arXiv:0704.0621v1  [math.CV]  4 Apr 2007
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS
MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Abstract. If µ is a finite complex measure in the complex plane C we denote by Cµ its
Cauchy integral defined in the sense of principal value. The measure µ is called reflectionless
if it is continuous (has no atoms) and Cµ = 0 at µ-almost every point. We show that if
µ is reflectionless and its Cauchy maximal function C
∗ is summable with respect to |µ|
then µ is trivial. An example of a reflectionless measure whose maximal function belongs
to the ”weak” L1 is also constructed, proving that the above result is sharp in its scale. We
also give a partial geometric description of the set of reflectionless measures on the line and
discuss connections of our results with the notion of sets of finite perimeter in the sense of
De Giorgi.
1. Introduction
This article discusses uniqueness theorems for Cauchy integrals of complex measures in
the plane. We consider the spaceM =M(C) of finite complex measures µ in C. The Cauchy
integral of a measure fromM is defined in the sense of principal value. First, for any µ ∈M ,
ε > 0 and any z ∈ C consider
Cµε (z) :=
ζ:|ζ−z|>ε
dµ(ζ)
ζ − z
Consequently, the Cauchy integral of µ can be defined as
Cµ(z) := lim
Cµε (z) ,
if the limit exists.
Unlike the Cauchy transform on the line, Cµ can vanish on a set of positive Lebesgue
measure: consider for example µ = dz on a closed curve, whose Cauchy transform is zero
at all points outside the curve. It is natural to ask if Cµ can also vanish on large sets with
respect to µ. If µ = δz is a single point mass, its Cauchy transform will be zero µ-a.e. due
to the above definition of Cµ in the sense of principal value. Examples of infinite discrete
measures with vanishing Cauchy transforms can also be constructed with little effort.
After that one arrives at the following corrected version of the question: Is it true that
any continuous µ ∈ M , such that Cµ(z) = 0 at µ-a.e. point, is trivial? As usual, we call
a measure continuous if it has no point masses. We denote the space of all finite complex
continuous measures by Mc(C).
This problem can also be interpreted in terms of uniqueness. Namely, if f and g are two
functions from L1(|µ|) such that C(f−g)µ = 0, µ-a.e., does it imply that f = g, µ-a.e.? This
way it becomes a problem of injectivity of the planar Cauchy transform.
The first author is supported by grants No. MTM2004-00519 and 2001SGR00431.
The second author is supported by N.S.F. Grant No. 0500852.
The third author is supported by N.S.F. Grant No. 0501067 .
http://arxiv.org/abs/0704.0621v1
2 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
First significant progress towards the solution of this problem was achieved by X. Tolsa
and J. Verdera in [14]. It was established that the answer is positive in two important
particular cases: when µ is absolutely continuous with respect to Lebesgue measure m2 in
C and when µ is a measure of linear growth with finite Menger curvature. The latter class
of measures is one of the main objects in the study of the planar Cauchy transform, see for
instance [11], [12] or [13].
As to the complete solution to the problem, it seemed for a while that the answer could
be positive for any µ ∈Mc, see for example [14]. However, in Section 5 of the present paper
we show that there exists a large set of continuous measures µ satisfying Cµ(z) = 0, µ-a.e.
Following [2], we call such measures reflectionless. This class seems to be an intriguing new
object in the theory.
On the positive side, we prove that if the maximal function associated with the Cauchy
transform is summable with respect to |µ| then µ cannot be reflectionless, see Theorem 2.1.
This result is sharp in its scale because the simplest examples of reflectionless measures
produce maximal functions that lie in the ”weak” L1(|µ|). We prove this result in Section 2
In view of this fact, we believe that the class of continuous measures with summable Cauchy
maximal functions also deserves attention.
A full description of this class and the (disjoint) class of reflectionless measures remains
an open problem.
Let us mention that if µ is a measure with linear growth and finite Menger curvature then
its Cauchy maximal function belongs to L2(|µ|), see [12, 13], and therefore is summable.
This fact relates Theorem 2.1 to the beforementioned result from [14]. The latter can also
be deduced in a different way, see Section 2.
From the point of view of uniqueness, our results imply that any bounded planar Cauchy
transform is injective, see corollary 2.5. This property is a clear analogue of the uniqueness
results for the Cauchy integral on the line or the unit circle.
In Section 3 we discuss other applications of Theorem 2.2. They involve structural theo-
rems of De Giorgi and his notion of a set of finite perimeter, see [5].
In Section 4 we study asymptotic behavior of the Cauchy transform near its zero set.
The results of this section imply that the Radon derivative of µ with respect to Lebesgue
measure m2 vanishes a.e. on the set {Cµ = 0}. In particular the set {Cµ = 0} must be a
zero set with respect to the variation of the absolutely continuous part of µ which is a slight
generalization of the first result of [14]. It is interesting to note that the most direct analogue
of this corollary on the real line is false: it is easy to construct an absolutely continuous (with
respect to m1 = dx) measure µ ∈M(R) such that |µ|({Cµ = 0}) > 0.
Finally, in Section 5 we attempt a geometric description of the set of reflectionless mea-
sures. We give a partial description of reflectionless measures on the line in terms of so-called
comb-like domains. We also provide tools for the construction of various examples of such
measures. In particular, we show that the harmonic measure on any compact subset (of
positive Lebesgue measure) of R is reflectionless.
Acknowledgments. The authors are grateful to Fedja Nazarov for his invaluable comments
and insights. The second author would also like to thank the administration and staff of
Centre de Recerca Matemática in Barcelona for the hospitality during his visit in the Spring
of 2006.
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 3
2. Measures with summable maximal functions
If µ ∈M we denote by Cµ∗ (z) its Cauchy maximal function
Cµ∗ (z) := sup
|Cµε (z)|.
Our first result is the following uniqueness theorem.
Theorem 2.1. Let µ ∈Mc. Assume that Cµ∗ (z) ∈ L1(|µ|) and that Cµ(z) exists and vanishes
µ-a.e. Then µ ≡ 0.
We first prove
Theorem 2.2. If Cµ∗ ∈ L1(|µ|) and Cµ(z) exists µ-a.e. then
µdµ(z) = 2
Cµ(t)dµ(t)
= [Cµ(z)]
for m2-a.e. point z ∈ C . (1)
Proof. Put
F := {z ∈ C :
d|µ|(t)
|t− z|
<∞} .
As |µ| is a finite measure,
m2(C \ F ) = 0 . (2)
Let z ∈ F . Then the integral
|t−ζ|>ε
dµ(t)dµ(ζ)
ζ − z
is absolutely convergent for any ε > 0.
Using the identity
(t− z)(z − ζ)
(z − ζ)(ζ − t)
(ζ − t)(t− z)
we obtain
|t−ζ|>ε
z − ζ
ζ − t
ζ − t
dµ(t)dµ(ζ) =
dµ(ζ)
ζ − z
|t−ζ|>ε
dµ(t)
dµ(t)
|ζ−t|>ε
dµ(ζ)
ζ − t
dµ(t) · Cµε (t) ·
dµ(ζ) · Cµε (ζ) ·
ζ − z
Cµε (t)dµ(t)
E := {z ∈ C :
Cµ∗ (t)d|µ|(t)
|t− z|
<∞} .
By assumption, the numerator Cµ∗ (t)d|µ|(t) is a finite measure. Therefore
m2(C \ E) = 0 . (3)
If z ∈ E then
Cµε (t)dµ(t)
Cµ(t)dµ(t)
. (4)
4 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
This formula is true as long as Cµ∗ ∈ L1(|µ|) and the principal value Cµ exists µ-a.e. by
the dominated convergence theorem. Thus
I = 2CC
µdµ(z) if z ∈ E . (5)
It is left to show that, since z ∈ F ,
I = [Cµ(z)]2 . (6)
Since z ∈ F , the following integral converges absolutely:
φε(t, z) :=
ζ∈C,|ζ−t|>ε
dµ(ζ)
ζ − z
φε(t, z)
dµ(t) .
Since the point z is fixed in F , we have that 1|ζ−z| ∈ L
1(|µ|), and therefore
|ζ−z|d|µ|(ζ) is
small if |µ|(A) is small. Denoting the disc centered at t and of radius ε by B(t, ε) we notice
1) φε(t, z) =
dµ(ζ)
ζ − z
B(t,ε)
dµ(ζ)
ζ − z
2) lim
|µ|(B(t, ε)) = 0.
uniformly in t. Otherwise µ would have an atom.
We conclude that, as ε → 0, the functions φε(t, z) converge uniformly in t ∈ C to φ(z) =∫
ζ−z . Hence for any z ∈ F and any t ∈ C \ z
φε(t, z)
→ φ(z)
, as ε → 0 .
Since φε(t, z) converge uniformly and z ∈ F ,
dµ(t)φε(t, z)
→ φ(z)
dµ(t)
= [Cµ(z)]2 .
We have verified (6).
Combining (5) and (6) we conclude that for z ∈ E ∩ F (so for m2-a.e. z ∈ C) we have
µdµ(z) = 2
Cµ(t)dµ(t)
= lim
I = [Cµ(z)]2 for m2-a.e. point z ∈ C . (7)
This formula is true as long as Cµ∗ ∈ L1(|µ|) and the principal value Cµ exists µ-a.e.
To deduce Theorem 2.1 suppose that Cµ vanishes µ-a.e. Then the left-hand side in (7) is
zero form2-a.e. point z. The same must hold for [C
µ(z)]2. But if Cµ(z) = 0 for Lebesgue-a.e.
point z ∈ C then µ = 0, see for example [6]. Theorem 2.1 is completely proved.
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 5
Remark. In the statement of Theorem 2.2 the condition Cµ∗ ∈ L1(|µ|) can be replaced
with the condition that Cµε converge in L
1(|µ|). The proof would have to be changed as
follows.
Like in the above proof one can show that at Lebesgue-a.e. point z
I = [Cµ(z)]2 . (8)
The relation
I = 2
Cµε (t)dµ(t)
for a.e. z can also be established as before. Since Cµε converge in L
1(|µ|), the last integral
converges to CC
µdµ(z) in the ”weak” L2(dxdy), which concludes the proof.
Hence we arrive at the following version of Theorem 2.1:
Theorem 2.3. Let µ ∈Mc. Assume that Cµε → 0 in L1(|µ|). Then µ ≡ 0.
This version has the following corollary:
corollary 2.4 ([14]). Let µ ∈M be a measure of linear growth and finite Menger curvature.
If Cµ = 0 at µ-a.e. point then µ ≡ 0.
Proof. The conditions on µ imply that the L2(|µ|)-norms of the functions Cµε are uniformly
bounded, see for instance [11]. Since Cµε also converge µ-a.e., they must converge in L
1(|µ|).
Remark As was mentioned in the introduction, Corollary 2.4 also follows from Theo-
rem 2.1. However, the above version of the argument allows one to obtain it without the
additional results of [12, 13] on the maximal function.
We also obtain the following statement on the injectivity of any bounded planar Cauchy
transform. As usual, we say that the Cauchy transform is bounded in L2(µ) if the functions
Cfdµε are uniformly bounded in L
2(µ)-norm for any f ∈ L2(µ). If Cµ is bounded, then Cfdµε
converge µ-a.e as ε → 0 and the image Cfdµ exists in a regular sense as a function in L2(µ),
see [13].
corollary 2.5. Let µ ∈ M be a positive measure. If Cµ is bounded in L2(µ) then it is
injective (has a trivial kernel).
Proof. Suppose that there is f ∈ L2(µ) such that Cfdµ = 0 at µ-a.e. point. Since both f
and Cfdµ∗ are in L
2(µ), Cfdµ∗ is in L
1(|f |dµ). Hence f is a zero-function by Theorem 2.1 �
Remark We have actually obtained a slightly stronger statement: If Cµ is bounded
in L2(µ) then for any f ∈ L2(µ) the functions f and Cfdµ cannot have disjoint essential
supports, i.e. the product fCfdµ cannot equal to 0 at µ-a.e. point.
