Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 29, 2004, 99–120

ASYMPTOTIC GROWTH OF CAUCHY TRANSFORMS

P. W. Jones and A. G. Poltoratski

Yale University, Department of Mathematics New Haven, CT 06520, U.S.A.; jones@math.yale.edu Texas A&M University, Department of Mathematics College Station, TX 77843, U.S.A.; alexei@math.tamu.edu

Abstract. Let μ be a complex measure on the real line. We denote by P μ and Qμ the Poisson and the conjugate Poisson integrals of μ in the upper half-plane. In this note we study the relative asymptotic growth of P μ and Qμ near the support of μ. In particular, we show that on μ almost every vertical line Qμ grows no slower than P μ. We also discuss applications to the theory of Cauchy transform in the plane and related questions on Riesz transforms in R^{n}.

1. Introduction

Let M(R) be the space of all complex measures μ on the real line satisfying

∫_{R} \frac{d|μ|(t)}{1 + |t|} < ∞.

We will denote by P μ and Qμ the Poisson and the conjugate Poisson integrals of μ ∈ M(R) in the upper half-plane C_{+} respectively:

P μ(x + iy) = ∫_{R} \frac{y}{(x − t)^{2} + y^{2}} dμ(t)

and

Qμ(x + iy) = ∫_{R} \frac{x − t}{(x − t)^{2} + y^{2}} dμ(t).

The Poisson kernel is an example of the so-called approximate unity, whereas the conjugate Poisson kernel is a typical singular kernel. Hence the boundary behavior of the Poisson and the conjugate Poisson integrals reflects different properties of the measure. The growth of P μ(z) as z → x ∈ R depends on the concentration of the mass near x. The behavior of Qμ depends on the “symmetry” of μ around x.

Nonetheless, as we will find out, the growth of P μ and Qμ must be almost the same near “most” points on the boundary. Roughly speaking, we show that the fast growth of mass near a point “usually” implies the lack of symmetry around it.

More precisely, we prove the following result. If μ ∈ M(R) we denote by μ_{s} its singular part with respect to the Lebesgue measure.

2000 Mathematics Subject Classification: Primary 30E20.

The first author is supported by grants NSF-DMS 9983403 and AFOSR-MIT 571 000164.

The second author is supported in part by N.S.F. grant DMS 0200699.

Theorem 1.1. Let μ ∈ M(R) and let Σ be a Borel subset of R. Then the following conditions are equivalent:

(1) Qμ(x + iy) = o(P μ(x + iy)) as y → 0+ for μ_{s} -a.e. x ∈ Σ;

(2) the restriction of μ_{s} on Σ is discrete.

Note that if a singular measure μ is such that

P μ(x + iy) = o(Qμ(x + iy))

as y → 0+ for μ-almost all x ∈ E then μ(E) = 0, see [7]. Hence, if μ is singular continuous, neither P μ nor Qμ can dominate its counterpart except near a zero set of points.

In Section 3 we consider the Riesz transform in R^{n}. Let M(R^{n−1}) be the space of all measures in the hyperplane R^{n−1} satisfying

∫_{R^{n−1}} \frac{1}{1 + |x|^{n−1}} dμ(x) < ∞.

For each μ ∈ M(R^{n−1}) the Riesz transform Rμ(x) is defined on the half-space

Rn_{+} = {(x_{1}, …, x_{n}) | x_{n} > 0} as Rμ(x) = ⟨R_{1}μ(x), …, R_{n}μ(x)⟩,

R_{i}μ(x) = ∫_{R^{n−1}} \frac{x_{i} − y_{i}}{|x − y|^{n}} dμ(y), i = 1, 2, …, n.

It is well known, that Riesz transforms can be viewed as generalizations of the Poisson integral (the nth transform R_{n}) and the conjugate Poisson integral (all other transforms R_{1}, …, R_{n−1} ) to higher dimensions.

The natural question that arises after Theorem 1.1 is whether an analogous statement holds for Riesz transforms in R^{n}. However, if one lets μ be the (n−2)- dimensional Lebesgue measure on an (n − 2)-dimensional cube in R^{n−1} and considers its Riesz transform in Rn_{+} , one obtains an example of a singular continuous measure satisfying

|⟨R_{1}μ((x_{1}, …, x_{n})), …, R_{n−1}μ((x_{1}, …, x_{n}))⟩| = o(R_{n}μ((x_{1}, …, x_{n})))

as x_{n} → 0+ for μ-a.e. (x_{1}, …, x_{n−1}, 0) ∈ R^{n−1} .

Nonetheless an analogue of Theorem 1.1 for Riesz transforms exists. One just has to replace the “radial” convergence with non-tangential. If x ∈ R^{n−1} and 0 < ϕ < \frac{1}{2}π we denote by Γ_{x}^{ϕ} the truncated cone:

Γ_{x}^{ϕ} = {y = ⟨y_{1}, …, y_{n} >∈ Rn_{+} | y_{n}/|y − x|⟩ sin ϕ, y_{n} < 1}.

As usual we write y →_{∢} x if y → x and there exists ϕ = ϕ(x) such that y ∈ Γ_{x}^{ϕ}.

