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Let’s work on for some fixed dimension . In the following I’ll write for the maximal function on uncentered cubes. In the following setting it won’t make a difference whether to work with this or with the usual one (defined on the centered balls), except in the constants; which I’ll throughout ignore thanks to my friend *the wiggle notation* and the fact that the dimension is fixed. In the proofs one has to resort alternatively to the dyadic, Hardy-Littlewood, centered versions of but it really just amounts to technicalities. I’ll ignore these minor issues as the reader can easily see how they are overcome. When in doubt, refer to the excellent [Duo], which is where all of the following is taken from.

I’ve split the post in two for convenience, you can find part II here.

**1. Definition of and weighted inequalities for the maximal function **

Definition 1 ( weights)A locally integrable non negative function is said to be in the weight class if:

- in case , it holds that for any and any cube
(here is the measure of w.r.t. measure , so );

- in case , for any cube it must be

It is useful to refer to the quantity

as the constant of .

Notice , so that the second factor in the last expression is some norm average of . To appreciate the symmetry, notice the condition is

A further and more substantial symmetry to appreciate is the fact that , since .

Finally, notice that the condition is equivalent to

Where do these conditions come from? Quite simply, one would like to extend boundedness results for operators to weighted spaces, like , and even viceversa. Starting from the maximal function , one would like to see for which weights and a given it still holds that

or its weak counterpart. Indeed, the condition above is an equivalent statement to the boundedness of , and this is where it comes from

Actually, the condition implies the strong boundedness too (if though), but we’ll see this later.

Let me state the weak boundedness clearly:

let’s see how to deduce from this the condition for . Choose a cube s.t. , then for any it is , thus

or in other words the ratio is bounded from below by , and in particular must be a.e.. Anyhow, we’d like some cancellation in the condition (4) and therefore we ask , that is . If we substitute, we get precisely the condition (2). When instead, for any take to be , then

and the condition (1) follows by the arbitraryness of .

Now we have to do the opposite, and prove the condition implies the weak inequality (3). The fundamental tool here is the **Calderón-Zygmund decomposition**. Recall that given a function and , we can decompose in and its complement , s.t.

- , where is a collection of disjoint dyadic cubes;
- for each it is
- a.e. on .

One constructs such a decomposition by the usual stopping time argument (I’ll talk about stopping time arguments in a future post, probably). How does this help us? It helps because all the cubes in are critical, in that . Moreover, is exactly the set , and by dilating the cubes by a constant factor, they’ll cover as well. Notice that by (5) for it is uniformly in the cubes that , for . This is an example of the technicalities I mentioned in the introduction; from now on we’ll just pretend , keeping in mind this isn’t 100% rigourous. We split in two cases:

- . In this case we rely on an inequality for which is true for
**any**positive :notice that Marcinkiewicz interpolation works on the as well (the case can be verified by hand), and thus the former implies for any

Anyway, if we substitute to then in the above can be relaxed to by the condition itself, and the weak (1,1) inequality follows.

I’ll prove (6) for the sake of completeness: make a Calderón-Zygmund decomposition of w.r.t. , then

- . Here we claim that the condition (2) is actually equivalent to (4), and with this we can take a Calderón-Zygmund decomposition (from now on CZ-decomposition) as done for above, and write
(we went from the 1st line to the 2nd thanks to (4)). This is the weak boundedness (3); it remains to prove the claim. This is easy, by Hölder and the condition (the opposite deduction has been made above):

and since we have exactly the factor in the condition, and thus the expression is

which is just (4) indeed.

Remark 1Of course, by Marcinkiewicz interpolation on the , it follows that if then

for all .

