After having studied the Hilbert transform in detail we now move to the study of general Calderón-Zygmund operators, that is operators given formally as

for an appropriate kernel . Let us quickly review what we used in order to show that the Hilbert transform is of weak type and strong type . First of all we essentially used the fact that the linear operator is defined on and bounded, that is, that it is of strong type . This information was used in two different ways. First of all, the fact that is defined on means that it is defined on a dense subspace of for every . Furthermore, the boundedness of the Hilbert transform on allowed us to treat the set where is the `good part’ in the Calderón-Zygmund decomposition of a function . Secondly, we used the fact that there is a specific representation of the operator of the form

whenever and has compact support and . For the Hilbert transform we had that the kernel is given as

We used the previous representation and the formula of to prove a sort of restricted boundedness of on functions which are localized and have mean zero, which is the content of Lemma 7 of Notes 6. This, in turn, allowed us to treat the `bad part’ of the Calderón-Zygmund decomposition of . From the proof of that Lemma it is obvious that what we really need for is a Hölder type condition. Note as well that for the Hilbert transform we first proved the bounds for and then the corresponding boundedness for followed by the fact that is essentially self-adjoint.

**1. Singular kernels and Calderón-Zygmund operators **

We will now define the class of Calderón-Zygmund operators in such a way that we will be able to repeat the schedule used for the Hilbert transform. We begin by defining an appropriate class of kernels , name the singular (or standard) kernels.

Definition 1 (Singular or Standard kernels)Asingular(orstandard) kernel is a function , defined away from the diagonal , which satisfies the decay estimate

Example 1Let be given as for with . Then is a singular kernel. Observe that is the singular kernel associated with the Hilbert transform.

where is a Hölder-continuous function:

for some . Then is a singular kernel.

Exercise 1Prove that the kernel of example 2 is a singular kernel.

Example 3Let satisfy the size estimateand the regularity estimates

away from the diagonal . Then is a singular kernel. In particular, the kernel given as

is a singular kernel since the gradient of is of the order . Thus the estimates (2) and (3) are consistent with (1) but of course do not follow from it.

Remark 1The constant appearing in (2), (3) is inessential. The conditions are equivalent with the corresponding conditions where is replaced by any constant between zero and one.

We are now ready to define Calderón-Zygmund operators.

Definition 2 (Calderón-Zygmund operators)ACalderón-Zygmund operator(in shortCZO) is a linear operator which is bounded on :and such that there exists a singular kernel for which we have

for all with compact support and .

Remark 2Note that the integral converges absolutely whenever has compact support and lies outside the support of . Indeed,by (1), for some . Observe that the integral in the last estimate converges.

Remark 3For any singular kernel one can define by means offor with compact support and . It is not necessary however that is a CZO since it might fail to be bounded on .

Remark 4It is not hard to see thatuniquelydetermines the kernel . That is iffor all with compact support, then almost everywhere (why?). The opposite is not true. Indeed, for any bounded function the operator defined as is a Calderón-Zygmund kernel with kernel zero. A more specific example is the identity operator which also falls in the previous class, and is CZO with kernel 0. However, this is the only ambiguity. See Exercise 2.

Exercise 2Let be two CZOs with the same singular kernel . Show that there exists a bounded function such thatfor all .

If is a CZO, the definition already contains the fact that is defined and bounded on , so we don’t need to worry about that. The next step is to establish the restricted boundedness for functions with mean zero. The following lemma is the analogue of Lemma 7 of Notes 6.

Lemma 3Let be a Euclidean ball in and denote by the ball with the same center and twice the radius, that is . Let have mean zero, that is . Then we have thatfor all . We conclude that

*Proof:*Using the fact that has zero mean on , for we can estimate

Integrating throughout we also get the second estimate in the lemma.

The only thing missing in order to conclude the proof of the bounds for CZOs is the the fact that they are self adjoint *as a class*. In particular, we need the following.

Lemma 4Let be a CZO. Consider the adjoint defined by means of

*Proof:* It is immediate from (4) and the fact that is bounded on that is also bounded on with the same norm. Now let have disjoint compact supports. We have

Now let and have support inside with . For , the functions are supported in so, for small enough, the support of is disjoint from the support of . By (5)we conclude that

Letting we get

for almost every . Since the conditions defining singular kernels are symmetric in the variables , the kernel is again a singular kernel so we are done.