Theorem 1.2. Let μ ∈ M(R^{n−1}) and let Σ ⊂ R^{n−1} be such that
(1)  |⟨R_{1}μ(y), …, R_{n−1}μ(y)⟩| = o(R_{n}μ(y))

as y →_{∢} x for μ-a.e. x ∈ Σ. Then the restriction of μ on Σ is absolutely continuous

with respect to the Lebesgue measure in R^{n−1} .

For μ ∈ M(R) we denote by Hμ(x, ε) its Hilbert transform: The kernel H_{x,ε}(t) is defined as 0 on (x − \frac{1}{2}ε, x + \frac{1}{2}ε) and as 1/(x − t) elsewhere and

Hμ(x, ε) = ∫_{R} H_{x,ε}(t) dt.

The standard argument shows (see Lemma 2.1) that the relation (1) in Theorem 1.1 is equivalent to

(2)  Hμ(x, y) = o(P μ(x + iy))

as y → 0 for μ-a.e. x.

An analog of Theorem 1.1 can be applied in the theory of the Cauchy transform in the plane. If μ ∈ M(R^{2}) we denote by Cμ the convolution of μ with 1/z in the sense of principal value: denote

Cμ(z, ε) = ∫_{\{|z−ξ|>ε\}} \frac{1}{z − ξ} dμ(ξ)

and put

Cμ(z) = lim_{ε→0} Cμ(z, ε).

It is a well-known phenomenon that under various conditions on the Cauchy transform a large part of the measure lies on a collection of smooth curves, see for instance [3], [4], [5] or [9]. The Cauchy transform of the restriction of the measure to one of such curves is similar to the Hilbert transform on the line. This may allow one to apply an analog of Theorem 1.1 and conclude that the measure does not have a singular continuous part on any of those curves.

To show an example of such an application, let us consider the following result by P. Mattila. We denote by B(a, r) the ball of radius r centered at a. We say that \underline{D}μ(a) > 0 if

lim inf_{r→0+} \frac{μ(B(a, r))}{r} > 0.

Theorem 1.3 ([3]). Let μ ∈ M(C) be a non-negative measure. If \underline{D}μ(z) > 0 and Cμ(z) exists for μ-a.e. z ∈ C then μ is concentrated on a countable set of C^{1} -curves, i.e. there exist C^{1} -curves γ_{1}, γ_{2}, … such that

μ(C \ ∪γ_{i}) = 0.

For this particular situation one can prove the following version of Theorem 1.1, see Section 4. We denote by H^{1} the one-dimensional Hausdorff measure.

Theorem 1.4. Let μ ∈ M(C), μ ≥ 0 and let γ be a C^{1} -curve in C. Suppose that Cμ exists μ-a.e. on γ. Then the restriction of μ on γ is the sum of a discrete measure and a measure absolutely continuous with respect to H^{1} .

(Instead of the existence of Cμ on γ one can actually require a slightly weaker condition, like C_{ε}μ(ξ) = o(μ(B(ξ, ε)/ε) as ε → 0+ for a.e. ξ with respect to the singular part of μ on γ, see Section 4.)

Together Theorems 1.4 and 1.3 give the following result:

Corollary 1.5. If μ ∈ M(C), μ ≥ 0 is such that \underline{D}μ(z) > 0 and Cμ(z) exists for μ-a.e. z, then μ is the sum of a discrete measure and a measure absolutely continuous with respect to H^{1} , concentrated on a countable union of C^{1} curves.

In particular, note that the continuous part of μ will automatically satisfy the so-called linear growth condition:

lim sup_{ε→0} \frac{μ_{c}(B(x, ε))}{ε} < ∞

for μ_{c} -a.e. x.

Another application of Theorem 1.1 concerns inner functions in the unit disk. It is well known that certain geometric properties of a conformal mapping may imply the existence of its derivative on the boundary of the domain. Our next theorem shows a similar property in a non-conformal situation.

Let Σ be a subset of T. We say that an inner function θ in the unit disk D is radial near Σ if it maps almost every radius that ends at Σ into a curve that is tangent to a radius. More precisely, θ is radial near Σ if for a.e. ξ ∈ Σ

Im θ(rξ)θ̄(ξ) = o(1 − |θ(rξ)|)

as r → 1−. Here θ(ξ) stands for the non-tangential limit of θ at ξ.

Theorem 1.6. An inner function in the unit disk is radial near Σ if and only if it has non-tangential (angular) derivatives almost everywhere on Σ.

I.e. an inner function is radial near Σ if and only if its zeros a_{n} and the singular measure σ corresponding to its singular factor satisfy

∑_{n} \frac{1 − |a_{n}|^{2}}{|a_{n} − ξ|^{2}} < ∞  and  ∫_{T} \frac{dσ(z)}{|z − ξ|^{2}} < ∞  for a.e. ξ ∈ Σ.

We say that an inner function θ has an angular derivative at ξ ∈ T if |θ(ξ)| = 1 and there exists a finite limit limz →_{∢} ξ(θ(z) − θ(ξ))/(z − ξ). (Note that the

condition |θ(ξ)| = 1 is redundant in the situation of Theorem 1.6.) There are several other equivalent definitions of the angular derivative related to each other by the Carathéodory theorem, see [8] for a detailed discussion.

Acknowledgement. The authors are grateful to the administration and staff of Institut Mittag-Leffler for their hospitality.

