BV Continuity for the uncentered
Hardy–Littlewood maximal operator
Abstract.
We prove the continuity of the map from to itself, where is the uncentered Hardy–Littlewood maximal operator. This answers a question of Carneiro, Madrid and Pierce.
Key words and phrases:
Maximal operators; continuity; bounded variation;2010 Mathematics Subject Classification:
26A45, 42B25, 39A12, 46E35, 46E39, 05C12.1. Introduction
1.1. A brief historical perspective and background
The study of regularity properties for maximal operators has been an important topic of research over the last years. The most classical object in this context is the centered Hardy–Littlewood maximal operator, that for every and is defined as
where and is the –dimensional Lebesgue measure. The uncentered Hardy–Littlewood maximal operator, denoted by , is defined analogously by taking the supremum over balls that contain the point but that are not necessarily centered at .
The program of studying the regularity of maximal functions was initiated by Kinnunen [15], who proved that the map
(1.1) |
is bounded on if Since then, several authors have made many contributions in this direction, including connections with potential theory and partial differential equations (see, e.g., [4, 5, 7, 13, 16, 22, 23]).
An important direction has been the study of the continuity of the map (1.1) in several settings. Since it is not necessarily a sublinear map at the derivative level, its continuity does not follow from the boundedness. The first result answering a question of this kind was due to Luiro [18], who proved that the map (1.1) is in fact continuous from to itself for each
1.2. The endpoint case
This case is much more subtle. In fact, Kinnunen’s result certainly does not hold because for every non trivial we have Nevertheless, since we are interested at the derivative level of the maximal functions, the natural version of the problem at is to study the map
(1.2) |
for In fact, the boundedness of the map (1.2) from to is an important open problem for In dimension the answer is positive, as has been established for the uncentered version in [24] and for the centered version in [17]. In [19], Luiro proved that the map (1.2) is bounded when restricted to radial data in the uncentered setting. Other boundedness results in the radial setting have been achieved in [2, 3, 8, 14, 20].
An important work for our purposes is [1], where the boundedness
for every was proved. Moreover, the authors proved that is absolutely continuous for providing then an improvement over the original function.
1.3. Continuity at the endpoint case
In this article we are particularly interested in the continuity of the map (1.2) at the endpoint We notice that the methods outlined by Luiro [18] cannot be adapted to our case, since they depend on the fact that the map
is bounded on for . In [12] the first continuity results at the endpoint were obtained. The authors proved, among other results, that the map
is continuous from to (cf. [12, Theorem 1]). This result was recently extended to the setting (the subspace of consisting of radial functions) in [9], where a more general approach, that can be applied in the context of other maximal functions, is proposed.
In [12] the authors also consider the space endowed with the norm About this, they asked the following question:
Question 1.
(Question B in [12]) Is the map continuous?
This question, in case of being answered affirmatively, would provide a generalization of [12, Theorem 1] (since embeds isometrically in ). It is important to notice that, in general, the continuity in the setting is more delicate that in the setting. An example of this is that in the fractional setting the analogue of [12, Theorem 1] holds (see [21]) but the answer to the analogue of the previous question is negative (see [12, Theorem 3]).
The main goal of the present manuscript is to answer this question. We prove the following.
Theorem 1.
The map is continuous.
Several of the arguments in [12] (also the arguments in [9]) rely on the regularity of the original function, therefore they are not enough to conclude Theorem 1. In fact, the authors (also the authors of [9]) used in their work a reduction of the problem to the analogous question stated for “lateral” maximal operators, but in our case that reduction causes several problems. For instance, the one-sided maximal function of a function in is not necessarily continuous, which brings additional difficulties. Therefore, a different idea is required in order to achieve the result. Our methods, although inspired by [12], provide a new approach to this kind of problems.
Having taken into account the main result obtained here, we summarize the situation of the endpoint continuity program (originally proposed in [12, Table 1]) in the table below. The word YES in a box means that the continuity of the corresponding map has been proved, whereas the word NO means that it has been shown that it fails. The remaining boxes are marked as OPEN problems. We notice that after this work the only open problems in this program are the ones related to the centered Hardy–Littlewood maximal operator in a continuous setting.
———— | continuity; continuous setting | continuity; continuous setting | continuity; discrete setting | continuity; discrete setting |
---|---|---|---|---|
Centered classical maximal operator | OPEN | OPEN | YES2 | YES4 |
Uncentered classical maximal operator | YES1 | YES: Theorem 1 | YES2 | YES1 |
Centered fractional maximal operator | YES5 | NO1 | YES3 | NO1 |
Uncentered fractional maximal operator | YES4 | NO1 | YES3 | NO1 |
2. Proof of Theorem 1
2.1. Preliminaries
In this subsection we develop the main tools required in our work. We start by stating the following result which describes the behavior of the maximal function at infinity.