We’ve mentioned before that is actually equivalent to the strong boundedness of . The CZ-decomposition works really well on weak inequalities, as it’s known from standard CZ theory of singular integrals; what about the strong inequalities though? Well, in this particular case we are able to somewhat refine the proof above, by using infinitely many CZ-decompositions indeed. Let me illustrate the strategy first. Suppose ; by the remark 1, the strong inequality would immediately follow if were to belong to some class for . It turns out this is always the case! Every class is made of the union of the previous classes:

This is quite an interesting property. First of all, by Hölder’s inequality\footnote{The spaces are not necessarily normed spaces, we’re using the general version of Hölder’s inequality.} we already know that is at least as large as : if and we assume holds, then we can write

since , and this is the condition. Notice in particular that we’ve proved

How to prove a property such as (7) ? We’re gonna use the following fundamental observation, which will be used multiple times in the following

Lemma 3[Reverse Hölder inequality] For any weight with there is a small enough s.t.

and the implied constants can depend on , on and the constant of , but nothing else.

This is called reverse for the obvious reason: indeed, Hölder goes the opposite direction

With Lemma 3 our claim that is in some for holds: indeed, remember , thus by the lemma

choose s.t. , then and therefore ; by this choice and the condition then

which is nothing but the condition for . Notice we can express the same fact by saying that there exists an s.t. .

So, we’ve laid out all the steps from Lemma 3 to the strong weighted inequality for when . It remains to prove the lemma, and we’ll be done with this section. As mentioned before, we have to build a series of CZ-decompositions. They’ll be CZ-decompositions of the function – of course – w.r.t. a collection of parameters rapidly increasing. The ordering of forces the resulting sets to be nested and in particular we’ll see they shrink rapidly. We’ll then use the shells to partition the whole domain, and use the fact that because a shell is outside of some , the value of on the shell is controlled by the corresponding . This is similar in spirit to decomposing a function according to its dyadic size

(indeed, we’ll see we have to take the ‘s lacunary). So, let’s start right away by fixing a cube : we want to prove that there exists an independent of the cube and s.t.

Fix and apply CZ-decompositions to w.r.t. every , by iterated bisection of the cube . This yields collections of cubes and sets

(the superscript is just for convenience) that are indeed equal to [technicalities alert]. Thus it is immediate that

This holds on the level of cubes too: every is contained in some cube . So, let’s denote by the collection of cubes in that are contained in ; then for any , thanks to the properties of CZ-decomposition we have the chain of inequalities

This inequality is meaningful only if , thus we want the ‘s to increase at least geometrically. Assume this is the case; then we want to invoke some sort of continuity of the measure : that is, if and for fixed , then uniformly in it must be

where . This is true and can be seen as follows: as seen at the beginning, for some constant

and therefore

Notice depends on the constant of too. Anyway, set ; we can fix for our purposes, i.e. . Then (8) implies

and by summing over all cubes in also

which means decreases geometrically and by nestedness

This pleasant property allows us to split all of the starting cube in

without missing any relevant bits. It remains therefore to estimate by exploiting this partition of . Here’s where the nested CZ-decompositions most come in handy: by construction, for it is ! We have control over the magnitude of and used the CZ-decompositions to achieve this control and give us structure as well. It’s now very easy to estimate

and the sum can be made finite by choosing sufficiently small. This yields

as we wanted, and the proof of the lemma is thus concluded.

The proofs in this section should make one appreciate the power of CZ-decomposition.

**2. weights characterization **

One first thing to notice is that is indeed the smallest of all the weight classes: if then

and therefore too. Nevertheless, combining weights in a simple way we can create weights for any : if , then

Indeed, this follows simply by the definition, since

What’s more surprising is perhaps the fact that *any* weight can be factored as for some weights (so weights are quotients of weights!). Let me show how: we have the identity , and we’d be done if it were . It isn’t necessarily so, but tweaking this factorization a little allows us to make it so the factors are provably in . Indeed, we prove that for some function we have , , and since that’s it. The function is constructed as follows: consider the operator

This maps to itself boundedly: indeed, if then , then by Theorem 2 too, and therefore . Moreover, it is positive and, by Minkowski’s inequality, sublinear. Now if define

which has norm dominated by ; it holds pointwise, by positivity,

and therefore

which is the condition for . As for the other factor, one has to use the analogous operator , and repeating the above you get s.t. . Now the only problem is that , but this can be solved by repeating the above directly on operator , which has the same properties as .