2. The growth of Cauchy integrals in the upper half-plane

We begin this section with the lemma, which illustrates the well-known fact that the conjugate Poisson integral is “close” to the Hilbert transform. For convenience we will often write Hμ(x + iy) instead of Hμ(x, y) (thus stressing the fact that Hμ(x, y) is close to Qμ(x + iy)). The notation H_{x+iy} will be used for the corresponding Hilbert kernel H_{x,y}. We will also use notations P_{z} and Q_{z} for the Poisson and conjugate Poisson kernels.

Lemma 2.1. Let μ ∈ M(R) be a positive measure.

(1) Suppose that |Qμ(iy)| < α(y)P μ(iy) for some positive function α: R_{+} → R_{+} , α(y) → 0 as y → 0. Then |Hμ(iy)| < β(y)P μ(iy) for some positive function β: R_{+} → R_{+} , β(y) → 0 as y → 0 depending only on α;

(2) Conversely, if |Hμ(iy)| < α(y)P μ(iy) for some positive function α: R_{+} → R_{+} , α(y) → 0 as y → 0 then |Qμ(iy)| < β(y)P μ(iy) for some positive function β: R_{+} → R_{+} , β(y) → 0 as y → 0 depending only on α.

Proof. (1) Denote H^{ε}μ(x + iy) = ∫_{(1+ε)^{−1}y}^{(1+ε)y} Hμ(x + is) ds. Then H^{ε}μ is an

integral transform of μ with the continuous kernel H_{x+iy}^{ε} = ∫_{(1+ε)^{−1}y}^{(1+ε)y} H_{x+is} ds. Linear combinations of functions

\frac{Q_{iy} − Q_{i}}{P_{i}} (t) = (y^{2} − 1)Q_{iy}(t)

are dense in the space of odd continuous functions on R̂ . Therefore for any ε > 0 one can choose constants c_{k} and points iy_{k} so that

|\frac{H_{i}^{ε} − Q_{i}}{P_{i}} − ∑_{1}^{n} c_{k} \frac{Q_{iy} − Q_{i}}{P_{i}} | < ε

on R. Then

|H_{i}^{ε} − ∑_{1}^{n+1} c_{k}Q_{iy_{k}} | < εP_{i}

on R where c_{n+1} = 1 − ∑_{k=1}^{n} c_{k}, y_{n+1} = 1. From the properties of Poisson, conjugate Poisson and Hilbert kernels this implies that for any 0 < s < 1

|H_{is}^{ε} − ∑_{1}^{n} c_{k}Q_{isy_{k}} | < εP_{is} on R.

Therefore, |H^{ε}μ(iy)| < β^{ε}(y)P μ(iy) for some positive function β^{ε}: R_{+} → R_{+} , β^{ε}(y) → 0 as y → 0, depending only on α.

To pass from H^{ε} to H notice that

|\frac{∂Hμ(iy)}{∂y}| < C_{1} \frac{P μ(iy)}{y}.

Hence, |Hμ(iy)| ≤ C_{2}(H^{ε}μ(iy) + εP μ(iy)).

(2) In a similar way, one can notice that Q_{i} can be approximated by linear combinations of H_{iy} (such an approximation can actually be constructed much easier than in part (1)).

Note: the lemma shows that the conjugate Poisson transform can be replaced with the Hilbert transform in the statement of Theorem 1.1 in the case when the measure is positive. To pass to the general case one can represent the measure as a linear combination of mutually singular positive measures and use Theorem 2.4 below.

Now we can proceed with the proof of Theorem 1.1. We start with the definition of a porous set. After that we prove a lemma, showing that any set E, such that the condition |Hμ(x, y)| = o(P μ(x + iy)) as y → 0+ is satisfied uniformly for x ∈ E, has to be porous. The proof of Theorem 1.1 will then be completed by showing that a singular continuous measure μ, satisfying Qμ = o(P μ) on μ-a.e. vertical line, cannot see a porous set, thus obtaining a contradiction.

Definition 2.2. We say that a set K ⊂ R is porous if for any x ∈ K and any ε > 0 there exists 0 < δ < ε such that (x − δ, x − δ/100) ∪ (x + δ/100, x + δ) does not intersect K.

Note that by the density theorem a porous set has to have zero Lebesgue measure.

Lemma 2.3. Let μ ∈ M(R) be a positive measure and let E be a closed set such that

(1) μ(E) > 0,

(2) dμ/dm(x) = ∞ for any x ∈ E and

(3) |Hμ(x + iy)| < α(y)P μ(x + iy) for some positive function α: R_{+} → R_{+} , α(y) → 0 as y → 0, at any x ∈ E. Then E is porous.

Here dμ/dm denotes the Radon derivative of μ with respect to the Lebesgue measure m.

Proof. By the last lemma, we can assume that Qμ grows uniformly slower than P μ with some function β replacing α in the corresponding estimate. By Theorem 1.2 (see next section for the proof), for μ-a.e. x ∈ E for any sector Γ_{x}^{ϕ}, ϕ > 0,

lim sup_{z→x, z∈Γ_{x}^{ϕ}} \frac{|Qμ(z)|}{P μ(z)} > 0.

Hence, for μ-a.e. x ∈ E one can consider the balls B_{n} = B(x + iε_{n}, \frac{1}{2}ε_{n}) such that

(3)  lim inf_{n→∞} \frac{max_{B_{n}} |Qμ|}{max_{B_{n}} P μ} > 0  where ε_{n} → 0+.

