I’ve been accepted to a Summer School in Bonn organized by prof. Christoph Thiele and prof. Diogo Oliveira e Silva. The subjects are and theorems, which give a characterization of boundedness for Calderon-Zygmund operators. I’m currently learning the basics so I can’t say much more, but for the summer school each student has been given an article to read which will have to be lectured to the other participants. My article is about boundedness of the Cauchy integral operator for a Lipschitz curve (a lipschitz graph, really), here’s the reference:

Two elementary proofs of the L2 boundedness of Cauchy integrals on Lipschitz curves.by Coifman, R. R.; Jones, Peter W.; Semmes, Stephen

J. Amer. Math. Soc. 2 (1989), no. 3, 553-564.

The article features two nice proofs of the same result which use different instruments (the first one relies on complex analysis, the second one on the construction of a suitable Haar-like basis), but are similar in spirit. Interesting parallels can be made between the two.

As I’m working out all the details in the paper, I thought that it might be good to share it. This first blog post is about the first proof of the theorem. A second blog post will include the second proof, as well as a comparison between the two.

**1. Main theorem **

Here is a Lipschitz graph in , that is

with . The epigraph is called and the hypograph .

One has the -norm on the curve

and by lipschitz property , so that

The Cauchy-integral operator is defined by

when , and when

Notice that the Cauchy integral is missing a factor of from Cauchy’s integral formula for holomorphic functions, but isn’t holomorphic and the factor is only a burden we don’t want to carry on here.

Remark 1There’s an issue in the definition of , namely that the (pointwise) limit along the imaginary axis might not be defined. I’ll deal with this later on, for now suppose it does indeed exist.

The main theorem is

Theorem 1 (-boundedness of )

and the constant depends on only, rather than .

We introduce a norm on the measurable -valued functions on :

where . Notice that , the Lebesgue volume element. Norm comes from an hermitian product, obviously, and the associated Hilbert space is denoted by .

**2. First proof **

We’ll suppose and to be in , since the conclusion for general (and ) can then be derived by taking the limit over a approximation.

First proof requires two lemmas, whose nature is that of Littlewood-Paley theory:

Lemma 2Let be a holomorphic function in such that at infinity. Then

Then is bounded:

With these two lemmas the proof is formally easy:

*Proof:* (of the Main Theorem)

Applying Lemma 2,

and applying Lemma 3

where we used Fubini to pass from 2nd to 3rd line and Cauchy-Schwarz in the end.

Remark 2is well defined, we can take the (complex) derivative inside the integral thanks to the hypothesis that is of compact support (and thus all integrability issues disappear).

*Proof:* (of Lemma 2)

The following will be needed in the proof: let be holomorphic functions in , such that and . Then

(). Notice the factor in the -integral that makes formulas `dimensionally consistent’.

To prove the first one it’s enough to use Green’s identity. But first, notice that writing ,

where we exploited the Cauchy-Riemann relations.

Now, by Green’s identity applied to functions and on the domain one has

and since is harmonic, , on , so that

Apply the same identity with in place of , and by the previous relations summing the two of them we get (3). Notice we applied Green’s identity to an unbounded domain, which is justified by the fact that at infinity.

As for (4), apply (3) to (which satisfies the suitable conditions),

since , and then (we use triangle inequality together with )

where we used (3) again.

Let’s turn to the proof of Lemma 2. Observe how, by the Riemann mapping theorem, there exists a conformal (see footnote [1]) map that maps into holomorphically. We’ll use it to change variables. Writing the norms explicitly, one has to prove

for some constant . Change variables using , writing to get

since (here is the Jacobian of interpreted as a function from the upper half-plane of to the real domain ). The factor is a bit annoying, but we can find a substitute for it using Köbe’s theorem, which we briefly state:

Theorem 4 (Köbe’s one-quarter theorem)Let be a holomorphic injective function (see footnote [2]) on the disk . Then the image of the function contains a ball of radius proportional to ; more precisely

Notice that if is defined on , that is a disk (see footnote [3]) of radius , then the image contains the ball of radius centered in . In our case, ; if we consider the restriction of to the ball , which is injective since is, then we have that the image contains , and thus . On the other hand, it follows from Schwarz’s lemma (see footnote [4]) that , thus .

So, what we’re interested in proving is

Before going further, observe that an inequality of the form

is equivalent to one of the form

This is because implies that either or , both equivalent to (if we don’t care about constants).

Observe that since is angle-preserving and is a Lipschitz graph, one has on the boundary for some . This means that there isn’t much cancellation in phase and so

Now we apply Green’s identity as we did before, to the functions and (unboundedness of domain is again justified by the decay at infinity)

Since and both and are holomorphic, a lot of terms get canceled (see footnote [5]):

Thus

and we recognize that the first term in the right hand side is the one we’re looking for: it’s the in . As for the second one, since it has a well defined logarithm, and we can write . Then . Moreover , which implies . By Cauchy-Schwarz

and we recognize the first term on the rhs as the in . As for the second term,

We can therefore apply (4) with , , , and obtain

in which we recognize term , and we’re done.

*Proof:* (of Lemma 3)

The result will be a consequence of Lemma 2 and Schur’s test. First notice that for to be in is equivalent to to be in . Moreover, since is the common boundary of and , Lemma 2 is true with in place of . Then, applying it to ,

and

Thus

where , . Now we introduce the linear operator given by

that is the operator given by integration against kernel ,

so that one has

To conclude from we need to prove , with constants depending on only. From the above discussion, this latest statement is implied (and equivalent to) by

and we claim that this is indeed true. To see this it is sufficient to apply Schur’s test to the kernel . So,

and since (they belong to opposite sides of )

By symmetry the integral in is bounded by the same quantity, and thus by Schur’s lemma is bounded.

The alternative proof will come soon, maybe with some related facts about , theorems.

[1] Conformal is equivalent to everywhere.

[2] Such functions are often referred to as univalent, in the literature.

[3] Ball and disk are synonyms here.

[4] Schwarz’s lemma states that for a holomorphic function it holds . Since is conformal, is holomorphic. Put and the inequality follows applying Schwarz’s lemma to .

[5] Here is the ordinary differentiation in , i.e. , and is the formal conjugate of . It is . For an holomorphic function .