The discussion above leads to the main theorem for CZOs:

Theorem 5Let be a Calderón-Zygmund operator. Then extends to a linear operator which is of weak type and of strong type for all where the corresponding norms depend only on and and .

**2. Pointwise convergence and maximal truncations **

Let be a CZO. The example of the Hilbert transform suggests that we should have the almost everywhere convergence

at least for nice functions . The truncated operators

certainly make sense for because of (1). However, the limit need not even exist in general or may exist and be different from . Here we can use the trivial example of the operator . As we have already observed this is a CZO operator with kernel . Thus for all but clearly in general.

The following lemma clears out the situation as far as the existence of the limit is concerned:

Lemma 6The limitexists almost everywhere for all if and only if the limit

exists almost everywhere.

*Proof:*First suppose that the limit exists for all and let with on . Then

Observe that by (1)the second integral on the right hands side converges absolutely. Since the limit on the left hand side exists we conclude that the limit on the right hand side exists as well. Conversely, suppose that the limit

exists and let . We have that

By the same considerations are before is a positive number that does not depend on . By the hypothesis we also have that . Finally for observe that we have

by (1). Since

dominated convergence implies that exists as well.

Thus, for specific kernels one has an easy criterion to establish whether the limit exists a.e. for `nice’ functions . For example, for the kernel of the Hilbert transform, the existence of the limit

is obvious. In order to extend the almost everywhere convergence to the class we need to consider the corresponding maximal function.

Definition 7Let be a CZO and define the truncations of as beforeThe

maximal truncationof is the sublinear operator defined as

The maximal truncation of a CZO has the same continuity properties as itself.

Theorem 8Let be a CZO and denote its maximal truncation. Then is of weak type and strong type for .

The proof of Theorem 8 depends on the following two results.

Lemma 9Let be an operator of weak type and . Then for every set with we have that

The proof of this lemma is a simple application of the representation of the norm in terms of level sets and is left as an exercise.

Exercise 3Prove Lemma 9 above.

The second result we need is the following lemma that gives a pointwise control of the maximal truncations of the CZO by an expression that involves the maximal function of and the maximal function of .

*Proof:*Let us fix a function and and consider the balls and its double . We decompose in the form

Since and obviously has compact support we can write

Also every is not contained in the support of thus

by (3), since for in the area of integration above. By this estimate we get that

Combining the previous estimates we conclude that for any

for some constant depending only on and .

If then we are done. If then there is such that . Let

and

Let . Then either or or . In the last case so in every case we conclude that thus . However we have that

Also, by the type of we get

Finally, if then . Otherwise so

Thus in every case we get that

Since the previous estimate is true for any we conclude that

which gives the desired estimate in the case .

For estimate (7)implies that

and integrate in to get

and thus

Note that

and by Lemma 9the last term is controlled by

since is of weak type . Gathering these estimates we get

as we wanted to show.

We can now give the proof of the fact that maximal truncation of a CZO is of weak type and strong type for .

*Proof:* Proof of Theorem 8. By Lemma 10 for we immediately get that is of strong type for since both and are. In order to show that is of weak type we argue as follows. By Lemma 10we have that

Thus the proof will be complete if we show that

As we have seen in Corollary 18 of Notes 5 we have that

where is the dyadic maximal function. Furthermore, using the Calderón-Zygmund decomposition it is not hard to see (see Exercise 4) that

Applying the last estimate to we get

For the set has finite measure. Thus by Lemma 9we conclude that

and thus by (8)that

This concludes the proof.

**3. Singular integral operators on and . **

The theory of Calderón-Zygmund operators developed so far is pretty satisfactory except for one point, the action of a CZO on . Exercise 4 from Notes 6 shows for example that in general a CZO cannot be bounded on . Furthermore, it is at the moment unclear how to define the action of on a general bounded function or even on a dense subset of . With a little effort however this can be achieved.