Let ζ_{n} be the “rescaling” maps: ζ_{n}(z) = (x + ε_{n}i) + \frac{1}{2}ε_{n}z. Then

u_{n} = \frac{1}{max_{B_{n}} P μ}Kμ ∘ ζ_{n}

is a sequence of functions in the unit disk whose real part is bounded by 1. The derivative of Qμ in B_{n} is bounded by Cε_{n}^{−1}P μ. Thus the derivatives of the imaginary parts of u_{n} are uniformly bounded in the unit disc. Also, by the definition of the set E

Im u_{n}(0) ≤ C \frac{Qμ(x + iε_{n})}{P μ(x + iε_{n})} → 0.

Thus the imaginary parts of u_{n} are also uniformly bounded. By the normal families argument, one can choose a subsequence u_{n_{k}} converging to an analytic function u pointvise in the disk. Let u = p + iq. By Harnak’s lemma, the real parts of u_{n} are bounded away from zero, and so p is non-zero. By (3) | Im u_{n}| > c > 0 on a subdisk of a fixed hyperbolic radius and hence q is non zero. By the definition of the set E, q = 0 on the vertical diameter d = (−i, i). Hence the partial derivative q_{y} is 0 on d. If q_{x} = 0 on d then by the Cauchy– Riemann equations q ≡ 0 and we have a contradiction. Therefore, for k large enough in each ball B_{n_{k}} there is at least one point z_{k} = x + iy_{k} on the vertical diameter where y_{k}|(Qμ)_{x}| > c max_{B_{n_{k}}} P μ for some positive c. Then the inequality y_{k}|(Qμ)_{x}| > \frac{1}{2}c max_{B_{n_{k}}} P μ must hold in the ball D_{k} = B(z_{k}, εy_{k}) for some ε > 0 and for k = 1, 2, …. Since Qμ is continuous, WLOG we can assume that

(4)  y_{k}(Qμ)_{x} > \frac{1}{2}cP μ in D_{k}.

Now if we choose n large, so that

(5)  β(y_{n}) ≪ ε

then (4) implies that for any

x_{1} ∈ (x − εy_{n}, x − εy_{n}/100) ∪ (x + εy_{n}/100, x + εy_{n})

we have

|Qμ(x_{1} + iy_{n})| > \frac{εc}{100}P μ(x_{1} + iy_{n}).

Together with (5) the last inequality implies that x_{1} ∉ E.

We will also need the following

Theorem 2.4 ([6]). Let μ, ν ∈ M(R), μ = fν + η where f ∈ L^{1}(|ν|) and η ⊥ ν. Then the limit

{lim \atop z →_{∢} x} \frac{Kμ(x)}{Kν(x)}

exists ν -a.e. It is equal to f(x) ν_{s} -a.e.

Lemma 2.5. Let μ and E be the same as in Lemma 2.3. Denote by ν the restriction of μ on E. For ν -a.e. x and any ε > 0 there exists δ, 0 < δ < ε such that

(1) (x − δ, x − δ/100) ∪ (x + δ/100, x + δ) does not intersect E;

(2) for any x_{0} + iy_{0} such that x_{0} ∈ (x − δ/100, x + δ/100) ∩ E, |x − x_{0}| < y_{0} < 2|x − x_{0}|,

|Hμ(x_{0} + iy_{0})| + |Hν(x_{0} + iy_{0})| < \frac{1}{1000}P μ(x_{0} + iy_{0}).

(Note: the condition |x − x_{0}| < y_{0} < 2|x − x_{0}| means that x_{0} + iy_{0} ∈ Γ_{x}^{π/4} \ Γ_{x}^{π/6} .)

Proof. The constant δ satisfying (1) exists by the last lemma.

Since x_{0} + iy_{0} ∈ Γ_{x}^{π/4} , by Theorem 2.4 (note that ν is a singular measure)

lim_{δ→0} \frac{Kμ(x_{0} + iy_{0})}{Kν(x_{0} + iy_{0})} = lim_{δ→0} \frac{P μ(x_{0} + iy_{0}) + iQμ(x_{0} + iy_{0})}{P ν(x_{0} + iy_{0}) + iQν(x_{0} + iy_{0})} = 1

for ν -a.e. x. Since x_{0} ∈ E, for small enough δ,

|Qμ(x_{0} + iy_{0})| < \frac{1}{1000}P μ(x_{0} + iy_{0}),

and therefore

|Qν(x_{0} + iy_{0})| < \frac{1}{1000}P ν(x_{0} + iy_{0}) ≤ \frac{1}{1000}P μ(x_{0} + iy_{0}).

Now one can use Lemma 2.1.

Proof of Theorem 1.1. (2) ⇒ (1). The implication is easy to verify if μ_{s} has just one point mass. If μ_{s} has more than one point mass, Theorem 2.4 implies that near each of those point masses the contribution of other point masses is negligeably small.

(1) ⇒ (2). By Theorem 2.4, one can reduce the statement to the case when μ is positive.