In the rest of this section we will discuss what other kernels could replace the Cauchy
kernel in the statement of Theorem 2.1.
If K(x) is a complex-valued function in Rn, bounded outside of any neighborhood of the
origin, and µ is a finite measure on Rn, one can define Kµ and Kµ∗ in the same way as C
and Cµ∗ were defined in the introduction.
6 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
The proof of Theorem 2.2 relied on the fact that the Cauchy kernel K(z) = 1/z is odd,
satisfies the symmetry condition (3), i.e.
K(x− y)K(y − z) +K(y − z)K(z − x) +K(z − x)K(x− y) ≡ 0, (9)
and is summable as a function of z for any t with respect to Lebesgue measure. Any K(x)
having these three properties could be used in Theorem 2.1. Out of these three conditions
the symmetry condition (9) seems to be most unique. However, other symmetry conditions
may result in formulas similar to Theorem 2.2 that could still yield Theorem 2.1.
Here is a different example. It shows that much less symmetry can be required from the
kernel if the measure is positive.
Theorem 2.6. Let µ be a positive measure in Rn. Suppose that the real kernel K(x) satisfies
the following properties:
1) K(−x) = −K(x) for any x ∈ Rn;
2) K(x) > 0 for any x from the half-space Rn+ = {x = (x1, x2, ..., xn) | x1 > 0}.
If Kµ∗ ∈ L1(µ) and Kµ(x) = 0 for µ-a.e. x then µ ≡ 0.
Note that real and imaginary parts of the Cauchy kernel, Riesz kernels in Rn, as well as
many other standard kernels satisfy the conditions of the theorem.
We will need the following
Lemma 2.7. Let K be an odd kernel. and let µ, ν ∈ M . Then
Kµε (z)dν(z) = −
Kνε (z)dµ(z) (10)
for any ε > 0.
Suppose that Kµ∗ ∈ L1(|ν|). If Kµ(z) exists ν-a.e. then
Kµ(z)dν(z) = − lim
Kνε (z)dµ(z).
In particular, suppose that both Kµ∗ ∈ L1(|ν|) and Kν∗ ∈ L1(|µ|). If Kµ(z) exists ν-a.e.
and Kν(z) exists µ-a.e. then
Kµ(z)dν(z) =
Kν(z)dµ(z).
Proof. Since K is odd, the first equation can be obtained simply by changing the order of
integration. The second and third equations now follow from the dominated convergence
theorem. �
Proof of Theorem 2.6. There exists a half-plane {x1 = c} in Rn such that µ({x1 = c}) = 0
but both µ({x1 > c}) and µ({x1 < c}) are non-zero. Denote by ν and η the restrictions of
µ onto {x1 > c} and {x1 < c} respectively. Then∫
Kνε (z)dµ(z) =
Kνε (z)dν(z) +
Kνε (z)dη(z).
The first integral on the right-hand side is 0 because of the oddness of K (apply the first
equation in the last lemma with µ = ν). The second condition on K and the positivity of
the measure imply that the second integral is positive and increases as ε → 0. Therefore∫
Kνε (z)dµ(z) cannot tend to zero. This contradicts the fact that K
µ = 0, ν-a.e. and the
second equation from the last lemma. �
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 7
3. Sets of finite perimeter
In this section we give another example of an application of Theorem 2.2. It involves the
notion of a set of finite perimeter introduced by De Giorgi in the 50’s, see [5]. We say that
a set G ⊂ R2 has finite perimeter (in the sense of De Giorgi) if the distributional partial
derivatives of its characteristic function χG are finite measures. Such sets have structural
theorems. For example, if G is such a set then the measure ∇χG is carried by a set E,
rectifiable in the sense of Besicovitch, i. e. a subset of a countable union of C1 curves and an
H1-null set, where H1 is the one-dimensional Hausdorff measure. Also the measure ∇χG is
absolutely continuous with respect to H1 restricted to E and its Radon-Nikodym derivative
is a unit normal vector H1-a.e. (notice that ∇χG is a vector measure). At H1-almost all
points of E the function χG has approximate “one-sided”’ limit. For more details we refer
the reader to [5].
The general question we consider can be formulated as follows: What can be said about µ
if Cµ coincides at µ-a.e. point with a ”good” function f? To avoid certain technical details,
all measures in this section are compactly supported. Furthermore, we will only discuss the
two simplest choices of f . As we will see, even in such elementary situations Theorem 2.2
yields interesting consequences.
As usual, when we say that Cµ = f at µ-a.e. point, we imply that the principal value
exists µ-almost everywhere.
Theorem 3.1. Let µ ∈ Mc be compactly supported. Assume that Cµ(z) = 1, µ-almost
everywhere and Cµ∗ ∈ L1(|µ|). Then µ = ∂̄χG, where G is a set of finite perimeter. In
particular, µ is carried by a set E, H1(E) < ∞, rectifiable in the sense of Besicovitch, and
µ is absolutely continuous with respect to the restriction of H1 to E.
Remark. The most natural example of such a measure is dz on a C1 closed curve. The
theorem says that, by the structural results of De Giorgi, this is basically the full answer.
Proof. By Theorem 2.2 we get that for Lebesgue-almost every point in C
[Cµ(z)]2 = 2Cµ(z) . (11)
In other words for m2-a.e. point z we have C
µ(z) = 0 or = 2. Let G denote the set where
Cµ(z) = 2. Since the Cauchy transform of any compactly supported finite measure must
tend to zero at infinity, this set is bounded. Consider the following equality
χG = C
understood in the sense that the two functions are equal as distributions. Taking distribu-
tional derivatives on both sides we obtain
∂̄χG = µ/2 and ∂χ̄G = µ̄/2.
Hence G has finite perimeter and the rest of the statement follows from the results of [5]. �
We say that a set G has locally finite perimeter (in the sense of De Giorgi) if the distribu-
tional derivatives of χG are locally finite measures. Our second application is the following
8 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Theorem 3.2. Let µ ∈ Mc be compactly supported. Assume that Cµ(z) = z, µ-almost
everywhere and Cµ∗ ∈ L1(|µ|). If µ(C) = 0 then µ = 2z∂̄χG, where G is a set with locally
finite perimeter. Whether µ(C) = 0 or not, µ is carried by a set E, H1(E) <∞, which is a
rectifiable set in the sense of Besicovitch, and µ is absolutely continuous with respect to the
restriction of H1 to E.
Remark. The most natural example of such a measure is zdz on a C1 closed curve. Our
statement shows that this is basically one-half of the answer. The other half is given by√
z2 − cdz as will be seen from the proof.
Proof. Again, from Theorem 2.2 we get that for Lebesgue-almost every point in C
[Cµ(z)]2 = 2Cζdµ(ζ)(z) . (12)
Notice that
Cζdµ(ζ)(z) =
ζ − z
dµ(ζ) = µ(C) + zCµ(z)
and we get a quadratic equation
[Cµ(z)]2 = 2zCµ(z)− p ,
where p := −2µ(C).
First case p = 0. Here we get
[Cµ(z)]2 = 2zCµ(z) .
We conclude that Cµ(z) = 0 or z for Lebesgue-a.e. point z ∈ C.
Again a bounded set G appears on which
Cµ = 2zχG(z)
in terms of distributions. Therefore
∂̄χG = dµ/2z ,
and the right hand side is a finite measure on any compact set avoiding the origin. Therefore,
G is a (locally) De Giorgi set.
Let us consider the case p 6= 0. For simplicity we assume p = 1, other p’s are treated in
the same way. Then we have to solve the quadratic equation
Cµ(z)2 − 2zCµ(z) + 1 = 0
for Lebesgue-a.e. point in C. Let us make the slit [−1, 1] and consider two holomorphic
functions in C \ [−1, 1]
r1(z) = z −
z2 − 1, r2(z) = z +
z2 − 1 ,
where the branch of the square root is chosen so that
r1(z) → 0, z → ∞ .
In other words we have the sets E1 and E2 such that m2(C \ E1 ∪ E2) = 0 and
z ∈ E1 ⇒ Cµ(z) = r1(z) ,
z ∈ E2 ⇒ Cµ(z) = r2(z) .
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 9
Obviously it is E1 that contains a neighborhood of infinity. The function z −
z2 − 1
outside of [−1, 1] can be written as Cµ0(z) where dµ0(x) = 1π
1− x2dx. Consider ν = µ−µ0.
z ∈ E1 ⇒ Cν(z) = 0 ,
z ∈ E2 ⇒ Cν(z) = 2
z2 − 1 := R(z) .
Therefore,
Cν(z) = R(z)χE2 . (13)
Notice that if R was analytic in an open domain compactly containing E2 we would conclude
from the previous equality that
ν = R(z)∂̄χE2 .
If, in addition, |R| was bounded away from zero on E2, we would obtain that ∂̄χE2 and ∂χE2
are measures of finite variation, and hence E2 is a set of finite perimeter. Notice that our
R(z) = 2
z2 − 1 is analytic in O := C \ [−1, 1] and is nowhere zero. We will conclude that
E2 is a set of locally finite perimeter. More precisely we will establish the following claim:
For every open disk V ⊂ O the set O ∩ E2 has finite perimeter.
Indeed, let W be a disk compactly containing V , W ⊂ O. Let ψ be a smooth function,
supported in W , ψ|V = 1. Multiply (13) by ψ and take a distributional derivative (against
smooth functions supported in V ). Then we get (using the fact that R is holomorphic on V )
ν|V = ∂̄(ψRχE2∩V )|V = ∂̄(RχE2∩V )|V = R∂̄(χE2∩V )|V .
We conclude immediately that E2 ∩ V is a set of finite perimeter. Therefore, E2 ∩D is a set
of finite perimeter, where D is a domain whose closure is contained compactly in O.
Recalling that µ = ν + µ0 we finish the proof. �
Remark 3.3. In is interesting to note that, as follows from the proof, if µ is the measure
from the statement of the theorem then one of the connected components of supp µ must
contain both roots of the equation z2 + 2µ(C) = 0.
We conclude this section with the following examples of measures µ whose Cauchy trans-
form coincides with z at µ-a.e. point
Examples. 1. Let Ω be an open domain with smooth boundary Γ. Suppose that [−1, 1] ⊂ Ω.
Let {Dj}∞j=1 be smoothly bounded disjoint domains in O := Ω \ [−1, 1], γj = ∂Dj . Assume
H1(γj) <∞ . (14)
LetR(z) be an analytic branch of 2
z2 − 1 inO. Consider the measure ν on Γ∪(∪γj)∪[−1, 1]
defined as
ν = R(z)dz|Γ − R(z)dz|∪γj −
1− x2dx|[−1,1].
10 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Cν(z) =
0 if z ∈ C \ Ō ,
0 if z ∈ ∪jDj ,
R(z) if z ∈ O \ ∪jD̄j .
Recall that R(z) = z +
z2 − 1 − (z −
z2 − 1) and that Cµ0(z) = z −
z2 − 1 for µ0 =
1− x2dx|[−1,1]. We conclude that for µ = ν + µ0 one has
Cµ(z) =
z2 − 1 if z ∈ C \ Ō ,
z2 − 1 if z ∈ ∪jDj ,
z2 − 1 if z ∈ O \ ∪jD̄j .
2. The second example is exactly the same as the first one but Dj,k = B(xj,k,
), xj,k =
j , 1 ≤ k ≤ j, j = 1, 2, 3.... Here the assumption (14) fails. But ν, defined as above,
will still be a measure of finite variation (and so will be µ): |ν|(C) ≤ C
In both examples Cµ(z) = z for µ-a.e. z.
