconditions: the LH0 condition is necessary only on the set where f is large, and theLH condition on the set where f is close to zero.

Third, our proof gives a unified treatment of the Hardy-Littlewood maximal op-erator and the fractional maximal operator. The proofs in [2] for the case > 0required first proving that the Hardy-Littlewood maximal operator is bounded onLp(), and then using this fact to prove that the fractional maximal operator mappedLp() into Lq(). In the classical Lebesgue spaces, however, it is possible to give asingle proof that works simultaneously for all , 0 < n. This is well-known; theproof is sketched in Duoandikoetxea [11]. This proof uses weak type inequalities andMarcinkiewicz interpolation, and so cannot be used in the variable Lebesgue spaces.In [2] the authors conjectured a generalization of (1.2) that would yield unified proof:if p()(f) 1, then there exists constant C and S L1 such that

Mf(x)q(x) CM(|f()|p()/p)(x)q + S(x).

However, this conjecture remains open.

The rest of this paper is organized as follows. In Section 2 we gather together somepreliminary results about variable Lebesgue spaces and about Calderon-Zygmundcubes. For complete information about these spaces we refer the reader to the papersby Kovacik and Rakosnk [15] or Fan and Zhao [12]. In Sections 3 and 4 we proveTheorem 1.2. The proof of the full result contains many technical details that obscurethe overall argument, so we first prove it in the special case that p+ < n/. Doingso results in some repetition, but it allows us to make clear the basic ideas of ourargument and to highlight the relative simplicity of this proof compared to earlierproofs. Finally, in Section 5, we prove Theorem 1.6. Throughout the paper, C


will denote a constant that may depend on n and p() but which may otherwisechange value at each appearance. In order to emphasize that we are dealing withvariable exponents, we will always write p() and q() for exponent functions; p andq will denote constants. Occasionally there will be minor differences in the argumentdepending on whether = 0 or > 0. We will highlight these but will generally givefull details only for the latter case, as the former case is usually easier.

2. Preliminary Results

The following lemmas are some key technical results needed in our proof. We havegathered them here to make our overall approach in the proofs clearer.

The first is the analogue of the monotone convergence theorem; in the more generalcontext of Banach function spaces it is referred to as the Fatou property of the norm.(See Bennett and Sharpley [1].)

Lemma 2.1. Given a non-negative function f Lp(), suppose the sequence {fN}of non-negative functions increases pointwise to f almost everywhere. Then fNp()increases to fp().

Proof. We may assume fp() > 0 since otherwise there is nothing to prove. Fix ,0 < < fp(); then by the definition of the norm,

1 < p()(f/) =

Rn\,p()

(f(x)

)p()dx+ 1fL(,p())

= limN

Rn\,p()

(fN(x)

)p()dx+ 1fNL(,p())

= limN

p()(fN/).

Therefore, for all N sufficiently large, p()(fN/) > 1, so fNp() > . Since we cantake any such , the desired conclusion follows at once.

To apply this lemma we need the following fact which is part of the folklore ofharmonic analysis. We include its short proof.

Lemma 2.2. Given , 0 < n, and a sequence of non-negative functions fNincreasing pointwise a.e. to a function f , then the functions MfN increase to Mfpointwise.

Proof. It follows at once from the definition that the sequence MfN is increasingand MfN(x) Mf(x) for all x. Now fix x and K such that K < Mf(x). Then


there exists a cube Q 3 x such that

K < |Q|/nQ

f(y) dx = limN

|Q|/nQ

fN(y) dx limN

MfN(x).

The desired conclusion follows immediately.

The next two lemmas are the only places we need to use the assumption that theexponent function is log-Holder continuous. The first appeared in [6] with balls inplace of cubes. The second is a special case of a result that appeared in [2] (seealso [4, 5]). For the convenience of the reader we include their short proofs.

Lemma 2.3. Given r() : Rn [0,) such that r() LH0 and r+ < , thereexists a constant C depending on n and the LH0 constant of r() such that given anycube Q and x Q,

|Q|r(x)r+(Q) C and |Q|r(Q)r(x) C.