Let us denote by μ_{sc} the singular continuous part of μ. We need to show that μ_{sc}(Σ) = 0. Suppose it is not so. Then one can choose a subset E ⊂ Σ, μ_{sc}(E) > 0 such that

|Hμ(x, y)| < α(y)P μ(x + iy)

for some positive function α: R_{+} → R_{+} , α(y) → 0 as y → 0, at any x ∈ E. Let ε be such that α(ε) < 1/1000. By Lemma 2.5 for any x ∈ E there exists δ < ε satisfying conditions (1) and (2) from the statement of the lemma. Let us cover E with such δ -neighborhoods. Let I be one of the intervals of this covering: I = (x − δ, x + δ) for some x ∈ E, δ = δ(x) < ε. Denote a = min E ∩ I, b = max E ∩ I (note: a, b ∈ (x − δ/100, x + δ/100)). Let Δ = b − a, z_{1} = a + 2Δi, z_{2} = b + 2Δi. Then

|Hν(z_{1}) − Hν(z_{2})| ≤ |Hμ(z_{1}) − Hμ(z_{2})| + |Hν(z_{1}) − Hμ(z_{1})| + |Hν(z_{2}) − Hμ(z_{2})|.

The first summand is small relative to P μ because a, b ∈ E and the other two are small because of part (2) of the last lemma. Altogether we get

|Hν(z_{1}) − Hν(z_{2})| ≤ \frac{5}{1000}P ν(z_{1}).

On the other hand, the difference of kernels H_{z_{1}} −H_{z_{2}} is 0 on (a, b) (both kernels are 0 there) and satisfies H_{z_{1}} − H_{z_{2}} < −1/2P_{z_{1}} outside of (x − 100δ, x + 100δ). Since ν is absent on (x − 100δ, x + 100δ) \ (x − δ, x + δ),

|Hν(z_{1}) − Hν(z_{2})| ≥ \frac{1}{2} ∫_{R\I} P_{z_{1}} dν.

Therefore

\frac{5}{1000}P ν(z_{1}) ≥ \frac{1}{2} ∫_{R\I} P_{z_{1}} dν

and

(6)  \frac{ν((a, b))}{b − a} ≥ P ν(z_{1}) − ∫_{R\I} P_{z_{1}} dν > \frac{1}{2}P ν(z_{1}).

To obtain the final contradiction consider points w_{1} = a + Δi, w_{2} = b + Δi. Note that, in the same way as before, the choice of I (part (2) of Lemma 2.5) and the fact that a, b ∈ E imply

|Hν(w_{1}) − Hν(w_{2})| ≤ \frac{5}{1000}P ν(w_{1}).

On the other hand, (H_{w_{1}} − H_{w_{2}}) > 1/2(b − a) on (a, b) and |H_{w_{1}} − H_{w_{2}}| < 2P_{z_{1}} on R \ I. Hence by (6),

\begin{aligned}|Hν(w_{1}) − Hν(w_{2})| & ≥ |∫_{(a,b)}(H_{w_{1}} − H_{w_{2}}) dν| − |∫_{R\I}(H_{w_{1}} − H_{w_{2}}) dν| \\ & ≥ \frac{1}{2} \frac{ν((a, b))}{b − a} − \frac{10}{1000}P ν(z_{1}) ≥ \frac{1}{10}P ν(z_{1}) ≥ \frac{1}{100}P ν(w_{1}). \end{aligned}

3. Non-tangential growth of Riesz transforms in the half-space

In this section we prove Teorem 1.2. We start with the following lemmas. For μ ∈ M(R^{n−1}) we denote by μ_{s} the part of μ that is singular with respect

to the (n − 1)-dimensional Lebesgue measure on R^{n−1} .

Lemma 3.1. Let μ ∈ M(R^{n−1}), 0 < ϕ < \frac{1}{2}π and let E ⊂ R^{n−1} be a set of zero (n − 1)-dimensional Lebesgue measure satisfying |μ_{s}|(E) > 0. Denote Γ = ⋃_{x∈E} Γ_{x}^{ϕ}. Then

(7)  ∫_{∂Γ} |Rμ| ds = ∞.

(In the last integral and throughout this section ds corresponds to the standard integration with respect to the surface area.)

Proof. Here we only give the proof for the case of the upper half-plane (n = 2). For this case the proof seems especially natural. The same ideas could be modified to obtain the general proof.

Let μ ∈ M(R) and Γ = ⋃_{t∈E} Γ_{t}^{ϕ}, where |E| = 0, |μ_{s}|(E) > 0. Suppose that Kμ is summable on ∂Γ with respect to the arclength. Define the function f on R in the following way:

f(x) = \{ \begin{matrix}|Kμ(x + iy)| & if x + iy ∈ ∂Γ for some 0 < y < 1, \\ 0 & otherwise. \end{matrix}

Then f ∈ L^{1}(R). Consider the Poisson integral P f of f in the upper half-plane. One can show that then

(8)  |Kμ| < C_{1}P f

a.e. on ∂Γ with respect to the arclength. Indeed, let x + iy ∈ ∂Γ for some y < 1. Then at least on one half of the interval (x−\frac{1}{2}y, x+\frac{1}{2}y) we have f > C_{3}|Kμ(x+iy)| because of the Lipschitz properties of Kμ in B(x + iy, \frac{1}{2}y). Since the Poisson kernel P_{x+iy} is larger than C_{4} on (x − \frac{1}{2}y, x + \frac{1}{2}y), we get (8) at x + iy.

Since Kμ is a function of the Smirnov class in Γ and P f is harmonic, (8) holds inside Γ as well. Since fdx ⊥ μ_{s}, we obtain a contradiction with Lemma 3.2 below (just put fdx = ν).

