This article is based on the paper of Coifman-Jones-Semmes (JAMS, 1989).

One of the fundamental problem is the solvability of the Dirichlet problem for the Poisson equation in a bounded domain(open and connected)

in :

Many authors have been studied the solvability of the problem in a various settings. One possible way to solve the problem is to use the method of layer potential. In this talk, we assume for simplicity. The fundamental solution for the Laplacian is defined by

where is the volume of the unit ball in . Suppose first that has -boundary. For , we define the *double layer potential* of by

where denotes the outward unit normal to the boundary . Then it is easy to check that is well-defined on and is harmonic in . Also, can be extended as a continuous function from to up to the boundary. Set for , . Since is of , one can easy to show that

(1)

for some constant .

For , we define

Due to (1), the above one is well-defined. We have the jump relation formula for the double layer potential:

for fixed , we have

Also, it can be shown that is compact, (and also , is compact). Hence by the Riesz-Schauder theory, the operator is invertible if and only if it is injective. Using some argument, one can show that it is injective

and so the Dirichlet problem has a solution. Uniqueness follows from the maximum principle. One can also obtain the solvability of the Dirichlet problem when the boundary data is in . We interpret the boundary condition via nontangential limits. We also note that the above argument holds when is a -domain.

The difficulty in extending these results to the case of and Lipschitz domains is that, in these cases, we have

So even the -boundedness, much less the compactness of is far from obvious. Another difficulty is that the operator we considered is of non-convolution type operators. So we cannot use the Fourier transform method to guarantee the -boundedness of the operator.

In Calder\’on’s remarkable paper, he proved that the Cauchy integral operator along a Lipschitz curve is bounded from to itself when the Lipschitz constant is small. This leads to the solvability of the Dirichlet problem for the Poisson equation in -domain due to Fabes-Rievere-Joedit. Later, Coifman-McIntosh-Meyer resolved the restrictive condition of Caderon’s result.

The first proof of theorem of Coifman-McIntosh-Meyer uses some inductive argument from Calder\’on’s result. One famous proof is to use the variant of -theorem, the -theorem due to David-Journ\’e-Semmes. But we do not pursuit in this direction. In 1989, Coifman-Jones-Semmes provide another proof using the theory of complex variables and some variant of the Littlewood-Paley theory. We also note that there is a geometric proof of theorem of Coifman-McIntosh-Meyer given by Melnikov and Verdera. Recently, Muscalu gives a new proof of Coifman-McIntosh-Meyer theorem whose methods can be extended to the case of multilinear operators.

In Section 1, we construct Haar system on the real line . In Section 2, we construct Haar system associated to an accretive function, which will be defined in later. This leads to the proof of Coifman-McIntosh-Meyer theorem. For those who are interested in, see the paper of Coifman-Jones-Semmes.

### 1. Haar systems on the real lines

In this section, we construct the Haar system on the real lines. Recall that a dyadic interval in is an interval of the form

where and are integers. For we denote by the set of all dyadic intervals in whose side length is . We also denote by the set of all dyadic intervals in Then we have

Moreover, the -algebra of measurable subset of formed by countable unions and

complements of elements of is increasing as increases, i.e., is a filtration. For simplicity, we write . Observe that any two dyadic intervals of the same side length either coincide or their interior are disjoint. Moreover, either two given dyadic interval contains the other or their interiors are disjoint.

Given a locally integrable function on , we let

denote the* average* of over an interval . The* conditional expectation* of a locally integrable function on with respect to the increasing family of -algebras generated by is defined as

We also define the *dyadic martingale difference operator* as follows:

also for .

*Remark (Justification of the terminology). *First, we justify the terminology “conditional expectation”. We first show that for any , we have

Since , for any , either or . So

The identity holds for all . Hence the identity holds for all . This shows that

which justify the terminology “conditional expectation”.

For a dyadic interval and write , where and are the left and right halves of . The function

is called the *Haar function associated with the interval* . By construction, the Haar functions have norm equal to . Moreover, if , either or not. Clearly, if , then

(2)

If , we may assume . Then either is contained in the left or in the right half of , on either of which is contant. Hence (2) follows. We introduce the notation

for . Then

The following proposition shows that the dyadic difference operator is a projection to the space which is spanned by Haar functions.

Proposition 1.Let . For all , we have

and also

Proof. Observe that every interval in is either an or an for some unique . Thus, we write

Also,

So

which is easily checked to be equal to

This coplemtes the first part. Now from the orthogonality of , we get

This completes the proof of Proposition 1.

Using the dyadic difference operator, we can decompose a function in as follows:

Theorem 2.Every function can be written as(3)

where the series converges almost everywhere and in . We also have

(4)

Proof. It follows from the Lebesgue differentitation theorem that there exists a set of measure zero on such that for all , we have

whenever is a sequence of decreasing intervals such that . Given in , there exists a unique sequence of dyadic intervals such that . Then for all , we have

Hence a.e. as . Since

where denotes the dyadic maximal function, we have that . Hence due to the -boundedness of the dyadic maximal operator, it follows from the dominated convergence theorem that in . For a given and as before, we have

which tends to zero as , since the side length of each is . Since , by the dominated convergence theorem, we conclude that in as . Observe that

as and a.e. and in . By Proposition ??,

This proves identity (3). To prove (4), we rewrite

For , we have

Since the last integral is nonzero only when . If this is the case, then for some dyadic interval . Then the function is supported in the interval and the function is constant on any dyadic subinterval of . Then

since . Hence, whenever . This shows that

Now the identity (4) follows from Proposition 1. This completes the proof of Theorem 2.

### 2. Modified haar systems with an accretive functions

The purpose of this section is to extend Proposition 1 and Theorem 2. This extension leads to a proof of the theorem of Coifman-McIntosh-Meyer. We introduce the following notion.

Definition.A bounded complex-valued function on is said to beaccretiveif there is a constant such that for almost all .

Let be an accretive function. For a measurable set in with , we define

Observe that

since is accretive. For each , we define

with a fixed choice of the square roots. If , then . For , we introduce a pseudo-inner product

on . By definition, each is supported on and is constant on and . Moreover,

The following theorem is a generalization of Theorem 2.

Theorem 3.Let . Then

where the sum converges in . Moreover,

for some constant .

Proof. We define

and

Then we first show that

Since is accretive, for any . Since , it follows from the Lebesgue differentiation theorem that

where is the unique dyadic interval such that for all and .

Since is accretive, we have

where with . Hence by the dominated convergence theorem, we get in as . Following the exactly same argument as in the proof of Theorem 2, we can also show that a.e. and in as . Also,

in and almost everywhere. Following the exact same argument in the proof of Proposition ??, we can show that

So the first conclusion of the lemma is verified.

Expanding out , we see that

since is accretive. So

Since , it follows from Theorem 1 that

It remains to show that

Recall that for some constant .

For , it follows from Theorem 1 that

Hence for any set , we have

(5)

(6)

Since

by (5),we have

Since

we get

This completes the first part of the proof. For the converse, let and set , .

Then

Since

Since is accretive, we get the desired result. This completes the proof.