Proof. We prove the first inequality; the proof of the second is identical. If `(Q) (2n)1, then

|Q|r(x)r+(Q) (2n)n(r+r) = C(n, r()).

If `(Q) < (2n)1, then for all y Q, |x y| < 1/2. In particular, since r() is

continuous, there exists y Q such that r(y) = r+(Q). Therefore, by the definitionof LH0,

|Q|r(x)r+(Q) (n1/2|x y|

)n|r(x)r(y)| exp

(C0(log(n

1/2) log |x y|) log |x y|

) C(n, r()).

Lemma 2.4. Let r() : Rn [0,) be such that r() LH, and let R(x) =(e+ |x|)N , N > n. Then there exists a constant C depending on n, N and the LHconstant of r() such that given any set E and any function F such that 0 F (y) 1for y E,

E

F (y)r(y) dy CE

F (y)r() dy + C

E

R(y)r() dy,(2.1) E

F (y)r() dy CE

F (y)r(y) dy + C

E

R(y)r() dy.(2.2)

Proof. We will prove (2.1); the proof of the second inequality is essentially the same.By the LH condition,

R(y)|r(y)r()| = exp(N log(e+ |y|)|r(y) r()|

) exp(NC).


Write the set E as E1 E2, where E1 = {x E : F (y) R(y)} and E2 = {x E :R(y) < F (y)}. Then

E1

F (y)r(y) dy E1

R(y)r(y) dy

E1

R(y)r()R(y)|r(y)r()| dy exp(NC)E1

R(y)r() dy.

Similarly, since F (y) 1,E2

F (y)r(y) dy E2

F (y)r()F (y)|r(y)r()| dy

E2

F (y)r()R(y)|r(y)r()| dy exp(NC)E2

F (y)r() dy.

The last two lemmas give some basic properties of cubes. The first defines the socalled Calderon-Zygmund cubes. This result is well-known for = 0for a proof seeDuoandikoetxea [11] or Garca-Cuerva and Rubio de Francia [13]. The same proofsgo through without significant changes for the case > 0. (For details, see [3].)Hereafter, given a cube Q and r > 0, let rQ denote the cube with the same center asQ and such that `(rQ) = r`(Q).

Lemma 2.5. Fix , 0 < n. Given a function f such that Q|f(y)| dy 0 as

|Q| , then for each > 0 there exists a set of pairwise disjoint dyadic cubes{Qj } such that

{x Rn : Mf(x) > 22n} j

3Qj ,

and

|Qj |/nQj

|f(x)| dx > .

Remark 2.6. The hypothesis on f is satisfied if it is bounded and has compact support.

The final lemma is a clever application of Holders inequality.

Lemma 2.7. Given , 0 < n, and p, q, such that 1 < p < n/ and 1/p 1/q =/n, then for every cube Q and non-negative function f ,

|Q|/nQ

f(x) dx (

Q

f(x)p dx

) 1p 1q(Q

f(x) dx

)p/q.


Proof. When = 0 this reduces to an identity, so we only need to consider the case > 0. By Holders inequality with exponent n/p > 1 and then with exponent p,

|Q|/nQ

f(x) dx = |Q|/nQ

f(x)p/nf(x)1p/n dx

|Q|/n(Q

f(x) dx

)p/n(Q

f(x) dx

)1p/n(

Q

f(x)p dx

)/n(Q

f(x) dx

)1p/n=

(Q

f(x)p dx

) 1p 1q(Q

f(x) dx

)p/q.

Remark 2.8. As a corollary to Lemma 2.7 we have that

Mf(x)q fqpp Mf(x)p.

Hence, the fact that M : Lp Lq follows immediately from the fact that the

Hardy-Littlewood maximal operator is bounded on Lp. In some sense, our proof ofTheorem 1.2 is a generalization of this approach.

3. Proof of Theorem 1.2: The case p+ < n/

Since p+ < n/, we have that p+, q+ 1} and f2 = f{x:f(x)1}; then p()(fi) fip() 1. Further, since Mf Mf1 + Mf2, it will suffice to show that for


i = 1, 2, that Mfiq() C(n, p()); since q+ < it will in turn suffice to showthat

q()(Mfi) =

RnMfi(x)

q(x) dx C.