Lemma 3.2 ([6]). Let μ, ν ∈ M(R) and ν ⊥ μ_{s}. Then

{lim \atop z →_{∢} x} \frac{|P μ|}{|P ν|} = ∞ for μ_{s} -a.e. x.

Proof. The statement presents a version of the Lebesgue theorem saying that for a summable function almost every point is its Lebesgue point. The classical proof can be easily modified to work in our case.

Lemma 3.3. Let u be a harmonic function in a domain in R^{n}. Denote by u_{1}, …, u_{n} its partial derivatives. Then the function u_{n}^{2} − C_{n} ∑_{i=1}^{n−1} u_{i}^{2} is superharmonic in the same domain for any C_{n} ≥ n − 1.

Proof. Note that since ∑_{i=1}^{n} u_{ii} = 0,

△(u_{j})^{2} = ∑_{i=1}^{n} ∂_{i}∂_{i}(u_{j})^{2} = ∑_{i=1}^{n} 2(∂_{i}u_{j})^{2} + 2u_{j} ∑_{i=1}^{n} ∂_{i}^{2}u_{j} = 2 ∑_{i=1}^{n}(u_{ij})^{2}.

Therefore

△(u_{n}^{2} − C_{n} ∑_{i=1}^{n−1} u_{i}^{2}) = 2(∑_{i=1}^{n}(u_{ni})^{2} − C_{n} ∑_{i=1}^{n}(u_{i(n−1)})^{2} ⋯ − C_{n} ∑_{i=1}^{n}(u_{i1})^{2})

= 2((u_{nn})^{2} − C_{n} ∑_{i=1}^{n−1}(u_{i(n−1)})^{2} … − C_{n} ∑_{i=1}^{n−1}(u_{i1})^{2})

≤ 2((u_{nn})^{2} − C_{n}(u_{(n−1)(n−1)})^{2} ⋯ − C_{n}(u_{11})^{2}).

Now using again the fact that u_{nn} = − ∑_{i=1}^{n−1} u_{ii} one can conclude that the last expression is negative.

Proof of Theorem 1.2. Let (1) be satisfied for μ-a.e. x ∈ Σ but |μ_{s}|(Σ) > 0. Denote by u the harmonic function in Rn_{+} such that ∇u = Rμ, i.e. R_{k}μ = u_{k}. Let \frac{1}{4}π < ϕ < \frac{1}{2}π. Then there exists a closed set E ⊂ Σ, |E| = 0, |μ|(E) > 0 and ε > 0 such that

(9)  |⟨u_{1}, …, u_{n−1}⟩| < \frac{1}{2\sqrt{(n − 1)}} |u_{n}|

on

Γ = {(y_{1}, …, y_{n}) ∈ ⋃_{x∈E} Γ_{x}^{ϕ}∩, 0 < y_{n} < ε}.

By Lemma 3.3 the function u_{n}^{2} − (n − 1) ∑_{1≤k<n} u_{k}^{2} is superharmonic on Γ and by (9) it is positive. Hence u_{n}^{2} − (n − 1) ∑_{1≤k<n} u_{k}^{2} , and therefore (by (9)) u_{n}^{2} is summable on the boundary of Γ with respect to the harmonic measure there.

Notice that Γ is a Lipschitz domain. The harmonic measure on ∂Γ can be written as w ds for some positive w. Since the angle ϕ is greater than \frac{1}{4}π, the density w satisfies

∫_{∂Γ} w^{−1} ds < ∞.

(One will need to “smooth-out” the upper part of the boundary to make the integral over the whole ∂Γ finite; we are actually only interested in the lower part of the boundary.) But now, by the Cauchy–Schwarz inequality,

∫_{∂Γ} |u_{n}| ds ≤ (∫_{∂Γ} u_{n}^{2}w ds)^{1/2}(∫_{∂Γ} w^{−1} ds)^{1/2} < ∞.

By (9) this implies that ∫_{∂Γ} |u| ds < ∞ which contradicts Lemma 3.1.

Let now ϕ satisfy 0 < ϕ ≤ \frac{1}{4}π. Suppose that for some x = (x_{1}, …, x_{n−1}) ∈ R^{n−1} we have

(10)  |⟨u_{1}, …, u_{n−1}⟩| = o(u_{n})

as z → x, z ∈ Γ_{x}^{ϕ} but

(11)  |⟨u_{1}, …, u_{n−1}⟩| ≠ o(u_{n})

in Γ_{x}^{3π/8} (the latter holds μ-a.e. by the first part of the proof). Suppose also that u_{n} → ∞ in Γ_{x}^{3π/8} (this holds a.e. with respect to the positive component of μ_{s}). Then there exist points z_{k} = ⟨x_{1}, …, x_{n−1}, y_{k}⟩, y_{k} → 0+ such that in each ball B_{k} = B(z_{k}, y_{k} sin \frac{3}{8}π) (B_{k} is the maximal ball centered at z_{k} that lies in Γ_{x}^{3π/8} ) there is a point w_{k} where

|⟨u_{1}(w_{k}), …, u_{n−1}(w_{k})⟩| > c|u_{n}(w_{k})|

for some fixed c > 0. We can choose B_{k} and w_{k} so that w_{k} ∈ (1 − ε)B_{k} = B(z_{k}, (1 − ε)y_{k} sin \frac{3}{8}π) for some small ε > 0. Consider the rescaling maps from the unit ball B = B(0, 1) to B_{k}:

ξ_{k}(z) = z_{k} + zy_{k} sin \frac{3π}{8} .

