Optimal bounds on the modulus of continuity of the uncentered Hardy-Littlewood maximal function
aa r X i v : . [ m a t h . C A ] S e p OPTIMAL BOUNDS ON THE MODULUS OF CONTINUITY OF THEUNCENTERED HARDY-LITTLEWOOD MAXIMAL FUNCTION
J. M. ALDAZ, L. COLZANI AND J. P´EREZ L ´AZARO
Abstract.
We obtain sharp bounds for the modulus of continuity of the uncentered maxi-mal function in terms of the modulus of continuity of the given function, via integral formulas.Some of the results deduced from these formulas are the following: The best constants for Lip-schitz and H¨older functions on proper subintervals of R are Lip α ( M f ) ≤ (1 + α ) − Lip α ( f ), α ∈ (0 , R , the best bound for Lipschitz functions is Lip( M f ) ≤ ( √ −
1) Lip( f ) . Inhigher dimensions, we determine the asymptotic behavior, as d → ∞ , of the norm of themaximal operator associated to cross-polytopes, euclidean balls and cubes, that is, ℓ p ballsfor p = 1 , , ∞ . We do this for arbitrary moduli of continuity. In the specific case of Lipschitzand H¨older functions, the operator norm of the maximal operator is uniformly bounded by2 − α/q , where q is the conjugate exponent of p = 1 ,
2, and as d → ∞ the norms approach thisbound. When p = ∞ , best constants are the same as when p = 1. Introduction.
The constants appearing in the weak and strong type inequalities satisfied by the Hardy-Littlewood maximal operator, in its different variants, have been subject to considerablescrutiny. We mention, for instance, [CF], [Al1], [Me1], [Me2], [GMM], [GM], [GK], [BD],[CLM], [St1], [St2], [St3], [Bou1], [Bou2], [Bou3], [Ca], [Mu], [StSt], [Al2], [Al3], [AlPe4],[AlPe5], [NaTa], and the references contained therein. Interest lies not only in determiningsharp inequalities, which in general are hard to come by (in fact, no best constants are knownfor dimensions larger than one) but also in finding out how constants change as certainparameters (for instance, the dimension) vary, or when the type of set one is averaging overis modified, or the space of functions one is considering is changed.Here we study the issue of optimal inequalities satisfied by the uncentered maximal operator M , and also its asymptotic behavior, but from a different viewpoint: Instead of consideringweak and strong type inequalities, we analyze the properties of M in connection with themodulus of continuity of a function. The overall emerging pattern reveals that the uncenteredmaximal operator improves regularity, by preserving moduli and reducing constants. But ingeneral there is no “qualitative” improvement in the modulus. For instance, it may happen Mathematical Subject Classification. that there is no change in the H¨older exponent of a function, not even in a weak, almosteverywhere sense. More precisely, we will see that there are H¨older functions f with exponent α ∈ (0 , α on a set of positive measure, such that M f is also no betterthan H¨older ( α ) on a set of the same measure, cf. Example 4.8. We note that the preservationof regularity does not extend to C functions, see Remark 4.7.This article is part of a wider project, which attempts to find out under which condi-tions and to what extent the Hardy-Littlewood maximal operator improves the regularity offunctions, in different settings. If there is such an improvement, then one can try to provevariants of inequalities involving derivatives (for example, Gagliardo-Niremberg-Sobolev typeinequalities) with DM f replacing Df . In applications one often needs to consider functionsmore general than those belonging to Sobolev spaces, so it is natural to look for this kind ofinequalities under as little regularity as possible. It is also possible to consider other maximaloperators, associated to smoother approximations of the identity, but we do not pursue thisline of research here.The study of the Hardy-Littlewood maximal operator acting on spaces that measuresmoothness was initiated in [Ki] by J. Kinnunen , who proved its boundedness on W ,p ( R d )for 1 < p ≤ ∞ , and also on the Lipschitz and H¨older classes (without increasing the corre-sponding constants); see also [KiLi], [HaOn], [KiSa], [Ta], [Lu]. In [AlPe], part of the projectoutlined in the previous paragraph is carried out for functions of bounded variation and d = 1 (the situation when d > f isof bounded variation, then the uncentered Hardy-Littlewood maximal function
M f is abso-lutely continuous (however, the centered maximal function need not even be continuous), andfurthermore, the variation is not increased by M . As application, a Landau type inequalityunder less regularity is presented in Theorem 5.1 of [AlPe].Given ( R d , k · k ), where k · k is an arbitrary norm in R d , and f : R d → R , we presentoptimal integral formulas for the modulus of continuity of M f in terms of the modulus of f . Note that distances appearing in the moduli of continuity and balls defining the maximalfunction, are determined according to k · k . However, we always use the same underlying d dimensional Lebesgue measure, regardless of the norm under consideration, i. e., there areno different normalizations of the measure for different norms. We take the viewpoint thatthere is (essentially) only one norm in dimension one, the usual absolute value, and this fixesLebesgue measure in every dimension by the requirement that a cube of sidelength one hasmeasure one. This also forces us to utilize the standard definition (via the euclidean length)and the standard normalizations, when dealing with Hausdorff measures. Such a convention(or some analogous consistency condition) is needed, for instance, to use Fubini’s Theorem,and more generally, the coarea formula.A very brief exposition of the contents of this paper follows next. The main integralformulas of the paper, valid for an arbitrary norm in the global case, where the domainunder consideration is the whole space R d , appear in Theorem 2.7 and Corollary 2.14. Theseformulas are then specialized to the following three norms and their associated maximalfunctions: The ℓ ∞ norm (cf. Theorem 3.1), the ℓ norm (cf. Theorem 3.14), and the ℓ odulus of continuity of the maximal function 3 norm (cf. Theorem 3.16). Their associated maximal functions are respectively defined byaveraging over cubes, euclidean balls, and cross-polytopes. A few consequences of theseTheorems are the following: The norm of the maximal operator acting on Lipschitz functionsis k M k Op(1) = √ − d = 1, cf. Corollary 3.6. For the ℓ ∞ norm the best constantsin dimensions 2 and 3 are approximately 0 .
574 and 0 . d , k M k Op(1) > ( d − / ( d + 1), and the error inthis estimate is o (1 / ( d + 1)), cf. Corollary 3.4. For the ℓ norm all constants are exactlythe same as for the ℓ ∞ norm, see Theorem 3.17. By way of contrast, we note that in theeuclidean case and on the class of Lipschitz functions, k M k Op(1) ≤ − / in every dimension,and this bound is optimal. With respect to the preceding results, some open questions arementioned; we indicate one now: Since k M f k ∞ = k f k ∞ and (under the euclidean norm) k DM f k ∞ ≤ − / k Df k ∞ , the maximal operator is a contraction on W , ∞ ( R d ), the Sobolevspace of essentially bounded functions with essentially bounded derivatives (note that thecontraction is not strict, consider for instance the constant functions). Is there some analogousresult for p < ∞ sufficiently large?The last section of this paper deals with the maximal operator on proper subintervals of R . In this case, the operator norm of M when acting on the H¨older and Lipschitz classes is k M k Op( α ) = (1 + α ) − (cf. Corollary 4.5). The local, higher dimensional case is not studiedhere; we only point out that constants when d > M c associated to euclidean ballssatisfies k M c k Op( α ) = 1 in all dimensions, so Lipschitz and H¨older constants are in general notreduced (cf. Remark 2.4). Thus, the preceding results, together with Theorem 2.5 of [AlPe]mentioned above, suggest that from the viewpoint of regularity the uncentered maximaloperator is a more natural object of study than the centered one.2. Definitions and global results for arbitrary norms.
Definition 2.1.
Let U ⊂ R d be an open set, let k · k be a norm on R d , and let B be ageneric ball with respect to this norm. Given a locally integrable function f : U → R , thenoncentered Hardy-Littlewood maximal function M f is defined by
M f ( x ) := sup x ∈ B ⊂ U | B | Z B | f ( y ) | dy. Here | B | stands for the Lebesgue measure of B . Regarding the centered maximal function M c f ( x ), one requires that balls be centered at x rather than just containing it, but everythingelse is as in the uncentered case. Definition 2.2.
The modulus of continuity of a function f is ω ( f, δ ) := sup {| f ( x ) − f ( y ) | : k x − y k ≤ δ } . We point out that both definitions depend on the norm under consideration (while the un-derlying Lebesgue measure does not). Although maximal functions and moduli of continuity
J. M. Aldaz, L. Colzani, J. P´erez L´azaro obtained via different norms will always be pointwise comparable, from the viewpoint of bestconstants distinctions cannot be neglected. Indeed, when the euclidean ( ℓ ) and the max( ℓ ∞ ) norms are used in R d , d > ℓ ∞ norm and arbitrary moduli of continuity are exactly the same as thoseobtained under the ℓ norm. Let us emphasize, though, that our results say nothing, forinstance, about the maximal function associated to cubes when the length used is the eu-clidean distance. For us, fixing a norm fixes the balls we average over; once we select, say, theeuclidean norm, the maximal function considered in our theorems will be the one associatedto euclidean balls.The following theorem is due to Juha Kinnunen, cf. [Ki]. While not explicitly stated there,its proof appears in [Ki], pp. 120-121, Remark 2.2 (iii). A small variant of the argumentyields the boundedness in W ,p ( R ), 1 < p ≤ ∞ , of the maximal operator (Remark 2.2 (i) of[Ki]); an abstract version can be found in Theorem 1 of [HaOn], where it is applied to thespherical maximal operator.In the next theorem M denotes a generic Hardy-Littlewood maximal operator (it could becentered or noncentered, associated to euclidean or to other balls). Theorem 2.3. (J. Kinnunen) . Let f : R d → R be locally integrable. Then, for every δ > , (2.3.1) ω ( M f, δ ) ≤ ω ( | f | , δ ) . Proof.
Let h, x ∈ R d and set f h ( x ) := f ( x + h ). By commutativity of M with translations, M f ( x + h ) = M f h ( x ), and by subadditivity, |M f h − M f | ≤ M ( | f h | − | f | ). Since averagesnever exceed a supremum, we have sup x ∈ R d M ( | f h | − | f | )( x ) ≤ sup x ∈ R d ( || f h | − | f || )( x ), andnow (2.3.1) follows by taking the sup over k h k ≤ δ . (cid:3) Remark 2.4.
