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:
Remark 1 Trying 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.
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
Exercise 1 Let . 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.
for every .
Changing variables this is equivalent to
For we have that
while for we can calculate
The previous estimates obviously imply that is absolutely integrable on . Furthermore
Exercise 2 Show 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 4 The Hilbert transform extends to an isometry on . We have that
for all . Furthermore, for the Hilbert transform can be defined as
(ii) The Hilbert transform is skew-adjoint on
(iii) We have the identity on :
Exercise 4 Let . 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
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 :
Remark 2 Since 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 .
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.
for all . We conclude that
Remark 3 Here 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
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 5 Let 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 4 In fact, tracking the constants in the previous argument we see that
Overall we have proved that is of strong type with a norm bound of the order
Remark 5 We 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
The verification of this formula is left as an exercise. Formula (7) is also true when or with obvious modifications.
Exercise 6 Prove formula (7).
A simple corollary of the boundedness of the Hilbert transform is the corresponding statement for .
Lemma 8 The 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.
However the boundedness of control the convergence of partial Fourier integrals.
Corollary 11 For 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 7 Show that extends to an operator of weak type on and that the partial Fourier integrals converge to in measure for . 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.]