Sparse domination and the strong maximal function
SSPARSE DOMINATION AND THE STRONGMAXIMAL FUNCTION
ALEX BARRON, JOS´E M. CONDE-ALONSO, YUMENG OU, AND GUILLERMO REY
Abstract.
We study the problem of dominating the dyadic strong maximal func-tion by (1 , Introduction
Recent years have seen a great deal of work around the concept of sparse dom-ination . Perhaps the easiest domination result is the one for the dyadic maximalfunction M D f ( x ) = sup Q (cid:51) x | Q | (cid:90) Q | f | , where the supremum is taken over the family D of all dyadic cubes in R d . Onecan use the Calder´on-Zygmund decomposition at varying heights to obtain (for eachnon-negative locally integrable function f ) a collection S of dyadic cubes from thesame filtration D such that M D f ≤ C (cid:88) Q ∈S (cid:104) f (cid:105) Q Q , where (cid:104) f (cid:105) Q = | Q | (cid:82) Q f , and S is sparse : Definition 1.
A collection of sets S is η -sparse if for all Q ∈ S there exists a subset E ( Q ) ⊆ Q such that | E ( Q ) | ≥ η | Q | and the collection { E ( Q ) } is pairwise disjoint. It is worth mentioning that this notion was already used by Sawyer [23] and byChrist-Fefferman [4] to prove certain weighted estimates, although the language thatthey used was different. Moreover, the result readily extends to more general mar-tingale filtrations and a similar estimate allows one to sparsely dominate martingaletransforms as well (see [17]).The recent interest in sparse domination results was sparked by Lerner (see forexample [20, 21, 22]) in connection with the famous A conjecture as it gave a In what follows, we will usually omit the parameter η and just say “sparse” instead of “ η -sparse”. a r X i v : . [ m a t h . C A ] N ov PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 2 particularly simple proof of the –now– theorem (first settled by Hyt¨onen in [15] afternumerous efforts by the mathematical community). Lerner’s original result was a dual form of sparse domination:(1.1) (cid:107)
T f (cid:107) X (cid:46) sup S (cid:13)(cid:13)(cid:13)(cid:88) Q ∈S (cid:104)| f |(cid:105) Q Q (cid:13)(cid:13)(cid:13) X , where the supremum is taken over all sparse collections (with respect to a finitenumber of dyadic grids), and X is any Banach function space .The number of sparse domination results is by now very large and it would beimpossible to give a detailed survey of all of them in a reasonable amount of text.But let us note that (1.1) has been generalized in many ways, and the typical resultis that for every pair of (sufficiently nice) functions f and g there exists a sparsecollection of cubes S such that(1.2) |(cid:104) T f, g (cid:105)| (cid:46) (cid:88) Q ∈S (cid:104)| f |(cid:105) Q (cid:104)| g |(cid:105) Q | Q | or more generally(1.3) |(cid:104) T f, g (cid:105)| (cid:46) (cid:88) Q ∈S (cid:104)| f |(cid:105) Q,r (cid:104)| g |(cid:105) Q,s | Q | , where (cid:104)| f |(cid:105) Q,r = (cid:104)| f | r (cid:105) r Q and 1 ≤ r, s < ∞ . The ( r, s ) sparse forms of type (1.3) arisewhen studying operators that fall outside the scope of classical Calder´on-Zygmundtheory, and which may not satisfy the full range of strong and weak-type L p estimatesimplied by (1.2). The operators that have been studied include but are not limitedto rough singular integral operators, Bochner-Riesz multipliers, spherical maximalfunctions, singular integrals along manifolds, pseudodifferential operators, and thebilinear Hilbert transform. See for example [1, 6, 17, 18, 2, 19, 5, 9, 8]. Sparsedomination provides a fine quantification of the mapping properties of operatorsthat carries much more information than L p bounds. In particular, it is a veryeffective way to obtain sharp quantitative weighted estimates and has also been usedto study previously unknown endpoint behaviors (see for instance [16]).In the present work we study the problem of sparsely dominating the bi-parameteranalogue of the the dyadic maximal function. This operator is the dyadic strongmaximal function:(1.4) M S f ( x ) := sup R (cid:51) x (cid:104)| f |(cid:105) R , where the supremum is taken over all dyadic rectangles containing x with sidesparallel to the axes. The strong maximal function is one of the most important PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 3 operators in the theory of multi-parameter singular integrals, associated with which isan underlying non-isotropic dilation structure. This class of operators arise naturallyin the theory of summation of multiple Fourier series, several complex variables andcertain boundary value problems, and is a first step into the study of operators withmore complicated dilation structure (e.g. Zygmund dilations). However, much lessis known about these operators compared to the one-parameter case, for examplethe sharp weighted bound, thus it is natural to ask whether the sparse dominationtechnique can be introduced into the multi-parameter setting to help the study.Unfortunately, this seems to be very difficulty due to the fact that one of the keyingredients in standard proofs of sparse domination, the stopping time argument, ismissing in the multi-parameter setting.We first note that in the study of multi-parameter sparse domination, the naturalgeometric objects to consider are axes-parallel rectangles instead of cubes, since M S can be large on axis-parallel rectangles of arbitrary eccentricity. Then, one observesthat the strong maximal function is indeed dominated by sparse forms (based onrectangles) when restricted to a single point mass. Indeed, one can make sense of(1.4) when applied to finite positive measures, and then it is easy to see that(1.5) M S ( δ )( x, y ) ≤ | x || y | . And this function can actually be dominated by a sparse operator: if we define I m = [0 , m ) and J m = [2 m − , m ) then1 | x || y | = (cid:88) m,n ∈ Z | x || y | J m × J n ( x, y ) ≤ (cid:88) m,n ∈ Z − m − n J m × J n ( x, y )= 4 (cid:88) m,n ∈ Z (cid:104) δ (cid:105) I m × I n I m × I n ( x, y ) . Note that the collection S = { I m × I n : m, n ∈ Z } is sparse since for every R = I m × I n we can define E ( R ) = J m × J n and this satisfies the conditions of Definition 1 with η = 1 / M S and the Hardy-Littlewood maximal functionsnear the L endpoint. Moreover, we know from the one-parameter theory that thereis a connection between sparse bounds and weak-type endpoint estimates; for exam-ple, a (1 ,
1) sparse bound of type (1.2) implies that T maps L into weak L (seethe appendix in [5]). Taking the above discussion into account, it is natural to ask PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 4 whether or not M S admits a (1 ,
1) sparse bound of the type(1.6) |(cid:104)M S f, g (cid:105)| (cid:46) (cid:88) R ∈S (cid:104)| f |(cid:105) R (cid:104)| g |(cid:105) R | R | , where the collection S consists of dyadic axis-parallel rectangles. We also note thatthere is no immediate contradiction implied by a bound of type (1.6); indeed, anestimate of the type |(cid:104) T f, g (cid:105)| (cid:46) (cid:88) R ∈S (cid:104)| f |(cid:105) R (cid:104)| g |(cid:105) R | R | implies that T maps L log L into weak L , but does not imply that T maps L intoweak L (the proof is similar to the argument in the appendix of [5]).Our main result answers the question raised above in the negative: there can beno domination by positive sparse forms of the type (1.2) for the strong maximalfunction. Theorem A.