Let us first fix a function and look at the formula

As we have already mentioned several times, such a formula is not meaningful throughout . Indeed the integral above need not converge, both close to the diagonal , since is singular, as well as at infinity since only decays like , not fast enough to make the integral above absolutely convergent. The first problem we have dealt with so far by considering functions with compact support and requiring the validity of (9)only for . A similar solution could work now but we still have a problem at infinity. Note that we didn’t run into this problem yet since we only considered functions in which necessarily possess decay at infinity. This is not necessarily the case for bounded functions. However, looking at the difference of the values of at two points with , we can formally write

Using the regularity condition (3)we see that

when . This is enough to assure integrability in the previous integral, as long as Motivated by this heuristic discussion we define for :

for some Euclidean ball so that . First of all it is easy to see that the integrals above make sense. Indeed, is well defined since is in . On the other hand, the integral in the second summand converges absolutely since we integrate away from , is bounded and behaves like for . However, (10)only defines up to a constant. Indeed it is easy to see that if are two different balls containing the difference in the two definitions is equal to

which is a constant independent of . Thus we only define modulo constants. This definition of gives a linear operator which extends our previous definitions on or . To deal with the ambiguity in the definition, we have to define the appropriate space.

Definition 11We say that two functions areequivalent modulo a constantif there exists a constant such that almost everywhere on . This is an equivalence relationship. By abuse of language and notation we will oftentimes identify an equivalence class with a representative of the class, much like we do with measurable functions.

Definition 12 (Bounded Mean Oscillation)Let be a locally integrable function , defined modulo a constant. We setto be the average of on the Euclidean ball . The norm of is the quantity

where the supremum varies over all Euclidean balls . The space is the set of all locally integrable functions , defined modulo a constant, such that . Thus, an element of is only defined up to a constant.

First of all observe that this is a good definition since replacing a function by for any constant does not affect its BMO norm. Thus, all elements in the equivalence class of have the same BMO norm. The previous quantity actually defines a norm, always keeping in mind that we identify functions that differ by a constant. For example any constant is equivalent to the function in BMO and thus if and only if almost everywhere for some .

It is not hard to give the following alternative description of the BMO norm, which is maybe a bit more revealing:

Proposition 13(i) Let . We have that(ii) For any locally integrable function and a cube set . We set

where the supremum is taken over all cubes Then

as in . Moreover

*Proof:*For (i) observe that for any ball we have

On the other hand for any we have

which gives the opposite inequality as well by taking the infimum over . The proof of the first claim in is identical. For the second claim in let and be a cube. Consider the smallest ball with the same center as . Then

Thus,

for any cube . Taking also the supremum over cubes proves the one direction of the inequality. The proof of the opposite inequality is similar.

Thus a function in BMO has the property that for any ball there is a constant such that . That is, the values of oscillate around by at most in average. Locally, and in the *mean*, the function has bounded oscillation.

The space BMO contains but also contains unbounded functions.

Proposition 14(i) For every we have thatthus .

(ii) The function is in . Thus is a proper subset of .

Our interest in the space BMO mainly lies in the fact that it serves as a substitute endpoint for the boundedness of CZOs, namely a CZO is bounded from to BMO, where should be defined as in (10). Note here that even though (10) only defines `up to constants’, this is the only possible definition of a BMO function.

Theorem 15Let be a CZO. Then for every we have that

*Proof:*Let be some ball in . We need to show that

and denote . We set

Since is of strong type we have

Thus by Cauchy-Schwartz we have

On the other hand for , the ball certainly contains both and so

Remembering that (10)only defines up to a constant we get

By Proposition 13 this proves the theorem.

** 3.1. The John-Nirenberg Inequality **

We will now see that although the space BMO contains unbounded functions like , this is in a sense the maximum possible growth for a BMO function. Although such a claim is not precise in a pointwise sense, it can be rigorously proved in the sense of level sets. Indeed, assuming then

for all balls . Using Chebyshev’s inequality this implies

This estimate is interesting for large, and states that on any ball the function exceeds its average by only on a small fraction of the ball . In fact, this can be improved.

Theorem 16 (John-Nirenberg inequality)Let . Then for any Euclidean cube we have thatfor all , where the constant depends only on the dimension .