Lemma 2.
Given let and Then
where .
Proof.
Without loss of generality we assume that and . Observe that
holds for any . Fix and let be such that for . We choose satisfying
Consider satisfying and any interval . If , then clearly
On the other hand, if , then
Since is arbitrary, the claim follows. ∎
The next goal is to use the norm to control the difference between two functions or between their maximal functions at a given point . The following estimates, although very basic, will be extremely useful later on.
Lemma 3.
Let . Then
hold for any .
Proof.
The first inequality follows since
Now, assume . By the first part of the lemma for any we have
Thus, and the second part follows as well. ∎
Contrasting with the setting (see [12, Lemma 14]), in our context to make the reduction to the case is much more problematic. In order to deal with this issue we require several results describing the relations between and .
In the following, for given we define and . Also, for each we introduce the quantity
where the supremum is taken over all and all sequences (notice that if is continuous at and , then the sequences satisfying can be considered instead and the supremum will not change). For a given partition we denote . Finally, we write and
Lemma 4.
Fix Then for any we have
Proof.
Fix . We write and , assuming that both of these sets are infinite (the other cases can be treated very similarly). Given we choose a partition such that
and
Then for fixed we construct by adding to (if needed) some extra points. The procedure consists of the following three steps.
-
(i)
We set
-
(ii)
For each we choose such that
and Similarly, we choose such that
and Then we set
-
(iii)
For , we let be the set of all pairs satisfying with and which are not of the form or . Let . If there exists such that , then we just add to If not, then at least one of the sets
and
must be non-empty (here is the usual sign function taking the value of , , or , if , , or , respectively). Assume (the other case is similar) and choose . Then for some We find such that (in particular, we have ), and add both and to The above process terminates after steps and we denote the final collection of points by
Having constructed we see that
Also, we obtain
since the only terms that contributes to are those corresponding to the pairs and Letting and we obtain the claim. ∎
Now, we use Lemma 4 to show that the map is continuous from to .
Lemma 5.
Fix and let be such that Then for any we have
Proof.
It is possible to verify that implies Thus, it remains to show
We define and as in the previous lemma. Given we choose such that
We also denote and By Lemma 3 we have that for big enough
and
Moreover, we have
and, similarly,
Finally, we observe that and for big enough, by the uniform convergence. Letting (and thus ) and applying Lemma 4, we obtain the claim. ∎
Let us now take a closer look at the properties of the maximal operator. Recall that the total variation of can be controlled by the total variation of . There is also a local version of this principle, where we focus on an interval . However in this case some boundary terms must be included. Thus, to avoid the possibility that behaves badly at or , we use its adjusted version defined by
It is known that is upper semicontinuous and that (see [1, Lemma 3.3]).
Lemma 6.
Fix Given we have
Proof.
This follows by a slight modification of the proof of [1, Lemma 3.9]. ∎
The next result gives us the uniform control (with respect to ) on the behavior of near infinity, provided that is a converging sequence in . This, in turn, allows one to restrict the attention to a bounded interval, while dealing with the total variations of the maximal functions . We point out that it is also possible to proceed without this reduction, but then for all considered functions the extended domain should be used instead of .
Lemma 7.
Fix and let be such that Then for any there exist such that
and
for every big enough.
Proof.
We prove that there exists such that the symmetric case is treated analogously. First we deal with the case where . Assume that . Let us take big enough such that (we use here Lemma 3) we have and for every Therefore, for big enough such that and for each we have This is the case because any interval satisfying contains since if , in particular and then
Therefore
for big enough, from where we conclude this case.
Now, we deal with the case where By Lemma 6 and [1, Lemma 3.3], assuming that is a continuity point for , we obtain
The analogous is obtained for instead of . Let us assume that is a continuity point for and every , such that By Lemma 5 we have
(2.1) |
for big enough. Also,
If is big enough to have , then by Lemma 3 we get and for big enough. Combining this with (2.1) concludes the proof. ∎
Before we prove our key result regarding the variation of maximal functions, we need the following definition. A given partition with has property (V) with respect to if for each we have .
Proposition 8.
Fix and let be such that Then
Proof.
By Lemma 7 it is enough to prove that for every interval with both and being points of continuity for and every . In the following we fix satisfying such assumption. Observe that Lemma 3 and Fatou’s lemma imply
Now, we prove the remaining inequality, that is,
Given we show that
holds if is big enough. Let be a partition satisfying
and
Also, by the uniform convergence and Lemma 5 we conclude that
(2.2) |
and
(2.3) |
hold for big enough. Now, we take such that and
Without loss of generality we can assume that is such that for each the set satisfies property (V) with respect to unless it consists of two elements. For each such we denote and claim that it is possible to find a partition such that
Indeed, for we use the convention that all the variation terms above are equal to so the inequality holds (we set or if or , respectively). It remains to consider the case in which property is guaranteed. We assume that (the opposite case can be treated analogously). Then shall be chosen in such a way that given we have
for odd, and
for even. We describe in detail the procedure for selecting the points If is odd, then we find an interval such that
Clearly, and we can find satisfying For even we take if or otherwise. Since there exists satisfying We note that the appropriate configuration of the sets guarantees that the inequalities hold.