For every
C > and < η < there exist a pair of compactlysupported integrable functions f and g such that |(cid:104)M S f, g (cid:105)| ≥ C (cid:88) R ∈S (cid:104)| f |(cid:105) R (cid:104)| g |(cid:105) R | R | , for all η -sparse collections S of dyadic rectangles with sides parallel to the axes. The proof of theorem A is based on the construction of pairs of extremal functionsfor which the sparse bound cannot hold. These extremal examples take advantage ofthe behavior of M S when applied to sums of several point masses. Indeed, when weapply M S to a point mass we get level sets that look like dyadic stairs , as displayedin Figure 1. These stairs are sparse, in fact the measure of their union is proportionalto the number of rectangles that form them. We shall describe a way to place manypoint masses sufficiently far from each other in such a way that these stairs are allnecessary to dominate M S , but are packed too tightly to be sparse.In particular, we shall construct a set of points P which are maximally separatedwith respect to a quantity that captures the biparametric structure of the problem.Our first extremal function is then the sum of point masses at the points in P . Aswe said above, this will guarantee that M S applied to it is uniformly large on theunit square. We complement the construction of the set P with another set, Z , withthe following structure: for each point p ∈ P there exist a large amount of points z ∈ Z that are both near p and far away from all the other points in P . Moreover,the family of points in Z associated with a given p ∈ P are not clustered together,but are instead spread out over the boundary of a hyperbolic ball centered at p . Oursecond extremal function is then the normalized sum of point masses at the points in PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 5
Figure 1.
Dyadic stairs Z . We will show that the placement of the points in Z implies that if a form givenby a family S dominates the pair (cid:104)M S f, g (cid:105) then the rectangles in S need to coverthe portion of the stair centered at p that contains each z , for all z ∈ Z . But thenthe resulting collection of rectangles cannot be sparse.We remark that our result actually extends to the situation in which we takeslightly larger averaged norms of f . In particular our methods allow us to pushTheorem A to the case of Orlicz φ ( L )-norms with φ ( x ) = x log( x ) α for α < /
2. Wecarry out this extension in Theorem 4.6.We have learned that our construction of P is closely connected to the fields ofdiscrepancy, combinatorics, and computational learning. In particular it is a specialconstruction of a low-discrepancy sequence similar to those of Hammersley, see forexample Definition 3.44 of [10]. These low-discrepancy sequences are also closelyconnected to the notion of ε -nets, though our point of view is in a sense oppositeto the one usually taken in the theory of ε -nets: we are interested in lower boundson the cardinality of the intersection of these sets with rectangles. The notion of( t, m, s )-nets is closer to our purposes, see [10]. Our results show that these kindof sets are particularly well-suited to the study of the strong maximal function.However, the theory of discrepancy and ε -nets is much more general so it would bevery interesting to find further connections between other types of maximal operatorsand the fields above. In particular Zygmund-type maximal operators (defined like PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 6 the strong maximal function, but with rectangles with fewer “parameters” than theambient dimension, see [7, 24, 13, 25]) could be amenable to this connection.The structure of the rest of the paper is the following. In Section 2, we provelower bounds for M S and, in the opposite direction, we prove an upper bound forsparse forms when acting on the same extremal functions. We assume the existenceof the key sets of points P and Z , which are needed to construct the aforementionedextremal functions. The sets P and Z are constructed in detail in Section 3, com-pleting the proof of Theorem A. Finally, in Section 4 we show how our results extendto higher dimensions by a tensor product argument. We also show that the strongmaximal function is in a sense supercritical for L -sparse forms, which allows us toslightly strengthen our result. Acknowledgements.
The authors would like to thank Jill Pipher and DmitriyBilyk for useful discussions regarding the content of this paper. The third namedauthor is supported by NSF-DMS
Lower bounds for M S and upper bounds for sparse forms We start by introducing a quantity which is intimately related with the geometryof the bi-parameter setting: for every pair of points p, q in [0 , we setdist H ( p, q ) = inf {| R | / : R ∈ D × D is a dyadic rectangle containing p and q } . One can also define a distance between two (dyadic) rectangles:dist H ( R , R ) := inf p ∈ R ,q ∈ R dist H ( p, q ) . Note that dist H is not a distance function, and not even a quasidistance, as thetriangle inequality is completely false in general. We will, however, refer to it as thedistance between two points. Using this language, the strong maximal function of apoint mass (1.5) can now be written as M S ( δ )( p ) = 1dist H ( p, . We are going to study the precise behavior of M S when applied to uniform prob-ability measures concentrated on certain point sets: given a finite set of points F let(2.1) µ = µ F = 1 F ) (cid:88) p ∈F δ p . In general evaluating the strong maximal function of a sum of point masses is acomputationally difficult problem: for every point p one has to consider all possiblesubsets of F whose minimal enclosing rectangle contains p . However, if we know apriori that F is ε -separated (with respect to dist H ), and that it contains sufficiently PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 7 many points, then this forces a certain uniform distribution of µ at large scales,reducing the problem to treating only small rectangles. In particular, let ε > F = P is such that P ) ≥ Cε − , and that it is ε -separated:dist H ( p, q ) ≥ ε ∀ p (cid:54) = q in P . Then for any rectangle with | R | ≥ ε we have (cid:104) µ (cid:105) R = µ ( R ) | R | ≤ R ∩ P ) C | R | ε − (cid:46) | R | ε − C | R | ε − = 1 C .
If instead | R | < ε then R can contain at most one point from P and so for any p for which dist H ( p, P ) (cid:46) M S ( µ )( p ) ∼ P ) sup q ∈P H ( p, q ) . As we mentioned in the introduction, if one can find another set of points Z each of which is very close to exactly one point p from P —and which are not veryclose to one another, so that they do not form large clusters—, then we can force acontradiction that makes sparse domination fail. The rest of this section is devotedto giving lower bounds for M S assuming the existence of such sets of points P and Z . In Section 2.2 we exploit the intuition given above and use it to give upperbounds for sparse forms that, together with the lower bounds from this section, willultimately yield a contradiction.The following two theorems give the full description of P and Z . We postponetheir proofs to Section 3. Theorem 2.1.
For every m ≥ there exists a collection of points P ⊂ [0 , suchthat (P.1) dist H ( p, q ) ≥ − m , ∀ p (cid:54) = q in P , and (P.2) P ) = 2 m +1 . Theorem 2.2.
For every k (cid:28) m and every − m -separated set P ⊂ [0 , thereexists a set Z ⊂ [0 , satisfying the following properties: (Z.1) Z ) (cid:38) m m . (Z.2) For every z ∈ Z there exists exactly one point p ( z ) ∈ P such that dist H ( p ( z ) , z ) < − m − . (Z.3) For every z ∈ Z , dist H ( p ( z ) , z ) ∼ − m − k . PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 8 (Z.4)
For every dyadic rectangle R with | R | ≤ − m − that intersects P we have R ∩ Z ) (cid:46) k. (Z.5) For every dyadic rectangle R with | R | ≥ − m − we have R ∩ Z ) (cid:46) m mk | R | . The implied constants are independent of k and m . Remark 2.3.
A simple pigeonholing argument shows that if P is the collection fromTheorem 2.1 and R is a dyadic rectangle with | R | ≥ − m − , then R ∩ P ) =2 m +1 | R | . Remark 2.4.