The estimate for f1. Let A = 22n, and for each k Z let

k = {x Rn : Mf1(x) > Ak}.

Since f1 is bounded and has compact support, by Lemma 2.7, Mf1 L, soMf1(x) < a.e., and Rn =

k k \ k+1 (up to a set of measure 0). Further,

for each k we can apply Lemma 2.5 to form the pairwise disjoint cubes {Qkj} suchthat

k j

3Qkj and |Qkj |/nQkj

f1(x) dx > Ak1.

Define the sets Ekj inductively: Ek1 = (k\k+1)3Qk1, Ek2 = ((k\k+1)3Qk2)\Ek1 ,

Ek3 = ((k \ k+1) 3Qk3) \ (Ek1 Ek2 ), etc. The sets Ekj are pairwise disjoint for allj and k and k \ k+1 =

j E

kj .

We now estimate as follows:RnMf1(x)

q(x) dx =k

k\k+1

Mf1(x)q(x) dx

k

k\k+1

[Ak+1]q(x) dx

A2q+3q+(n)k,j

Ekj

(|3Qkj |/n

3Qkj

f1(y) dy

)q(x)dx.(3.1)

To estimate the integral in the last sum, we apply Lemma 2.7 with exponentspjk = p(3Q

kj ) and qjk = q(3Q

kj ):

|3Qkj |/n

3Qkj

f1(y) dy

(3Qkj

f1(y)pjk dy

) 1pjk 1qjk

(

3Qkj

f1(y) dy

)pjk/qjk.

Since f1 = 0 or f1 1 pointwise,3Qkj

f1(y)pjk dy

Rnf1(y)

p(y) dy 1.


Therefore,

k,j

Ekj

(|3Qkj |/n

3Qkj

f1(y) dy

)q(x)dx

k,j

Ekj

(

3Qkj

f1(y) dy

)pjkq(x)/qjkdx

k,j

Ekj

(

3Qkj

f1(y)pjk/p dy

)pq(x)/qjkdx.

Since q() LH0 and q+ 1, M is bounded on Lp . Hence,

C

Rnf1(x)

p(x) dx

C.


The estimate for f2. We argue exactly as we did above for f1, forming the sets kand Ekj using Lemma 2.5. We thus get

RnMf2(x)

q(x) dx Ck,j

Ekj

(|3Qkj |/n

3Qkj

f2(y) dy

)q(x)dx.

We claim that

(3.2) F = |3Qkj |/n

3Qkj

f2(y) dy 1.

If = 0, this is immediate since f2 1. If > 0, then by Holders inequality andsince p(y) n/,

F

(3Qkj

f2(y)n/ dy

)/n

(3Qkj

f2(y)p(y) dy

)/n 1.

Therefore, by Lemma 2.4 with R(x) = (e+ |x|)n.

k,j

Ekj

(|3Qkj |/n

3Qkj

f2(y) dy

)q(x)dx

Ck,j

Ekj

(|3Qkj |/n

3Qkj

f2(y) dy

)q()dx+ C

k,j

Ekj

R(x)q() dx.

We can immediately estimate the second term: since q() > 1 and the sets Ekj aredisjoint,

k,j

Ekj

R(x)q() dx

RnR(x)q() dx C.

To estimate the first term we apply Lemma 2.7 with exponents p() and q():

|3Qkj |/n

3Qkj

f2(y) dy

(3Qkj

f2(y)p()

) 1p()

1q()

(

3Qkj

f2(y) dy

)p()/q().

To estimate the first integral on the right-hand side, we again apply Lemma 2.4 withR(x) = (e+ |x|)n:

3Qkj

f2(y)p() C

3Qkj

f2(y)p(y) dy + C

3Qkj

R(y)p() dy C.


Since p() > 1, M is bounded on Lp(). Therefore,k,j

Ekj

(|3Qkj |/n

3Qkj

f2(y) dy

)q()dx C

k,j

Ekj

(

3Qkj

f2(y) dy

)p()dx

Ck,j

Ekj

Mf2(x)p() dx

C

RnMf2(x)

p() dx

C

Rnf2(x)

p() dx;

since f2 1 we can apply Lemma 2.4 again to conclude

C

Rnf2(x)

p(x) dx+ C

RnR(x)p() dx

C.