We now want to characterize the weights completely. It should be no surprise that the maximal function has its role in here. Let’s state a theorem directly

- Suppose locally integrable and finite a.e.. Then for any one has
The constant here depends on but not on .

- Suppose . Then there exists , s.t. too (so that ) and a s.t.

Thus, the weights are essentially of the form . *Proof:* In order to prove (1.) we have to prove that for any cube

The idea is that some part of the function already behaves well w.r.t. the inequality, and the other part can be dealt with. Write and , so that

because . Therefore it is enough to prove the inequality for and in the LHS. The term containing is already fine: by construction, if is such that , then there must exist a cube s.t. , and therefore it must be that (here is the side length). It follows that dilating by a constant (depending on ) then and thus

which implies pointwise on that . The inequality is thus satisfied by . As for , we resort to an inequality of Kolmogorov (proved in the remark following the end of the proof), that says that for a weak operator it holds

We now prove (2.). By the reverse Hölder inequality (Lemma 3) there exists some s.t.

and taking the supremum over , by the condition we have

(the inequality from below is a consequence of Lebesgue differentiation theorem). This is enough, because then and , and the latter size comparison means that .

Remark 2By the weak inequality, if

Thus Kolmogorov’s inequality (9) is proved.

**3. Extrapolation **

After we’ve built our weights, we might want to go back to the original question and study on which spaces typical operators (e.g. singular integrals) are bounded. One might expect a priori a very complicated behaviour, as there are many possibilities. There is though one strong property in action here: an operator can’t be bounded for all weights in a class without being bounded for *any* weigth in any class (w.r.t. the exponents of the corresponding classes). Formally

Theorem 5 (Extrapolation)Fix . Suppose is a bounded operator on all of for all . Then it is bounded on all of for for any .

This is indeed a very strong statement, as the hypothesis involves just one exponent and the conclusion holds for any. We can express the collection of all spaces with weights succintly as .

The proof of the theorem relies mainly on the properties of . Indeed, let’s prove a weaker statement first: that is bounded on , where and (i.e. bounded on every space in ). It is weaker in the fact that . By the characterization of weights (theorem 4), is an weight since , and therefore is an weight; then

because too, and finally since one has , and therefore the last expression is bounded by , as we wish. This proves is bounded on for . To pass to all the we have to play a little with the reverse Hölder inequality (Lemma 3). Notice if we had for any then we’d have automatically: every belongs to some for some lesser , and then will do. So we set to prove boundedness on , by exploiting the boundedness. To this extent, by duality there exists s.t.

and moreover has norm . Now, trivially for any you might choose. By the previous section, since is locally integrable, we have is an weight, and we can use boundedness, so that the norm above is bounded by

and by Hölder with exponents , this is further bounded by

The first factor is exactly what we were hoping for, so we have to prove the second factor doesn’t screw things up. , and remember , so that . By reverse Hölder inequality again, if is close enough to 1 then too, and therefore by theorem 2 we can bound the second factor above by

This concludes the proof. Thus, we’ve seen how boundedness for produces boundedness for every . I would like to stress the power of this fact by pointing out that the constants in the hypothesis are allowed to depend on without restriction.

Remark 3Take an operator bounded on . Fix some . What can we say about the range of ‘s for which is bounded on ? First of all, for some , and therefore by the extrapolation result is bounded on as well, and by Marcinkiewicz interpolation it’s bounded inbetween and . Next, for all , and therefore is bounded on for all . Thus boundedness on (along with all the other weights) implies boundedness on an open set of exponents of the form , where .

End of part I. Part II will be about applications of weights theory.

References:

[Duo] J. Duoandikoetxea, “*Fourier Analysis*“, Graduate Studies in Mathematics, vol. 29, AMS Providence, Rhode Island.