Above, we deliberately omitted the precise dependence of m on k . Itturns out that any m ≥ k + C is admissible as one can check from the proofs inSection 3. This justifies our choices of m and k in the next subsections. Lower bound for M S . For the rest of this section k (cid:28) m will be some fixedlarge numbers, and ( P , Z ) will be the sets given by Theorems 2.1 and 2.2. Also, µ = µ P and ν = ν Z will always denote the associated uniform probability measuresintroduced in (2.1). The proofs of Theorems 2.2 and 2.2 in the next section show that µ and ν can be arbitrarily well-approximated by L functions, in particular becausewe will be able to choose the points from P and Z in small cubes. Because of this wewill prove Theorem A with f and g replaced by the measures µ and ν , respectively,and the full result can then be recovered by a simple limiting argument. Proposition 2.5.
Under these conditions we have (2.3) (cid:104)M S ( µ ) , ν (cid:105) (cid:38) k . Proof.
The positivity of M S makes the proof almost trivial since we do not need tocare about possible interactions among the points in P . In particular for all z ∈ Z let p ( z ) be the point guaranteed by (Z.2) of Theorem 2.2. Then, by (2.2) and ((Z.3)), M S ( µ )( z ) ≥ P ) dist H ( p ( z ) , z ) ∼ m + k P ) . Now by (P.2) of Theorem 2.1, M S ( µ )( z ) (cid:38) k and the claim follows from the fact that ν is a uniform probability measure over theset Z . (cid:3) PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 9
Upper bound for sparse forms.
We now prove upper bounds for sparseforms when acting on µ and ν . Our results here are written in the language ofCarleson sequences: Definition.
Let α be a non-negative function defined on all rectangles that is zerofor all but a finite collection of rectangles. We say that α is Λ -Carleson if for allopen sets Ω we have (cid:88) R ⊆ Ω α R | R | ≤ Λ | Ω | , where the sum is taken over all rectangles contained in Ω . We call a collection of dyadic rectangles S a Λ-Carleson collection if the sequence α R = (cid:40) R ∈ S , − -sparse and vice-versa, as was shown in [14] (seealso [11]). Proposition 2.6.
For every Λ -Carleson collection S and all k and m we have (2.4) (cid:88) R ∈S (cid:104) µ (cid:105) R (cid:104) ν (cid:105) R | R | (cid:46) Λ k (cid:16) k m (cid:17) . Proof.
For any rectangle R let R be the intersection of R with the unit square [0 , .We will show that(2.5) (cid:104) µ (cid:105) R (cid:104) ν (cid:105) R (cid:46) k (cid:16) k m (cid:17)(cid:16) | R || R | (cid:17) . Assume first that R is large , i.e.: | R | ≥ − m − . Then by the 2 − m -separation of P(cid:104) µ (cid:105) R = 12 m +1 P ∩ R ) | R | = | R || R | . Similarly, for ν we have by (Z.5): (cid:104) ν (cid:105) R (cid:46) k | R || R | . Therefore, for large rectangles we have (cid:104) µ (cid:105) R (cid:104) ν (cid:105) R (cid:46) k (cid:16) | R || R | (cid:17) . PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 10
If instead R is small , i.e.: | R | < − m − , then we have to be more careful. If (cid:104) µ (cid:105) R (cid:104) ν (cid:105) R (cid:54) = 0then R must contain at least one point p ∈ P . In fact, since R is sufficiently small,there exists exactly one p ∈ P ∩ R . Hence (cid:104) µ (cid:105) R (cid:46) m | R | . Since (cid:104) ν (cid:105) R (cid:54) = 0 the rectangle R must contain at least one point from Z . Thisimplies a lower bound on the size of R . To see this observe that if z is any pointin R ∩ Z , then by the smallness of R and (Z.2) we must have p = p ( z ), and hencedist H ( p, z ) ∼ − m − k by (Z.3), therefore | R | (cid:38) − m − k . Now, by (Z.4) we have R ∩ Z ) (cid:46) k, so (cid:104) µ (cid:105) R (cid:46) k and (cid:104) ν (cid:105) R (cid:46) k k m , from which inequality (2.5) follows.Now we can finish the proof by splitting S into families S j as follows: S j = { R ∈ S : 2 − j − | R | ≤ | R | < − j | R |} . Note that the rectangles in S j are contained inΩ j = { p ∈ R : M S ( [0 , ) (cid:38) − j } , which, by the weak-type boundedness of the strong maximal function, satisfies(2.6) | Ω j | (cid:46) j j . Then by estimate (2.5), the Carleson condition, and (2.6), (cid:88) R ∈S (cid:104) µ (cid:105) R (cid:104) ν (cid:105) R | R | = ∞ (cid:88) j =0 (cid:88) R ∈S j (cid:104) µ (cid:105) R (cid:104) ν (cid:105) R | R | (cid:46) k (cid:16) k m (cid:17) ∞ (cid:88) j =0 − j (cid:88) R ∈S j | R |≤ k (cid:16) k m (cid:17) ∞ (cid:88) j =0 − j Λ | Ω j | PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 11 (cid:46) Λ k (cid:16) k m (cid:17) ∞ (cid:88) j =0 j − j (cid:46) Λ k (cid:16) k m (cid:17) . (cid:3) Remark 2.7.
An examination of the proof above shows that we do not use the fullpower of the Carleson condition. In particular, the collection S can be assumed tosatisfy the Λ -Carleson packing condition with respect to the unit cube and the levelsets Ω j , but it does not need to be Λ -Carleson with respect to any other open sets(and in particular need not be Λ -Carleson at small scales). We can now formally conclude the proof of Theorem A conditionally on Theorems2.1 and 2.2.
Proof of Theorem A.
Choose m = k k . Proposition 2.5 implies that (cid:104) M S ( µ ) , ν (cid:105) (cid:38) k , so Proposition 2.6 forces Λ → ∞ if we make k → ∞ , which leads to a contradiction. (cid:3) Construction of the extremal sets
In this section, we will sometimes need to work with the projections of a rectangle R onto the axes. To that end, if R = I × J we denote π ( R ) := I, π ( R ) := J. For dyadic intervals I we let (cid:98) I denote the dyadic parent of I . We also let I ( j ) denotethe j -fold dyadic dilation of I , so that I ( j ) is the dyadic interval containing I with | I | = 2 j | I | .3.1. Construction of P and the proof of Theorem 2.1. The following obser-vation will be useful in the construction:
Lemma 3.1.
Let R = I × J , R = I × J be two dyadic rectangles such that I ∩ I = J ∩ J = ∅ . Then dist H ( p , p ) = dist H ( q , q ) , ∀ p , q ∈ R , p , q ∈ R . Proof.
Given any p ∈ R , p ∈ R , it suffices to show that any dyadic rectangle R = I × J containing both p and p needs to contain R and R . Since I ∩ I i (cid:54) = ∅ , i = 1 ,
2, and I ∩ I = ∅ , one has I ⊃ I i , i = 1 ,
2. The same holds for J : J ⊃ J i for i = 1 , (cid:3) PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 12
Proof of theorem 2.1.