4. Proof of Theorem 1.2: The general case

The proof of the general case has much the same outline as the proof whenp+ < n/ given in the previous section, but it is made more complicated by thetechnicalities needed to deal with the fact that the exponent function q() is un-bounded and may in fact equal on a set of positive measure. In the proof thatfollows we attempt to strike a balance between brevity and completeness, and we willrefer back to the proof in Section 3 for those details which remain the same.

Arguing as we did before, we may assume without loss of generality that f isnon-negative, bounded, has compact support, and that fp() = 1. Then

p()(f) =

Rn\,p()

f(x)p(x) dx+ fL(,p()) 1.

Decompose f as f1 + f2 + f3, where

f1 = f{x:f(x)>1},

f2 = f{xRn\,p():f(x)1},

f3 = f{x,p():f(x)1}.

(Note that f3 6= 0 only if = 0.) Then supp(f1) Rn \ ,p() (up to a set ofmeasure zero), and p()(fi) fip() 1. We will show that there exist constantsi = i(n, p()) > 0 such that

q()(Mfi/i) 1.


In each case, we will write the constant i as a product of constants needed at differentstages of the proof.

The estimate for f1. Let 11 = 111. Then

q()(111Mf1) =

Rn\,q()

[111Mf1(x)]q(x) dx+ 111Mf1L(,q()).

We will show that each term on the right is bounded by 1/2. To estimate the first,let A = 22n, and define

k = {x Rn \ ,q() : Mf1(x) > Ak}.Since Mf1 L, Mf1(x) < a.e., and so Rn \ ,q() =

k k \ k+1 (up to a

set of measure 0). By Lemma 2.5 there exist pairwise disjoint cubes {Qkj} such that

k j

3Qkj and |Qkj |/nQkj

f1(x) dx > Ak1,

and we can form sets Ekj that are pairwise disjoint for all j and k and and such that

k \ k+1 =j E

kj . Now let 1 = A

23n and estimate as follows:Rn\,q()

[111Mf1(x)]q(x) dx =

k

k\k+1

[111Mf1(x)]q(x) dx

k

k\k+1

[111Ak+1]q(x) dx

k,j

Ekj

(11|3Qkj |/n

3Qkj

f1(y) dy

)q(x)dx.(4.1)

To estimate the integral in the last sum, we apply Lemma 2.7 with exponentspjk = p(3Q

kj ) and qjk = q(3Q

kj ). (Since E

kj k Rn \ ,q(), both of these

exponents are finite.) This yields

|3Qkj |/n

3Qkj

f1(y) dy

(3Qkj

f1(y)pjk dy

) 1pjk 1qjk

(

3Qkj

f1(y) dy

)pjk/qjkSince f1 = 0 or f1 1 pointwise and supp(f1) Rn \ ,p(),

3Qkj

f1(y)pjk dy

Rn\,p()

f1(y)p(y) dy p()(f1) 1.

Therefore,


k,j

Ekj

(11|3Qkj |/n

3Qkj

f1(y) dy

)q(x)dx

k,j

Ekj

11(3Qkj

f1(y) dy

)pjk/qjkq(x) dxk,j

Ekj

11(3Qkj

f1(y)pjk/p dy

)p/qjkq(x) dx.Define the exponent function r() = 1/q(). Then r() LH0, r+ 1, and

r+(3Qkj ) = 1/qjk. Therefore, by Lemma 2.3, we can choose 1 < 1 so that

1|3Qkj |p/qjk |3Qkj |p/q(x).

Further, arguing as before,3Qkj

f1(y)p(y)/p dy

3Qkj

f1(y)p(y) dy 1.

Therefore, since x Ekj 3Qkj , q(x) qjk, and assuming for the moment that1 < 1,

k,j

Ekj

11(3Qkj

f1(y)pjk/p dy

)p/qjkq(x) dxk,j

Ekj

|3Qkj |p(1

3Qkj

f1(y)p(y)/p dy

)q(x)p/qjkdx

k,j

Ekj

|3Qkj |p(1

3Qkj

f1(y)p(y)/p dy

)pdx

k,j

Ekj

p1 M(f1()p()/p)(x)p dx

Rnp1 M(f1()p()/p)(x)p dx.