Remark 5Obviously it doesn’t make any difference to work with balls instead of cubes so the the previous theorem remains valid with balls replacing cubes .

*Proof:*For let us denote by the best constant in the inequality

valid for any cube and with By Chebyshev’s inequality combined with the trivial bound we get

which is of course quite far from the desired estimate

This will be achieved by iterating a local Calderón-Zygmund decomposition as follows.

Let us fix a cube and consider the family of cubes inside which are formed by bisecting each side of . Then define the second generation by bisecting the sides of each cube in and so on. The family of all cubes in all generation will be denoted by . For a level to be chosen later let be the `bad’ cubes in , that is the cubes such that

where .

Finally let be the family of maximal bad cubes. Since for the original cube , every bad cube is contained in a maximal bad cube. As in the global Calderón-Zygmund decomposition we conclude that

for each cube where the constant depends only on the dimension . We also conclude that

if by the dyadic maximal theorem. Remembering the initial normalization we get

and for

Now consider . We have

However this means that

whenever . Suppose that . Since is non-increasing and the trivial estimate we get

for (say) and . On the other hand, for we have

so the proof is complete.

Corollary 17Consider the version of the BMO normThen

Exercise 5Use the John-Nirenberg and the description of norms in terms of level sets to prove Corollary 17

Finally, we show how we can use the space as a different endpoint in the Log-convexity estimates for the norms.

Lemma 18Let and . Then and

*Proof:*Obviously it is enough to assume that otherwise there is nothing to prove. Also by homogeneity we can normalize so that . Now form the Calderón-Zygmund decomposition of at level and denote by the family of bad cubes as usual. For each cube we then have

From the John-Nirenberg inequality we conclude that

for all the bad cubes . Since we have that for we get

for all . On the other hand, since we have

We conclude the proof by using the description of the norm in terms of level sets and using (12) for and (11) for .

Exercise 6 (The sharp Maximal function)For definethe sharp maximal functionObserve that if and only if and, in particular,

Show that for every we have

**4. Vector valued Calderón-Zygmund Singular integral operators **

We close this chapter on CZOs by describing a vector valued setup in which all our results on CZOs go through almost verbatim. We will see an application of these vector valued results in our study of *Littlewood-Paley* inequalities.

So let be a separable Hilbert space with inner product and norm and consider a function . All the well known facts about spaces of measurable scalar functions have almost obvious generalizations in this setup once we fix some analogies. For example, the function will be called measurable if for every the function is a measurable function of . If is measurable then is also measurable. We then denote the space of all measurable functions such that

and the usual corresponding definition for

It is not hard to check the duality relations for these spaces; for example

for all . Also our interpolations theorems, the Marcinkiewicz interpolation theorem and the Riesz thorin interpolation theorem go through in this setup as well.

Moreover, if a function is absolutely integrable, we can define its integral as an element of by defining the functional

Note here that is uniquely defined as a functional in . Indeed, is obviously linear and by the Cauchy-Schwartz inequality we have

By the Riesz representation theorem on Hilbert spaces, there is a unique element of , which we denote by , such that , that is

Finally, if are separable Hilbert spaces we denote by to be the space of bounded linear operators , equipped with the usual operator norm:

Again, a function will be called measurable if for every the function

is a measurable -valued function.

We are now ready to give the description of vector valued CZOs. We start with the definition of a singular kernel.

Definition 19 (Vector valued singular Kernel)Let be two separable Hilbert spaces and be a function defined away from the diagonal Then will be called a (vector-valued)singular kernelif it obeys the size estimate

Definition 20Let be separable Hilbert spaces. An linear operator is called a (vector valued)Calderón-Zygmundoperator (vector valued CZO) from to if it is bounded from tofor all , and there exists a vector valued singular kernel such that

whenever has compact support and .

Adjusting the proof of the scalar case to this vector valued setup we get the corresponding statement of Theorem 5.

Theorem 21Let be separable Hilbert spaces and be a vector valued Calderón-Zygmund operator from to .(i) The operator is of weak type

for all .

(ii) For all , is of strong type

for all .