Observe, also, that the partition either consists of elements or satisfies property (V) with respect to . Thus, regardless of which case occurs, we obtain
where for and for (the boundary values depend on and the parity of ). Similarly,
(we eventually change the sign of the second term on the left-hand side in order to get the boundary coefficients equal to and ). Consequently, the claim follows since for each we have
Now, we apply the claim in order to get the following estimate
where In particular, we note that consists of at most elements and thus
follows by Lemma 3 for big enough. Combining the above inequalities with (2.3), we arrive at
which, in view of (2.2), gives
provided that is big enough. Consequently, ∎
Having obtained Proposition 8 we continue with the remaining tools required. Our general purpose in the next few lemmas is to get more information about the derivative of maximal function. In particular, we are interested in studying the behavior of the sequence for a given point .
Lemma 9.
Fix and let be such that If with , and a.e. with , then we have
Proof.
Follows a slight modification in [12, Lemma 12]. ∎
Let us now define Since is absolutely continuous and is upper semicontinuous, we have that is open. We notice that if , then there exists a finite interval such that Indeed, there exists a sequence such that and Since , we have for some . Thus, by taking a subsequence (if required), we get and , with By the boundedness of we conclude that Also, let us observe that therefore or
The next result states that for a.e. the derivative of the maximal function can be described by an explicit formula.
Lemma 10.
Let . Assume that is differentiable and is continuous at (that happens a.e. because and have bounded variation). Let us suppose that is such that there exists an interval with such that and or . Then
Also, if , then we have .
Proof.
The last claim follows because is a local minimum of Assume without loss of generality that , (the other case is similar). We have, for , that
as Therefore Also, for we have
as This concludes the proof. ∎
Now, we can use the obtained formula to prove the following result regarding pointwise convergence.
Lemma 11.
Fix and let be such that Then
a.e. in
Proof.
The claim is trivial if has measure zero, so assume this is not the case. We define as the analogue of for Let us take such that and are differentiable at for every and and is continuous at . By the uniform convergence we have that for big enough. We also make the following observation. If there are intervals such that , then the quantities are bounded below uniformly. Indeed, if for a sequence we have , then we would have by the uniform convergence and continuity of at , contradicting the pointwise convergence of the maximal functions.
Assume first that and take such that Then for big enough we have Also, there exists such that for big enough and each we have . We can observe then that for big enough. Let us assume that we have and a sequence such that
(2.4) |
Without loss of generality assume that (the other case is treated analogously). Since , there exists a subsequence (that we also denote by ) such that Moreover, in view of the previous observation, we have . Thus, Lemma 9 gives and consequently, in view of Lemma 10, we obtain . Also, holds. However, by the uniform convergence we have
reaching a contradiction with (2.4). Thus, we conclude this case.
Now, if then by Lemma 10 we have Also, if for a subsequence we have , then Therefore, this subcase follows and we can assume that It is now enough to prove that Let us suppose that for some and a subsequence we have As before, we assume the case . We claim that there exists such that for big enough we have Indeed, in view of
we have and thus gives our claim. Now, since we have that for big enough. Consequently, there exists a subsequence (that we also denote by ) such that for some Then by Lemma 9 we have that Therefore, Lemma 10 gives
and the left-hand side must be equal to . Since we have
by the uniform convergence, we reach a contradiction. This concludes the proof. ∎
It remains to take a look at the set .
Lemma 12.
Let . Then for a.e. we have
Proof.
Assume that and are differentiable at (this happens a.e. because and have bounded variation). Then, since and , we have Now, assume, in order to get a contradiction, that (the other case is analogous). Then there exist such that for every Thus, for a.e. we have , which implies a contradiction. ∎
Combining the previous results we obtain the following.
Corollary 13.
Fix and let be such that Then
2.2. Proof of Theorem 1
Finally, we are ready to prove the main result. In what follows denotes the set analogous to defined for instead of
2.3. Concluding remarks
We end our discussion by showing that the assumptions are important, not only
Example.
Let and take
Then we have , while .
Indeed, the first claim is obvious and for the second one we argue as follows. We observe that and for . Moreover, if , then for any we have
which is due to the fact that for any interval we have Thus, for we have .
Acknowledgements.
Cristian González-Riquelme was supported by CAPES-Brazil. Dariusz Kosz was supported by the National Science Centre of Poland, project no. 2016/21/N/ST1/01496. The authors are thankful to Emanuel Carneiro for helpful discussions during the preparation of this manuscript.
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