4. Asymptotic behavior near the zero-set of Cµ
In this section we take a slightly different approach. We study asymptotic properties of
measures near the sets where the Cauchy transform vanishes. Theorem 4.2 below shows
that near the density points of such sets the measure must display a certain ”irregular”
asymptotic behavior.
As was mentioned in the introduction, one of the results of [14] says that an absolutely
continuous planar measure cannot be reflectionless. This result is not implied by our Theorem
2.1 because an absolutely continuous measure may not have a summable Cauchy maximal
function. It is, however, implied by Theorem 4.2, see Corollary 4.4 below.
When estimating Cauchy integrals one often uses an elementary observation that the
difference of any two Cauchy kernels 1/(z − a) − 1/(z − b) can be estimated as O(|z|−2)
near infinity. To obtain higher order of decay one may consider higher order differences.
Here we will utilize the following estimate of that kind, which can be verified through simple
calculations.
Lemma 4.1. If a, b, c ∈ B(0, r) be different points, |a − b| > r. Then there exist constants
A,B ∈ C such that |A|, |B| < 2
z − a
z − b
z − c
∣∣∣∣ <
outside of B(0, 2r).
(Namely, A = b−c
b−a , B =
a−b .)
If µ ∈M consider one of its Riesz transforms in R3, R1µ(x, y, z), defined as
R1µ(x, y, z) =
|(u, v, 0)− (x, y, z)|3
dµ(u+ iv).
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 11
This transform is the planar analogue of the Poisson transform. In particular,
R1µ(x, y, z) =
(x+ iy)
for all points w = x+ iy ∈ C where the Radon derivative
(w) = lim
µ(B(w, r))
|B(w, r)|
exists.
For measures on the line or on the circle their Poisson integrals and Radon derivatives
(with respect to the one-dimensional Lebesgue measure) are very much related but not always
equivalent. When the asymptotics of the Poisson integral and the ratio from the definition
of the Radon derivative are different near a certain point it usually means that the measure
is ”irregular” near that point. It is not difficult to show that if µ is absolutely continuous
then at a Lebesgue point of its density function the Radon derivative of µ and the Poisson
integral of |µ| (or R1|µ| if n > 1) behave equivalently. Even for singular measures on the
circle, if a measure possesses a certain symmetry near a point, then the same equivalent
behavior takes place, as follows for instance from [1], Lemma 4.1. In fact, it is not easy to
construct a measure so that its Poisson integral and Radon derivative behaved differently
near a large set of points. The same can be said about the Riesz transform and the Radon
derivative. Thus one may interpret our next result as an evidence that, for a planar measure
µ, most points where Cµ = 0 are ”irregular.”
Theorem 4.2. Let µ ∈ M and let w = x + iy be a point of density (with respect to m2) of
the set E = {Cµ = 0}. Then
µ(B(w, r))
= o (R1|µ|(x, y, r)) as r → 0 + .
In view of the above discussion this implies
corollary 4.3. If w is a point of density of the set E = {Cµ = 0}, such that there exists the
Radon derivative d|µ|/dm2(w) 6= 0, then
µ(B(w, r)) = o (|µ|(B(w, r))) as r → 0+ (16)
and dµ/dm2(w) = 0.
Since m2-almost every point of a set is its density point, we also obtain the following
version of the result from [14]:
corollary 4.4. The set E = {Cµ = 0} has measure zero with respect to the absolutely
continuous component of µ.
Proof of Theorem 4.2. without loss of generality w = 0. Choose a C∞0 test-function φ sup-
ported in B := B(0, r), and such that 0 ≤ φ ≤ D/r2, |∇φ| ≤ A/r3 and
φ dm2 = 1. Denote
the complement of E by Ec. Then
φdµ = 〈φ, ∂̄Cµ〉 = 〈∂̄φ, Cµ〉 = 〈χEc∂̄φ, Cµ〉 =
χEc ∂̄φ dm2(z)
ζ − z
dµ(ζ) (17)
All we need is to show that the last integral is small. Then, since the first integral in
(17) is similar to the right-hand side of (16) we will complete the proof. The main idea for
12 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
the rest of the proof is to make the function F (ζ) =
χEc ∂̄φ dm2(z)
ζ−z ”small” by subtracting a
linear combination of Cauchy kernels corresponding to points from E, which will not change
its integral with respect to µ.
Namely, let a, b ∈ B(0, r) ∩ E be any two points such that |a − b| > r. By the previous
lemma for any z ∈ B(0, r) there exist constants A = A(z), B = B(z), of modulus at most 2,
such that (15) holds with c = z. Integrating (15) with respect to χEc ∂̄φ dm2(z) we obtain
that ∣∣∣∣
χEc ∂̄φ dm2(z)
ζ − z
ζ − a
ζ − b
∣∣∣∣ < C
ε(r)r
|ζ |3
outside of B(0, 2r) for some constants A∗, B∗, where ε(r) = |B(0, r)∩Ec|/r2 = o(1) as r → 0.
The constants satisfy |A∗|, |B∗| < 2 ε(r)
Notice that if w ∈ E then
ζ−wdµ = 0 by the definition of the set E. Hence, since
a, b ∈ E,
χEc ∂̄φ dm2(z)
ζ − z
dµ(ζ) =
χEc ∂̄φ dm2(z)
ζ − z
ζ − a
ζ − b
dµ(ζ)
B(0,2r)
C\B(0,2r)
= I1 + I2.
For I2 we now have
C\B(0,2r)
χEc ∂̄φ dm2(z)
ζ − z
ζ − a
ζ − b
dµ(ζ)
C\B(0,2r)
ε(r)r
|ζ |3
d|µ|(ζ) ≤ Cε(r)R1|µ|(0, 0, r).
In I1 we estimate each summand separately. First,
B(0,2r)
χEc ∂̄φ dm2(z)
ζ − z
dµ(ζ)
∣∣∣∣ ≤
B(0,2r)
|ζ − z|
χEcdm2(z)d|µ|(ζ)
|µ|(B(0, 2r)) ≤ C
ε(r)R1|µ|(0, 0, r).
To estimate the second and third summands of I1, recall that the only restriction on the
choice of a, b ∈ B(0, r) ∩ E was that |a − b| > r. This condition will be satisfied, for
instance, if a ∈ B1 = B(−56r,
r) and b ∈ B2 = B(56r,
r). If we average the modulus of the
second summand over all choices of a ∈ B1 ∩ E, recalling that A∗ = A∗(a) always satisfies
|A∗| ≤ 2 ε(r)
, we get
|B1 ∩ E|
B(0,2r)
A∗(a)
ζ − a
dµ(ζ)
∣∣∣∣dm2(a) ≤
|B1 ∩ E|
B(0,2r)
|A∗(a)|
|ζ − a|
dm2(a)d|µ|(ζ)
r|µ|(B(0, 2r)) ≤ Cε(r)R1|µ|(0, 0, r).
It is left to choose a ∈ B1 ∩ E for which the modulus is no greater than its average. The
same can be done for b. The proof is finished. �
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 13
5. Reflectionless measures and Combs
As was mentioned in the introduction, following [2], we will call a non-trivial continuous
finite measure µ ∈M(C) reflectionless if Cµ(z) = 0 at µ-a.e. point z.
Perhaps the simplest example of a reflectionless measure is the measure µ = 1
(1−x2)−1/2dx
on [−1, 1], the harmonic measure of C \ [−1, 1] corresponding to infinity. The fact that
µ is reflectionless can be verified through routine calculations or via the conformal map
interpretation of the harmonic measure. It will also follow from a more general Theorem 5.4
below.
At the same time, since Cµ∗ ≍ (1 − x2)−1/2 on [−1, 1], this simple example complements
the statement of Theorem 2.1. Since the function (1−x2)−1/2 belongs to the ”weak” L1(|µ|),
the summability condition for the Cauchy maximal function proves to be exact in its scale.
In the rest of this section we discuss further examples and properties of positive reflec-
tionless measures on the line.
Let us recall that functions holomorphic in the upper half plane C+ and mapping it to
itself (having non-negative imaginary part) are called Nevanlinna functions. Let M+(R)
denote the class of finite positive measures compactly supported on R. The function f is a
Nevanlinna function if and only if it has a form
f(z) = az + b+
t2 + 1
]dρ(t) ,
where ρ is a positive measure on R such that
dρ(t)
< ∞, a > 0, b ∈ R are constants.
If the representing measure is from M+(R) and f(∞) = 0, the formula becomes simpler:
f(z) =
dµ(x)
x−z .
Definition. A simply connected domain O is comb-like if it is a subset of a half-strip
{w : ℑw ∈ (0, π),ℜw > q}, for some q ∈ R, contains another half-strip {w : ℑw ∈
(0, π),ℜw > r} for some r ∈ R and has the property that
for any w0 = u0 + iv0 ∈ O the whole ray {w = u+ iv0, u ≥ u0} lies in O . (18)
If in addition H1(∂O∩B(0, R)) <∞ for all finite R, we say that O is a rectifiable comb-like
domain.
Let O be a rectifiable comb-like domain, Γ = ∂O. Then by the Besicovitch theory we
know that for H1-a.e. pont w ∈ Γ there exists an approximate tangent line to Γ, see [3] for
details. We wish to consider rectifiable comb-like domains satisfying the following geometric
property:
for a.e. w ∈ Γ approximate tangent line is either vertical or horizontal. (19)
It is not difficult to verify that for any conformal map F : C+ → O, O is comblike if and
only if F ′ is a Cauchy potential of µ ∈ M+(R): F ′(z) =
dµ(x)
x−z . It is, therefore, natural to
ask the following
Question. Which comb-like domains correspond to reflectionless measures µ ∈M+(R)?
An answer would give a geometric description of reflectionless measures from M+(R). If,
in addition, a comb-like domain is rectifiable, then the answer is given by
14 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Theorem 5.1. 1) Rectifiable comb-like domains correspond exactly to those measures
µ ∈M+(R) that are absolutely continuous with respect to dx and satisfy
dµ(x)
∈ H1loc(C+). (20)
2) An absolutely continuous measure satisfying (20) is reflectionless if and only if the corre-
sponding comb-like domain has the property (19).
Remarks.
1) Of course not every comb-like domain gives rise to a reflectionless measure fromM+(R).
Just take any comb-like domain which appears as F (C+), where F =
∫ z ∫ dµ(x)
x−z for a singular
µ ∈M+(R). By a result from [9] singular measures cannot be reflectionless.
2) On the other hand, even if µ = g(x)dx is a reflectionless absolutely continuous measure,
the corresponding conformal map F =
∫ z ∫ dµ(x)
x−z : C+ → O can be onto a non-rectifiable
domain.
3) For non-rectifiable domains we have no criteria to recognize which ones correspond to
reflectionless measures.
4) It is well known, and not difficult to prove, that the antiderivative of a Nevanlinna
function is a conformal map, see for instance [4]. If F =
∫ z ∫ dµ(x)
x−z , µ ∈ M+(R) then ℑF (x)
is an increasing function on R whose derivative in the sense of distributions is µ. The image
F (C+) lies in the strip {ℑw ∈ (0, π‖µ‖)}.
Theorem 5.1 will follow from Theorems 5.2 and 5.3 below.
Theorem 5.2. Let F be a conformal map of C+ on a rectifiable comb-like domain O. Then
F (z) =
∫ z ∫ dµ(x)
x−z , µ ∈ M+(R), µ << dx. Also
dµ(x)
x−z ∈ H
loc(C+). If in addition O satisfies
(19) then µ is reflectionless.
