This week we come to the study of singular integral operators, that is operators of the form

defined initially for `nice’ functions . Here we typically want to include the case where has a singularity close to the diagonal

which is not locally integrable. Typical examples are

and in one dimension

and so on. Observe that these kernels have a non integrable singularity both at infinity as well as on the diagonal . It is however the local singularity close to the diagonal that is important and will lead us to characterize a kernel as a singular kernel. For example, the kernel

*is not* a singular kernel since its singularity is locally integrable. Observe that for Schwartz functions it makes perfect sense to define

and in fact the previous integral operator was already considered in the Hardy-Littlewood-Sobolev inequality of Exercise 12 in Notes 5 and can be treated via the standard tools we have seen so far.

Thus, if one insists on writing the representation formula (1) throughout then will not be a function in general. Indeed, the discussion in Notes 4 reveals that if the operator is translation invariant then the kernel must necessarily be of the form for an appropriate tempered distribution :

Bearing in mind that there are tempered distributions which do not arise from functions or measures we see that (1) does not make sense in general and it should be understood in a different way. To give a more concrete example, think of the principal value distribution and write

Here we would like to rewrite this in the form

but this does not make sense even for since the function is not locally integrable on the diagonal .

In fact, the representation (1) of the operator will not be true in general but we will satisfy ourselves with its validity for functions , of compact support, and whenever does not lie in the support of . Indeed, if has compact support and then in (1) and thus we are away from the diagonal. Indeed, returning to the principal value example, observe that the integral

makes perfect sense when has compact support and .

Eventually the theory of singular integral operators does not depend on translation invariance; singular kernels of the type can be viewed as a special case of the more general class of singular kernels which satisfy appropriate growth and regularity assumptions. It is however instructive to consider the translation invariant case first. In the Calderón-Zygmund theory of singular integral operators we will start with more or less assuming that the operator is well defined and bounded on and that its kernel satisfies certain growth and regularity conditions. Alternatively, assumptions on will allow us to show the -boundedness. We will see that under these conditions will extend to a bounded operator on for and of weak type .

**1. The Hilbert transform **

In order to illustrate the general ideas let us consider what is probably the primordial example of a singular integral operator, the *Hilbert transform*, given in the form

Remembering the principal value distribution we can rewrite this in the form

at least whenever . The previous formula makes sense just because the principal value of is a well defined tempered distribution. Alternatively, we can repeat the argument we used for to write for any and a Schwartz function

Observe that we heavily rely on the fact that the kernel has zero mean on symmetric intervals around (and away from) the origin:

The mean value theorem now shows that is uniformly bounded by thus the limit of the first summand as exists and we have that

Remark 1Trying to write the Hilbert transform as an integral operator with respect to a kernel ,

we immediately run into the problem that the principal value distribution does not arise from a function. The previous discussion allows us however to write

whenever is a compactly supported function in or and . This is essentially equivalent to the fact that the integrals

are absolutely convergent whenever and is fixed.

Thus we see that the Hilbert transform is a linear operator which is at least well defined on the Schwartz class . This is quite promising since we know that is dense in for . Of course, in order to extend the action of to say we need to exhibit the continuity of on the dense subclass . In our general theory this will be a `given’, that is that our operator is bounded on . To make this general assumption meaningful we have to exhibit that it is indeed satisfied in the model case of the Hilbert transform. We begin this investigation by first showing a simple asymptotic relationship.

Before giving the proof of this Lemma let us discuss its consequences. Already the expression (2) shows that is a bounded function whenever . Indeed, using the mean value theorem for the first term in (2) and Hölder’s inequality for the second term we have that

As a result, the integrability of for solely depends on the behavior of at infinity. Now the lemma just stated shows that

whenever with . Thus for a general with non-zero mean, fails to be in since it doesn’t decay fast enough at infinity. It is however in for any . As we shall see the failure of continuity of on has a weak substitute, namely that is of weak type and this is the typical behavior of all singular integral operators we want to consider.

*Proof of Lemma 1*: The proof is a variation of the idea used in (2). For any and large we can write

For observe that whenever thus we have that

as since is a Schwartz function. On the other hand, for we have that whenever . We get

as since is integrable, being a Schwartz function. Now consider the expression

thus

as .