This simple result already contains the sharp bound for the centered maximaloperator M c on the Lipschitz and H¨older classes. Let ψ ( x ) = max { − | x | , } be definedon R ; clearly, M c ψ ( x ) = ψ ( x ) for every x ∈ [ − / , / M c ψ ) = Lip( ψ ) = 1. Now, this example can be easily adaptedto higher dimensions (for instance, by making the corresponding function depend only onthe first coordinate), and it follows that the centered maximal operator on R d does not ingeneral reduce the Lipschitz constant of a function. By way of contrast, we mention thatthe uncentered maximal operator on R satisfies Lip( M f ) ≤ ( √ −
1) Lip( f ), cf. Corollary3.6 below, while on R d with the Euclidean norm, the bound Lip( M f ) ≤ − / Lip( f ) holdsuniformly in the dimension (Theorem 3.14).As for H¨older functions, the preceding example can be easily modified to yield the sameconclusion. In one dimension, set η ( x ) := max { − | x | α , } when x ≤
0, set η ( x ) := 1 + | x | α on [0 , / η to [1 / , ∞ ) by reflection about the x = 1 / η on 0 < x < / M c η ( x ) = η ( x ) for every x ∈ [0 , / odulus of continuity of the maximal function 5 Our aim in this paper is to find best inequalities in the spirit of (2.3.1) for the uncenteredmaximal operator, which, as noted in the introduction, has better properties regarding theregularization of functions than its centered relative.
Definition 2.5.
A function ω : [0 , ∞ ) → [0 , ∞ ) is a modulus of continuity if it is the mod-ulus of continuity of a uniformly continuous function, i.e., if there is a uniformly continuousfunction f such that for all δ ≥ ω ( δ ) = ω ( f, δ ). Remark 2.6.
Often a modulus of continuity ω : [0 , ∞ ) → [0 , ∞ ) is defined as a continuous,nondecreasing, subadditive function (so ω ( a + b ) ≤ ω ( a ) + ω ( b )), vanishing at zero. Note thatfrom continuity and subadditivity it follows that ω is uniformly continuous. It is well knownand not difficult to check that the modulus of continuity of a uniformly continuous function f has these properties, while given any ω satisfying the above conditions, there is a uniformlycontinuous f such that ω ( · ) = ω ( f, · ), namely ω itself. We do not assume that moduli ofcontinuity are bounded, so our results apply to general Lipschitz and H¨older functions. Toavoid trivialities, given a generic modulus of continuity, we shall assume it is not identicallyzero.The next theorem, and its corollary 2.14, contain the main integral formulas of the paper,valid (in the global case) for all norms and all dimensions. Note that on R d the value of M f ( x ) is the same regardless of whether the balls we average over are taken to be open orclosed. We shall assume whatever is more convenient at any given point. For instance, inthe next theorem and its proof we suppose that balls are closed. Additionally, we use thefollowing notation: a n ↑ b (resp. a n ↓ b ) means that the sequence { a n } converges to b in amonotone increasing (resp. decreasing) fashion. Theorem 2.7.
Let d ≥ and let f be a locally integrable function on R d . Then, for everynorm k · k on R d and every t ≥ , (2.7.1) ω ( M f, t ) ≤ sup { v ∈ R d : k v k =1 } inf { c ∈ R d ,R> k v − c k≤ R } | B (0 , | Z B (0 , ω ( | f | , t k c + Ru k ) du. The preceding inequality is optimal in the sense that given a modulus of continuity ω , thereexists a function ψ such that letting f = ψ in (2.7.1), the following equalities hold: For all t > , we have ω ( ψ, t ) = ω ( t ) , and furthermore, (2.7.2) ω ( M ψ, t ) = sup { v ∈ R d : k v k =1 } inf { c ∈ R d ,R> k v − c k≤ R } | B (0 , | Z B (0 , ω ( t k c + Ru k ) du. If v is a unit vector, to find its associated infimum in (2.7.2) it is enough to consider the set (2.7.3) (cid:8) c ∈ R d , R > k v − c k = R and R ≤ (cid:9) . As for the extremal functions ψ appearing in (2.7.2), they have the following form: If ω is bounded, we can take ψ ( x ) = k ω k L ∞ ([0 , ∞ )) − ω ( k x k ) , while if ω is unbounded, we set ψ ( x ) = P ∞ n =1 (Ω n − w ( k x − a n k )) + , where { Ω n } ∞ ↑ ∞ and { a n } ∞ diverges to infinity so fast, J. M. Aldaz, L. Colzani, J. P´erez L´azaro that on the support of each spike the maximal function does not depend on any of the otherspikes.
Observe that (2.7.1) is stronger, for the uncentered maximal operator, than Kinnunen’sinequality ω ( M f, t ) ≤ ω ( | f | , t ), as can be seen by letting R → c → v ). Notealso that the right hand side of (2.7.2) depends only on the modulus of continuity and on thenorm. Regarding the extremal functions, our strategy to find them is easy to explain (and theproof of the theorem shows that it works): Suppose ω is a bounded modulus of continuity,and suppose that a given function g has a global maximum at x . Then M g ( x ) = g ( x ),and in order to maximize M g ( x ) − M g ( y ) for each y = x , we want to minimize M g ( y ).Thus g should have the fastest possible decay allowed by ω in every direction (and hence theleast possible mass). This is precisely what ψ ( x ) := k ω k L ∞ ([0 , ∞ )) − ω ( k x k ) does. A similarobservation (in a local sense) can be made for unbounded moduli. Finally, we mention thatwhile it is natural to suspect that the centers c in (2.7.3) associated to v should be chosen sothat B ( c, R ) ⊂ B (0 , ℓ , ℓ and ℓ ∞ norms. Proof.
Given f ≥
0, we assume that the associated modulus of continuity ω ( f, t ) is finite forevery t >
0, for otherwise there is nothing to prove. Hence, we take f to be locally bounded.We may also assume that M f is not constant, and in particular, that it is not identically ∞ .Next, choose x, y ∈ R and suppose that M f ( y ) < M f ( x ). Set E := { b ∈ R d , T > k ( y − x ) − b k ≤ T } and F := { a ∈ R d , S > k x − a k ≤ S } . Then, for all ( a, S ) ∈ F we have E ⊂ { b ∈ R d , T > y ∈ B ( a + b, S + T ) } by the triangleinequality. Hence M f ( x ) − M f ( y ) = sup F | B ( a, S ) | Z B ( a,S ) f ( u ) du − M f ( y ) ≤ sup F inf E (cid:18) | B ( a, S ) | Z B ( a,S ) f ( u ) du − | B ( a + b, S + T ) | Z B ( a + b,S + T ) f ( u ) du (cid:19) ≤ sup F inf E (cid:18) | B (0 , | Z B (0 , | f ( a + Su ) − f ( a + b + ( S + T ) u ) | du (cid:19) ≤ sup F inf E | B (0 , | Z B (0 , ω ( f, k b + T u k ) du = inf E | B (0 , | Z B (0 , ω ( f, k b + T u k ) du, where the sup F has been deleted from the last line since neither a nor S appear in the integral.By symmetry of B (0 , | M f ( x ) − M f ( y ) | ≤ inf E | B (0 , | Z B (0 , ω ( f, k b + T u k ) du odulus of continuity of the maximal function 7 also holds when M f ( y ) > M f ( x ), and thus it always holds. Writing in (2.7.4) c k y − x k = b , R k y − x k = T , and y − x = k y − x k v , where k v k = 1, we have | M f ( x ) − M f ( y ) | ≤ inf { c ∈ R d ,R> k v − c k≤ R } | B (0 , | Z B (0 , ω ( f, k y − x kk c + Ru k ) du. Since the right hand side of the last inequality is increasing in k y − x k , it follows that forevery t > ω ( M f, t ) = sup { x,y ∈ R d : k x − y k≤ t } | M f ( x ) − M f ( y ) | (2.7.5) ≤ sup { v ∈ R d : k v k =1 } inf { c ∈ R d ,R> k v − c k≤ R } | B (0 , | Z B (0 , ω ( f, t k c + Ru k ) du. Next we prove that this inequality is sharp. Suppose first that ω is a bounded modulus ofcontinuity. Set Ω := k ω k L ∞ ([0 , ∞ )) , and write ψ ( x ) := Ω − ω ( k x k ). Fix t >
0. If x satisfies0 < k x k ≤ t , then we can express x = tv , where 0 < k v k ≤
1, and we get ω ( M ψ, t ) ≥ M ψ (0) − M ψ ( x ) = inf { c ∈ R d ,R> k v − c k≤ R } | B (0 , | Z B (0 , ω ( t k c + Ru k ) du. Now the result follows by taking the supremum over all v such that 0 < k v k ≤
1. If ω isunbounded, we modify ψ it as follows: Take a sequence of suitably chosen “spike” functions(Ω n − ω ( k x − a n k )) + , and then set ψ ( x ) = P ∞ n =1 (Ω n − w ( k x − a n k )) + . “Suitably chosen” inthe preceding sentence means that the heights Ω n tend to infinity and the different spikes areplaced so far apart (i.e., a n → ∞ so fast) that on the support of each spike, the others neednot be taken into account when computing M ψ ( x ) (in particular, different spikes will havedisjoint supports, so there are no convergence issues with the series defining ψ ).In order to apply formula (2.7.5), it is useful to narrow down as much as possible wherethe infimum occurs. First we show that it is enough to consider R ≤
1. Fix a unit vector v and suppose R >
1. Since k v − k = 1 ≤ R , the origin is an admissible center c associatedto v . But then averaging ω ( f, t k · k ) over B (0 ,
1) yields a value no larger than averaging overany other ball B ( c, R ) containing v , for every vector in B ( c, R ) \ B (0 ,
1) has norm largerthan one, hence larger than the norm of any vector in B (0 , \ B ( c, R ), and additionally | B ( c, R ) \ B (0 , | > | B (0 , \ B ( c, R ) | .Next we prove that it is enough to consider pairs ( c, R ) for which k v − c k = R . Suppose k v − c k < R . Since the continuous function r rR −k v − rc k changes sign on [0 , r ∈ [0 ,
1] such that k v − r c k = r R and, for such an r , we have ω ( f, t k r c + r Ru k ) ≤ ω ( f, t k c + Ru k ). Hence, we do no worse by using the pair ( r c, r R ) instead of ( c, R ), and(2.7.3) follows. (cid:3) Remark 2.8.
We mention that to obtain (2.7.3), instead of the selfcontained argument givenabove, we could have applied Anderson’s Theorem (cf., for instance, Theorem 1.11, pg. 376of [Ga], or Theorem 1 of [An]), which for nonnegative, integrable, symmetric and unimodal
J. M. Aldaz, L. Colzani, J. P´erez L´azaro functions f , and origin symmetric convex bodies K ⊂ R d , tells us that Z K f ( x + cy ) dx ≥ Z K f ( x + y ) dx, where 0 ≤ c ≤ y ∈ R d . Since we shall use Anderson’s Theorem later on, it is statedhere for easy reference. Definition 2.9.
