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This is the 2nd part of a post on basic weights theory. In the previous part I included definitions and fundamental facts, namely the nestedness of the classes, the weighted inequalities for and an extrapolation result for general operators. Refer to part I for notation and those results. This part includes instead a fundamental result on Calderón-Zygmund operators, together with inequalities relevant to the proof, and an elegant proof of Marcinkiewicz’s multiplier theorem as an application.

All of the following is taken from the excellent [Duo].

**4. Calderón-Zygmund operators**

This section is about the following result: if is now a Calderón-Zygmund operator, then is bounded on any for all ; moreover, if then is weakly bounded on . This is exactly what one would’ve hoped for, and suggests that the weights theory is indeed one worth pursuing.

Recall a CZ-operator is an bounded operator given by integration against a CZ-kernel , defined on at least, i.e.

with the properties (for some )

- ;
- for ;
- for .

Then, let’s state formally

Theorem 6Let be a Calderón-Zygmund operator. Then is bounded on all for :

Moreover,

The proof of the second part is a close adaptation of the standard proof of weak boundedness of a CZ-operator, using CZ-decomposition. The first part relies instead on some inequalities relating maximal functions to the sharp maximal function , the maximal function that controls oscillation around the average:

where . In particular, one needs to prove the two lemmas

Then the first part of theorem 6 follows at once if you can prove :

where the last inequality holds by the strong version of theorem 2 provided is close enough to for to belong in (again, by reverse Hölder inequality). Again, the reason we resort to maximal functions of the form is because equality (7) gives us some room to stretch our exponents. As for proving , it is enough to prove , and it is enough to prove the result for bounded and compactly supported since these functions are dense in every . Therefore, if , the mass of around the support of is bounded by

which is finite since for every and by reverse Hölder inequality the second factor is finite as long as is sufficiently small. As for the mass of away from the support of , we can estimate pointwise, for

and therefore

once again we have to use the reverse Hölder inequality: for some , therefore

and since the above series is convergent (we’ve used (5)).

It remains to prove the lemmas, and to prove the weak boundedness part of theorem 6. We prove the latter here and move the proofs of the lemmas to an appendix to ease the presentation.

So, we want to prove that for and with CZ-kernel, bounded,

Do a CZ-decomposition of w.r.t. , which yields and write

where

and

is defined as a consequence. Notice everywhere. Then it is enough to estimate

Since is bounded ( and thus too), one has

and since we can estimate

and thus is alright.

As for , we exploit cancellation on every . Write for (so and ) and for (i.e. points close to any ); we only need to estimate the contribution from outside this set, since it’s small

Thus, since for any , estimate

where is the center of cube , and by the hypotheses on the kernel this quantity is bounded by

Now and , thus this is bounded by

Since , the factor in front of the integral is

and as done before, by the condition we can bound the expression by

and since and , the last term is controlled by and we’re done.

Remark 4We’ve seen is bounded on for any , but how does the operator norm depend on ? for the case it was conjectured that

and this was known as conjecture. It was solved positively by Hyt\”{o}nen in 2010, and is now known as theorem. One of the presentations at last year’s Summer School in Bonn was exactly about the solution to the conjecture. I can’t remember the details but I’m planning to read the whole paper, hopefully I’ll get back on this.

**5. Quick recap **

- the condition is equivalent to strong boundedness of for , and to weak boundedness for ;
- For every one can find a smaller s.t. too; in particular, every class is exactly the union of the previous classes: . This property turns out to be fundamental;
- functions of the form for are weights (and all weights are like this, modulus a factor );
- Once an operator is bounded on all for just one exponent , it is bounded on all for every exponent ;
- if is Calderón-Zygmund in particular, then it is bounded on every for , and weakly bounded w.r.t. weights.

**6. Applications: an elegant proof of Marcinkiewicz multiplier theorem **

As a treat to the reader who made it so far, I want to show the usefulness of weights theory by a remarkable example. We’ll prove Marcinkiewicz multipliers are bounded for with very little effort. Remember is said to be a Marcinkiewicz multiplier if and the total variation of *on every dyadic interval* is uniformly bounded by a constant. Then we can state formally

Theorem 9 (Marcinkiewicz multipliers)Let be a Marcinkiewicz multiplier. Then the operator defined by

is bounded for every .

The proof goes through an easy lemma:

Lemma 10Let be a family of operators bounded on for any with uniform constants (possibly depending on ). Then for

*Proof:* The result is trivial for . Now, for , the square of the norm on the left is

for some s.t. . Now , and the latter is in for and thus is in too. Therefore the above is bounded by (remember the constant only depends on , not on )

and by Hölder’s inequality

and by boundedness of , if you choose close enough to s.t. ,

The remaining case follows by duality applying the previous part to the adjoints.

We know for sure by a previous section that Calderón-Zygmund operators are bounded for and in particular for . We want to prove that the localized operator with multiplier , where is the dyadic interval , is a family of bounded operators as required in the lemma. Then, by standard Littlewood-Paley theory it will follow (here is the operator associated to multiplier )

therefore proving the theorem.

Now, by the bounded variation condition we know there must exists s.t. for

and with this we can rewrite formally

and since ,

where is the operator with multiplier . Now, we know that both and can be written in terms of the Hilbert transform\footnote{In general , where is modulation, i.e. .}, and are thus bounded with comparable constants (independent of ). Therefore for

which by Minkowski’s inequality is bounded by

by the hypothesis of bounded variation of on dyadic intervals. This concludes the proof.

**7. Appendix: proofs of Lemma 7 and Lemma 8**

*Proof:* For a fixed cube and it is enough to show that for some

since this is an equivalent characterization\footnote{Indeed, because , and the viceversa (with constant 1) is trivial.}. The strategy is to estimate separately the contributions from around the cube and far from it. Thus write where . We claim that is a good constant. Indeed

By boundedness of ,

As for the second term, it is bounded by

and since is a CZ-kernel this is bounded by

Before proving the remaining lemma we show how to prove inequalities of the form

by using what’s known as a *good lambda inequality*, i.e. for some and for every

(of course we’ll be using a different measure, but it can be adapted easily). Heuristically, such an inequality is saying that can’t be much smaller than in a big proportion of the volume where is large – the portion must shrink as the ratio decreases. With (10), one has

and

thus for small enough one has

and the inequality follows by taking the limit for . One minor point: you have to verify the partial integrals are finite, and this is true for if (easy).

*Proof:* By the remark just made, we can prove a good lambda inequality:

The sets are the union of disjoint dyadic cubes (by definition of ). Take a maximal in such union, i.e. but if is the parent dyadic cube then . It is enough to prove the good lambda inequality on each , so our goal is

We can make a further reduction thanks to the reverse Hólder inequality: for any there exists a s.t. for

indeed, by Hólder’s inequality

and you can choose . Thus the good lambda inequality will be proved if we can prove that

Take in the set on the left side of the inequality. Then, since ,

This implies

and by weak boundedness of

which concludes the proof if the set is non-empty (if it is, there’s nothing to prove).

References:

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