Exercise 1Let . Show that if and only if

Hint:Examine the decay of for by using the identity .

** 1.1. The Hilbert transform on **

Having exhibited that whenever our next task is to show that is bounded as an operator , that is to show that

for all . Remember that since is dense in such an estimate will allow us to extend to a bounded linear operator on . There are several different approaches to such a theorem, most of them connected to the significance of the Hilbert transform in complex analysis and in the theory of holomorphic functions. First we exhibit the connection with Cauchy integrals.

Proposition 2Let be a function on such that is well defined, say and for large. Then

for every .

*Proof:* By translation invariance of and taking complex conjugate in both sides of the identity it suffices to show that

Changing variables this is equivalent to

Now let

For we have that

while for we can calculate

The previous estimates obviously imply that is absolutely integrable on . Furthermore

as can be seen by a direct calculation. Thus by the previous calculations it suffices to show that

which follows by dominated convergence since and is bounded.

Exercise 2Show that for satisfying for the Hilbert transform is indeed well defined. Furthermore, show that it indeed suffices to show (3) in the previous proposition. In particular exhibit how the full statement of the previous follows from (3).

*Proof:* Let us define the Cauchy-type integral

Then Proposition 2 shows that

Observe by the proof of the proposition applied to the function that

for all . Thus by Minkowski’s integral inequality we get that

By dominated convergence we conclude that converges to in as well. By Plancherel’s theorem we get that we must also have that

in , as . Note here that the Fourier transform is well defined since and in this case we have exhibited that . The problem now reduces to calculating the Fourier transform of for and see what happens in the limit. Consider the truncations

Let us write

Then as in by dominated convergence and thus

as . We now have that

However we have that

Now Cauchy’s theorem from Complex analysis shows that whenever .

The previous definitions allow us to conclude that the Fourier transform

whenever and thus that

whenever . We conclude that

Now not that the Hilbert transform satisfies

where remember that . So for we can write

In other words for we get that .

The previous theorem shows in particular that for all . This allows us to extend the Hilbert transform to a bounded linear operator on . In fact is an isometry by Plancherel’s theorem and the fact that . Furthermore, although at the current stage it is not clear that our original definition makes sense on , we can directly define the Hilbert transform on by means of

which is a good definition whenever . In fact, recalling the discussion on multiplier transformations it is clear that the operator on is the multiplier transformation associated with the multiplier which is obviously a bounded function. This is automatic from the definition

and the fact that . We also have that which is also obvious from the fact that is an isometry.

Corollary 4The Hilbert transform extends to an isometry on . We have that

for all . Furthermore, for the Hilbert transform can be defined as

Corollary 5Consider the Hilbert transform . Then we have the following properties (i) The Hilbert transform commutes with translations and dilations (but not modulations).

(ii) The Hilbert transform is skew-adjoint on

(iii) We have the identity on :

Exercise 3Prove Corollary 5 above.Hint:Use the formula of Theorem 3.

Exercise 4Let . Show that

Conclude that the Hilbert transform is not of strong type nor of strong type .

** 1.2. The Hilbert transform on **

So far we have defined our first singular integral operator, the Hilbert transform. This is an operator that is bounded on and that has the representation

whenever has compact support and . The function

is the singular kernel associated with the Hilbert transform. Although we have seen that the Hilbert transform can be described for all , at least for nice functions , the restricted representation just described is all we really need to execute our program. Furthermore, this approach will serve as a good introduction to the general case of Calderón-Zygmund operators. From the previous discussion we know that the Hilbert transform is not of type nor of type . The following theorem is the main result of the theory.

Theorem 6(i) The Hilbert transform is of weak type ; for we have that

(ii) For , the Hilbert transform is of strong type ; for we have

*Proof:* We will divide the proof in several steps. The most important one however is the proof of the weak type . All the rest really relies on exploiting the symmetries of the Hilbert transform, interpolation and duality. ** **

**step 1; the weak bound:** We fix a level and a function and write the Calderón-Zygmund decomposition of the function at level in the form

Recall that the `bad part’ is described as

where is a collection of disjoint dyadic intervals (since ) and each is supported on . Furthermore we have that

and

Recall also that

by the maximal theorem. On the other hand the `good part’ is bounded

and its norm is controlled by the norm of :

Observe that thus and by the log-convexity of the norm we have

Remark 2Since it follows that as well. Also, by the definition of the pieces it is easy to see that as well. However, we will not use the bounds on nor on , the fact that they belong to being merely a technical assumption that allows us to define their Hilbert transforms. Overall, the hypothesis that cannot be used in any quantitative way if we ever want to extend our results to for .