Given a modulus of continuity ω , define the Lipschitz space Lip( ω ) =Lip( ω, X ) via the seminorm k f k Lip( ω,X ) := sup { x,y ∈ X : x = y } | f ( x ) − f ( y ) | ω ( k x − y k ) = sup t> ω ( f, t ) ω ( t ) . Then f ∈ Lip( ω, X ) if (and only if) k f k Lip( ω,X ) < ∞ . In this section and the next we willalways have X = R d , so reference to X shall usually be omitted. Next, set(2.9.1) k M k Op( ω ) := sup k f k Lip( ω ) =0 k M f k Lip( ω ) k f k Lip( ω ) = sup k f k Lip( ω ) =1 k M f k Lip( ω ) . When ω ( t ) = t α and 0 < α ≤
1, we use Lip α ( X ) (or just Lip( α )) to denote the corre-sponding spaces of H¨older continuous functions and of Lipschitz functions on X , Lip α ( f )to denote k f k Lip( ω ) , and k M k Op( α ) to denote k M k Op( ω ) . If α = 1, we often omit it, simplywriting Lip( X ) and Lip( f ).While the notation does not make it explicit, since we are considering functions defined on( R d , k · k ), k M k Op( ω ) depends both on d and on k · k . But dimension and norm will always beclear from context. Remark 2.10.
Kinnunen’s Theorem (2.3) shows that kMk
Op( ω ) ≤ M , since for every t > f with k f k Lip( ω,X ) ≤
1, we have ω ( M f, t ) ≤ ω ( | f | , t ) ≤ ω ( f, t ) ≤ ω ( t ). Example 2.11.
Note that for f ∈ Lip( ω ), its norm k f k Lip( ω ) depends not only on the spaceLip( ω ), but also on the modulus ω used to define it (we write norm for short, even thoughwe mean seminorm). Consider, for instance, the functions on R that are both bounded andLipschitz. As a set, this space can be defined via many moduli, for example ω ( t ) := min { t, } and ω ′ ( t ) := min { t, / } . Then ψ ( x ) = (1 − | x | ) + = max { , − | x |} has norm 1 in Lip( ω ) andnorm 2 in Lip( ω ′ ). We also mention that since lim t →∞ ω ( M ψ, t ) = 1, we have k M k Op( ω ) = 1,so the upper bound 1 can be attained by the uncentered maximal operator. But we shall seethat for many standard moduli strict inequality holds. Remark 2.12.
Identifying some extremal functions, as we do in Theorem 2.7, allows us toimmediately improve the general bound 1 in the uncentered case, on Lipschitz functions, forevery dimension d , and all norms. Corollary 2.13.
Fix d ≥ . Given any norm k · k on R d , the associated maximal operatoracting on Lipschitz functions satisfies k M k Op(1) ≤ d/ ( d + 1) . odulus of continuity of the maximal function 9 Proof.
Let ψ ( x ) = max { − k x k , } (since we are in the Lipschitz case, it is immaterial whichheight Ω we select in Theorem 2.7, so we just pick Ω = 1). Given a vector v with k v k = 1,we estimate M ψ ( v ) by integrating over the support of ψ , that is, over the unit ball accordingto k · k . This gives a lower bound for M ψ on the unit sphere, and hence an upper bound for k M k Op(1) . Since for d ≥ / ( d + 1), we have k M k Op(1) ≤ − / ( d + 1) . (cid:3) Of course, the very general but otherwise rather crude estimates given by the precedingcorollary cannot be expected to be sharp. In dimension one there is essentially one norm,and the best constant is √ −
1, as will be seen below, rather than 1 /
2. We shall show thatin dimensions two and three the constants 2 / / ℓ and ℓ ∞ norms.Let f be a smooth, compactly supported function on R d , and recall that Lip( M f ) = k DM f k ∞ . Since by the previous corollary k DM f k ∞ ≤ d/ ( d + 1) k Df k ∞ , it is natural tosuspect that if p d is high enough, for every p ≥ p d one can find a c p ∈ (0 ,
1) such that k DM f k p ≤ c p k Df k p (with c p independent of f , see also question 2 below). But in this paperwe study the size of k DM f k p only when p = ∞ .Note also that asymptotically the above corollary does not improve the general upperbound 1. Thus, it is natural to enquire whether bounds strictly less than 1 and independentof the dimension can be obtained, by a more careful choice of averaging ball. It turns outthat for cubes (with sides parallel to the axes, that is, ℓ ∞ balls) and cross-polytopes ( ℓ balls)the constant 1 is the correct asymptotic value, and the trivial upper bound d/ ( d + 1) fromthe previous result is “essentially” optimal (actually, we shall see later that the lower bound( d − / ( d + 1) is a better asymptotic estimate). However, substantial improvement is possiblefor Euclidean ( ℓ ) balls, and thus, the same question on the p norm of the derivative arises:Can we have c p < p < ∞ sufficiently high? And can we take p to be independent of thedimension?We conclude this section by using the notation from Definition 2.9 to summarize the maincontents of Theorem 2.7. While formula (2.14.1) below does not look very promising, due tothe successive appearance of two suprema and one infimum, the fact is that it will allow usto obtain optimal asymptotic estimates for the norms considered in the next section, and insome cases we will be able to actually compute the number k M k Op( ω ) . Corollary 2.14.
Let k · k be a norm on R d , and let ω be a modulus of continuity. Then theassociated maximal operator M acting on Lip( ω ) has norm given by (2.14.1) k M k Op( ω ) = sup t> sup { v ∈ R d : k v k =1 } inf { c ∈ R d , Using one of the extremal functions ψ found in Theorem 2.7 we have k M k Op( ω ) = k M ψ k Lip( ω ) , and now the result follows from (2.7.2) together with (2.7.3). (cid:3) Bounds in the ℓ ∞ , ℓ and ℓ cases. Next we specialize Theorem 2.7 and Corollary 2.14 to the three norms in the title of thissection. This specialization will allow us to find explicit constants, at least for low dimensionsand in the case of cubes (where the needed arguments seem to be simpler). As we mentioned,however, all constants will turn out to be exactly the same when working with the ℓ ∞ and ℓ norms. Theorem 3.1. Let M be the uncentered maximal operator associated to balls defined by the ℓ ∞ norm on R d , i.e., to cubes with sides parallel to the coordinate axes, and let ω be a modulusof continuity. Then (3.1.1) k M k Op( ω ) = sup t> ( inf ≤ s ≤ ω ( t ) Z [ − s, d ω ( t k x k ∞ ) dx (1 + s ) d ) , or equivalently, (3.1.2) k M k Op( ω ) = sup t> (cid:26) inf ≤ s ≤ d (1 + s ) d ω ( t ) (cid:20) d Z s u d − ω ( tu ) du + Z s ( u + s ) d − ω ( tu ) du (cid:21)(cid:27) . Moreover, if we choose the same modulus ω for every dimension d , then k M k Op( ω ) is nonde-creasing in d , and lim d →∞ k M k Op( ω ) = 1 . Of the two expressions for k M k Op( ω ) given in the preceding theorem, (3.1.1) is the onewith a clearer geometric content: It says that an optimal choice for v is (1 , , . . . , − , d . On the otherhand, while harder to interpret, formula (3.1.2) turns out to be computationally much moreconvenient. Proof. From (2.14.1) it follows that(3.1.3) k M k Op( ω ) = sup t> sup { v ∈ R d : k v k ∞ =1 } inf { c ∈ R d ,R> k v − c k ∞ = R ≤ } ω ( t ) Z [ − , d ω ( t k c + Ru k ∞ ) du d , so if we write v := (1 , , . . . , k M k Op( ω ) ≥ sup t> inf { c ∈ R d ,R> k v − c k ∞ = R ≤ } ω ( t ) Z [ − , d ω ( t k c + Ru k ∞ ) du d . We claim that the infimum in (3.1.4) is attained when c = (1 − R, − R, . . . , − R ) for some R ∈ (0 , 1] (which may vary with the dimension). Let c = ( c , c , . . . , c d ) satisfy k v − c k ∞ = R ≤ 1. Then c , c , . . . , c d ≥ − R . Suppose c > − R (else, do nothing and consider c instead). Translate c to c := (1 − R, c , . . . , c d ), by moving it parallel to e . Now, for every x ∈ B ( c, R ) \ B ( c , R ), | x | > 1, while for every x ∈ B ( c , R ) \ B ( c, R ), | x | ≤ 1. Since paralleltransport in the direction of e does not change any of the other coordinates, the averagevalue of k · k ∞ is not increased. Then repeat this process with each coordinate. odulus of continuity of the maximal function 11 Next, note that the infimum is attained when 1 / ≤ R ≤ 1. This immediately follows fromthe fact that for each u ∈ B (0 , f ( R ) := k (1 − R ) v + Ru k ∞ is decreasing on(0 , / | − R + Ru i | = 1 − (1 − u i ) R is decreasing there.Thus(3.1.5) k M k Op( ω ) ≥ sup t> inf { / ≤ R ≤ } ω ( t ) Z [ − , d ω ( t k (1 − R ) v + Ru k ∞ ) du d . On the other hand, using (3.1.3) and, for each each unit vector v , taking the infimum over asmaller set of associated centers c , we get(3.1.6) k M k Op( ω ) ≤ sup t> sup { v ∈ R d : k v k ∞ =1 } inf { / ≤ R ≤ } ω ( t ) Z [ − , d ω ( t k (1 − R ) v + Ru k ∞ ) du d . We show that the supremum over unit vectors is attained on v , so in fact the right hand sidesof (3.1.5) and (3.1.6) are equal. From this, (3.1.1) follows by making the change of variable x = (1 − R ) v + Ru , and relabeling s = 2 R − 1. Fix R ∈ [1 / , v = ( v , . . . , v d ) satisfy k v k ∞ = 1. By symmetry considerations we may assume that v , . . . , v d ≥ 0. Now we argue asbefore. If v = 1 do nothing and move to v . Else, v < 1, so shift v to v := (1 , v , . . . , v d ) byparallel transport in the direction of e . Then for every x ∈ B ((1 − R ) v, R ) \ B ((1 − R ) v , R ),(1 − R ) v − R < x < 0, so | x | < | (1 − R ) v − R | , while if x ∈ B ((1 − R ) v , R ) \ B ((1 − R ) v, R ),we have | x | > (1 − R ) v + R . Since the average value of | x | increases after the shift andthe other coordinates do not change, the average value of k · k ∞ increases. Then repeat thisprocess with each coordinate.To obtain (3.1.2) from (3.1.1), break up the integral appearing in (3.1.1) into the regions {k x k ∞ < s } and {k x k ∞ ≥ s } , and