Put

v_{k} = \frac{1}{max_{B_{k}} |∇u|}(∇u) ∘ ξ_{k}.

Then {v_{k}} is a sequence of gradients of harmonic functions in the unit ball B whose magnitude is bounded by 1. By the normal families argument, one can choose a subsequence v_{n_{k}} converging to a gradient v, |v| < 1 pointwise in the disk. Since at the origin the last coordinates of v_{k} are bounded away from zero, the last coordinate of v is bounded away from zero at the origin. Since the last coordinates of v_{k} are positive (recall that we assumed u_{n} → ∞ in Γ_{x}^{3π/8} ) the last coordinate is bounded away from zero at the origin and positive in B. Hence it is bounded away from zero in (1 − ε)B. By the choice of w_{k}, for each k there is a point in B(0, (1 − ε)) where the magnitude of the first n − 1 coordinates of v_{k} is large in comparison to the last coordinate. Thus the first n − 1 coordinates of v cannot be all zero. But since |⟨u_{1}, …, u_{n−1}⟩| = o(u_{n}) in Γ_{x}^{ϕ}, the first n − 1 coordinates of v are zero on a large part of the ball (a set of nonzero volume) and we have a contradiction. It is left to notice that by the first part of the proof (11) holds in Γ_{x}^{3π/8} for μ_{s} -a.e. x and therefore (10) may not hold in Γ_{x}^{ϕ} except for a zero set of x with respect to the positive part of μ_{s}. Other parts of μ_{s} can be treated similarly.

4. Applications

4.1. Measures in the plane. The goal of this subsection is to prove Theorem 1.4.

Let μ be a finite positive measure in C and let γ be a C^{1} curve.

WLOG γ is a graph of a C^{1} function f on the interval [−1, 1]: γ = {x + if(x)}. Denote by Ω_{±} the open domains above and below γ, i.e. Ω_{+} = {x + iy | |x| ≤ 1, y > f(x)} and Ω_{−} = {x + iy | |x| ≤ 1, y < f(x)}.

First, we will “move” the whole μ under the graph γ to be able to consider the holomorphic function Cμ in Ω_{+} . This will allow us to apply complex methods like in Sections 3 and 4.

To do this, consider the map ϕ: Ω_{+} ↦ Ω_{−}, ϕ(x+iy) = x+i(f(x)−(y−f(x))) that maps points in Ω_{+} into points in Ω_{−} symmetric with respect to γ. Denote by ν the restriction of μ on γ. Let η = μ − ν and denote by η_{±} the restrictions of η on Ω_{±}. Consider the measure ϕ(η_{+}) on Ω_{−}: [ϕ(η_{+})](B) = η_{+}(ϕ^{−1}(B)) for any Borel B ⊂ Ω_{−}. Denote η^{∗} = η_{−} + ϕ(η_{+}) and μ^{∗} = ν + η^{∗}.

Now Cμ^{∗} is a holomorphic function in Ω_{+} . If z ∈ γ denote by α_{z} the arctangent of the slope of the tangent line at z. Then the point z + ie^{α_{z}}ε approaches z along the normal line from Ω_{+} .

Throughout this section, if μ ∈ M(C) is supported on a rectifiable curve, we denote by μ_{ac}, μ_{s} and μ_{sc} the absolutely continuous, singular and singular continuous parts of μ with respect to H^{1} on the curve.

We want to proceed as follows. Suppose that Cμ is finite μ-a.e. on γ. First we will show that Re e^{−α_{z}}C_{ε}μ^{∗}(z+ie^{α_{z}}ε), the analog of the conjugate Poisson integral from the line case considered in Section 2, grows slower than Im e^{−α_{z}}C_{ε}μ^{∗}(z + ie^{α_{z}}ε), the analog of the Poisson integral, as ε → 0+ on ν_{s} -a.e. normal line, see Claims 4.1–4.6 below. Then, using methods similar to those from Sections 2 and 3, we will show that this is possible only if ν_{s} is discrete.

Claim 4.1. For every z ∈ γ there exists a finite constant C such that for any ε > 0

Im e^{−α_{z}}C_{ε}μ^{∗}(z) ≥ Im e^{−α_{z}}C_{ε}μ(z) + C.

Proof. Let z = 0 ∈ γ and assume that the tangent line to γ at 0 is horisontal. Then e^{−α_{z}} = 1. For any δ > 0, γ lies inside {|y| < δ|x|} near 0. WLOG we can assume that the whole γ lies there. Note that

Im Cμ(w) = ∫ \frac{Im w − Im ξ}{|w − ξ|^{2}} dμ(ξ).

The kernel of Im C_{ε}(0) is negative in the upper half-plane and positive in the lower half-plane. The part of η_{+} that lies above y = 3δ|x| was mapped by ϕ from the upper to the lower half-plane, and therefore after replacing that part with its “image” under ϕ the integral could only increase. WLOG the part of η_{+} under y = 3δ|x| is pure point. Notice that each point mass moves down under ϕ. If δ is small enough, for any point mass that lies inside {|y| < 3δ|x|} such a motion increases its integral.