We are going to show the following claim by induction: forevery non-negative integer m there exist 2 m +1 dyadic squares Q m , . . . , Q m m +1 in[0 , satisfying(1) (cid:96) ( Q mi ) = 2 − m − for all i ,(2) dist H ( Q mi , Q mj ) ≥ − m for all i (cid:54) = j .Assuming the claim, it is enough to take P = { p Q : Q ∈ { Q m , . . . , Q m m +1 }} , where p Q denotes the center of the cube Q —in fact, any point of Q would work—.Then, by Lemma 3.1 we immediately get (P.1) and (P.2) and we end the proof.We turn to the proof of the claim. The case m = 0 is easy: it suffices to take Q = [0 , / and Q = [1 / , . Assume now by induction that the theorem istrue for m −
1. Scale the family obtained in that step by 1 / Q ∈ D ([0 , ). Denote thecubes so constructed by P Q , . . . , P Q m − . By the dilation invariance of dist H we have (cid:96) ( P Qi ) = 2 − · − m − − = 2 − m . Therefore, it is enough to choose a first-generation child from each of these squares { P Qi } i,Q in such a way that (2) is satisfied. Write D ([0 , ) = { Q , Q , Q , Q } ,where the children are listed in the order of upper left, upper right, lower left, andlower right. We first consider Q and Q . For each P Qi with Q ∈ { Q , Q } we choosean arbitrary first-generation child and call it Q Qi . By induction we havedist H ( Q Qi , Q Qj ) ≥ dist H ( P Qi , P Qj ) ≥ − · − ( m − = 2 − m for all i (cid:54) = j, while the distance between any square in Q and any other square in Q must beexactly 1 according to Lemma 3.1.Now we must choose the children from the squares in the other diagonal. Thischoice is more delicate since the squares from the first diagonal could be much closer.By the pigeonhole principle, for each P = P Qi , Q ∈ { Q , Q } , there exist exactly twoother squares, P Q j and P Q k , which are at distance at most 2 − m . Indeed, accordingto Lemma 3.1, the distance condition forces P Q j , P Q k to be in the same row (orcolumn) as P Qi , while each row (or column) of Q (or Q ) contains exactly one squareby induction hypothesis. In fact, the distance between P Qi and P Q j is precisely equalto 0, and the same holds for P Q k .We now project Q Q j and Q Q k , the children chosen from P Q j and P Q k onto P .Note that their projection leaves exactly one first-generation child of P untouched,which we select as Q Qi . It suffices to show that the distance from Q Qi to any square in Q is larger than 2 − m , since the case of Q is symmetric and any other combination PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 13 can be dealt with in the same way as above. To see this, if the square in Q doesnot come from the P Q j chosen above, the distance has to be larger than 2 − m byLemma 3.1. Otherwise, the square is exactly Q Q j . By the construction above, bothits horizontal projection and its vertical projection and the respective ones of Q Qi aredisjoint, which allows one to apply Lemma 3.1 once more to conclude the proof. (cid:3) Construction of Z . In the construction, there are two aspects that will requirespecial care: each point in Z has to be close to exactly one point from P (but not tooclose), while two different points in Z cannot be too close to each other. We begin byconstructing certain special rectangles inside which we shall place the points forming Z . Definition.
We say that a rectangle R is a standard rectangle if it belongs to thecollection E = { R ⊆ [0 , : R dyadic, | R | = 2 − m − , and R ∩ P (cid:54) = ∅} . If R ∈ E we let p R denote the unique point in R ∩ P . Also let E p denote the collectionof standard rectangles containing p ∈ P . One can obtain E p by area-preserving dilations of one fixed R ∈ E p . Indeed, forany dyadic R = I × J and any j ∈ Z defineDil j R = I (cid:48) × J (cid:48) , where I (cid:48) × J (cid:48) is the unique dyadic rectangle in E p of dimensions 2 j | I | × − j | J | . Thenit is easy to see that for each standard rectangle R , E p = { Dil j R ⊆ [0 , } . Inparticular, for any p ∈ P there are approximately m distinct standard rectangles in E p . Given x, y ∈ R , let δ ( x, y ) = inf {| Q | : Q ∈ D is a dyadic interval containing x and y } . Also, for dyadic intervals K define δ ( x, K ) := inf y ∈ K δ ( x, y ) . Now suppose R = I × J ∈ E , with p R = p = ( x, y ). Given k ≥
1, we define I ( k ) to bethe largest dyadic interval I (cid:48) such that δ ( x, I (cid:48) ) = 2 − k +1 | I | , and define J ( k ) similarly. Note that I (1) is the child of I that does not contain x ,and for all k ≥ I ( k ) is the unique dyadic offspring of I of generation k satisfying x / ∈ I ( k ) and x ∈ (cid:99) I ( k ) (the dyadic parent of I ( k ) ).The following result is the main technical lemma needed for our construction. PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 14
Figure 2. log-log sketch of some standard rectangles. In blue are thesubrectangles I (1) × J ( k ) with k = 3. Lemma 3.2.
Fix an integer k ≥ with k (cid:28) m . For any dyadic rectangle R = I × J ∈ E p there exists a sub-rectangle R ∗ ⊂ I (1) × J ( k ) such that for all q (cid:54) = p in P : dist H ( q, R ∗ ) ≥ − m − . We will apply this lemma to construct the points Z used in the proof of the maintheorem as follows: given R = I × J ∈ E , we choose one point z R in R ∗ ⊂ I (1) × J ( k ) and let Z = (cid:83) R ∈E z R . Figure 2 gives some intuition about the situation. Note thatthis immediately guarantees properties (Z.1), (Z.2) and (Z.3) of Theorem 2.2. Thisplacing also allows us to prove the remaining two properties. We start by provingdirectly (Z.4): Lemma 3.3.
For every R ∈ E we have R ∩ Z ) (cid:46) k. Proof.
First observe that if z R (cid:48) ∈ R with R, R (cid:48) ∈ E then by Lemma 3.2 we must have p R (cid:48) = p R = p . So let Z R be defined by Z R = { T ∈ E p : z T ∈ R } . It suffices to show that Z R ) (cid:46) k .Observe that any T ∈ Z R must have π ( T ) ⊆ π ( R ). Indeed, if π ( T ) (cid:41) π ( R ),by definition z T ∈ π ( T ) (1) × π ( T ) ( k ) and one has π ( R ) ⊂ π ( T ) (1) . But this isimpossible since π ( T ) (1) ⊆ π ( T ) \ π ( R ). In addition, there must hold | π ( T ) | ≤ k − | π ( R ) | , as the smallest interval containing both π ( p T ) and π ( z T ) is (cid:92) π ( T ) ( k ) . The desiredestimate then follows from the fact that there are O ( k ) such standard rectangles. (cid:3) PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 15
Proof of lemma 3.2 and (Z.5).
In this subsection, given two dyadic rect-angles R , R that are not contained in one another but which do intersect at somenontrivial set, we say that R intersects R horizontally if π ( R ) (cid:40) π ( R ) , and similarly R intersects R vertically if π ( R ) (cid:40) π ( R ) . One can see that these two options are mutually exclusive for R and R . Also,one can see that if R intersects R horizontally then π ( R ) (cid:40) π ( R ), and if R intersects R vertically then π ( R ) (cid:40) π ( R ). We start by recording the followinginformation that will be used later: Lemma 3.4.
Let R = I × J ∈ E be such that I × J ( (cid:96) +1) ⊆ [0 , , (cid:96) ≥ . Then I ( (cid:96) ) × J ( (cid:96) +1) contains exactly one q ∈ P , q (cid:54) = p R , and furthermore q ∈ I ( (cid:96) ) × ( J ( (cid:96) +1) \ J ( (cid:96) ) ) . Proof.