then separate these into the sets where | x i | = k x k ∞ , for i = 1 , . . . , d . Note, for instance, that if we are working over {k x k ∞ ≥ s } ∩ { x = k x k ∞ } , then,for a fixed value of x , the coordinates x , . . . , x d of the associated vertical section E x satisfy − s ≤ x i ≤ x , so | E x | = ( s + x ) d − . A similar remark can be made about the integral over {k x k ∞ < s } , so applying Fubini’s Theorem we obtain (3.1.2).Next we show that the norm of M does not decrease when the dimension changes from d to d + 1 if we keep the same modulus ω . For notational simplicity, we suppose that d = 1.The argument for arbitrary d is the same. Let s d ∈ [0 , 1] be the minimizing value of s indimension d . Then1(1 + s ) Z [ − s , ω ( t max {| x | , | y |} ) dxdy ≥ s ) Z − s Z − s ω ( t | x | ) dxdy = 11 + s Z − s ω ( t | x | ) dx ≥ 11 + s Z − s ω ( t | x | ) dx. To finish, we show that as the dimension d → ∞ , for a fixed modulus ω we have k M k Op( ω ) → 1. This is simply a consequence of the fact that for every s ∈ [0 , − s, d concentrates near the norm one vectors. More precisely, let X be a random vector, chosen uniformly from [ − s, d . Then its coordinate functions X i areindependent random variables, uniformly distributed over [ − s, we use P to denote both uniform probabilities on [ − s, d and on [ − s, < ε < 1, set t = 1, and choose δ ∈ (0 , ε ) so that ω (1 − δ ) /ω (1) > − ε . Then, for any 0 ≤ s ≤ d , we have P ( k X k ∞ > − δ ) ≥ P (max { X , . . . , X d } > − δ ) = 1 − Π d P ( X i ≤ − δ )= 1 − (cid:18) s − δ s (cid:19) d ≥ − (cid:18) − δ (cid:19) d > − ε. Thus, if the dimension d is large enough (depending of ω and ε ) it follows that1 ≥ k M k Op( ω ) ≥ inf ≤ s ≤ ω (1) Z [ − s, d ω ( k x k ∞ ) dx (1 + s ) d ≥ (1 − ε ) ω (1 − δ ) ω (1) > (1 − ε ) . (cid:3) Observe that in the above proof we did not need to establish how the optimal value s d of s ∈ [0 , 1] behaves as d → ∞ . Intuition suggests that since the measure of [ − s d , d concentrates near its border as d grows, in order to minimize the average value of the normover this cube, the origin should be increasingly closer to the boundary of [ − s d , d . Or, inother words, s d should approach 0 as d → ∞ . This intuition is, in fact, completely erroneous,as formula (3.2.2) below shows (see also the proof of Corollary 3.4): When d → ∞ , theoptimal s d tends to 1, and thus the optimal averaging cube has ℓ ∞ diameter approaching2. Nevertheless, we shall show that in the euclidean case the above intuition is correct: As d → ∞ the origin must indeed be increasingly closer to the boundary of the optimal ball, inorder to minimize the average value of the norm, and the ℓ diameter of the optimal averagingball must approach 1 rather than 2. This helps to understand why the asymptotic behaviorof k M k Op(1) is so different in the ℓ ∞ and ℓ cases.Next we specialize the preceding theorem to the Lipschitz and H¨older functions, obtainingthe following corollary. Corollary 3.2. Let α ∈ (0 , , and consider the space ( R d , k · k ∞ ) . Then, on Lip α ( R d , k · k ∞ ) , (3.2.1) k M k Op( α ) = min ≤ s ≤ ( d (1 + s ) d " d s α + d α + d + d − X j =0 (cid:18) d − j (cid:19) s d − − j − s d + α α + j + 1 . In particular, when α = 1 , that is, for Lipschitz functions on ( R d , k · k ∞ ) we have (3.2.2) k M k Op(1) = dd + 1 − d + 1 max To obtain (3.2.1), use (3.1.2) in Theorem 3.1 with ω ( t ) = t α , and integrate. As for(3.2.2), it does not seem to be easy to derive it from (3.2.1) by substituting α = 1. Instead,use (3.1.2) again, evaluating the integral R s ( u + s ) d − udu via the change of variable v = u + s .This yields (3.2.2) but with max ≤ s ≤ . To further refine this expression and obtain max Specializing formula (3.2.1) to d = 1 and d = 2 gives the following expressions.We mention that they can also be obtained easily and directly from (3.1.1).When d = 1, we find that for every f ∈ Lip α ( R ) and every α ∈ (0 , k M k Op( α ) = min 1) is shown, as above, by elementary calculus arguments:Fix α ∈ (0 , s ∈ [0 , 1] write g α ( s ) := 1 + s α +1 ( α + 1) (1 + s ) . Evaluating g α on 0 and 1we see (by inspection) that it achieves its maximum value at these points. Since g ′′ α > , k M k Op( α ) < (1 + α ) − , which is the sharpbound in the local case, that is, for H¨older functions on intervals (cf. Corollary 4.5 below).Thus, H¨older constants are smaller on the line than on proper subintervals. Note also that as α → k M k Op( α ) → 1, again by (3.3.1). Convergence to 1 as α → d = 2, formulas (3.2.1) and (3.1.1) become(3.3.2) k M k Op( α ) = min ≤ s ≤ (cid:18) α + 1 + ( α + 2) s + (2 α + 1) s α +2 ( α + 1)( α + 2) (1 + s ) (cid:19) . Here calculus arguments regarding extrema are more involved, and in fact, as d grows theformulas given by (3.2.1) become less manageable, though of course, numerical estimation ispossible. However, in the simpler Lipschitz case, where α = 1, there is still a good deal ofexplicit information that can be extracted from (3.2.2). Contrary to our usual notation, inthe next corollary we shall indicate the dependency of the maximal operator on the dimension d by writing M d . Corollary 3.4. On Lip( R d , k· k ∞ ) , k M d k Op(1) = ( d − / ( d +1)+ o (1 / ( d +1)) . More precisely, (3.4.1) d − d + 1 < k M d k Op(1) ≤ dd + 1 − d + 1 (cid:18) − √ d (cid:19) " − (cid:18) − √ d − (cid:19) d . Proof. The lower bound follows from (3.2.2) by noticing that g d ( s ) := s − d s d +1 / (1 + s ) d 1] to s . For instance, the choices s = 0 and s = 1 recover the general bound d/ ( d + 1) from Corollary 2.13 (so k M d k Op(1) = ( d − / ( d + 1) + O (1 / ( d + 1))), but of courseone can do better: The right hand side of (3.4.1) is obtained by taking s = 1 − / √ d . (cid:3) While sufficient to prove asymptotic equivalence, the choice s = 1 − / √ d made aboveis somewhat arbitrary and can easily be improved, at the cost of getting more complicated upper bounds (so the second inequality in (3.4.1) is also strict). The optimal choice is theunique solution to the polynomial equation given next. Lemma 3.5. The norm of the maximal operator M on Lip( R d , k·k ∞ ) is obtained by evaluating (3.5.1) h d ( s ) := dd + 1 − d + 1 (cid:18) s − d s d +1 (1 + s ) d (cid:19) on the unique solution s d inside (0 , of the polynomial equation (3.5.2) p d ( s ) := 2 d s d +1 − (1 + s ) d +1 + 2 d ( d + 1) s d = 0 . Proof. Formula (3.5.1) follows immediately from (3.2.2). Recall from the proof of Corollary3.2 that the function g d ( s ) := s − d s d +1 / (1 + s ) d has a unique maximum on (0 , g ′ d = 0, or equivalently, p d ( s ) := (1 + s ) d +1 − d ( d + 1) s d − d s d +1 = 0, onthe said interval. (cid:3) When d ≤ 3, deg p d ≤ 4, so its roots can be found explicitly using Cardan’s formula. Wedo this next, thereby obtaining the sharp constant in dimension 1, and for the ℓ ∞ norm, thesharp constants in dimensions 2 and 3. Details are included for the readers convenience. Corollary 3.6. On Lip( R ) we have k M k Op(1) = √ − . Proof. Solving p ( s ) = 1 − s − s = 0 on (0 , h ( s ) = h ( √ − 1) = √ − (cid:3) Corollary 3.7. On Lip( R , k · k ∞ ) we have (3.7.1) k M k Op(1) = 4 √ (cid:18) π (cid:19) + √ (cid:18) π (cid:19) − 14 sec (cid:18) π (cid:19) − . Proof. We use Lemma 3.5, finding first the unique root of s +3 s − s − / , s s − s − s +8 / . Since (8 / − / = − / < s = (cid:0) / √ (cid:1) y , toobtain 4 y − y = −√ / π/ πk ), and now we use the trigonometrical identity4 cos ( ϑ ) − ϑ ) = cos (3 ϑ ) to conclude that s = (cid:0) / √ (cid:1) cos (5 π/ 18 + 2 kπ/ k = 0 , , k = 0 belongs to (1 , s = √ cos π − ∈ (0 , h (cf. (3.5.1)) on s and simplifying once more we find that h ( s ) = 4 √ (cid:18) π (cid:19) + √ (cid:18) π (cid:19) − 14 sec (cid:18) π (cid:19) − . (cid:3) Remark 3.8. Therefore k M k Op(1) ≈ . 574 on Lip( R , k · k ∞ ). Remark 3.9. In dimension two, the results (3.3.2) and (3.7.1) hold verbatim if we use the ℓ norm instead of the ℓ ∞ norm, since in dimension two ℓ balls are just rotated cubes. Infact, this phenomenon repeats itself in every dimension, despite the fact that the geometry isvery different when d ≥ 3. Thus, the next result holds also for the ℓ norm. odulus of continuity of the maximal function 15 Corollary 3.10. On Lip( R , k · k ∞ ) we have (3.10.1) k M k Op(1) = 1 − / (( √ √ / +( √ −√ / ) / (cid:18) (cid:16) / (( √ √ / +( √ −√ / ) / − (cid:17) / (cid:19) . Proof. By Lemma 3.5, it is enough to find the the unique root s of p ( s ) = 7 s + 28 s − s − s − , 1) and then evaluate h ( s ) (see (3.5.1)). Using the change of variable s = 1 / (2 t − < s < t > 1, and p ( s ) = 0 if and only if2 t − t + 3 = 0 . Now it can be checked by direct substitution that t = (cid:0) ( √ √ / + ( √ − √ / (cid:1) / / (cid:18) / (( √ √ / + ( √ − √ / ) / − (cid:19) / ! satisfies 2 t − t + 3 = 0, and furthermore, it is the unique t > s in (0 , s directly in h , it is moreconvenient to simplify h ( s ) first. Note that h ( s ) = 3 + 8 s + 6 s + 7 s s ) , and also 7 s = − s + 6 s + 4 s + 1, since p ( s ) = 0. Eliminating the fourth order termand simplifying we get k M k Op(1) = h ( s ) = 1 − (cid:18) s s (cid:19) . Using s = 1 / (2 t − k M k Op(1) = 1 − t − , and now (3.10.1)follows by substituting in the numerical value of t . (cid:3) Remark 3.11. k M k Op(1) ≈ . R , k · k ∞ ).Next we study the cases p = 2 and p = 1. Since in one dimension all the ℓ p unit ballscoincide with the interval [ − , d = 1 is covered by Theorem 3.1. Lemma 3.12. Let d ≥ , let f : [0 , ∞ ) → [0 , ∞ ) be an increasing function, and let ≤ p < ∞ . Then, for all R > , all c = ( c , . . . , c d ) ∈ R d , and all a ∈ R such that | a | ≤ | c | , we have Z B ( ae ,R ) f ( k x k pp ) dx ≤ Z B ( c,R ) f ( k x k pp ) dx. Proof. Write ˆ c = ( c , . . . , c d ) and ˆ x = ( x , . . . , x d ). Using Fubini’s Theorem and Anderson’sTheorem (cf. Remark 2.8) in d and d − Z B ( c,R ) f ( k x k pp ) dx = Z c + Rc − R (cid:18)Z B (ˆ c, ( R p −| c − x | p ) /p ) f ( | x | p + k ˆ x k pp ) d ˆ x (cid:19) dx ≥ Z c + Rc − R (cid:18)Z B (0 , ( R p −| c − x | p ) /p ) f ( | x | p + k ˆ x k pp ) d ˆ x (cid:19) dx = Z B ( c e ,R ) f ( k x k pp ) dx ≥ Z B ( ae ,R ) f ( k x k pp ) dx. (cid:3) To apply (2.14.1), it is useful to determine on which unit vectors the supremum is attained.We have seen that one such vector in the ℓ ∞ case is (1 , . . . , p norms, if e canbe selected, this usually leads to simplification of the formulas. The next lemma reduces thequestion of the optimality of e to the two dimensional case. Lemma 3.13. Let f : [0 , ∞ ) → [0 , ∞ ) be an increasing function, and let ≤ p < ∞ . If forall c ∈ R and all R ∈ (0 , we have Z B ( c,R ) f ( k x k pp ) dx ≤ Z B ( k c k p e ,R ) f ( k x k pp ) dx, then for all d ≥ , all c ∈ R d , and all R ∈ (0 , , Z B ( c,R ) f ( k x k pp ) dx ≤ Z B ( k c k p e ,R ) f ( k x k pp ) dx. Proof. Assume the result is true for d ≥ 2. Let c ∈ R d +1 and let R > 0. We write c = ( c, b ),where c ∈ R d , x = ( x, y ), x ∈ R d , and x = ( x , ˆ x ) , ˆ x = ( x , . . . , x d ) ∈ R d − . From Fubini’sTheorem, induction, and the assumption for d = 2, we get Z B ( c,R ) f ( k x k pp ) dx = Z b + Rb − R Z | x − c | p + ... + | x d − c d | p ≤ R p −| y − b | p f ( k x k pp + | y | p ) dxdy ≤ Z b + Rb − R Z | x −k c k p | p + | x | p + ... + | x d | p ≤ R p −| y − b | p f ( k x k pp + | y | p ) dxdy = Z k ˆ x k pp ≤ R p Z | x −k c k p | p + | y − b | p ≤ R p −k ˆ x k pp f ( k ˆ x k pp + | x | p + | y | p ) dx dy ! d ˆ x ≤ Z k ˆ x k pp ≤ R p Z | x −k c k p | p + | y | p ≤ R p −k ˆ x k pp f ( k ˆ x k pp + | x | p + | y | p ) dx dy ! d ˆ x = Z B ( k c k p e ,R ) f ( k x k pp ) dx. (cid:3) For euclidean balls, optimality of e , or any other vector (in every dimension), follows fromsymmetry. Now let d = 2. For ℓ balls, optimality of e follows from the optimality of (1 , ℓ ∞ balls, since each unit ball and its corresponding norm can be obtained from the othervia a rotation and a dilation. And in this case e is strictly better than nearby vectors. Wehave not been able to prove in a direct way the optimality of e for other ℓ p balls, 1 < p < odulus of continuity of the maximal function 17 It is well known, and we use it below in the cases p = 1 , 2, that when 1 ≤ p < ∞ the measureof the unit ball concentrates near the vectors with norm one and first coordinate equal to zero,that is, near the “vertical equator” perpendicular to e : { x ∈ R d : x = 0 and k x k p = 1 } .Indeed, the sections of the unit ball perpendicular to e are balls in R d − , and a small decreasein the radius r causes a large decrease in mass whenever d is high, since Lebesgue measure in R d − scales like r d − . It follows that the measure of the unit ball concentrates on the sectionsof maximal radius, i.e., when x ≈ 0. Likewise, the measure of the unit ball concentrates nearthe unit sphere. Thus we have concentration near the vertical equator, since the intersectionof two very large subsets of the unit ball must be large. Of course, when p = ∞ thisconcentration near the vertical equator does not takes place.Let B dp be the ℓ p unit ball { x ∈ R d : k x k p ≤ } (for convenience here we take balls to beclosed), let S d − p be the corresponding unit sphere, given by { x ∈ R d : k x k p = 1 } , and let | B dp | and | S d − p | be their respective d and d − r not necessarilyequal to one, we write B dp ( r ) and S d − p ( r ). Theorem 3.14. Let d ≥ , let M be the uncentered maximal operator associated to ballsdefined by the ℓ norm, i.e., to euclidean balls, and let ω be a modulus of continuity. Then (3.14.1) k M k Op( ω ) = sup t> inf / ≤ R ≤ ( d − d/ ω ( t ) √ π Γ(1 / d/ × Z − Z (1 − u ) / ω (cid:0) t ((1 − R + Ru ) + R ρ ) / (cid:1) ρ d − dρdu . Furthermore, if we select the same modulus ω in all dimensions, we have (3.14.2) sup t> ω (cid:16) − t (cid:17) ω ( t ) ≤ lim inf d →∞ k M k Op( ω ) and (3.14.3) lim sup d →∞ k M k Op( ω ) ≤ inf r> sup t> ω (cid:16) − rt (cid:17) ω ( t ) . Under the additional assumption that ω is concave, the limit exists, it is equal to the left handside of (3.14.2), and bounds k M k Op( ω ) uniformly in d : Given d ≥ , (3.14.4) k M k Op( ω ) ≤ sup t> ω (cid:16) − t (cid:17) ω ( t ) = lim d →∞ k M k Op( ω ) . In particular, for the H¨older and Lipschitz classes we obtain k M k Op( α ) ≤ − α for all d ≥ , and lim d →∞ k M k Op( α ) = 2 − α . In order to give an idea about the size of the constant term (for d fixed) in (3.14.1), wepoint out that(3.14.5) (cid:18) d (cid:19) / ≤ Γ(1 + d/ / d/ ≤ (cid:18) d + 12 (cid:19) / . This is an easy consequence of the log-convexity of the Γ function (cf. Exercise 5, pg. 216of [Web]). Note also that if instead of taking the infimum in the right hand side of (3.14.1)we just set R = 1, we are averaging over the whole unit ball, that is, we are acting as inthe proof of Corollary 2.13. And indeed, if we change to polar coordinates and integrate werecover the bound d/ ( d + 1). So here, a better choice of R leads to lower asymptotic bounds.We shall see that in fact, when d → ∞ the optimal R = R ( d ) approaches 1 / 2, as intuitionsuggests. The s -dimensional Hausdorff measure on R d is denoted by H s . Proof. As in the case of Theorem 3.1, it is enough to consider balls contained in B d . To see this,fix any unit vector v , and let B ( c, R ) be a minimizing ball for v in (2.14). Suppose B ( c, R )has points outside B d . Translating B ( c, R ) towards the origin along the ray { tc : t ≥ } determined by the vector c , so that the displaced ball B ( c ′ , R ) is fully contained in B d andtangent to the unit sphere, leads to Z B (0 , ω ( t k c ′ + Ru k ) du ≤ Z B (0 , ω ( t k c + Ru k ) du by Anderson’s Theorem (see Remark 2.8). It may well happen that after the translation v / ∈ B ( c ′ , R ). If so, rotate B ( c ′ , R ) about the origin to make its new center lie in the segment[0 , v ]. Since this does not change the value of the integral, we conclude that it suffices toconsider balls contained in B d . Again by rotational symmetry we may take v to be e , so itis enough to consider centers c = (1 − R ) e and radii R , with 0 ≤ R ≤ 1. Hence, by (2.14.1)we have(3.14.6) k M k Op( ω ) = sup t> inf ≤ R ≤ ω ( t ) Z B d ω ( t k (1 − R ) e + Ru k ) du (cid:12)(cid:12) B d (cid:12)(cid:12) . Note that the infimum in (3.14.6) is attained when R ∈ [1 / , f ( R ) := k (1 − R ) e + Ru k is decreasing when R ≤ / 2, so the minimum mustindeed occur on 1 / ≤ R ≤ 1. Differentiating f ( R ) = (1 − R + Ru ) + R P di =1 u i − R u ,and using P di =1 u i ≤ R ≤ / 2, the claim follows.Given u ∈ B d , we write u = ( u , y ), where u ∈ R and y ∈ R d − . Denote by P d the uniformprobability on B d , so dP d ( u ) = du/ | B d | . To prove (3.14.3) and (3.14.2), fix t > 0, and notethat by concentration of measure near the vertical equator, for each δ > d →∞ P d ( {| u | < δ, − δ < k y k ≤ } ) = 1 . Of course, the weaker assertion(3.14.8) lim d →∞ P d ( {| u | < δ } ) = 1 odulus of continuity of the maximal function 19 also holds. Since k (1 − R ) e + Ru k ≤ u ∈ B d and all R ∈ [1 / , / ≤ R ≤ Z B d \{| u | <δ } ω ( t k (1 − R ) e + Ru k ) ω ( t ) dP d ( u ) ≤ P d ( {| u | ≥ δ } ) . Next, note that on B d ∩ {| u | < δ } , for every d and every R ∈ [1 / , 1] we have(3.14.10) k (1 − R ) e + Ru k ≤ (cid:0) (1 − R + δ ) + R (cid:1) / . The unique minimum of h ( R ) := (1 − R + δ ) + R on [1 / , 1] is attained at R δ = (1 + δ ) / h ( R δ ) = (1 + δ ) / 2. Thus,(3.14.11) k (1 − R δ ) e + R δ u k ≤ − / (1 + δ ) , so splitting B d into the regions where | u | < δ and | u | ≥ δ we obtain(3.14.12)inf / ≤ R ≤ Z B d ω ( t k (1 − R ) e + Ru k ) ω ( t ) dP ( u ) ≤ ω (cid:0) t (cid:0) − / (1 + δ ) (cid:1)(cid:1) ω ( t ) + P d ( {| u | ≥ δ } ) . Taking on both sides of the preceding inequality first sup t> , second, lim sup d →∞ , and third,inf δ> , from (3.14.6) (with R ∈ [1 / , d →∞ k M k Op( ω ) ≤ inf δ> sup t> ω (cid:16) − (1 + δ ) t (cid:17) ω ( t ) . This proves (3.14.3).The argument used to obtain (3.14.2) is similar. Fix u = ( u , y ) ∈ R d and note that on R the function h ( R ) := (1 − R + Ru ) + R k y k = k (1 − R ) e + Ru k achieves its unique minimum at R u = (1 − u ) / [(1 − u ) + k y k ], where it takes the value h ( R u ) = k y k / [(1 − u ) + k y k ]. Thus, for 0 < δ < u ∈ B d ∩{| u | < δ, − δ < k y k ≤ } we have 1 − δ p (1 + δ ) + 1 ≤ k y k p (1 − u ) + k y k ≤ k (1 − R ) e + Ru k , from which it follows that(3.14.14) sup t> P d ( {| u | < δ, − δ < k y k ≤ } ) ω (cid:16) t ((1 + δ ) + 1) − / (1 − δ ) (cid:17) ω ( t )(3.14.15) ≤ sup t> inf / ≤ R ≤ Z B d ω ( t k (1 − R ) e + Ru k ) ω ( t ) dP d ( u ) = k M k Op( ω ) . Now (3.14.14) increases with d , so taking the limit inferior in (3.14.14) and (3.14.15) as d → ∞ , we get(3.14.16) sup t> ω (cid:16) t ((1 + δ ) + 1) − / (1 − δ ) (cid:17) ω ( t ) ≤ lim inf d →∞ k M k Op( ω ) . Likewise, the left hand side of the preceding inequality is decreasing in δ ∈ (0 , δ , interchanging it with the supremum over t , and letting δ ↓ u = ( u , y ) ∈ R d and y = ρη , where ρ = k y k and k η k = 1. By Fubini’sTheorem(3.14.17) Z B d ω ( t k (1 − R ) e + Ru k ) du = Z − Z B d − ((1 −| u | ) / ) ω ( t k (1 − R ) e + R ( u , y ) k ) dydu . Using polar coordinates on the vertical sections, or equivalently, by the coarea formula, weget Z B d − ((1 − u ) / ) ω ( t k (1 − R ) e + R ( u , y ) k ) dy = Z (1 − u ) / Z S d − ( ρ ) ω (cid:0) ( | (1 − R ) + Ru | + R k y k ) / (cid:1) d H d − ( y ) dρ = ( d − | B d − | Z (1 − u ) / ω (cid:0) ( | (1 − R ) + Ru | + R ρ ) / (cid:1) ρ d − dρ. Since | B d | = π d/ Γ(1+ d/ , (3.14.1) follows from (3.14.6) and the preceding equalities.Suppose next that ω is concave. We show that the inequality in (3.14.4) holds for every d ≥ 1. When d = 1 the result follows from (3.1.1) and the concavity of ω :(3.14.18) k M k Op( ω ) ≤ sup t> ω ( t ) Z ω ( tx ) dx ≤ sup t> ω ( t ) ω (cid:18) t Z xdx (cid:19) = sup t> ω ( t/ ω ( t ) . Let d > t > ω ( z ) is concave, then so is ω (cid:0) tz / (cid:1) . Thus, given x ≥ y ≥ 0, we have ω (cid:0) ( x − y ) / (cid:1) + ω (cid:0) ( x + y ) / (cid:1) ≤ ω (cid:0) x / (cid:1) . We write the integralappearing in (3.14.1), together with its constant term, as I = Z − Z (1 − u ) / ω (cid:0) t ((1 − R ) + R ( u + ρ ) + 2 R (1 − R ) u ) / (cid:1) dµ where dµ = | ( d − B d − | (cid:12)(cid:12) B d (cid:12)(cid:12) ρ d − dρdu odulus of continuity of the maximal function 21 defines a probability on B ∩ { ρ ≥ } . Actually we get a probability µ d for each dimension d , but since this is not relevant in the following argument we omit the reference to d in thenotation. Now from concavity and the fact that u + ρ ≤ I ≤ Z Z (1 − u ) / ω (cid:0) t ((1 − R ) + R ( u + ρ )) / (cid:1) dµ ≤ ω (cid:0) t ((1 − R ) + R ) / (cid:1) . Using inf / ≤ R ≤ ω (cid:0) t ((1 − R ) + R ) / (cid:1) = ω (cid:16) − t (cid:17) we obtain (3.14.4).Regarding the assertion about the H¨older and Lipschitz classes, for d > ω ( t ) = t α . And for d = 1, the upper bound is immediate from (3.14.18). (cid:3) Remark 3.15. Consider the Lipschitz functions on R d . By the preceding theorem, for every d we have k M k Op(1) ≤ − / on Lip( R d , k · k ). However, on Lip( R d , k · k ∞ ), k M k Op(1) > ( d − / ( d + 1), by Corollary 3.4. Since 2 − / < / any optimal constant in the ℓ ∞ case, with d ≥ 6, is strictly larger than all the optimal constants in the Euclidean case ( d = 1 , , , . . . ).This illustrates the fact that using different norms on R d may result in obtaining very differentbest constants. But the opposite can also happen: Best constants are identical for the ℓ and ℓ ∞ norms, as we shall prove by showing that the corresponding integral formulas for k M k Op( ω ) are actually the same.Even though we are using different norms to define maximal operators and their associatedmoduli of continuity, we adopt the convention that the s -dimensional Hausdorff measure H s is always defined via the Euclidean ℓ metric, and normalized by the factor π s/ / Γ(1 + s/ s = d is a positive natural number, the cube of sidelength 1 has Hausdorff d measure 1.A consistent definition is required in order to use Fubini’s Theorem, or more generally, thecoarea formula. For example, by our convention the length of the polygonal curve S is 4 √ ℓ distance instead ofthe Euclidean distance. Theorem 3.16. Let d ≥ , let M be the uncentered maximal operator associated to ballsdefined by the ℓ norm, i.e., to cross-polytopes, and let ω be a modulus of continuity. Then (3.16.1) k M k Op( ω ) = sup t> inf / ≤ R ≤ d ( d − ω ( t ) Z − Z −| u | ω ( t ( | (1 − R ) + Ru | + Rρ )) ρ d − dρdu . If we choose the same modulus ω in all dimensions, then we have lim d →∞ k M k Op( ω ) = 1 .Proof. Setting c = (1 − R ) v in (2.14.1), we get k M k Op( ω ) ≤ sup t> sup { v ∈ R d : k v k =1 } inf 2, by Lemma 3.13. Thus(3.16.2) k M k Op( ω ) ≤ sup t> inf 1. Thus,this is the case on almost all R d − . Set g ( y ) = 1 / | J f ( y ) | and use (3.16.5) to obtain | B d − | = Z B d − dy = Z Z S d − ( ρ ) √ d − d H d − ( y ) dρ == | S d − |√ d − Z ρ d − dρ = | S d − | ( d − √ d − . From this equality and the coarea formula (once more), with g ( y ) = ω ( t ( | (1 − R ) + Ru | + R k y k )) | J f ( y ) | , we get Z B d − (1 −| u | ) ω ( t k (1 − R ) e + R ( u , y ) k ) dy = Z −| u | Z S d − ( ρ ) ω ( t ( | (1 − R ) + Ru | + Rρ )) d H d − ( y ) √ d − dρ = 1 √ d − Z −| u | ω ( t ( | (1 − R ) + Ru | + Rρ )) Z S d − ρ d − d H d − ( y ) dρ odulus of continuity of the maximal function 23 = ( d − | B d − | Z −| u | ω ( t ( | (1 − R ) + Ru | + Rρ )) ρ d − dρ. It is well known (and easy to compute) that | B d | = d d ! , so | B d − || B d | = d and we obtain k M k Op( ω ) = sup t> inf ≤ R ≤ d ( d − ω ( t ) Z − Z −| u | ω ( t ( | (1 − R ) + Ru | + Rρ )) ρ d − dρdu . To see that the infimum is attained when R ∈ [1 / , , / 2] the function f ( R ) = 1 − R + Ru + Rρ has a negative derivative for every ρ ∈ [0 , − | u | ). Finally, 1is a uniform upper bound by Kinnunnen’s Theorem, and concentration of measure near thevertical equator { u ∈ R d : u = 0 and k u k = 1 } entails that lim d →∞ k M k Op( ω ) = 1. (cid:3) Observe, for instance, that unlike the case of cubes and the ℓ ∞ norm, it is not clear fromformula (3.16.1) that k M k Op( ω ) increases with the dimension. However, this must be thecase, since given an arbitrary modulus, constants for the ℓ and the ℓ ∞ norms are equal ineach dimension. Of course this is trivial for d = 1, and clear for d = 2, since in this casecross-polytopes are rotated squares. But when d > Theorem 3.17. Fix d ≥ and select a modulus of continuity ω . Then k M k Op( ω ) has ex-actly the same value regardless of whether distances and the maximal operator are computedaccording to the ℓ norm, or to the ℓ ∞ norm. Recall that in the case of the ℓ ∞ norm the asymptotic value 1 was obtained by usingconcentration near the boundary of the cube [ − s, d , and more precisely, near the norm1 vectors, while in the ℓ case we utilized concentration near the vertical equator. By thepreceding theorem one of these (different) arguments is redundant. Proof. Fix t > 0. To see that the values of k M k Op( ω ) given by (3.1.1) and (3.16.1) are indeedthe same, it is enough to show thatinf / ≤ R ≤ d ( d − Z − Z −| u | ω ( t ( | (1 − R ) + Ru | + Rρ )) ρ d − dρdu (3.17.1) = inf ≤ s ≤ d (1 + s ) d (cid:18) d Z s z d − ω ( tz ) dz + Z s ( z + s ) d − ω ( tz ) dz (cid:19) . Write I := Z − Z −| u | ω ( t ( | (1 − R ) + Ru | + Rρ )) ρ d − dρdu = I + I , where I := Z R − R Z −| u | ω ( t ((1 − R ) + R ( u + ρ ))) ρ d − dρdu and I := Z R − R − Z u ω ( t ( R − R ( ρ − u ))) ρ d − dρdu . To compute I , we use the change of variables v = 1 − R + R ( u + ρ ), y = Ru and Fubini’sTheorem: I = Z RR − Z −| y | + y − R + y ω ( tv ) ( v + R − − y ) d − R d dvdy = 1 R d Z Z v + R − { R − , v − } ω ( tv )( v + R − − y ) d − dydv = 1 R d ( d − (cid:18)Z R − ω ( tv ) ( v + 2 R − d − d − dv + Z R − ω ( tv ) v d − dv (cid:19) . Likewise, to compute I we set v = R − R ( ρ − u ), y = Ru , and interchange the orderof integration: I = Z R − − R Z R − R − − y ω ( tv ) ( v + 1 − R + y ) d − R d dvdy = 1 R d ( d − Z R − ω ( tv ) v d − dv. Adding up we obtain I = 1 R d ( d − (cid:18)Z R − ω ( tv ) ( v + 2 R − d − d − dv + 2 Z R − ω ( tv ) v d − dv (cid:19) . To finish, set s = 2 R − 1, multiply by d ( d − / 2, and take the corresponding infima to get(3.17.1). (cid:3) We conclude this section with four questions and some variants of these. We conjecturethat the following version of Theorems 3.14 and 3.16 holds for all p ∈ (1 , 2) and d ≥ 2: Forevery modulus of continuity ω we have(3.17.2) k M k Op( ω ) = sup t> inf ≤ R ≤ ( d − d/p )2 ω ( t ) Γ(1 + 1 /p )Γ(1 + ( d − /p ) × Z − Z (1 −| u | p ) /p ω (cid:0) t ( | (1 − R ) + Ru | p + R p ρ p ) /p (cid:1) ρ d − dρdu . Let q = p/ ( p − 1) be the conjugate exponent of p . Choose the same modulus of continuity ω in every dimension d . Thenlim inf d →∞ k M k Op( ω ) ≥ sup t> ω (cid:16) − q t (cid:17) ω ( t ) . and lim sup d →∞ k M k Op( ω ) ≤ inf r> sup t> ω (cid:16) − q rt (cid:17) ω ( t ) . odulus of continuity of the maximal function 25 Suppose additionally that ω is concave. Then for every d ≥ k M k Op( ω ) ≤ sup t> ω (cid:16) − q t (cid:17) ω ( t ) . In particular, for the H¨older and Lipschitz classes we obtain k M k Op( α ) ≤ − αq for all d ≥ , and lim d →∞ k M k Op( ω ) = 2 − αq . The main obstacle for proving this result would be removed by a positive answer to thenext question. Question 1. Is e a maximizing vector for 1 < p < d = 2.It is plausible that best bounds increase with the dimension for 1 ≤ p < ∞ , as it happens inthe case p = ∞ . This would immediately imply that the asymptotic bounds are also uniformupper bounds.As we mentioned before, it seems likely that v p := d − /p (1 , , . . . , 1) is an optimizing vectorwhen 2 < p < ∞ . But while using e leads to simplification of the integral