Proof. without loss of generality O ⊂ {ℜz > 0}. Put Φ = eF . Then the image Φ(O) is
the subdomain of the complement of the unit half-disk in C+ which is the union of rays
(R(θ)eiθ,∞). Consider the subdomain of the upper half-disk D := {z : 1/z ∈ Φ(O)}. Define
G as the smallest open domain containing D and its reflection D := {z̄ : z ∈ D}. Then G is
a star-like domain inside the unit disk. The preimage of G ∩ R under Φ is the union of two
Infinite rays R1 = [−∞, a), R2 = (b,∞], a < b. Therefore, by reflection principle C \ [a, b] is
mapped conformally (by the extension of Φ which we will also denote by Φ) onto star-like
Since Φ : C+ → G, where G is star-like, it is well-known that argΦ(x+ iδ) is an increasing
function of x, see [7].
We conclude that the argument of Φ is monotone. Therefore, ℑF (x + iδ) is monotone,
and so ℑf(x+ iδ) is positive, where f = F ′. We see that f = F ′ is a Nevanlinna function.
From the structure of our comb-like domain, we conclude immediately that its representing
measure µ has compact support, so we are in M+(R). Also, let us prove that µ << dx. The
boundary of our comb is locally rectifiable. So f = F ′ belongs locally to the Hardy class
H1(C+), [16]. Since ℑf is the Poisson integral of µ,
ℑf = Pµ =
(x− t)2 + y2
dµ(t),
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 15
and f is in H1(C+) locally, we conclude that µ = ℑfdx,ℑf ≥ 0 a.e., [16].
Now suppose that, in addition, O = F (C+) has the property (19). Let us recall that
for a simply connected domain with rectifiable boundary Γ the restriction of the Hausdorff
measure H1|Γ is equivalent to the harmonic measure ν on O. Therefore the tangent lines
to Γ are either vertical or horizontal a.e. with respect to ν. The measure ν is the image
of the harmonic measure λ of C+ which is equivalent to the Lebesgue measure on the line.
We have a conformal map F (a continuous function up to the boundary of C+ because it
is an anti-derivative of an H1loc-function) which pushes forward λ to ν. Call a point w0 ∈ Γ
accessible from O if there exists a ray x0 + iy, 0 < y < 1, such that w0 = limy→0 F (x0 + iy).
Almost every point of Γ (w.r. to ν) is accessible from O. For ν-a.e. accessible w0 ∈ Γ
where the tangent line is vertical (horizontal) we can say that ℜF ′(x0) = 0 (ℑF ′(x) = 0).
So R = E1 ∪ E2 ∪ E3, where |E3| = 0, |E1 ∩ E2| = 0, and E1 = {x ∈ R : ℜF ′(x) = 0},
E2 = {x ∈ R : ℑF ′(x) = 0}. We already know that the measure µ = ℑF ′(x)dx represents
f(z) = F ′(z) =
dµ(t)
t−z . Notice that
R\E2 · =
·. But we also know that boundary
values exist dx-almost everywhere, i.e.
dµ(t)
t− x− iy
= ℜF ′(x) = 0
for a.e. x ∈ E1 and therefore for µ-a.e. x ∈ E1. This means (see [16]) that
dµ(x)
= 0 µ-a.e.
Definition. A simply connected rectifiable comb-like domain O is called a comb if its “left”
boundary consists of countably many horizontal and vertical segments.
A comb is called a straight comb if O = {w : ℑw ∈ (0, π),ℜw > 0} \ S, where the set S
is relatively closed with respect to the strip {w : ℑw ∈ (0, π),ℜw > 0} and is the union of
countably many horizontal intervals Rn = (iyn, ln + iyn]. We require also that
ln <∞ .
Example. Let F be a conformal map of C+ on a comb O. By our last theorem F ′(z) =∫
dµ(x)
x−z , where µ ∈M+(R) is reflectionless: C
µ(x) = 0 for µ-a.e. x.
Definition. Let E be a compact subset of the real line. Let E have positive logarithmic
capacity, so Green’s function G of C \E exists. The domain C \E is called Widom domain
G(c) <∞ ,
where the summation goes over all critical points of G (we assume that G is a Green’s
function with pole at infinity.
Example. Let E be a compact subset of the real line of the positive length. We assume
that every point of E is regular in the sense of Dirichlet for the domain C \ E, and we
also assume that C \ E is not a Widom domain. Such E exist in abundance. We will
see below, that the harmonic measure ω of C \ E (with pole at infinity) is reflectionless.
Consider F (z) =
∫ z ∫ dω(x)
z−x for z ∈ C+. It is easy to see that F (z) = G(z) + iG̃(z) + const,
16 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
where G̃ is the harmonic conjugate of G. This F is a conformal map (see [4]) of C= onto
a domain D lying in the strip {w : ℑw ∈ (0, π)}. It is easy to see that complementary
intervals of E will be mapped by F onto straight horizontal segments on the boundary of
D. Each finite complementary interval contains exactly one critical point of G, and clearly
the length of the corresponding straight horizontal segment is G(c) (this follows from the
formula F (z) = G(z) + iG̃(z) + const).
As the domain C \ E was not a Widom domain, we have that the sum of lengths of
abovementioned straight horizontal segment is infinite. So domain D is not rectifiable.
Therefore the reflectionless property of µ alone does not say anything about the rectifiability
of the domain, which is the target domain of the conformal map F (z) =
∫ z ∫ dµ(x)
z−x .
Theorem 5.3. Let µ be absolutely continuous positive measure on R and let Cµ ∈ H1loc(C+).
Then F (z) =
∫ z ∫ dµ(x)
x−z is a conformal map of C+ onto a rectifiable comb-like domain O. If
µ is reflectionless then O has the property (19).
Proof. Consider F (z) =
∫ z ∫ dµ(x)
x−z . Since µ is positive, it is a conformal map. If µ is such
that f(z) = Cµ ∈ H1loc(C+) then F (z) =
f maps C+ onto a domain with locally rectifiable
boundary (see [16]).
If, in addition, µ = ℑfdx is reflectionless, then for a.e. point of P := {x ∈ R : ℑf(x) > 0}
we have ℜf(x) = 0. Conformal map F (z) is continuous up to the boundary of C+ and its
boundary values F (x) form a (locally) absolutely continuous function, F ′(x) = f(x) a.e. As
at almost every point we have either ℑF ′(x) = 0 or ℜF ′(x) = 0 we conclude that O = F (C+)
has the property (19).
We also need the following definition.
Definition. A compact subset E in R is called homogeneous if there exist r, δ > 0 such that
for all x ∈ E, |E ∩ (x− h, x+ h)| ≥ δh for all h ∈ (0, r).
Example. Let E ⊂ R be a compact set of positive length. Let µ be a reflectionless
measure supported on E, µ = g(x)dx. Let in addition E be a homogeneous set. Then
F (z) =
∫ z ∫ dµ(x)
x−z is a conformal map from C+ on a rectifiable comb-like domain satisfying
(19).
Proof. The Cauchy integral Cgdx considered in C\E will be in the Hardy class H1(C\E). In
fact the reflectionless property of gdx implies that its limits from C± will be both integrable
with respect to dx|E .
Now we use homogenuity of E and Zinsmeister’s theorem [15] to conclude that f(z) =
Cgdx(z) is in the usual H1loc(C). Then the conformal map F (z) =
f maps C+ onto a
rectifiable subdomain of a strip. We use Theorem 5.3 to get the rest of our example’s
claims. �
The simple example of a reflectionless measure mentioned at the beginning of this section,
as well as many other explicit examples, are given by our next statement.
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 17
Theorem 5.4. Let E be a compact set of positive lenght, E ⊂ R. Let ω be a harmonic
measure of C \ E with pole at infinity. Then ω is reflectionless.
Example. The simplest comb is a strip {w : ℑw ∈ (0, π),ℜw > 0}. Consider F (z) =
log(z +
z2 − 1). It maps conformally C+ onto the strip. Its derivative f(z) = 1√z2−1 is
x−z and dµ =
1−x2 is the harmonic measure of C \ [−1, 1].
Proof of Theorem 5.4. We need to show that Cω = 0 at ω-a.e point. From our definitions
it can be seen, that Cω on the line coincides with the Hilbert transform of ω, which in its
turn is asymptotically equivalent to the conjugate Poisson transform Qω. Thus all we need
to establish is that
Qω(x+ ih) =
(x− y)2 + h2
dω(y) = ℜ
dω(y)
x− ih− y
→ 0 as h→ 0+ (21)
for almost every x. Instead, we have that the Green’s function F (x) defined as
F (x) =
log |x− y|dω(y) + C∞,
where C∞ is a real constant (Robin’s constant), is equal to 0 at every density point of
E, see for example [8]. The idea of the proof is to show that Qω(x + iε) behaves like
(F (x+ ε) + F (x− ε))/ε near almost every x. The technical details are as follows.
Introduce
φ(y) :=
|1− y|
|1 + y|
y2 + 1
, (22)
φx,h(y) :=
y − x
The function φ(y) decreases as 1/y2 at infinity, hence it is in L1(R, dx) and so are φx,h(y)
with a uniform bound on the norm. However, these functions are not bounded, which makes
it difficult to use them in our estimates. To finish the proof we will first obtain a bounded
version of φx,h(y) through the following averaging procedure.
Let ω = g(x)dx. Choose x to be a Lebesgue point of g and a density point of E. Fixing
sufficiently small h > 0 we can find the set A(x, h) ⊂ (x−h, x−h/2)∪ (x+h/2, x+h) such
• A(x, h) consists of density points of E,
• |A(x, h)| ≥ h/2,
• A(x, h) is symmetric with respect to x.
Let Tx,h := T := {t ∈ (0, h) : x+ t ∈ A(x, h)}. Then |T | ≥ h/4. Now put
ψx,h(y) :=
φx,t(y) dt .
By (22) one can see immediately that
|ψx,h| ≤
for some M > 0 and |ψx,h(y)| ≤ C
, for |y| > h . (23)
18 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Also, since ∫
φ dy = 0 .
we have that ∫
ψx,h dy = 0 .
Therefore,
g(y)ψx,h(y) dy| = |
(g(y)− g(x))ψx,h(y) dy| ≤
|g(y)− g(x)||ψx,h|(y) dy.
Now notice that (23) implies that |ψx,h| is majorated by an approximate unity (for instance,
by a constant multiple of the Poisson kernel corresponding to z = x + ih). Since x is a
Lebesgue point for g(x), this means that the last integral tends to 0 as h→ 0.
Looking at the definitions of Tx,h and ψx,h(y) we can see that
g(y)ψx,h(y) dy =
|Tx,h|
(F (x+ t)− F (x− t))−ℜ
g(y)dy
x− it− y
where F (x) is the Green’s function. As we mentioned before, F is zero at the density points
of E. We conclude that
|Tx,h|
g(y)dy
x− it− y
→ 0, h→ 0 + .
for a.e. x on the Borel support of g. Since the Cauchy integral of g has a limit a.e. we obtain
g(y)dy
x− ih− y
→ 0, h→ 0 + .
Remark. All reflectionless measures on R discussed in this section, including those provided
by Theorem 5.4 are absolutely continuous with respect to Lebesgue measure. One may
wonder if there exist singular reflectionless measures. The answer is negative. More generally,
as follows from a theorem from [9], if principal values of the Hilbert transform exist µ-a.e.
for a continuous µ ∈M(R) then µ << dx .
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Leningrad, 1950. 336 pp.