We have that | (cid:99) I ( (cid:96) ) × J ( (cid:96) +1) | = 2 − (cid:96) +1 (cid:96) +1 | I || J | = 2 − m , so there are exactly twopoints from P in (cid:99) I ( (cid:96) ) × J ( (cid:96) +1) . One of these points must be p R , so the other one is q . Toprove the second part of the assertion, note that we cannot have q ∈ ( (cid:99) I ( (cid:96) ) \ I ( (cid:96) ) ) × J ( (cid:96) +1) or q ∈ I ( (cid:96) ) × J ( (cid:96) ) , since in each case we would have p R and q contained in a dyadicrectangle of area smaller or equal than 2 − m − , violating (P.1). (cid:3) Lemma 3.5.
Let R = I × J ∈ E . For any (cid:96) ≥ with (cid:96) +1 | J | ≤ , there existsexactly one T = W × H ∈ E intersecting R vertically such that p T (cid:54) = p R , W ⊂ I ( (cid:96) ) ,and | W | = 2 − | I ( (cid:96) ) | .Proof. By Lemma 3.4 the rectangle I ( (cid:96) ) × J ( (cid:96) +1) contains exactly one q ∈ P and q (cid:54) = p R . Moreover, we know that q ∈ I ( (cid:96) ) × ( J ( (cid:96) +1) \ J ( (cid:96) ) ). We define W as the child of I ( (cid:96) ) that contains π ( q ), and choose H so that π ( q ) ∈ H and T = W × H has area2 − m − . In particular, we have H = J ( (cid:96) +1) . By construction, T ∈ E .We now claim that T is the unique standard rectangle intersecting R verticallywith p T (cid:54) = p R , W ⊂ I ( (cid:96) ) , and | W | = | I ( (cid:96) ) | . By contradiction, suppose there existedanother T (cid:48) = W (cid:48) × H (cid:48) satisfying the same properties. Then since W, W (cid:48) ⊂ I ( (cid:96) ) with | W | = | W (cid:48) | we know that T and T (cid:48) are disjoint and therefore p T (cid:54) = p T (cid:48) . Since T (cid:48) is a standard rectangle and T, T (cid:48) both intersect R vertically, we also know that H (cid:48) = H = J ( (cid:96) +1) . But this contradicts Lemma 3.4: that there can only be one q ∈ P in I ( (cid:96) ) × J ( (cid:96) +1) with q (cid:54) = p R , so p T = p T (cid:48) , which is a contradiction. (cid:3) PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 16
Lemma 3.5 holds for rectangles intersecting R horizontally with exactly the sameproof, which we omit for brevity. Assume now that the rectangle R = I × J has beenfixed. Let E (cid:96),vR = (cid:8) T ∈ E : T intersects R vertically and π ( T ) ⊂ I ( (cid:96) ) (cid:9) . and E (cid:96),hR = (cid:8) T ∈ E : T intersects R horizontally and π ( T ) ⊂ J ( (cid:96) ) (cid:9) . We will analyze the vertical rectangles E (cid:96),vR ; essentially the same arguments apply tothe horizontal collection E (cid:96),hR .If S and T are in E (cid:96),vR , we say that S (cid:22) T if π ( S ) ⊂ π ( T ) (an analogous ordercan be defined in E (cid:96),vR ). Denote by E (cid:96),v, ∗ R = { T , T , . . . } the set of maximal elementswith respect to the ordering (cid:22) . Assume they are ordered so that the sequence a i := | π ( T i ) | is non-increasing. Lemma 3.6.
The sequence a i satisfies a i = 2 − i | I ( (cid:96) ) | , ≤ i ≤ m − (cid:96). Proof.
By Lemma 3.5 we find that T satisfies π ( T ) ⊂ I ( (cid:96) ) and a = 2 − | I ( (cid:96) ) | .Moreover, we know that there can be no other T j with a j = a , and so in particular a ≤ − | I ( (cid:96) ) | .We assume inductively that the desired result holds for T , . . . , T i − , and aim toshow that a i = 2 − i | I ( (cid:96) ) | . Note that { π ( T ) , . . . , π ( T i − ) } are pairwise disjoint. Thenby induction π ( T i ) ⊂ I ( (cid:96) ) \ i − (cid:71) j =1 π ( T j ) , and we also have that a j = 2 − a j − and a = 2 − | I ( (cid:96) ) | . Therefore, π ( T i ) must becontained in some interval K with | K | = 2 − i +1 | I ( (cid:96) ) | . Moreover, by induction K and π ( T i − ) must have the same dyadic parent.We first show that a i < | K | . To see this, suppose by contradiction that π ( T i ) = K .Then a i = a i − and (cid:92) π ( T i ) = (cid:92) π ( T i − ) . Since both T i − and T i must also have the same height, their union is a dyadicrectangle of area 2 − m − which contains both p T i − and p T i , which is impossible by(P.1).Finally, we have to show that a i ≥ − a i − or, equivalently, that there exists T ∈ E (cid:96),v, ∗ R with | π ( T ) | = 2 − a i − . To that end, let ˜ T i − ∈ E p Ti − be the uniquerectangle with π ( ˜ T i − ) = (cid:98) K . If 2 | π ( ˜ T i − ) | ≤
1, one can apply Lemma 3.5 with
PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 17 (cid:96) = 1 to ˜ T i − to conclude that there is exactly one T ∈ E intersecting ˜ T i − verticallywith p T (cid:54) = p T i − and | π ( T ) | = 2 − | (cid:98) K (1) | = 2 − i | I ( (cid:96) ) | .T i = T by maximality, and this closes the induction. If on the other hand 2 | π ( ˜ T i − ) | >
1, then we claim that i > m − (cid:96) and we have finished. Indeed, if T i exists in thiscase, one has | π ( T i ) | < | K | = 2 − | π ( ˜ T i − ) | . Therefore, | π ( T i ) | > | π ( ˜ T i − ) | ≥ | π ( ˜ T i − ) | > , which means it is impossible for T i to be contained in [0 , . (cid:3) Proof of lemma 3.2.
It is enough to show the following slightly stronger claim: forany pair of integers k, (cid:96) ≥
1, any R = I × J ∈ E contains a non-empty rectangle E × F such that E × F ⊆ I ( (cid:96) ) × J ( k ) \ (cid:0) (cid:91) R (cid:48) ∈E p R (cid:48) (cid:54) = p R R (cid:48) (cid:1) . Any R (cid:48) ∈ E which intersects I ( (cid:96) ) × J ( k ) must be in one of the two collections E (cid:96),vR , E k,hR , when p R (cid:54) = p R (cid:48) .We start by constructing E , so now we only need to take rectangles belonging to E (cid:96),v, ∗ R into account. From Lemma 3.6 we know that a i = | π ( T i ) | satisfies a i = 2 − i | I ( (cid:96) ) | .It thus follows that | π ( T i ) | must increase in size exponentially, and therefore thetotal number of T i ∈ E (cid:96),v, ∗ R is bounded by some constant that depends on R (since allrectangles are contained in [0 , ). Since there are only finitely many π ( T i ) ⊂ I ( (cid:96) ) ,the estimate on their sizes from Lemma 3.6 implies that there must be some dyadicsubinterval E ⊂ I ( (cid:96) ) with π ( T i ) ∩ E = ∅ for all i, as desired. The same procedure yields an interval F ⊂ J ( k ) which no rectangles T (cid:48) ∈ E k,h, ∗ R can intersect. (cid:3) Finally, we turn to the proof of (Z.5), which is the content of the next lemma.