formulas, using v p does not. Furthermore, e fits well with Fubini, in the sense that sections of d -balls yield d − e . This does not happen with v p , so a changeof coordinates would not help on that respect. In any case, we suspect that on the range2 < p < ∞ , as p → ∞ constants become increasingly worse and approach the bounds thathold for p = ∞ . Question 2. We have seen that if f is Lipschitz and M is the maximal function associatedto euclidean balls, then for every dimension d , Lip( M f ) ≤ − / Lip( f ), or equivalently, k DM f k ∞ ≤ − / k Df k ∞ . It is natural to expect a similar behavior for p “close” to ∞ . Moreprecisely, given c ∈ (2 − / , p c such that for all p ≥ p c , if Df ∈ L p ,then k DM f k p ≤ c k Df k p ? Or given p >> 1, is it possible to find such a c ∈ (2 − / , L p inequalities satisfied by M to obtain W ,p results.Since k M f k p ≥ k f k p always, this type of argument will never yield constants below 1. Onthe other hand, if we fix d , by Corollary 2.13 we have k DM f k ∞ ≤ (1 + d ) − d k Df k ∞ forthe maximal function associated to any ball, so it is natural to seek inequalities of the form k DM f k p ≤ c p k Df k p , where c p < p is high enough, and M is defined via an arbitrarynorm. At the other extreme, given f ∈ W , ( R d ) with d ≥ 2, it is not known whetherthere is a constant c (independent of f ) such that k DM f k ≤ c k f k W , ( R d ) . We mentionthat for some related maximal operators, such as, for instance, the strong maximal operator(where averages are taken over rectangles with sides parallel to the axes) such constants donot exist, cf. Theorem 2.21 of [AlPe2]. It follows from Theorem 2.5 of [AlPe] that if d = 1then k DM f k ≤ c k Df k , and c = 1 is sharp. For d ≥ f to be radial, for instance). Whether or not c < ∞ for d ≥ 2, one would expect c p > p , c p < values of p , and c p = 1 for some “crossing” p , which will likely depend on d and the ball usedto define M . A motivation to obtain detailed information on DM f comes not only from thepossible use of DM f as a substitute for Df , but also because it will yield new informationabout the maximal function itself.On a more speculative mood, we note that where balls of arbitrarily small radii have tobe taken into account, the function is “large” and the maximal function coincides with it,so maximal functions with smaller Lipschitz norms, and hence lower rate of decay “from thetop”, will tend to be larger in an L p sense. Since asymptotically constants are smaller for ℓ balls than for cubes or ℓ balls (and we believe this is also the case for other balls), it istempting to conjecture that the maximal function associated to Euclidean balls is at least“as efficient” at capturing mass as maximal functions associated to other balls. Our resultssuggest this only in the weakest possible sense, since different norms are used to computedistances, the directions we consider when measuring the modulus of continuity are those offastest decay and not some “average direction”, and L p norms of extremal functions vary with p (at least in the bounded case). Nevertheless, it seems worthwhile to try to find out whetherthe Euclidean maximal function, both in the centered and uncentered versions, has largeroperator norm on L p spaces than maximal functions associated to other balls (needless tosay, weak type results in this line would also be interesting). In fact, the weaker “comparisontheorem” suggested next would already have many consequences. Question 3. Let M e denote the maximal operator, either centered or uncentered, as-sociated to Euclidean balls, and M b the corresponding (centered or uncentered) operatorassociated to some other ball (defined by a different norm). Prove or refute the followingstatement: For every p ∈ (1 , ∞ ),10 kM e k L p ( R d ) → L p ( R d ) ≥ kM b k L p ( R d ) → L p ( R d ) . Of course, 10 is not important here, any other constant c would do. But we do mean toemphasize that 10 does not depend on anything, in particular not on p or d . “Constants”that depend on the dimension are trivial to obtain by the equivalence of all norms in R d . Thepreceding conjecture, if true in the centered case, would imply that the uniform bounds in thedimension proved by E. M. Stein for Euclidean balls also hold for all other balls (includingfor instance, cubes) and all p > 1. An opposing viewpoint can be found in [Mu], pg. 298,where it is suggested that this result may be false for cubes and p ≤ / ℓ norm means that we are outside the odulus of continuity of the maximal function 27 scope of Theorem 2.7. However, the maximal operator associated to cubes fits well with theproduct structure of R d , so it is natural, and common, to use it together with the naturaldistance on R d , the euclidean length. The next question seems therefore interesting to us: Question 4. If in Theorem 3.1 we keep the maximal operator associated to cubes butconsider the ℓ instead of the ℓ ∞ norm, how do the conclusions change?4. Local results in one dimension. Next we study the local case in one dimension, that is, when the domain is a propersubinterval of R . While the idea of the proof is the same as in the previous results, formallythe next theorem does not follow from them, so we include the full argument. In fact, notationis considerably simplified by the fact that only intervals with x as one endpoint need to beconsidered when computing M f ( x ). In order to define any such interval it is enough to specifythe other extreme, there is no need to talk about centers, radii and the relations betweenthem.We note that constants are worse in the local case than in the global case due, to the factthat a proper subinterval I of R has at least one boundary point, say, for instance a leftendpoint a . Then the decay restrictions imposed by the modulus of continuity only hold tothe right of a , and so the level sets of an extremal function will be in general smaller thanthey would be if the function were defined over the whole real line. This is reflected in thedifferent integral formulas; recall that for R , Theorem 3.1 tells us that(4.0.4) ω ( M f, t ) ≤ min ≤ s ≤ 11 + s Z − s ω ( | f | , tu ) du. The existence of a boundary point entails that we must take s = 0 in the local case, cf. (4.3.1)and (4.3.2) below. Remark 4.1. If f is uniformly continuous on a bounded interval I , then its modulus ofcontinuity is constant on [ | I | , ∞ ), with value ω ( f, | I | ). Thus, if we are given a modulus ω ,and | I | < ∞ , we cannot expect to find a function ψ ∈ Lip( ω, I ) with ω ( ψ, t ) = ω ( t ) for all t > 0. In general, the best we can do is to find ψ so that ω ( ψ, t ) = ω ( t ) on (0 , | I | ]. Remark 4.2. By an interval I we always mean a nondegenerate interval, so the emptyset ( a, a ] and points [ a, a ] are excluded. Also, the requirement that subintervals be properleaves out the already studied case I = R . Other than that, there are no restrictions on thesubintervals of R considered in Theorem 4.3. Nevertheless, for convenience we shall assume in the proof that intervals are not open. The open case is handled in essentially the sameway, via a limit argument. So I will contain at least one endpoint. To simplify notation, weshall assume that this endpoint is the origin, and furthermore, that it is the left endpoint of I . Theorem 4.3. Let I ⊂ R be a proper subinterval, and let f : I → R be locally integrable.Then, for every t > , (4.3.1) ω ( M f, t ) ≤ Z ω ( | f | , tu ) du. Inequality (4.3.1) is sharp in the sense that for every modulus of continuity ω and every δ > with δ ≤ | I | , there exists a nonnegative, uniformly continuous function ψ : I → R such thatfor all t ∈ (0 , δ ) , we have ω ( ψ, t ) = ω ( t ) and (4.3.2) ω ( M ψ, t ) = Z ω ( tu ) du. Moreover, if ω is bounded, then the function ψ can be chosen so that it satisfies ω ( ψ, t ) = ω ( t ) and (4.3.2) for all t > with t < | I | .Proof. Assume f ≥ 0, and note that when we evaluate M f ( x ), taking the supremum overall intervals containing x yields the same value as taking the supremum over intervals having x as a boundary point (in other words, M f is the maximum of the right and left one sidedmaximal functions). Let x, y ∈ I be such that M f ( y ) < M f ( x ). Then M f ( x ) − M f ( y ) = sup S ∈ I S − x Z Sx f ( u ) du − M f ( y ) ≤ sup S ∈ I (cid:18) S − x Z Sx f ( u ) du − S − y Z Sy f ( u ) du (cid:19) (4.3.3) ≤ sup S ∈ I Z | f ( x + ( S − x ) u ) − f ( y + ( S − y ) u ) | du ≤ Z ω ( | f | , | x − y | u ) du. Now using a symmetry argument between x and y , and taking the supremum over | x − y | ≤ t ,(4.3.1) follows.We prove the optimality of (4.3.1) on the interval I = [0 , ∞ ). The argument can be easilyadapted to other proper subintervals of R . If ω is bounded, set ψ := k ω k L ∞ ([0 , ∞ )) − ω .It is immediate from the definitions that ω ( ψ, t ) = ω ( t ). Since ψ is decreasing, M ψ ( t ) = t − R t ψ ( u ) du = R ψ ( tu ) du , so(4.3.4) ω ( M ψ, t ) ≥ M ψ (0) − M ψ ( t ) = k ω k L ∞ ([0 , ∞ )) − Z ψ ( tu ) du = Z ω ( tu ) du. Thus, by (4.3.1) we have equality. If ω is unbounded, the result follows from the previouscase by fixing δ > ω δ ( t ) := min { ω ( t ) , ω ( δ ) } . (cid:3) Remark 4.4. In (4.3.4) we are crucially using that t < | I | . Suppose I = [0 , ω ( t ) = ω ( ψ, t ) = min { t, } , where ψ ( x ) = 1 − x . Then for all t ≥ ω ( M ψ, t ) = 1 / < t →∞ R ω ( tu ) du. For the same reason, the restriction t < | I | is also needed in (4.5.1)below. odulus of continuity of the maximal function 29 Recall that k f k Lip( ω,I ) = sup t> ω ( f, t ) /ω ( t ). In the next result we use the following no-tation: ψ T : [0 , ∞ ) → [0 , ∞ ) is defined by ψ T := ( T − ω ) + . We shall see that only thefunctions ψ T need to be taken into account when computing k M k Op( ω ) . If ω is bounded weset T = k ω k L ∞ ([0 , ∞ )) (as usual) and just write ψ . Corollary 4.5. Let ω be a modulus of continuity and let I be a proper subinterval of R . On Lip( ω, I ) , (4.5.1) k M k Op( ω ) = sup The sharp bounds given above yield information about related inequalities.For instance, in Theorem 5.1 of [AlPe] the following Landau type inequality is proven: If u : [0 , ∞ ) → R is an absolutely continuous function such that its derivative is of boundedvariation, then(4.6.1) k u ′ k ∞ ≤ k u k ∞ (cid:0) k DM ( u ′ + ) k ∞ + k DM ( u ′− ) k ∞ (cid:1) . It is natural to seek a lower bound on the best possible constant c that can replace 48 in(4.6.1). Recall that the classical (sharp) Landau inequality assumes more regularity on thepart of u ′ : If it is absolutely continuous on [0 , ∞ ), then(4.6.2) k u ′ k ∞ ≤ k u k ∞ k u ′′ k ∞ . As noted in Remark 5.5, of [AlPe], (4.6.1) implies Landau’s inequality, save for the issue ofbest constants. Using Corollary 4.5 and (4.6.1) we have that k DM ( u ′ + ) k ∞ + k DM ( u ′− ) k ∞ ≤ − k Du ′ + k ∞ + 2 − k Du ′− k ∞ ≤ k Du ′ k ∞ , so 4 ≤ c ≤ 48. On the whole real line, the constantfrom Theorem 5.1 of [AlPe] appearing in (4.6.1) is 24 instead of 48, and the best constant inthe sharp Landau’s inequality (4.6.2) is 2 instead of 4. Arguing as before and using Corollary3.6, we have k DM ( u ′ + ) k ∞ + k DM ( u ′− ) k ∞ ≤ √ − k Du ′ k ∞ , so 1 / ( √ − ≤ c ≤ R the best constant appearing in the generalized Landau inequality (4.6.1) is strictlylarger than the best constant 2 in the classical Landau inequality. Remark 4.7. By Corollary 4.5 the functions ψ α ( x ) := max { − x α , } are extremal in theclasses Lip( α )([0 , ∞ )). Now M ψ α ( x ) = 1 − x α α on 0 ≤ x < M ψ α ( x ) = α (1 + α ) x on 1 ≤ x < ∞ ). Thus, the H¨older or Lipschitz exponent of M ψ α is no better than that of ψ α . But in this example there is only one “bad point”: For every ε > 0, both ψ α and M ψ α are Lipschitz on [ ε, ∞ ). In view of the regularizing properties of M , one may wonder, forinstance, whether M f must have a better H¨older exponent than f save for small sets, or insome “almost everywhere” sense. We shall see below that the answer is negative.We also note that the preservation of regularity does not extend to the C r classes. To seethis, let f be the sum of two smooth bump functions with disjoint supports and recall that M f is the maximum of the right and left one sided maximal functions. This entails thatbetween the bumps of f , M f achieves its minimum value at a point of non-differentiability,since the right and left derivatives of M f are nonzero there and have opposite signs.Next we show that a H¨older condition on f is not sufficient to ensure the differentiabilityalmost everywhere of M f , so in particular M f can fail to be absolutely continuous. Thefollowing example also shows that the H¨older exponent of M f may be as bad as that of f on a large set. To prove this we suitably modify the fat Cantor set defined in Example 4.2 of[AlPe], and the functions used there. Example 4.8. For every α ∈ (0 , there exists a function f ∈ Lip α ([0 , , with Lip α ( f ) = 1 ,such that M f is not differentiable on a set E of positive measure. Furthermore, given any β ∈ ( α, , M f is not locally H¨older ( β ) at any point of E . More precisely, given any x ∈ E ,and any interval I , relatively open in [0 , and with x ∈ I , we have M f / ∈ Lip β ( I ). Proof: As in Example 4.2 of [AlPe], and unlike the usual construction of the Cantor set,instead of removing the “central part” of every interval at each stage, we remove several parts.Let F = [0 , 1] and let F n be the finite union of closed subintervals of [0 , 1] obtained at step n of the construction, to be described below. As usual C := ∩ n F n . Denote by ℓ ( J ) and r ( J )the left and right endpoints of an interval J . At stage n we remove 2 − n of the mass in the odulus of continuity of the maximal function 31 preceding set, so | C | > − | ∪ ∞ F cn | ≥ − P ∞ − n = 2 / 3. Let I n − be a component of F n − ,and let O ( n, x ) the open interval centered at x of length 2 − n | I n − | . Select the unique 2 n points x , x . . . , x n ∈ I n − such that i) ℓ ( I n − ) < x < · · · < x n < r ( I n − ), and ii) after theremoval of the 2 n open intervals O ( n, x i ) from I n − , the 2 n + 1 remaining components haveequal length, which therefore must be (1 − − n ) | I n − | / (2 n + 1). Thus, x − x = x i +1 − x i for i < n , while r ( I n − ) − x n = x − ℓ ( I n − ) < x − x . Note that the extremes of I n − areleft untouched.Repeat the same process with the other components of F n − to obtain F n , which by con-struction is a disjoint union of closed intervals, all of length (1 − − n ) | I n − | / (2 n + 1).Next we define the function f . Given t > 0, set g t ( x ) = 1 − ( t − x ) α on [0 , t ) and extend itto ( − t, t ) as an even function. Then let f be identically 1 on C , and let f ( x ) := g t ( x − s ) oneach interval of type O ( j, s ), where t is half the length of O ( j, s ). To see why f ∈ Lip α ([0 , α ( f ) = 1, note that if x ∈ C and y ∈ [0 , \ C , say, with y < x , then there is aunique O ( n, w ) containing y , so we can replace x by r ( O ( n, w )) and then f ( x ) − f ( y ) = f ( r ( O ( n, w ))) − f ( y ), while r ( O ( n, w )) − y ≤ x − y ; the case where x and y belong to differentcomponents of [0 , \ C can be reduced to the previous one: Let y ∈ O ( n, u ), x ∈ O ( m, w ),and suppose y < x . If f ( x ) < f ( y ), replace y with ℓ ( O ( m, w )); otherwise, replace x with r ( O ( n, u )). So it is enough to verify that f has the desired properties on the closure of anarbitrary O ( j, s ), and this is true since it holds for | x | α , its translates, reflections, etc. Next,let f n := 1 on F n and f n := f on [0 , \ F n . Then f ≤ f n , so M f ≤ M f n , and of coursethe latter function is easier to work with. Let x i and O ( n, x i ) be as above. For notationalconvenience, suppose we are taking i = 2; the same reasoning applies to the other x j . Toevaluate M f n ( x ) we only need to consider intervals J that have x as an endpoint and, weclaim, omit both x and x . To see this, assume that x = ℓ ( J ). Since f n ( x ) < M f n ( x ) < J must satisfy f n ( r ( J )) = M f n ( x ). Otherwise we could improve the average byeither shortening or enlarging J . Thus, r ( J ) ∈ O ( n, x ) ∩ { x < x } . Of course, the situationis completely symmetrical if x = r ( J ) (here ℓ ( J ) ∈ O ( n, x ) ∩ { x > x } ), so we keep theassumption x = ℓ ( J ). Next, define h ≥ f n by setting h ≡ O ( n, x ) ∩ { x < x } and h ≡ f n elsewhere. Clearly M f n ( x ) < x − x Z x x h = ( − − n n +1 + 2 − n − ) | I n − | + R − n − | I n − | − x α dx ( − − n n +1 + 2 − n ) | I n − | = 1 − ( − n − α +1) ( − − n n +1 + 2 − n )( α + 1) | I n − | α < − − αn − n − α − | I n − | α . Fix z ∈ C ∩ { M f = 1 } . For each n , let I n,z be the component of F n that contains z , and let w n be midpoint of the nearest interval O ( n ) deleted at step n (if there are two such midpoints,pick any). Then | z − w n | < − n | I n − | , and furthermore | I n − | < − n ( n − , since the number ofcomponents of the set F n − is Q n − i =1 (2 i + 1) > Q n − i =1 i = 2 n ( n − . Thus, for every β ∈ ( α, we havelim sup w → z | M f ( z ) − M f ( w ) || z − w | β ≥ lim sup n →∞ − M f n ( w n ) | z − w n | β ≥ lim n →∞ − αn − n − α − − βn (2 − n ( n − ) β − α = ∞ . Since M f ≥ f a.e., the sets E := C ∩ { M f = 1 } and C have the same measure, so taking β = 1 it follows that M f is not differentiable on a set of measure at least 2 / 3, while for β ∈ ( α, 1) we obtain the failure of the local H¨older condition with exponent β . (cid:3) Remark 4.9. The following example shows that local, higher dimensional generalizationsof the preceding results may fail. Recall that in Definition 2.1 the balls in R d we av-erage over, are assumed to be fully contained in the (open) domain U of f . Let U = (cid:8) ( x, y ) : x ∈ R d , y ∈ R , y > exp ( − / | x | ) (cid:9) be a domain with a cusp and let f ( x, y ) = y .Then lim y ↓ M f (0 , y ) = 0, while M f (0 , 1) = ∞ . References [Al1] Aldaz, J. M. Remarks on the Hardy-Littlewood maximal function. Proceedings of the Royal Societyof Edinburgh A, 128 (1998), 1–9.[Al2] Aldaz, J. M. The weak type (1 , bounds for the maximal function associated to cubes grow to infinitywith the dimension. To appear in Ann. Math. 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E-mail address : [email protected] Dipartimento di Matematica, Universit`a di Milano-Bicocca, Edificio U5, via R. Cozzi 53,20125 Milano, Italia. E-mail address : [email protected] Departamento de Matem´aticas y Computaci´on, Universidad de La Rioja, 26004 Logro˜no,La Rioja, Spain. E-mail address ::