Department de Matematiques, Uneversitat Autonoma de Barcelona, 08193 Bellaterra,
Barcelona, Spain
E-mail address : mark.melnikov@gmail.com
Texas A& M University, Department of Mathematics, College Station, TX 77843, USA
E-mail address : alexeip@math.tamu.edu
Dept. Math., Michigan State Univ., East Lansing MI 48823, USA, and, School of Math.,
University of Edinburgh, Edinburgh UK EH9 EJ6
E-mail address : sashavolberg@yahoo.com

  If $\mu$ is a finite complex measure in the complex plane $\C$ we denote by
$C^\mu$ its Cauchy integral defined in the sense of principal value. The
measure $\mu$ is called reflectionless if it is continuous (has no atoms) and
$C^\mu=0$ at $\mu$-almost every point. We show that if $\mu$ is reflectionless
and its Cauchy maximal function $C^\mu_*$ is summable with respect to $|\mu|$
then $\mu$ is trivial. An example of a reflectionless measure whose maximal
function belongs to the "weak" $L^1$ is also constructed, proving that the
above result is sharp in its scale. We also give a partial geometric
description of the set of reflectionless measures on the line and discuss
connections of our results with the notion of sets of finite perimeter in the
sense of De Giorgi.

Introduction
This article discusses uniqueness theorems for Cauchy integrals of complex measures in
the plane. We consider the spaceM =M(C) of finite complex measures µ in C. The Cauchy
integral of a measure fromM is defined in the sense of principal value. First, for any µ ∈M ,
ε > 0 and any z ∈ C consider
Cµε (z) :=
ζ:|ζ−z|>ε
dµ(ζ)
ζ − z
Consequently, the Cauchy integral of µ can be defined as
Cµ(z) := lim
Cµε (z) ,
if the limit exists.
Unlike the Cauchy transform on the line, Cµ can vanish on a set of positive Lebesgue
measure: consider for example µ = dz on a closed curve, whose Cauchy transform is zero
at all points outside the curve. It is natural to ask if Cµ can also vanish on large sets with
respect to µ. If µ = δz is a single point mass, its Cauchy transform will be zero µ-a.e. due
to the above definition of Cµ in the sense of principal value. Examples of infinite discrete
measures with vanishing Cauchy transforms can also be constructed with little effort.
After that one arrives at the following corrected version of the question: Is it true that
any continuous µ ∈ M , such that Cµ(z) = 0 at µ-a.e. point, is trivial? As usual, we call
a measure continuous if it has no point masses. We denote the space of all finite complex
continuous measures by Mc(C).
This problem can also be interpreted in terms of uniqueness. Namely, if f and g are two
functions from L1(|µ|) such that C(f−g)µ = 0, µ-a.e., does it imply that f = g, µ-a.e.? This
way it becomes a problem of injectivity of the planar Cauchy transform.
The first author is supported by grants No. MTM2004-00519 and 2001SGR00431.
The second author is supported by N.S.F. Grant No. 0500852.
The third author is supported by N.S.F. Grant No. 0501067 .
http://arxiv.org/abs/0704.0621v1
2 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
First significant progress towards the solution of this problem was achieved by X. Tolsa
and J. Verdera in [14]. It was established that the answer is positive in two important
particular cases: when µ is absolutely continuous with respect to Lebesgue measure m2 in
C and when µ is a measure of linear growth with finite Menger curvature. The latter class
of measures is one of the main objects in the study of the planar Cauchy transform, see for
instance [11], [12] or [13].
As to the complete solution to the problem, it seemed for a while that the answer could
be positive for any µ ∈Mc, see for example [14]. However, in Section 5 of the present paper
we show that there exists a large set of continuous measures µ satisfying Cµ(z) = 0, µ-a.e.
Following [2], we call such measures reflectionless. This class seems to be an intriguing new
object in the theory.
On the positive side, we prove that if the maximal function associated with the Cauchy
transform is summable with respect to |µ| then µ cannot be reflectionless, see Theorem 2.1.
This result is sharp in its scale because the simplest examples of reflectionless measures
produce maximal functions that lie in the ”weak” L1(|µ|). We prove this result in Section 2
In view of this fact, we believe that the class of continuous measures with summable Cauchy
maximal functions also deserves attention.
A full description of this class and the (disjoint) class of reflectionless measures remains
an open problem.
Let us mention that if µ is a measure with linear growth and finite Menger curvature then
its Cauchy maximal function belongs to L2(|µ|), see [12, 13], and therefore is summable.
This fact relates Theorem 2.1 to the beforementioned result from [14]. The latter can also
be deduced in a different way, see Section 2.
From the point of view of uniqueness, our results imply that any bounded planar Cauchy
transform is injective, see corollary 2.5. This property is a clear analogue of the uniqueness
results for the Cauchy integral on the line or the unit circle.
In Section 3 we discuss other applications of Theorem 2.2. They involve structural theo-
rems of De Giorgi and his notion of a set of finite perimeter, see [5].
In Section 4 we study asymptotic behavior of the Cauchy transform near its zero set.
The results of this section imply that the Radon derivative of µ with respect to Lebesgue
measure m2 vanishes a.e. on the set {Cµ = 0}. In particular the set {Cµ = 0} must be a
zero set with respect to the variation of the absolutely continuous part of µ which is a slight
generalization of the first result of [14]. It is interesting to note that the most direct analogue
of this corollary on the real line is false: it is easy to construct an absolutely continuous (with
respect to m1 = dx) measure µ ∈M(R) such that |µ|({Cµ = 0}) > 0.
Finally, in Section 5 we attempt a geometric description of the set of reflectionless mea-
sures. We give a partial description of reflectionless measures on the line in terms of so-called
comb-like domains. We also provide tools for the construction of various examples of such
measures. In particular, we show that the harmonic measure on any compact subset (of
positive Lebesgue measure) of R is reflectionless.
Acknowledgments. The authors are grateful to Fedja Nazarov for his invaluable comments
and insights. The second author would also like to thank the administration and staff of
Centre de Recerca Matemática in Barcelona for the hospitality during his visit in the Spring
of 2006.
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 3
2. Measures with summable maximal functions
If µ ∈M we denote by Cµ∗ (z) its Cauchy maximal function
Cµ∗ (z) := sup
|Cµε (z)|.
Our first result is the following uniqueness theorem.
Theorem 2.1. Let µ ∈Mc. Assume that Cµ∗ (z) ∈ L1(|µ|) and that Cµ(z) exists and vanishes
µ-a.e. Then µ ≡ 0.
We first prove
Theorem 2.2. If Cµ∗ ∈ L1(|µ|) and Cµ(z) exists µ-a.e. then
µdµ(z) = 2
Cµ(t)dµ(t)
= [Cµ(z)]
for m2-a.e. point z ∈ C . (1)
Proof. Put
F := {z ∈ C :
d|µ|(t)
|t− z|
<∞} .
As |µ| is a finite measure,
m2(C \ F ) = 0 . (2)
Let z ∈ F . Then the integral
|t−ζ|>ε
dµ(t)dµ(ζ)
ζ − z
is absolutely convergent for any ε > 0.
Using the identity
(t− z)(z − ζ)
(z − ζ)(ζ − t)
(ζ − t)(t− z)
we obtain
|t−ζ|>ε
z − ζ
ζ − t
ζ − t
dµ(t)dµ(ζ) =
dµ(ζ)
ζ − z
|t−ζ|>ε
dµ(t)
dµ(t)
|ζ−t|>ε
dµ(ζ)
ζ − t
dµ(t) · Cµε (t) ·
dµ(ζ) · Cµε (ζ) ·
ζ − z
Cµε (t)dµ(t)
E := {z ∈ C :
Cµ∗ (t)d|µ|(t)
|t− z|
<∞} .
By assumption, the numerator Cµ∗ (t)d|µ|(t) is a finite measure. Therefore
m2(C \ E) = 0 . (3)
If z ∈ E then
Cµε (t)dµ(t)
Cµ(t)dµ(t)
. (4)
4 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
This formula is true as long as Cµ∗ ∈ L1(|µ|) and the principal value Cµ exists µ-a.e. by
the dominated convergence theorem. Thus
I = 2CC
µdµ(z) if z ∈ E . (5)
It is left to show that, since z ∈ F ,
I = [Cµ(z)]2 . (6)
Since z ∈ F , the following integral converges absolutely:
φε(t, z) :=
ζ∈C,|ζ−t|>ε
dµ(ζ)
ζ − z
φε(t, z)
dµ(t) .
Since the point z is fixed in F , we have that 1|ζ−z| ∈ L
1(|µ|), and therefore
|ζ−z|d|µ|(ζ) is
small if |µ|(A) is small. Denoting the disc centered at t and of radius ε by B(t, ε) we notice
1) φε(t, z) =
dµ(ζ)
ζ − z
B(t,ε)
dµ(ζ)
ζ − z
2) lim
|µ|(B(t, ε)) = 0.
uniformly in t. Otherwise µ would have an atom.
We conclude that, as ε → 0, the functions φε(t, z) converge uniformly in t ∈ C to φ(z) =∫
ζ−z . Hence for any z ∈ F and any t ∈ C \ z
φε(t, z)
→ φ(z)
, as ε → 0 .
Since φε(t, z) converge uniformly and z ∈ F ,
dµ(t)φε(t, z)
→ φ(z)
dµ(t)
= [Cµ(z)]2 .
We have verified (6).
Combining (5) and (6) we conclude that for z ∈ E ∩ F (so for m2-a.e. z ∈ C) we have
µdµ(z) = 2
Cµ(t)dµ(t)
= lim
I = [Cµ(z)]2 for m2-a.e. point z ∈ C . (7)
This formula is true as long as Cµ∗ ∈ L1(|µ|) and the principal value Cµ exists µ-a.e.
To deduce Theorem 2.1 suppose that Cµ vanishes µ-a.e. Then the left-hand side in (7) is
zero form2-a.e. point z. The same must hold for [C
µ(z)]2. But if Cµ(z) = 0 for Lebesgue-a.e.
point z ∈ C then µ = 0, see for example [6]. Theorem 2.1 is completely proved.
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 5
Remark. In the statement of Theorem 2.2 the condition Cµ∗ ∈ L1(|µ|) can be replaced
with the condition that Cµε converge in L
1(|µ|). The proof would have to be changed as
follows.
Like in the above proof one can show that at Lebesgue-a.e. point z
I = [Cµ(z)]2 . (8)
The relation
I = 2
Cµε (t)dµ(t)
for a.e. z can also be established as before. Since Cµε converge in L
1(|µ|), the last integral
converges to CC
µdµ(z) in the ”weak” L2(dxdy), which concludes the proof.
Hence we arrive at the following version of Theorem 2.1:
Theorem 2.3. Let µ ∈Mc. Assume that Cµε → 0 in L1(|µ|). Then µ ≡ 0.
This version has the following corollary:
corollary 2.4 ([14]). Let µ ∈M be a measure of linear growth and finite Menger curvature.
If Cµ = 0 at µ-a.e. point then µ ≡ 0.
Proof. The conditions on µ imply that the L2(|µ|)-norms of the functions Cµε are uniformly
bounded, see for instance [11]. Since Cµε also converge µ-a.e., they must converge in L
1(|µ|).
Remark As was mentioned in the introduction, Corollary 2.4 also follows from Theo-
rem 2.1. However, the above version of the argument allows one to obtain it without the
additional results of [12, 13] on the maximal function.
We also obtain the following statement on the injectivity of any bounded planar Cauchy
transform. As usual, we say that the Cauchy transform is bounded in L2(µ) if the functions
Cfdµε are uniformly bounded in L
2(µ)-norm for any f ∈ L2(µ). If Cµ is bounded, then Cfdµε
converge µ-a.e as ε → 0 and the image Cfdµ exists in a regular sense as a function in L2(µ),
see [13].
corollary 2.5. Let µ ∈ M be a positive measure. If Cµ is bounded in L2(µ) then it is
injective (has a trivial kernel).
Proof. Suppose that there is f ∈ L2(µ) such that Cfdµ = 0 at µ-a.e. point. Since both f
and Cfdµ∗ are in L
2(µ), Cfdµ∗ is in L
1(|f |dµ). Hence f is a zero-function by Theorem 2.1 �
Remark We have actually obtained a slightly stronger statement: If Cµ is bounded
in L2(µ) then for any f ∈ L2(µ) the functions f and Cfdµ cannot have disjoint essential
supports, i.e. the product fCfdµ cannot equal to 0 at µ-a.e. point.