Since and is linear, we have the following basic estimate

The part that corresponds to is the easy one to estimate. This is not surprising since is the good part. Since we already know that is of strong type it’s certainly of weak type thus we have

by (5). Thus this estimate for the good part is exactly what we want. Let’s move now to the estimate for the bad part. The main ingredient for the estimate of the bad part is the following statement which we formulate as a lemma for future reference.

Lemma 7Let be any interval in and denote by the interval with the same center as and twice its length. For support in and with zero mean on , , we have

for all . We conclude that

Remark 3Here we require that is also in just in order to make sure that is well defined. Note that in the case of the Hilbert transform it can be verified directly that is well defined for and . However we prefer this formulation since for more general Calderón-Zygmund operators we will only have a formula available to us for with compact support and .

*Proof:* Using the zero mean value hypothesis for we can write for

Now since we have that

so we can write

as we wanted to show. The second claim of the lemma follows easier by integrating this estimate.

We now go back to the estimate of . First of all note that

for almost every . Indeed, if we enumerate the cubes in as then we have that for every thus in . Since is an isometry on it follows that converges to in as well. Taking subsequences we then have that almost everywhere. Thus

almost everywhere and we get the claim by letting .

For each let denote the cube with the same center and twice the side-length. We now estimate the `bad part’ as follows

By the Calderón-Zygmund decomposition we have that

which takes care of the first summand. For the second we use Lemma 7 to write

again by the Calderón-Zygmund decomposition. Observe that each and has mean zero on so the appeal to Lemma 7 is legitimate. Summing up the estimates for all the bad cubes in we get

By Chebyshev’s inequality we thus get

Summing up the estimates for the bad part we conclude that

By (6) now we conclude that

whenever .

We have a priori assumed that in order to have a good definition of . However, the weak inequality on allows us to extend the Hilbert transform to a linear operator on which is also of weak type . The details are left as an exercise.

Exercise 5Let be a linear operator which is of weak type . Show that extends to a linear operator on which is of weak type , with the same constant.

**step 2; the strong bound:** As promised, the difficult part of the proof was the weak bound. The rest is routine. first of all observe that since is of weak type and strong type , the Marcinkiewicz interpolation theorem allow us to show that is of strong type for any . To treat the interval we argue by duality, exploiting the fact that is almost self-adjoint (in fact it is skew adjoint as we have seen in Corollary 5). Indeed, let and . Now for any we have

using the fact that is of strong type since . Taking the supremum over all with we get

for as well, whenever . Using standard arguments again this shows that extends to a bounded linear operator on , .

Remark 4In fact, tracking the constants in the previous argument we see that

and

Overall we have proved that is of strong type with a norm bound of the order

Remark 5We have exhibited that extends to a bounded linear operator to for and that it is of weak type . However, for a general , , there is no reason why should by given by the same formula by which it was initially defined; remember that

Thus the question whether a.e., for , is very natural. Since we know this convergence is true for the dense subset , the study of the pointwise convergence amounts to studying the boundedness properties of the corresponding maximal operator

Thus if one can show that is of weak type for example, the pointwise convergence of to would follow by Proposition 1 of Notes 5. Such an estimate is actually true and thus this formula extends to all functions for . We will however see this in the general theory of Calderón-Zygmund operators of which the Hilbert transform is a special case and so we postpone the proof until then.

** 1.3. The Hilbert transform and the boundary values of holomorphic functions **

In this section we briefly discuss the connection of the Hilbert transform with the boundary values of holomorphic functions in the upper half plane. Let us write

for the upper half plane. Two function on are called *conjugate harmonic functions* if they are the real and imaginary part respectively of a holomorphic function in the upper half plane, where . Thus we have that

By definition both are real and harmonic. Moreover, they satisfy the *Cauchy-Riemann* equations (since is holomorphic). Now assume that has a boundary value on the real line . Then

Of course, some technical assumptions are needed to make all these claims rigorous as for example assuming that the holomorphic function F has some decay of the form in the upper half plane.