Since p > 1, M is bounded on Lp , so we can choose 1 < 1 such that

Rnp1 M(f1()p()/p)(x)p dx

1

2

Rnf1(y)

p(y) dy 12.


We will now show that 111Mf1L(,q()) 1/2. Since 11 1/4, it willsuffice to show (after possibly taking 1 smaller than the value chosen above) that

(4.2) 1Mf1L(,q()) 2.

Fix x ,q(). Since supp(f1) Rn \ ,q(), when computing Mf1(x) we canrestrict ourselves to cubes Q 3 x such that |Q \,q()| > 0. In particular, thereexists such a cube that satisfies

Mf1(x) 2|Q|/nQ

f1(y) dy.

Fix r, q(Q) < r 0sufficiently small (but not depending on our choice of r),(

1|Q|/nQ

f1(y) dy

)q(xr) Q

f1(y)p(y) dy 1

|Q|.

Therefore, we have that

1|Q|/nQ

f1(y) dy |Q|1/r 1.

Since this is true for all r large, we can take the limit as r to get

1Mf1(x) 21|Q|/nQ

f1(y) dy 2.

Since this estimate holds for almost all x, we have proved inequality (4.2). Thus wehave proved that q()(111Mf1) 1.

The estimate for f2. Let 12 = 2222. Then

q()(2222Mf2)

=

Rn\,q()

[2222Mf2(x)]q(x) dx+ 2222Mf2L(,q()).

We will again show that each term is bounded by 1/2. The second is very easy toestimate. Since M : L

n/ L with constant 1, and since f2 1,

2222Mf2L(,q()) 2222(

Rnf2(y)

n/ dy

)/n 2222

(Rnf2(y)

p(y) dy

)/n 2222[p()(f2)]/n 2222.


As we will see below, 2222 1/2.To estimate the first term, we form the sets k, Q

kj and E

kj as before using

Lemma 2.5. If we set 2 = A23n, and argue as we did for f1, we get

Rn\,q()[2222Mf2(x)]

q(x) dx k,j

Ekj

(222|3Qkj |/n

3Qkj

f2(y) dy

)q(x)dx.

At this point we consider two cases: q() = and q() 0 sufficiently close to 0.

We now consider the more difficult case when q()


where R(y) = (e+ |x|)N , and N is chosen so that the second integral converges andin fact so that

RnR(y)1/p() dy

RnR(y)1/q() dy 1.

(The reason for this choice will be made clear below.) Since we also have that3Qkj

g2(y) dy

Rnf2(y)

p(y) dy 1,

we can choose 2 > 0 so that

k,j

Ekj

(222|3Qkj |/n

3Qkj

g2(y)1/p(y) dy

)q(x)dx

k,j

Ekj

22

[(

3Qkj

g2(y)1/p(y) dy

)p() ]q(x)/q()dx.

Since the quantity in square brackets is less than 1, and since 1/q() LH, we canagain apply Lemma 2.4 to conclude that we can choose 2 > 0 such that

k,j

Ekj

22

[(

3Qkj

g2(y)1/p(y) dy

)p() ]q(x)/q()dx

k,j

Ekj

2

(

3Qkj

g2(y)1/p(y) dy

)p()dx+

1

6

RnR(x)1/q() dx

Rn2M(g2()1/p())(x)p() dx+

1

6.

The maximal operator is bounded on Lp() since p() p > 1, and so we canapply Lemma 2.4 a third time to conclude that

RnM(g2()1/p())(x)p() dx C

Rng2(x)

p()/p(x) dx

C

Rng2(x) dx+ C

RnR(x)1/p() dx C.

Therefore, we can choose 2 > 0 so thatRn2M(g2()1/p())(x)p() dx+

1

6 1

3+

1

6=

1

2.

This completes the estimate for f2.


The estimate for f3. Recall that by the definition of f3 we only need this estimatewhen = 0, so p() = q(). Let 1 = 3 1/2. Then

p()(3Mf3) =

Rn\,p()

[3Mf3(x)]p(x) dx+ 3Mf3L(,p()).