In the rest of this section we will discuss what other kernels could replace the Cauchy
kernel in the statement of Theorem 2.1.
If K(x) is a complex-valued function in Rn, bounded outside of any neighborhood of the
origin, and µ is a finite measure on Rn, one can define Kµ and Kµ∗ in the same way as C
and Cµ∗ were defined in the introduction.
6 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
The proof of Theorem 2.2 relied on the fact that the Cauchy kernel K(z) = 1/z is odd,
satisfies the symmetry condition (3), i.e.
K(x− y)K(y − z) +K(y − z)K(z − x) +K(z − x)K(x− y) ≡ 0, (9)
and is summable as a function of z for any t with respect to Lebesgue measure. Any K(x)
having these three properties could be used in Theorem 2.1. Out of these three conditions
the symmetry condition (9) seems to be most unique. However, other symmetry conditions
may result in formulas similar to Theorem 2.2 that could still yield Theorem 2.1.
Here is a different example. It shows that much less symmetry can be required from the
kernel if the measure is positive.
Theorem 2.6. Let µ be a positive measure in Rn. Suppose that the real kernel K(x) satisfies
the following properties:
1) K(−x) = −K(x) for any x ∈ Rn;
2) K(x) > 0 for any x from the half-space Rn+ = {x = (x1, x2, ..., xn) | x1 > 0}.
If Kµ∗ ∈ L1(µ) and Kµ(x) = 0 for µ-a.e. x then µ ≡ 0.
Note that real and imaginary parts of the Cauchy kernel, Riesz kernels in Rn, as well as
many other standard kernels satisfy the conditions of the theorem.
We will need the following
Lemma 2.7. Let K be an odd kernel. and let µ, ν ∈ M . Then
Kµε (z)dν(z) = −
Kνε (z)dµ(z) (10)
for any ε > 0.
Suppose that Kµ∗ ∈ L1(|ν|). If Kµ(z) exists ν-a.e. then
Kµ(z)dν(z) = − lim
Kνε (z)dµ(z).
In particular, suppose that both Kµ∗ ∈ L1(|ν|) and Kν∗ ∈ L1(|µ|). If Kµ(z) exists ν-a.e.
and Kν(z) exists µ-a.e. then
Kµ(z)dν(z) =
Kν(z)dµ(z).
Proof. Since K is odd, the first equation can be obtained simply by changing the order of
integration. The second and third equations now follow from the dominated convergence
theorem. �
Proof of Theorem 2.6. There exists a half-plane {x1 = c} in Rn such that µ({x1 = c}) = 0
but both µ({x1 > c}) and µ({x1 < c}) are non-zero. Denote by ν and η the restrictions of
µ onto {x1 > c} and {x1 < c} respectively. Then∫
Kνε (z)dµ(z) =
Kνε (z)dν(z) +
Kνε (z)dη(z).
The first integral on the right-hand side is 0 because of the oddness of K (apply the first
equation in the last lemma with µ = ν). The second condition on K and the positivity of
the measure imply that the second integral is positive and increases as ε → 0. Therefore∫
Kνε (z)dµ(z) cannot tend to zero. This contradicts the fact that K
µ = 0, ν-a.e. and the
second equation from the last lemma. �
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 7
3. Sets of finite perimeter
In this section we give another example of an application of Theorem 2.2. It involves the
notion of a set of finite perimeter introduced by De Giorgi in the 50’s, see [5]. We say that
a set G ⊂ R2 has finite perimeter (in the sense of De Giorgi) if the distributional partial
derivatives of its characteristic function χG are finite measures. Such sets have structural
theorems. For example, if G is such a set then the measure ∇χG is carried by a set E,
rectifiable in the sense of Besicovitch, i. e. a subset of a countable union of C1 curves and an
H1-null set, where H1 is the one-dimensional Hausdorff measure. Also the measure ∇χG is
absolutely continuous with respect to H1 restricted to E and its Radon-Nikodym derivative
is a unit normal vector H1-a.e. (notice that ∇χG is a vector measure). At H1-almost all
points of E the function χG has approximate “one-sided”’ limit. For more details we refer
the reader to [5].
The general question we consider can be formulated as follows: What can be said about µ
if Cµ coincides at µ-a.e. point with a ”good” function f? To avoid certain technical details,
all measures in this section are compactly supported. Furthermore, we will only discuss the
two simplest choices of f . As we will see, even in such elementary situations Theorem 2.2
yields interesting consequences.
As usual, when we say that Cµ = f at µ-a.e. point, we imply that the principal value
exists µ-almost everywhere.
Theorem 3.1. Let µ ∈ Mc be compactly supported. Assume that Cµ(z) = 1, µ-almost
everywhere and Cµ∗ ∈ L1(|µ|). Then µ = ∂̄χG, where G is a set of finite perimeter. In
particular, µ is carried by a set E, H1(E) < ∞, rectifiable in the sense of Besicovitch, and
µ is absolutely continuous with respect to the restriction of H1 to E.
Remark. The most natural example of such a measure is dz on a C1 closed curve. The
theorem says that, by the structural results of De Giorgi, this is basically the full answer.
Proof. By Theorem 2.2 we get that for Lebesgue-almost every point in C
[Cµ(z)]2 = 2Cµ(z) . (11)
In other words for m2-a.e. point z we have C
µ(z) = 0 or = 2. Let G denote the set where
Cµ(z) = 2. Since the Cauchy transform of any compactly supported finite measure must
tend to zero at infinity, this set is bounded. Consider the following equality
χG = C
understood in the sense that the two functions are equal as distributions. Taking distribu-
tional derivatives on both sides we obtain
∂̄χG = µ/2 and ∂χ̄G = µ̄/2.
Hence G has finite perimeter and the rest of the statement follows from the results of [5]. �
We say that a set G has locally finite perimeter (in the sense of De Giorgi) if the distribu-
tional derivatives of χG are locally finite measures. Our second application is the following
8 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Theorem 3.2. Let µ ∈ Mc be compactly supported. Assume that Cµ(z) = z, µ-almost
everywhere and Cµ∗ ∈ L1(|µ|). If µ(C) = 0 then µ = 2z∂̄χG, where G is a set with locally
finite perimeter. Whether µ(C) = 0 or not, µ is carried by a set E, H1(E) <∞, which is a
rectifiable set in the sense of Besicovitch, and µ is absolutely continuous with respect to the
restriction of H1 to E.
Remark. The most natural example of such a measure is zdz on a C1 closed curve. Our
statement shows that this is basically one-half of the answer. The other half is given by√
z2 − cdz as will be seen from the proof.
Proof. Again, from Theorem 2.2 we get that for Lebesgue-almost every point in C
[Cµ(z)]2 = 2Cζdµ(ζ)(z) . (12)
Notice that
Cζdµ(ζ)(z) =
ζ − z
dµ(ζ) = µ(C) + zCµ(z)
and we get a quadratic equation
[Cµ(z)]2 = 2zCµ(z)− p ,
where p := −2µ(C).
First case p = 0. Here we get
[Cµ(z)]2 = 2zCµ(z) .
We conclude that Cµ(z) = 0 or z for Lebesgue-a.e. point z ∈ C.
Again a bounded set G appears on which
Cµ = 2zχG(z)
in terms of distributions. Therefore
∂̄χG = dµ/2z ,
and the right hand side is a finite measure on any compact set avoiding the origin. Therefore,
G is a (locally) De Giorgi set.
Let us consider the case p 6= 0. For simplicity we assume p = 1, other p’s are treated in
the same way. Then we have to solve the quadratic equation
Cµ(z)2 − 2zCµ(z) + 1 = 0
for Lebesgue-a.e. point in C. Let us make the slit [−1, 1] and consider two holomorphic
functions in C \ [−1, 1]
r1(z) = z −
z2 − 1, r2(z) = z +
z2 − 1 ,
where the branch of the square root is chosen so that
r1(z) → 0, z → ∞ .
In other words we have the sets E1 and E2 such that m2(C \ E1 ∪ E2) = 0 and
z ∈ E1 ⇒ Cµ(z) = r1(z) ,
z ∈ E2 ⇒ Cµ(z) = r2(z) .
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 9
Obviously it is E1 that contains a neighborhood of infinity. The function z −
z2 − 1
outside of [−1, 1] can be written as Cµ0(z) where dµ0(x) = 1π
1− x2dx. Consider ν = µ−µ0.
z ∈ E1 ⇒ Cν(z) = 0 ,
z ∈ E2 ⇒ Cν(z) = 2
z2 − 1 := R(z) .
Therefore,
Cν(z) = R(z)χE2 . (13)
Notice that if R was analytic in an open domain compactly containing E2 we would conclude
from the previous equality that
ν = R(z)∂̄χE2 .
If, in addition, |R| was bounded away from zero on E2, we would obtain that ∂̄χE2 and ∂χE2
are measures of finite variation, and hence E2 is a set of finite perimeter. Notice that our
R(z) = 2
z2 − 1 is analytic in O := C \ [−1, 1] and is nowhere zero. We will conclude that
E2 is a set of locally finite perimeter. More precisely we will establish the following claim:
For every open disk V ⊂ O the set O ∩ E2 has finite perimeter.
Indeed, let W be a disk compactly containing V , W ⊂ O. Let ψ be a smooth function,
supported in W , ψ|V = 1. Multiply (13) by ψ and take a distributional derivative (against
smooth functions supported in V ). Then we get (using the fact that R is holomorphic on V )
ν|V = ∂̄(ψRχE2∩V )|V = ∂̄(RχE2∩V )|V = R∂̄(χE2∩V )|V .
We conclude immediately that E2 ∩ V is a set of finite perimeter. Therefore, E2 ∩D is a set
of finite perimeter, where D is a domain whose closure is contained compactly in O.
Recalling that µ = ν + µ0 we finish the proof. �
Remark 3.3. In is interesting to note that, as follows from the proof, if µ is the measure
from the statement of the theorem then one of the connected components of supp µ must
contain both roots of the equation z2 + 2µ(C) = 0.
We conclude this section with the following examples of measures µ whose Cauchy trans-
form coincides with z at µ-a.e. point
Examples. 1. Let Ω be an open domain with smooth boundary Γ. Suppose that [−1, 1] ⊂ Ω.
Let {Dj}∞j=1 be smoothly bounded disjoint domains in O := Ω \ [−1, 1], γj = ∂Dj . Assume
H1(γj) <∞ . (14)
LetR(z) be an analytic branch of 2
z2 − 1 inO. Consider the measure ν on Γ∪(∪γj)∪[−1, 1]
defined as
ν = R(z)dz|Γ − R(z)dz|∪γj −
1− x2dx|[−1,1].
10 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Cν(z) =
0 if z ∈ C \ Ō ,
0 if z ∈ ∪jDj ,
R(z) if z ∈ O \ ∪jD̄j .
Recall that R(z) = z +
z2 − 1 − (z −
z2 − 1) and that Cµ0(z) = z −
z2 − 1 for µ0 =
1− x2dx|[−1,1]. We conclude that for µ = ν + µ0 one has
Cµ(z) =
z2 − 1 if z ∈ C \ Ō ,
z2 − 1 if z ∈ ∪jDj ,
z2 − 1 if z ∈ O \ ∪jD̄j .
2. The second example is exactly the same as the first one but Dj,k = B(xj,k,
), xj,k =
j , 1 ≤ k ≤ j, j = 1, 2, 3.... Here the assumption (14) fails. But ν, defined as above,
will still be a measure of finite variation (and so will be µ): |ν|(C) ≤ C
In both examples Cµ(z) = z for µ-a.e. z.