Lemma 3.7.
For every dyadic rectangle R ⊂ [0 , with | R | ≥ − m − we have R ∩ Z ) (cid:46) m mk | R | . Proof.
First, we may assume that | R | = 2 − m − and show R ∩ Z ) (cid:46) mk. Indeed, if it is larger we can just write it as a disjoint union of dyadic rectangles ofarea 2 − m − and use the estimate for rectangles of that size. Since | R | = 2 − m − we PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 18 know that it contains exactly one point p ∈ P , and from Lemma 3.3 we know R contains about k points z ∈ Z such that p ( z ) = p For every z ∈ R ∩ Z such that p ( z ) (cid:54) = p , there exists T ∈ E for which z T = z ,which we denote by T z . Since both R and T z are dyadic rectangles and | T z | < | R | we must have Z ∩ R = V (cid:116) H (cid:116) O , where V consists of those points z ∈ Z ∩ R such that T z intersects R vertically, H consists of those points z ∈ Z ∩ R such that T z intersects R horizontally, and O is thecollection of points z ∈ Z whose associated point p ( z ) = p . We will only estimatethe size of V , since the argument for H is similar.By the construction of Z and the previous arguments, for each z ∈ V there exists T ∈ E (cid:96),v, ∗ R containing z . By Lemma 3.3 it suffices to show that there are no morethan some constant times m such rectangles.Let ˜ R = ˜ I × ˜ J be the unique rectangle in E satisfying p R ∈ ˜ R ⊂ R and π ( ˜ R ) = π ( R ) . Observe that any rectangle T ∈ E (cid:96),v, ∗ R intersecting R vertically and whose z T is inside R must intersect ( ˜ I ) (1) × ˜ J vertically (since othwerwise z T would be too close to p ).According to Lemma 3.5 with (cid:96) = 1, the maximal rectangle with shortest heightmust have height at least 4 | ˜ J | = | π ( R ) | and the heights of these maximal rectanglesincrease exponentially. Therefore the number of maximal rectangles going through( ˜ I ) (1) is at most log (2 m | π ( R ) | ) ≤ m + log ( | π ( R ) | ) ≤ m. (cid:3) The proof of Theorem 2.2 is complete.4.
Extensions and open questions
Failure of sparse bound for bi-parameter martingale transform.
In thissubsection, we extend our main theorem to the bi-parameter martingale transform,which is a 0-complexity dyadic shift that resembles the behavior of the bi-parameterHilbert transform. In general, given a sequence σ = { σ R } R ∈D×D satisfying | σ R | ≤ T σ ( f ) := (cid:88) R ∈D σ R (cid:104) f, h R (cid:105) h R . We have the following lower bound result.
PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 19
Theorem 4.1.
Let measures µ, ν be the same as above. Then there exists a martin-gale transform T σ such that |(cid:104) T σ ( µ ) , ν (cid:105)| (cid:38) k . Recalling the upper estimate of sparse forms obtained in Subsection 2.2, one im-mediately gets the following corollary:
Corollary 4.2.
There cannot be any (1 , sparse domination for all bi-parametermartingale transforms.Proof of Theorem 4.1. We first assume that σ R = 0 unless R ⊂ [0 , . In general,for any R = I × J ∈ E that gives rise to some point z ∈ Z , denote T R := (cid:99) I ( (cid:96) ) × (cid:100) J ( k ) ,which is the smallest rectangle that contains both z and p ( z ) ∈ P . Also, denote thestandard rectangle that gives rise to z ∈ Z by R z . Now for any fixed z ∈ Z , T σ ( µ )( z ) = 1 P (cid:88) p ∈P (cid:88) R ⊂ [0 , σ R h R ( p ) h R ( z ) . Set σ T = 0 unless T = T R for some R ∈ E . In particular, { T R } R ∈E and { R } R ∈E areone-to-one, and each T R contains exactly one p R ∈ P . Hence, T σ ( µ )( z ) ∼ − m (cid:88) R ∈E σ T R h T R ( p R ) h T R ( z ) . Note that z cannot be contained in T R unless R = R z according to Lemma 4.3below, therefore, T σ ( µ )( z ) ∼ − m σ T Rz h T Rz ( p R z ) h T Rz ( z ) = 2 − m σ T Rz | T R z | − ξ T Rz , where ξ T Rz = ±
1. Choosing σ T Rz = ξ T Rz , one has T σ ( µ )( z ) ∼ − m | T R z | − = 2 − m · (cid:0) − k +1 | R z | (cid:1) − ∼ k , and the desired estimate follows immediately. (cid:3) In the proof above, we have used the key observation that even though R ∈ E cancontain many points z ∈ Z with p R = p R z , T R can only contain one z R ∈ Z . This isjustified by the following result. Lemma 4.3.
For any integers k, (cid:96) ≥ and R = I × J ∈ E , let z R ∈ Z be the pointchosen in R as in Subsection 3.2, and T R = (cid:99) I ( (cid:96) ) × (cid:100) J ( k ) . Then z T / ∈ T R , ∀ R, T ∈ E , R (cid:54) = T. Proof.
Fix R = I × J ∈ E and denote z = z R . Given any other R (cid:48) = I (cid:48) × J (cid:48) ∈ E ,our goal is to show z R (cid:48) / ∈ T R . Obviously, if p R (cid:48) (cid:54) = p R , then by Lemma 3.2 z R (cid:48) is noteven contained in R . It thus suffices to assume p R (cid:48) = p R = p . Suppose | I | > | I (cid:48) | , i.e. PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 20 I = I (cid:48) ( j ) for some j ≥
1, then | J | = 2 − j | J (cid:48) | and J ⊂ J (cid:48) . Since both (cid:100) J ( k ) , (cid:100) J (cid:48) ( k ) contain π ( p ), one has (cid:100) J (cid:48) ( k ) (cid:41) (cid:100) J ( k ) .We claim that π ( z R (cid:48) ) / ∈ (cid:100) J ( k ) , hence z R (cid:48) / ∈ T R . Indeed, π ( z R (cid:48) ) ∈ J (cid:48) ( k ) , and (cid:100) J ( k ) is either contained in J (cid:48) ( k ) or its dyadic sibling. But it is impossible for (cid:100) J ( k ) to becontained in J (cid:48) ( k ) as it would imply π ( p ) ∈ J (cid:48) ( k ) , which is an obvious contradiction.Therefore, π ( z R (cid:48) ) / ∈ (cid:100) J ( k ) and the proof for the case | I | > | I (cid:48) | is complete.The case | I | < | I (cid:48) | can be treated symmetrically, as one has | J | < | J (cid:48) | and canshow in the same way as above that π ( z R (cid:48) ) / ∈ (cid:99) I ( (cid:96) ) . (cid:3) Remark 4.4.
Using the same method, one can easily show that there exists a bi-parameter dyadic shift of any given complexity which cannot have a (1 , sparsebound. We omit the details. Failure of sparse bound in higher dimensions.
For n ≥
1, define the n -parameter strong maximal function M n ( f )( x ) := sup R (cid:51) x | R | (cid:90) R | f ( y ) | dy, f : R n → C , where R is any n -dimensional dyadic rectangle. Clearly, M S equals M . We havethe following theorem. Theorem 4.5.