We will show that each term is less that 1/2. To estimate the second, since M isbounded on L with constant 1,

3Mf3L(,p()) 3f3L 3 1

2.

To estimate the first term, we consider two cases: p() = and p() < .In the first case we argue exactly as we did before in the estimate for f2. Since1/p() LH, p(x) C1 log(e + |x|). Since f3 1, Mf3 1. Therefore, for 3sufficiently close to 0,

Rn\,p()[3Mf3(x)]

p(x) dx

RnC1 log(e+|x|)3 dx

1

2.

Now suppose p() 1, M is bounded on Lp(), and since f3 1 we can again applyLemma 2.4 (with the same function R) to conclude that

Rn\,p()Mf3(x)

p() dx

Rnf3(x)

p() dx

C

Rnf3(x)

p(x) dx+ C

Rn\,p()

R(x)1/p() dx C.

Combining these estimates, we see that we can choose 3 > 0 such thatRn\,p()

[3Mf3(x)]p(x) dx 1

2.


5. Proof of Theorem 1.6

The proof is nearly identical to the proof of Theorem 1.2 and we sketch the details.We begin by making the same reductions as before, and writing f = f1 + f2 + f3.Then for fixed t > 0,

{x Rn : Mf(x) > t} i

{x Rn : Mfi(x) > t/3} =i

i

Therefore, it will suffice to show that for each i, tiq() C, and in turn it willsuffice to show that for some i > 0,

q()(iti) =

i\,q()

[it]q(x) dx+ itiL(,q()) 1.

As before, for each i we will show that each term on the right is bounded by 1/2 forsuitable choice of i.

The estimates for the second term are immediate. For all i,

itiL(,q()) iMfiL(,q()),

and the bounds on the right-hand side given above did not depend on the fact thatp > 1. Since the other hypotheses hold, the same proofs yield the desired estimates.

To estimate i\,q()

[it]q(x) dx

we have to avoid using the Hardy-Littlewood maximal operator, since our hypothesesno longer guarantee that it is bounded. We apply Lemma 2.5 to find disjoint dyadiccubes {Qj} such that

i j

3Qj and |Qj|/nQj

fi(y) dy > 22n t

3.

Then we can find disjoint sets Ej such that Ej 3Qj and i \ ,q() =j Ej.

Now if i = 1, we argue as we did in the estimate of f1 in the previous section,replacing 3Qkj by Q

kj and using the fact that p = 1 to get

1\,q()[1t]

q(x) dx j

Ej

1Qj

f1(y)p(y) dy dx.

Since Ej 3Qj and the cubes Qj are disjoint, we can choose 1 > 0 so thatj

Ej

1Qj

f1(y)p(y) dy dx

j

1

2

Qj

f1(y)p(y) dy 1

2

Rnf1(y)

p(y) dy 12.


If i = 2, the estimate when q() = is exactly the same as in the estimate forf2 above. When q() 0 such that right-hand side is bounded by 1/2.

Finally, if i = 3, we may assume as before that = 0. In this case, since f3 1,Mf3 1, so 3 is non-empty only if t < 3. If p() =, then

3\,p()[3t]

p(x) dx

3\,p()[33]

p(x) dx,

and we can repeat the argument for the estimate of f3 given above.

If p() 0 such that3\,p()

[3t]p(x) dx 1

2.

References

[1] C. Bennett and R. Sharpley. Interpolation of operators, volume 129 of Pure and Applied Math-ematics. Academic Press Inc., Boston, MA, 1988.

[2] C. Capone, D. Cruz-Uribe, and A. Fiorenza. The fractional maximal operator and fractionalintegrals on variable Lp spaces. Revista Math. Iberoamericana, To appear.

[3] D. Cruz-Uribe. New proofs of two-weight norm inequalities for the maximal operator. GeorgianMath. J., 7(1):3342, 2000.


[4] D. Cruz-Uribe, A. Fiorenza, and C. J. Neugebauer. The maximal function on variable Lp spaces.Ann. Acad. Sci. Fenn. Math., 28(1):223238, 2003. See errata [5].