4. Asymptotic behavior near the zero-set of Cµ
In this section we take a slightly different approach. We study asymptotic properties of
measures near the sets where the Cauchy transform vanishes. Theorem 4.2 below shows
that near the density points of such sets the measure must display a certain ”irregular”
asymptotic behavior.
As was mentioned in the introduction, one of the results of [14] says that an absolutely
continuous planar measure cannot be reflectionless. This result is not implied by our Theorem
2.1 because an absolutely continuous measure may not have a summable Cauchy maximal
function. It is, however, implied by Theorem 4.2, see Corollary 4.4 below.
When estimating Cauchy integrals one often uses an elementary observation that the
difference of any two Cauchy kernels 1/(z − a) − 1/(z − b) can be estimated as O(|z|−2)
near infinity. To obtain higher order of decay one may consider higher order differences.
Here we will utilize the following estimate of that kind, which can be verified through simple
calculations.
Lemma 4.1. If a, b, c ∈ B(0, r) be different points, |a − b| > r. Then there exist constants
A,B ∈ C such that |A|, |B| < 2
z − a
z − b
z − c
∣∣∣∣ <
outside of B(0, 2r).
(Namely, A = b−c
b−a , B =
a−b .)
If µ ∈M consider one of its Riesz transforms in R3, R1µ(x, y, z), defined as
R1µ(x, y, z) =
|(u, v, 0)− (x, y, z)|3
dµ(u+ iv).
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 11
This transform is the planar analogue of the Poisson transform. In particular,
R1µ(x, y, z) =
(x+ iy)
for all points w = x+ iy ∈ C where the Radon derivative
(w) = lim
µ(B(w, r))
|B(w, r)|
exists.
For measures on the line or on the circle their Poisson integrals and Radon derivatives
(with respect to the one-dimensional Lebesgue measure) are very much related but not always
equivalent. When the asymptotics of the Poisson integral and the ratio from the definition
of the Radon derivative are different near a certain point it usually means that the measure
is ”irregular” near that point. It is not difficult to show that if µ is absolutely continuous
then at a Lebesgue point of its density function the Radon derivative of µ and the Poisson
integral of |µ| (or R1|µ| if n > 1) behave equivalently. Even for singular measures on the
circle, if a measure possesses a certain symmetry near a point, then the same equivalent
behavior takes place, as follows for instance from [1], Lemma 4.1. In fact, it is not easy to
construct a measure so that its Poisson integral and Radon derivative behaved differently
near a large set of points. The same can be said about the Riesz transform and the Radon
derivative. Thus one may interpret our next result as an evidence that, for a planar measure
µ, most points where Cµ = 0 are ”irregular.”
Theorem 4.2. Let µ ∈ M and let w = x + iy be a point of density (with respect to m2) of
the set E = {Cµ = 0}. Then
µ(B(w, r))
= o (R1|µ|(x, y, r)) as r → 0 + .
In view of the above discussion this implies
corollary 4.3. If w is a point of density of the set E = {Cµ = 0}, such that there exists the
Radon derivative d|µ|/dm2(w) 6= 0, then
µ(B(w, r)) = o (|µ|(B(w, r))) as r → 0+ (16)
and dµ/dm2(w) = 0.
Since m2-almost every point of a set is its density point, we also obtain the following
version of the result from [14]:
corollary 4.4. The set E = {Cµ = 0} has measure zero with respect to the absolutely
continuous component of µ.
Proof of Theorem 4.2. without loss of generality w = 0. Choose a C∞0 test-function φ sup-
ported in B := B(0, r), and such that 0 ≤ φ ≤ D/r2, |∇φ| ≤ A/r3 and
φ dm2 = 1. Denote
the complement of E by Ec. Then
φdµ = 〈φ, ∂̄Cµ〉 = 〈∂̄φ, Cµ〉 = 〈χEc∂̄φ, Cµ〉 =
χEc ∂̄φ dm2(z)
ζ − z
dµ(ζ) (17)
All we need is to show that the last integral is small. Then, since the first integral in
(17) is similar to the right-hand side of (16) we will complete the proof. The main idea for
12 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
the rest of the proof is to make the function F (ζ) =
χEc ∂̄φ dm2(z)
ζ−z ”small” by subtracting a
linear combination of Cauchy kernels corresponding to points from E, which will not change
its integral with respect to µ.
Namely, let a, b ∈ B(0, r) ∩ E be any two points such that |a − b| > r. By the previous
lemma for any z ∈ B(0, r) there exist constants A = A(z), B = B(z), of modulus at most 2,
such that (15) holds with c = z. Integrating (15) with respect to χEc ∂̄φ dm2(z) we obtain
that ∣∣∣∣
χEc ∂̄φ dm2(z)
ζ − z
ζ − a
ζ − b
∣∣∣∣ < C
ε(r)r
|ζ |3
outside of B(0, 2r) for some constants A∗, B∗, where ε(r) = |B(0, r)∩Ec|/r2 = o(1) as r → 0.
The constants satisfy |A∗|, |B∗| < 2 ε(r)
Notice that if w ∈ E then
ζ−wdµ = 0 by the definition of the set E. Hence, since
a, b ∈ E,
χEc ∂̄φ dm2(z)
ζ − z
dµ(ζ) =
χEc ∂̄φ dm2(z)
ζ − z
ζ − a
ζ − b
dµ(ζ)
B(0,2r)
C\B(0,2r)
= I1 + I2.
For I2 we now have
C\B(0,2r)
χEc ∂̄φ dm2(z)
ζ − z
ζ − a
ζ − b
dµ(ζ)
C\B(0,2r)
ε(r)r
|ζ |3
d|µ|(ζ) ≤ Cε(r)R1|µ|(0, 0, r).
In I1 we estimate each summand separately. First,
B(0,2r)
χEc ∂̄φ dm2(z)
ζ − z
dµ(ζ)
∣∣∣∣ ≤
B(0,2r)
|ζ − z|
χEcdm2(z)d|µ|(ζ)
|µ|(B(0, 2r)) ≤ C
ε(r)R1|µ|(0, 0, r).
To estimate the second and third summands of I1, recall that the only restriction on the
choice of a, b ∈ B(0, r) ∩ E was that |a − b| > r. This condition will be satisfied, for
instance, if a ∈ B1 = B(−56r,
r) and b ∈ B2 = B(56r,
r). If we average the modulus of the
second summand over all choices of a ∈ B1 ∩ E, recalling that A∗ = A∗(a) always satisfies
|A∗| ≤ 2 ε(r)
, we get
|B1 ∩ E|
B(0,2r)
A∗(a)
ζ − a
dµ(ζ)
∣∣∣∣dm2(a) ≤
|B1 ∩ E|
B(0,2r)
|A∗(a)|
|ζ − a|
dm2(a)d|µ|(ζ)
r|µ|(B(0, 2r)) ≤ Cε(r)R1|µ|(0, 0, r).
It is left to choose a ∈ B1 ∩ E for which the modulus is no greater than its average. The
same can be done for b. The proof is finished. �
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 13
5. Reflectionless measures and Combs
As was mentioned in the introduction, following [2], we will call a non-trivial continuous
finite measure µ ∈M(C) reflectionless if Cµ(z) = 0 at µ-a.e. point z.
Perhaps the simplest example of a reflectionless measure is the measure µ = 1
(1−x2)−1/2dx
on [−1, 1], the harmonic measure of C \ [−1, 1] corresponding to infinity. The fact that
µ is reflectionless can be verified through routine calculations or via the conformal map
interpretation of the harmonic measure. It will also follow from a more general Theorem 5.4
below.
At the same time, since Cµ∗ ≍ (1 − x2)−1/2 on [−1, 1], this simple example complements
the statement of Theorem 2.1. Since the function (1−x2)−1/2 belongs to the ”weak” L1(|µ|),
the summability condition for the Cauchy maximal function proves to be exact in its scale.
In the rest of this section we discuss further examples and properties of positive reflec-
tionless measures on the line.
Let us recall that functions holomorphic in the upper half plane C+ and mapping it to
itself (having non-negative imaginary part) are called Nevanlinna functions. Let M+(R)
denote the class of finite positive measures compactly supported on R. The function f is a
Nevanlinna function if and only if it has a form
f(z) = az + b+
t2 + 1
]dρ(t) ,
where ρ is a positive measure on R such that
dρ(t)
< ∞, a > 0, b ∈ R are constants.
If the representing measure is from M+(R) and f(∞) = 0, the formula becomes simpler:
f(z) =
dµ(x)
x−z .
Definition. A simply connected domain O is comb-like if it is a subset of a half-strip
{w : ℑw ∈ (0, π),ℜw > q}, for some q ∈ R, contains another half-strip {w : ℑw ∈
(0, π),ℜw > r} for some r ∈ R and has the property that
for any w0 = u0 + iv0 ∈ O the whole ray {w = u+ iv0, u ≥ u0} lies in O . (18)
If in addition H1(∂O∩B(0, R)) <∞ for all finite R, we say that O is a rectifiable comb-like
domain.
Let O be a rectifiable comb-like domain, Γ = ∂O. Then by the Besicovitch theory we
know that for H1-a.e. pont w ∈ Γ there exists an approximate tangent line to Γ, see [3] for
details. We wish to consider rectifiable comb-like domains satisfying the following geometric
property:
for a.e. w ∈ Γ approximate tangent line is either vertical or horizontal. (19)
It is not difficult to verify that for any conformal map F : C+ → O, O is comblike if and
only if F ′ is a Cauchy potential of µ ∈ M+(R): F ′(z) =
dµ(x)
x−z . It is, therefore, natural to
ask the following
Question. Which comb-like domains correspond to reflectionless measures µ ∈M+(R)?
An answer would give a geometric description of reflectionless measures from M+(R). If,
in addition, a comb-like domain is rectifiable, then the answer is given by
14 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Theorem 5.1. 1) Rectifiable comb-like domains correspond exactly to those measures
µ ∈M+(R) that are absolutely continuous with respect to dx and satisfy
dµ(x)
∈ H1loc(C+). (20)
2) An absolutely continuous measure satisfying (20) is reflectionless if and only if the corre-
sponding comb-like domain has the property (19).
Remarks.
1) Of course not every comb-like domain gives rise to a reflectionless measure fromM+(R).
Just take any comb-like domain which appears as F (C+), where F =
∫ z ∫ dµ(x)
x−z for a singular
µ ∈M+(R). By a result from [9] singular measures cannot be reflectionless.
2) On the other hand, even if µ = g(x)dx is a reflectionless absolutely continuous measure,
the corresponding conformal map F =
∫ z ∫ dµ(x)
x−z : C+ → O can be onto a non-rectifiable
domain.
3) For non-rectifiable domains we have no criteria to recognize which ones correspond to
reflectionless measures.
4) It is well known, and not difficult to prove, that the antiderivative of a Nevanlinna
function is a conformal map, see for instance [4]. If F =
∫ z ∫ dµ(x)
x−z , µ ∈ M+(R) then ℑF (x)
is an increasing function on R whose derivative in the sense of distributions is µ. The image
F (C+) lies in the strip {ℑw ∈ (0, π‖µ‖)}.
Theorem 5.1 will follow from Theorems 5.2 and 5.3 below.
Theorem 5.2. Let F be a conformal map of C+ on a rectifiable comb-like domain O. Then
F (z) =
∫ z ∫ dµ(x)
x−z , µ ∈ M+(R), µ << dx. Also
dµ(x)
x−z ∈ H
loc(C+). If in addition O satisfies
(19) then µ is reflectionless.