Let ( D n ) denote the following statement: there exists C > suchthat for all compactly supported integrable functions f, g on R n , there exists a sparsecollection of dyadic rectangles S such that |(cid:104)M n ( f ) , g (cid:105)| ≤ C (cid:88) R ∈S (cid:104)| f |(cid:105) R (cid:104)| f |(cid:105) R | R | . Then for all n ≥ , there holds ( D n ) = ⇒ ( D n − ) . Proof.
The desired result follows from a tensor product argument. For the sakeof simplicity, we will only prove ( D ) = ⇒ ( D ). Our goal is to show ( D ): forany given functions f, g on R , we would like to find a sparse dominating form for (cid:104)M ( f ) , g (cid:105) . Define (cid:101) f = f ⊗ χ [0 , and (cid:101) g = g ⊗ χ [0 , , both are compactly supportedintegrable functions on R . By the assumption ( D ), there exists a sparse collection (cid:101) S of rectangles in R such that |(cid:104) M ( (cid:101) f ) , (cid:101) g (cid:105)| ≤ C (cid:88) (cid:101) R = R × J ∈ (cid:101) S (cid:104)| (cid:101) f |(cid:105) (cid:101) R (cid:104)| (cid:101) g |(cid:105) (cid:101) R | (cid:101) R | , where R ∈ R , J ∈ R . PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 21
By definition, for all (cid:101) x = ( x, x ) ∈ R with x ∈ [0 , M ( (cid:101) f )( (cid:101) x ) = sup R × J (cid:51) (cid:101) x | R | · | J | (cid:90) R × J | (cid:101) f ( (cid:101) y ) | d (cid:101) y = M ( f )( x ) M ( [0 , )( x ) = M ( f )( x ) . Therefore, (cid:104) M ( (cid:101) f ) , (cid:101) g (cid:105) R = (cid:104) M ( f ) , g (cid:105) R . It thus suffices to show that the sparse form in R can be dominated by a sparseform in R .To see this, rewrite (cid:88) (cid:101) R = R × J ∈ (cid:101) S (cid:104)| (cid:101) f |(cid:105) (cid:101) R (cid:104)| (cid:101) g |(cid:105) (cid:101) R | (cid:101) R | = (cid:88) (cid:101) R = R × J ⊂ R α (cid:101) R (cid:104)| (cid:101) f |(cid:105) (cid:101) R (cid:104)| (cid:101) g |(cid:105) (cid:101) R | (cid:101) R | where α (cid:101) R = 1 if (cid:101) R ∈ (cid:101) R , and 0 otherwise. The sparsity of (cid:101) R thus means for someconstant Λ there holds (cid:88) (cid:101) R ⊂ (cid:101) Ω α (cid:101) R | (cid:101) R | ≤ Λ | (cid:101) Ω | , ∀ open set (cid:101) Ω ⊂ R . One can further simplify the expression (cid:88) (cid:101) R = R × J ⊂ R α (cid:101) R (cid:104)| (cid:101) f |(cid:105) (cid:101) R (cid:104)| (cid:101) g |(cid:105) (cid:101) R | (cid:101) R | = (cid:88) R ⊂ R (cid:104)| f |(cid:105) R (cid:104)| g |(cid:105) R | R | (cid:88) J ⊂ R α R × J | J | · (cid:18) | [0 , ∩ J || J | (cid:19) =: (cid:88) R ⊂ R (cid:104)| f |(cid:105) R (cid:104)| g |(cid:105) R | R | β R . It suffices to show that { β R } R ⊂ R is a Carleson sequence.Indeed, decomposing β R = ∞ (cid:88) j =0 − j (cid:88) J ⊂ R | J ∩ [0 , |∼ − j | J | α R × J | J | , one has for all open set Ω ⊂ R that (cid:88) R ⊂ Ω β R | R | = ∞ (cid:88) j =0 − j (cid:88) R ⊂ Ω (cid:88) J ⊂ R | J ∩ [0 , |∼ − j | J | α R × J | R × J |≤ ∞ (cid:88) j =0 − j (cid:88) R × J ⊂ (cid:101) Ω α R × J | R × J | , PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 22 where (cid:101)
Ω := Ω × { x ∈ R : M ( χ [0 , )( x ) > C − j } . By the weak L boundedness of M , one has | (cid:101) Ω | (cid:46) j | Ω | , hence (cid:88) R ⊂ Ω β R | R | (cid:46) Λ ∞ (cid:88) j =0 − j | Ω | (cid:46) Λ | Ω | . The proof is complete. (cid:3)
Similar results hold true for multi-parameter martingale transforms as well. Onecan define (cid:101) f = f ⊗ h [0 , and (cid:101) g = g ⊗ h [0 , . Then one observes that |(cid:104) T σ ( h [0 , ) , h [0 , (cid:105)| ≥ C for some martingale transform T σ in the third variable and therefore can arguesimilarly as above. The details are left to the reader.4.3. Other sparse forms.
As mentioned in the introduction, it is also interestingto consider the possibility of domination of the bi-parameter operators by biggersparse forms, for instance those of type ( φ, (cid:88) R ∈S (cid:104)| f |(cid:105) R,φ (cid:104)| g |(cid:105) R | R | , where (cid:104) f (cid:105) R,φ = inf (cid:40) λ > | R | (cid:90) R φ (cid:32) f ( x ) λ (cid:33) ≤ (cid:41) dx, and interesting norms include for example φ ( x ) = x [log( e + x )] α , α >
0, and φ ( x ) = x p , p >
1. We can actually extend our main theorem to the following:
Theorem 4.6.
Let φ ( x ) = x [log( e + x )] α , < α < . Then there cannot be any ( φ, -sparse domination for the strong maximal function or the martingale transform T σ .Proof. Fix k (cid:28) m . We need to first slightly modify the measure µ to get a functionthat lies in the correct normed space. Let ν be the same as before and define f = 1 P (cid:88) p ∈P Q p | Q p | , where for each p ∈ P , Q p (cid:51) p is the little cube (of side-length ∼ − m ) that isconstructed in the proof of Theorem 2.1. By the exact same argument in Section2, it is easy to check that the same lower bound for the bilinear forms associated to M S and the special martingale transform T σ still hold true, i.e. (cid:104) M S ( f ) , ν (cid:105) (cid:38) k , (cid:104) T σ ( f ) , ν (cid:105) (cid:38) k . PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 23
We shall now focus on showing the upper bound. For every Λ-Carleson collection S we will show that(4.1) (cid:88) R ∈S (cid:104) f (cid:105) R,φ (cid:104) ν (cid:105) R | R | (cid:46) Λ km α (cid:16) k m (cid:17) . If (4 .