[5] D. Cruz-Uribe, A. Fiorenza, and C. J. Neugebauer. Corrections to: The maximal function onvariable Lp spaces [Ann. Acad. Sci. Fenn. Math. 28 (2003), no. 1, 223238; ]. Ann. Acad. Sci.Fenn. Math., 29(1):247249, 2004.

[6] L. Diening. Maximal function on generalized Lebesgue spaces Lp(). Math. Inequal. Appl.,7(2):245253, 2004.

[7] L. Diening. Maximal function on Musielak-Orlicz spaces and generalized Lebesgue spaces. Bull.Sci. Math., 129(8):657700, 2005.

[8] L. Diening. Habilitation, Universitat Freiburg, 2007.[9] L. Diening, P. Harjulehto, P. Hasto, Y. Mizuta, and T. Shimomura. Maximal functions in

variable exponent spaces: limiting cases of the exponent. Preprint, 2007.[10] L. Diening, P. Hasto, and A. Nekvinda. Open problems in variable exponent Lebesgue and

Sobolev spaces. In FSDONA04 Proceedings (Drabek and Rakosnik (eds.); Milovy, Czech Re-public, pages 3858. Academy of Sciences of the Czech Republic, Prague, 2005.

[11] J. Duoandikoetxea. Fourier analysis, volume 29 of Graduate Studies in Mathematics. AmericanMathematical Society, Providence, RI, 2001.

[12] X. Fan and D. Zhao. On the spaces Lp(x)() and Wm,p(x)(). J. Math. Anal. Appl., 263(2):424446, 2001.

[13] J. Garca-Cuerva and J.L. Rubio de Francia. Weighted norm inequalities and related topics,volume 116 of North-Holland Mathematics Studies. North-Holland Publishing Co., Amsterdam,1985.

[14] V. Kokilashvili and S. Samko. On Sobolev theorem for Riesz-type potentials in Lebesgue spaceswith variable exponent. Z. Anal. Anwendungen, 22(4):899910, 2003.

[15] O. Kovacik and J. Rakosnk. On spaces Lp(x) and W k,p(x). Czechoslovak Math. J.,41(116)(4):592618, 1991.

[16] A.K. Lerner. Some remarks on the Hardy-Littlewood maximal function on variable Lp spaces.Math. Z., 251(3):509521, 2005.

[17] B. Muckenhoupt and R. Wheeden. Weighted norm inequalities for fractional integrals. Trans.Amer. Math. Soc., 192:261274, 1974.

[18] J. Musielak. Orlicz spaces and modular spaces, volume 1034 of Lecture Notes in Mathematics.Springer-Verlag, Berlin, 1983.

[19] A. Nekvinda. Hardy-Littlewood maximal operator on Lp(x)(Rn). Math. Inequal. Appl.,7(2):255265, 2004.

[20] A. Nekvinda. A note on maximal operator on `{pn} and Lp(x)(R). J. Funct. Spaces Appl.,5(1):4988, 2007.

[21] L. Pick and M. Ruzicka. An example of a space Lp(x) on which the Hardy-Littlewood maximaloperator is not bounded. Expo. Math., 19(4):369371, 2001.

[22] S. Samko. On a progress in the theory of Lebesgue spaces with variable exponent: maximaland singular operators. Integral Transforms Spec. Funct., 16(5-6):461482, 2005.


Dept. of Mathematics, Trinity College, Hartford, CT 06106-3100, USA

E-mail address: [email protected]

Section of Applied Mathematics, University of Freiburg, Eckerstr. 1, 79104 Freiburg,Germany


Dipartimento di Costruzioni e Metodi Matematici in Architettura, Universita diNapoli, Via Monteoliveto, 3, I-80134 Napoli, Italy, and Istituto per le Applicazionidel Calcolo Mauro Picone, sezione di Napoli, Consiglio Nazionale delle Ricerche,via Pietro Castellino, 111, I-80131 Napoli, Italy


1. Introduction2. Preliminary Results3. Proof of Theorem 1.2: The case p+

Documents

A NEW PROOF OF THE BOUNDEDNESS OF MAXIMAL …diening/archive/newmax6.pdf · a new proof of the boundedness of maximal operators on variable lebesgue spaces d. cruz ... are simpler