Proof. without loss of generality O ⊂ {ℜz > 0}. Put Φ = eF . Then the image Φ(O) is
the subdomain of the complement of the unit half-disk in C+ which is the union of rays
(R(θ)eiθ,∞). Consider the subdomain of the upper half-disk D := {z : 1/z ∈ Φ(O)}. Define
G as the smallest open domain containing D and its reflection D := {z̄ : z ∈ D}. Then G is
a star-like domain inside the unit disk. The preimage of G ∩ R under Φ is the union of two
Infinite rays R1 = [−∞, a), R2 = (b,∞], a < b. Therefore, by reflection principle C \ [a, b] is
mapped conformally (by the extension of Φ which we will also denote by Φ) onto star-like
Since Φ : C+ → G, where G is star-like, it is well-known that argΦ(x+ iδ) is an increasing
function of x, see [7].
We conclude that the argument of Φ is monotone. Therefore, ℑF (x + iδ) is monotone,
and so ℑf(x+ iδ) is positive, where f = F ′. We see that f = F ′ is a Nevanlinna function.
From the structure of our comb-like domain, we conclude immediately that its representing
measure µ has compact support, so we are in M+(R). Also, let us prove that µ << dx. The
boundary of our comb is locally rectifiable. So f = F ′ belongs locally to the Hardy class
H1(C+), [16]. Since ℑf is the Poisson integral of µ,
ℑf = Pµ =
(x− t)2 + y2
dµ(t),
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 15
and f is in H1(C+) locally, we conclude that µ = ℑfdx,ℑf ≥ 0 a.e., [16].
Now suppose that, in addition, O = F (C+) has the property (19). Let us recall that
for a simply connected domain with rectifiable boundary Γ the restriction of the Hausdorff
measure H1|Γ is equivalent to the harmonic measure ν on O. Therefore the tangent lines
to Γ are either vertical or horizontal a.e. with respect to ν. The measure ν is the image
of the harmonic measure λ of C+ which is equivalent to the Lebesgue measure on the line.
We have a conformal map F (a continuous function up to the boundary of C+ because it
is an anti-derivative of an H1loc-function) which pushes forward λ to ν. Call a point w0 ∈ Γ
accessible from O if there exists a ray x0 + iy, 0 < y < 1, such that w0 = limy→0 F (x0 + iy).
Almost every point of Γ (w.r. to ν) is accessible from O. For ν-a.e. accessible w0 ∈ Γ
where the tangent line is vertical (horizontal) we can say that ℜF ′(x0) = 0 (ℑF ′(x) = 0).
So R = E1 ∪ E2 ∪ E3, where |E3| = 0, |E1 ∩ E2| = 0, and E1 = {x ∈ R : ℜF ′(x) = 0},
E2 = {x ∈ R : ℑF ′(x) = 0}. We already know that the measure µ = ℑF ′(x)dx represents
f(z) = F ′(z) =
dµ(t)
t−z . Notice that
R\E2 · =
·. But we also know that boundary
values exist dx-almost everywhere, i.e.
dµ(t)
t− x− iy
= ℜF ′(x) = 0
for a.e. x ∈ E1 and therefore for µ-a.e. x ∈ E1. This means (see [16]) that
dµ(x)
= 0 µ-a.e.
Definition. A simply connected rectifiable comb-like domain O is called a comb if its “left”
boundary consists of countably many horizontal and vertical segments.
A comb is called a straight comb if O = {w : ℑw ∈ (0, π),ℜw > 0} \ S, where the set S
is relatively closed with respect to the strip {w : ℑw ∈ (0, π),ℜw > 0} and is the union of
countably many horizontal intervals Rn = (iyn, ln + iyn]. We require also that
ln <∞ .
Example. Let F be a conformal map of C+ on a comb O. By our last theorem F ′(z) =∫
dµ(x)
x−z , where µ ∈M+(R) is reflectionless: C
µ(x) = 0 for µ-a.e. x.
Definition. Let E be a compact subset of the real line. Let E have positive logarithmic
capacity, so Green’s function G of C \E exists. The domain C \E is called Widom domain
G(c) <∞ ,
where the summation goes over all critical points of G (we assume that G is a Green’s
function with pole at infinity.
Example. Let E be a compact subset of the real line of the positive length. We assume
that every point of E is regular in the sense of Dirichlet for the domain C \ E, and we
also assume that C \ E is not a Widom domain. Such E exist in abundance. We will
see below, that the harmonic measure ω of C \ E (with pole at infinity) is reflectionless.
Consider F (z) =
∫ z ∫ dω(x)
z−x for z ∈ C+. It is easy to see that F (z) = G(z) + iG̃(z) + const,
16 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
where G̃ is the harmonic conjugate of G. This F is a conformal map (see [4]) of C= onto
a domain D lying in the strip {w : ℑw ∈ (0, π)}. It is easy to see that complementary
intervals of E will be mapped by F onto straight horizontal segments on the boundary of
D. Each finite complementary interval contains exactly one critical point of G, and clearly
the length of the corresponding straight horizontal segment is G(c) (this follows from the
formula F (z) = G(z) + iG̃(z) + const).
As the domain C \ E was not a Widom domain, we have that the sum of lengths of
abovementioned straight horizontal segment is infinite. So domain D is not rectifiable.
Therefore the reflectionless property of µ alone does not say anything about the rectifiability
of the domain, which is the target domain of the conformal map F (z) =
∫ z ∫ dµ(x)
z−x .
Theorem 5.3. Let µ be absolutely continuous positive measure on R and let Cµ ∈ H1loc(C+).
Then F (z) =
∫ z ∫ dµ(x)
x−z is a conformal map of C+ onto a rectifiable comb-like domain O. If
µ is reflectionless then O has the property (19).
Proof. Consider F (z) =
∫ z ∫ dµ(x)
x−z . Since µ is positive, it is a conformal map. If µ is such
that f(z) = Cµ ∈ H1loc(C+) then F (z) =
f maps C+ onto a domain with locally rectifiable
boundary (see [16]).
If, in addition, µ = ℑfdx is reflectionless, then for a.e. point of P := {x ∈ R : ℑf(x) > 0}
we have ℜf(x) = 0. Conformal map F (z) is continuous up to the boundary of C+ and its
boundary values F (x) form a (locally) absolutely continuous function, F ′(x) = f(x) a.e. As
at almost every point we have either ℑF ′(x) = 0 or ℜF ′(x) = 0 we conclude that O = F (C+)
has the property (19).
We also need the following definition.
Definition. A compact subset E in R is called homogeneous if there exist r, δ > 0 such that
for all x ∈ E, |E ∩ (x− h, x+ h)| ≥ δh for all h ∈ (0, r).
Example. Let E ⊂ R be a compact set of positive length. Let µ be a reflectionless
measure supported on E, µ = g(x)dx. Let in addition E be a homogeneous set. Then
F (z) =
∫ z ∫ dµ(x)
x−z is a conformal map from C+ on a rectifiable comb-like domain satisfying
(19).
Proof. The Cauchy integral Cgdx considered in C\E will be in the Hardy class H1(C\E). In
fact the reflectionless property of gdx implies that its limits from C± will be both integrable
with respect to dx|E .
Now we use homogenuity of E and Zinsmeister’s theorem [15] to conclude that f(z) =
Cgdx(z) is in the usual H1loc(C). Then the conformal map F (z) =
f maps C+ onto a
rectifiable subdomain of a strip. We use Theorem 5.3 to get the rest of our example’s
claims. �
The simple example of a reflectionless measure mentioned at the beginning of this section,
as well as many other explicit examples, are given by our next statement.
UNIQUENESS THEOREMS FOR CAUCHY INTEGRALS 17
Theorem 5.4. Let E be a compact set of positive lenght, E ⊂ R. Let ω be a harmonic
measure of C \ E with pole at infinity. Then ω is reflectionless.
Example. The simplest comb is a strip {w : ℑw ∈ (0, π),ℜw > 0}. Consider F (z) =
log(z +
z2 − 1). It maps conformally C+ onto the strip. Its derivative f(z) = 1√z2−1 is
x−z and dµ =
1−x2 is the harmonic measure of C \ [−1, 1].
Proof of Theorem 5.4. We need to show that Cω = 0 at ω-a.e point. From our definitions
it can be seen, that Cω on the line coincides with the Hilbert transform of ω, which in its
turn is asymptotically equivalent to the conjugate Poisson transform Qω. Thus all we need
to establish is that
Qω(x+ ih) =
(x− y)2 + h2
dω(y) = ℜ
dω(y)
x− ih− y
→ 0 as h→ 0+ (21)
for almost every x. Instead, we have that the Green’s function F (x) defined as
F (x) =
log |x− y|dω(y) + C∞,
where C∞ is a real constant (Robin’s constant), is equal to 0 at every density point of
E, see for example [8]. The idea of the proof is to show that Qω(x + iε) behaves like
(F (x+ ε) + F (x− ε))/ε near almost every x. The technical details are as follows.
Introduce
φ(y) :=
|1− y|
|1 + y|
y2 + 1
, (22)
φx,h(y) :=
y − x
The function φ(y) decreases as 1/y2 at infinity, hence it is in L1(R, dx) and so are φx,h(y)
with a uniform bound on the norm. However, these functions are not bounded, which makes
it difficult to use them in our estimates. To finish the proof we will first obtain a bounded
version of φx,h(y) through the following averaging procedure.
Let ω = g(x)dx. Choose x to be a Lebesgue point of g and a density point of E. Fixing
sufficiently small h > 0 we can find the set A(x, h) ⊂ (x−h, x−h/2)∪ (x+h/2, x+h) such
• A(x, h) consists of density points of E,
• |A(x, h)| ≥ h/2,
• A(x, h) is symmetric with respect to x.
Let Tx,h := T := {t ∈ (0, h) : x+ t ∈ A(x, h)}. Then |T | ≥ h/4. Now put
ψx,h(y) :=
φx,t(y) dt .
By (22) one can see immediately that
|ψx,h| ≤
for some M > 0 and |ψx,h(y)| ≤ C
, for |y| > h . (23)
18 MARK MELNIKOV, ALEXEI POLTORATSKI, AND ALEXANDER VOLBERG
Also, since ∫
φ dy = 0 .
we have that ∫
ψx,h dy = 0 .
Therefore,
g(y)ψx,h(y) dy| = |
(g(y)− g(x))ψx,h(y) dy| ≤
|g(y)− g(x)||ψx,h|(y) dy.
Now notice that (23) implies that |ψx,h| is majorated by an approximate unity (for instance,
by a constant multiple of the Poisson kernel corresponding to z = x + ih). Since x is a
Lebesgue point for g(x), this means that the last integral tends to 0 as h→ 0.
Looking at the definitions of Tx,h and ψx,h(y) we can see that
g(y)ψx,h(y) dy =
|Tx,h|
(F (x+ t)− F (x− t))−ℜ
g(y)dy
x− it− y
where F (x) is the Green’s function. As we mentioned before, F is zero at the density points
of E. We conclude that
|Tx,h|
g(y)dy
x− it− y
→ 0, h→ 0 + .
for a.e. x on the Borel support of g. Since the Cauchy integral of g has a limit a.e. we obtain
g(y)dy
x− ih− y
→ 0, h→ 0 + .
Remark. All reflectionless measures on R discussed in this section, including those provided
by Theorem 5.4 are absolutely continuous with respect to Lebesgue measure. One may
wonder if there exist singular reflectionless measures. The answer is negative. More generally,
as follows from a theorem from [9], if principal values of the Hilbert transform exist µ-a.e.
for a continuous µ ∈M(R) then µ << dx .
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Department de Matematiques, Uneversitat Autonoma de Barcelona, 08193 Bellaterra,
Barcelona, Spain
E-mail address : mark.melnikov@gmail.com
Texas A& M University, Department of Mathematics, College Station, TX 77843, USA
E-mail address : alexeip@math.tamu.edu
Dept. Math., Michigan State Univ., East Lansing MI 48823, USA, and, School of Math.,
University of Edinburgh, Edinburgh UK EH9 EJ6
E-mail address : sashavolberg@yahoo.com