1) holds true, then by taking m = 2 k and by noting that α < , one immediatelysees that Λ blows up to infinity as k → ∞ , which completes the proof.We now prove (4.1). Compared to the estimate of (cid:104) µ (cid:105) R in the proof of Theorem 2.6,it suffices to show that there is at most an extra factor of m α in the upper estimateof the new average form (cid:104) f (cid:105) R,φ . Specifically, in the case | R | = | R ∩ [0 , | ≥ − m − ,we would like to show that(4.2) (cid:104) f (cid:105) R,φ (cid:46) | R || R | · max (cid:18) m α , (cid:20) log (cid:0) | R || R | (cid:1)(cid:21) α (cid:19) . Note that if this estimate holds true, then one can proceed as in the proof of Theorem2.6 to conclude that contribution to (cid:80) R ∈S (cid:104) f (cid:105) R,φ (cid:104) ν (cid:105) R | R | from rectangles of this typeis controlled by Λ k (1 + m α ). Similarly, in the case | R | < − m − , we will show that(4.3) (cid:104) f (cid:105) R,φ (cid:46) m α m | R | , which, combined with the estimates for large rectangles, completes the proof of (4.1).To see why (4.2) holds, let λ > | R | (cid:90) R φ (cid:18) f ( x ) λ (cid:19) dx = 1 | R | (cid:88) p ∈P∩ R (cid:90) R ∩ Q p φ (cid:80) q ∈P Qq ( x ) | Q q | λ · P dx = 1 | R | · {P ∩ R } · | R ∩ Q p | · φ (cid:18) λ · P · | Q p | (cid:19) ≤ | R || R | · − m · φ (cid:18) m λ (cid:19) . Take λ = | R || R | · max (cid:16) m α , (cid:104) log (cid:0) | R || R | (cid:1)(cid:105) α (cid:17) , it suffices to show that the expression abovewith this λ value is (cid:46)
1. Indeed, note that e (cid:28) m λ , so | R || R | · − m · φ (cid:18) m λ (cid:19) (cid:46) (cid:104) m + log (cid:0) | R || R | (cid:1) − α log (cid:0) max (cid:0) m, log( | R | / | R | ) (cid:1)(cid:1)(cid:105) α max (cid:16) m α , (cid:104) log (cid:0) | R || R | (cid:1)(cid:105) α (cid:17) (cid:46) . PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 24
To see (4.3), again let λ > | R | (cid:90) R φ (cid:18) f ( x ) λ (cid:19) dx ≤ − m φ (cid:18) m λ (cid:19) = 1 λ (cid:20) log (cid:0) e + 2 m λ (cid:1)(cid:21) α . Take λ = m α m | R | , then it suffices to show that the expression above with this λ valueis (cid:46)
1. Indeed, the lower bound | R | (cid:38) − m − k guarantees that m λ (cid:29) e , hence1 λ (cid:20) log (cid:0) e + 2 m λ (cid:1)(cid:21) α (cid:46) m | R | m α (cid:2) m + log | R | − α log m (cid:3) α (cid:46) m | R | (cid:46) . (cid:3) The argument in the theorem above is not strong enough to produce a contra-diction when α ≥ . In other words, it could be possible that the bi-parameteroperators that are considered have a ( φ, α ≥ .Moreover, it is easy to check that a very similar argument as in Theorem 4.6 barelyfails for concluding a contradiction when φ ( x ) = x p , p >
1. We think that positiveor negative results for those types of bounds would be very interesting.
Problem.
Determine whether ( φ, -type sparse bounds for M S are possible for φ ( x ) = x p , p > or φ ( x ) = x [log( e + x )] α ( α ≥ ). References [1] D. Beltran and L. Cladek. Sparse bounds for pseudodifferential operators.
ArXiv e-prints , Nov.2017.[2] C. Benea, F. Bernicot, and T. Luque. Sparse bilinear forms for Bochner Riesz multipliers andapplications.
Trans. London Math. Soc. , 4(1):110–128, 2017.[3] L. Carleson. A counterexample for measures bounded on h p for the bi-disc. Number 7 inMittag-Leffler Report. 1974.[4] M. Christ and R. Fefferman. A note on weighted norm inequalities for the Hardy-Littlewoodmaximal operator. Proc. Amer. Math. Soc. , 87(3):447–448, 1983.[5] J. M. Conde-Alonso, A. Culiuc, F. Di Plinio, and Y. Ou. A sparse domination principle forrough singular integrals.
Anal. PDE , 10(5):1255–1284, 2017.[6] J. M. Conde-Alonso and G. Rey. A pointwise estimate for positive dyadic shifts and someapplications.
Math. Ann. , 365(3-4):1111–1135, 2016.[7] A. C´ordoba. Maximal functions: a problem of A. Zygmund. In
Euclidean harmonic analysis(Proc. Sem., Univ. Maryland, College Park, Md., 1979) , volume 779 of
Lecture Notes in Math. ,pages 154–161. Springer, Berlin, 1980.[8] A. Culiuc, F. Di Plinio, and Y. Ou. Domination of multilinear singular integrals by positivesparse forms.
Journal of the London Mathematical Society , 0(0).[9] A. Culiuc, F. Di Plinio, and Y. Ou. Uniform sparse domination of singular integrals via dyadicshifts.
Math. Res. Lett. , 2(1), 2018.[10] J. Dick and F. Pillichshammer.
Digital nets and sequences . Cambridge University Press, Cam-bridge, 2010. Discrepancy theory and quasi-Monte Carlo integration.
PARSE DOMINATION AND THE STRONG MAXIMAL FUNCTION 25 [11] L. E. Dor. On projections in L . Ann. of Math. (2) , 102(3):463–474, 1975.[12] R. Fefferman. Harmonic analysis on product spaces.
Ann. of Math. (2) , 126(1):109–130, 1987.[13] R. Fefferman and J. Pipher. A covering lemma for rectangles in R n . Proc. Amer. Math. Soc. ,133(11):3235–3241, 2005.[14] T. S. Hanninen. Equivalence of sparse and Carleson coefficients for general sets.
ArXiv e-prints ,Sept. 2017.[15] T. P. Hyt¨onen. The sharp weighted bound for general Calder´on-Zygmund operators.
Ann. ofMath. (2) , 175(3):1473–1506, 2012.[16] B. Krause and M. T. Lacey. Sparse Bounds for Maximally Truncated Oscillatory SingularIntegrals.
ArXiv e-prints , Jan. 2017.[17] M. T. Lacey. An elementary proof of the A bound. Israel J. Math. , 217(1):181–195, 2017.[18] M. T. Lacey. Sparse Bounds for Spherical Maximal Functions.
ArXiv e-prints , Feb. 2017.[19] M. T. Lacey and D. Mena Arias. The sparse T1 theorem.
Houston J. Math. , 43(1):111–127,2017.[20] A. K. Lerner. A pointwise estimate for the local sharp maximal function with applications tosingular integrals.
Bull. Lond. Math. Soc. , 42(5):843–856, 2010.[21] A. K. Lerner. On an estimate of Calder´on-Zygmund operators by dyadic positive operators.
J.Anal. Math. , 121:141–161, 2013.[22] A. K. Lerner. A simple proof of the A conjecture. Int. Math. Res. Not. IMRN , (14):3159–3170,2013.[23] E. T. Sawyer. A characterization of a two-weight norm inequality for maximal operators.
StudiaMath. , 75(1):1–11, 1982.[24] F. Soria. Examples and counterexamples to a conjecture in the theory of differentiation ofintegrals.
Ann. of Math. (2) , 123(1):1–9, 1986.[25] A. Stokolos. Properties of the maximal operators associated with bases of rectangles in R . Proc. Edinb. Math. Soc. (2) , 51(2):489–494, 2008.
Alexander Barron, Department of Mathematics, Brown University, 151 KassarHouse, Providence, RI, USA.
E-mail address : [email protected] Jos´e M. Conde-Alonso, Department of Mathematics, Brown University, 151 Kas-sar House, Providence, RI, USA.
E-mail address : [email protected] Yumeng Ou, Department of Mathematics, Box B6-230, One Bernard Baruch Way,New York, NY, USA
E-mail address : [email protected] Guillermo Rey, 127 Vincent Hall, 206 Church St. SE Minneapolis, MN, USA
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