Conversely, Let be a real function and be the Poisson kernel for the upper half plane

As we have seen, the convolution is a harmonic function in the upper half plane . Observe that

Consider now the *conjugate Poisson kernel*

The name comes from the fact that both are both real harmonic functions and writing we have

which is holomorphic in the upper half plane. Thus , are conjugate harmonic functions which is what makes the functions conjugate harmonic functions as well. We conclude that the function

is harmonic in the upper half plane and that

is holomorphic in the upper half plane.

Finally observe that according to the previous formulae we have

In this language, Proposition 2 just states that converges to its boundary value as . We also see that the imaginary part of converges to the Hilbert transform:

both in and almost everywhere.

** 1.4. Frequency cut-off multipliers and partial Fourier integrals **

Remember that for a bounded function the operator

is a multiplier operator (associated to the multiplier ) and that . We also say that is a multiplier on if extends to a bounded linear operator . Thus we see that the Hilbert transform is a multiplier operator on associated with the multiplier

which is obviously a bounded function with . A very closely related multiplier is the *frequency cutoff multiplier*. Given an interval in the frequency space, where , we define the operator by means of the formula

Thus the operator applied to , localizes the function in frequency, in the interval . Such operators as well as their multidimensional analogues turn out to be very important in harmonic analysis as well as in the theory of partial differential operators. Obviously is bounded on , since . However, the corresponding estimate in is far from obvious. After all the work we have done for the Hilbert transform though, we can get the bounds for as a simple corollary. This is based on the observation that

where the equality should be understood as an equality of operator in . Here remember that

The verification of this formula is left as an exercise. Formula (7) is also true when or with obvious modifications.

Exercise 6Prove formula (7).

A simple corollary of the boundedness of the Hilbert transform is the corresponding statement for .

Lemma 8The operator is of strong type for :

Note that the operator norm of does not depend on .

Now for and define the partial Fourier integral operator

Observe that these integrals are the -means of the integral . We have seen that the Gauss-Weierstrass or Abel means of this integral converge to , both almost everywhere as well as in the sense. However the function is much rougher. We still have the following theorem as a consequence of the bound for the Hilbert transform.

Theorem 9For the operator has a unique extension to a bounded linear operator on for .

However the boundedness of control the convergence of partial Fourier integrals.

Lemma 10The partial Fourier integrals converge to in the norm for if and only if is of strong type uniformly in .

Now Theorem 9 and Lemma 10 immediately imply:

Corollary 11For the partial Fourier integrals converge to in the norm.

The question whether converges to almost everywhere is much harder. For the answer is positive and this is the content of the famous Carleson-Hunt theorem. This theorem was first proved by Carleson for and then extended to by Hunt. A counterexample by Kolmogorov shows that both the and the almost everywhere convergence of the partial Fourier integrals fail for .

Exercise 7Show that extends to an operator of weak type on and that the partial Fourier integrals converge toin measurefor . Conclude that for almost every there is a subsequence such that .

*[Update 15th May 2011: Equation (7) moved to the right place, Exercise 1 slightly changed.]
*

Great article – I like the motivation you provided! In particular Lemma 1, was a point I was unaware of when I first learned singular integrals, but it’s good to know because it tells you the most you can expect for L^p estimates. Also, it’s the same estimate for the Hardy–Little maximal inequality in 1 dimensions as you remarked in the previous lecture.

By the way, the link to Carleson’s theorem at the end doesn’t work.

[Corrected thanks! Y.]Thank you for the article, very pleasant to read.

I have one additional question : I read somewhere that the Hilbert transform of a Hölder continuous function, say on [-a ; a], is still Hölder continuous on [-a/2 ; a/2].

Do you have any idea how to prove it ?

Dear Florian,

to be honest I don’t know of this specific result from the top of my head. I would try to argue via the Fourier transform of H(f) though. I will try to come back with a more precise answer. To clear out one thing, do you mean that your function is also compactly supported on [-a,a]?

yannis

Thank you Yannis for this answer. I also had the idea of using the Fourier transform but I have not done the calculation so far (I have tried the naive way to show it, with some refinements, and it did not work…)

My question was general, I did not assume that the function is compactly supported, only that it is square integrable on R.

Yet, if it is simpler, let us assume this hypothesis at first.