If μ ∈ M(C) we will denote by 𝒫μ(z, ε) the integral

𝒫μ(z, ε) = ∫ \frac{ε}{|ξ − z|^{2} + ε^{2}} dμ(ξ).

Let ψ be the map from Clos Ω_{−} to the closed lower half-plane defined as ψ(f(x) − iy) = −iy for every y ≥ 0 (recall that γ is the graph of f). The map ψ projects γ on the real line and sends every curve γ − iy into the horisontal segment [−1 − iy, 1 − iy] below the real line. Denote by μ_{p}^{∗} the measure ψ(μ^{∗}), i.e. μ_{p}^{∗}(B) = μ^{∗}(ψ^{−1}(B)). Similarly, let ν_{p}, η_{p}^{∗} stand for the images of the corresponding measures.

Claim 4.2. We have

Im Cη_{p}^{∗}(ψ(z) + iε) = o(𝒫ν(z, ε)) for ν_{s} -a.e. z as ε → 0+.

Proof. The function Im Cη_{p}^{∗} is a positive harmonic function in the upper half-plane. Therefore it is equal to P σ for some positive measure σ on R. The measure σ is absolutely continuous. Indeed, denote by η_{ε} the restriction of η_{p}^{∗} on {Im z < −ε} and let σ_{ε} be the corresponding measure on R: Im Cη_{ε} = P σ_{ε}. Then σ_{ε} → σ in norm. Since all σ_{ε} are absolutely continuous, so is σ. Therefore by Lemma 4.8

Im Cη_{p}^{∗}(x + iε) = P σ(x + iε) = o(P ν_{p}(x + iε)) for (ν_{p})_{s} -a.e. x as ε → 0+.

It is left to notice that for any z ∈ γ there exists C > 0 such that 𝒫ν(z, ε) > CP ν_{p}(x + iε).

Let ξ ∈ γ and δ > 0. Near ξ, γ lies in Δ_{ξ} = e^{α_{ξ}}{| Im(z −ξ)| < δ| Re(z −ξ)|}. Our next claim shows that the part of μ^{∗} that lies outside of Δ_{ξ} has little influence on the asymptotics of e^{α_{ξ}} Im C_{ε}μ^{∗}ξ.

Claim 4.3. For any δ > 0 and ξ ∈ γ denote by η_{ξ}^{∗} the restriction of η^{∗} on C \ Δ_{ξ}. Then for ν_{s} -a.e. ξ

e^{α_{ξ}} Im C_{ε}η^{∗}ξ = o(𝒫μ^{∗}(ξ, ε)).

Proof. By comparing the kernels one can notice that

|e^{α_{ξ}} Im C_{ε}η^{∗}ξ| < C Im Cη_{p}^{∗}(ψ(ξ) + iε).

Now the statement follows from the previous claim.

Now we show that Im e^{−α_{z}}Cμ^{∗}(w) grows fast as w approaches z non-tan-gentially from Ω_{+} for ν_{s} -a.e. z. For the rest of this subsection for any ξ ∈ γ, 0 < ϕ < \frac{1}{2}π we denote by Γ_{ϕ}(ξ) the non-tangential sector \{Im e^{α_{ξ}}(z −ξ)/|z −ξ| > sin ϕ}. Note that near ξ the sector Γ_{ϕ}(ξ) lies entirely in Ω_{+} .

Claim 4.4. For any 0 < ϕ < \frac{1}{2}π

\frac{1}{L}𝒫μ^{∗}(0, ε) + C ≤ Im e^{−α_{z}}Cμ^{∗}(w) ≤ L𝒫μ^{∗}(0, ε) + C

as w → z, w ∈ Γ_{ϕ}(z), |w − z| = ε for some positive L for ν_{s} -a.e. z ∈ γ.

Proof. Again we can assume that z = 0 ∈ γ, the tangent line to γ at 0 is horisontal and γ lies in {|y| < δ|x|}, where δ is so small that γ does not intersect the sector Γ_{ϕ}(0). Let D be a large positive constant. Simple calculations show that for all ξ ∈ {|y| < δ|x|}, |ξ| > Dε we have

\frac{− Im ξ}{|ξ|^{2}} + C_{1} \frac{ε}{ε^{2} + |ξ|^{2}} ≤ \frac{Im w − Im ξ}{|w − ξ|^{2}} ≤ \frac{− Im ξ}{|ξ|^{2}} + C_{2} \frac{ε}{ε^{2} + |ξ|^{2}}

for some C_{1}, C_{2} > 0 (if δ is small and D is large enough). The part of the measure outside of {|y| < δ|x|} can be ignored by the previous claim. Therefore

Im Cμ^{∗}(w) = ∫_{|ξ|>Dε}\frac{Im w − Im ξ}{|w − ξ|^{2}} dμ^{∗}(xi) + ∫_{|ξ|≤Dε}\frac{Im w − Im ξ}{|w − ξ|^{2}} dμ^{∗}(xi)

≍ ∫_{|ξ|≤Dε}\frac{Im w − Im ξ}{|w − ξ|^{2}} dμ^{∗}(xi)+∫_{|ξ|≥Dε} \frac{ε}{ε^{2} + |ξ|^{2}} dμ^{∗}(xi) + Im C_{Dε}μ^{∗}(0).

Since μ^{∗} is concentrated under y = δ|x|

∫_{|ξ|≤Dε} \frac{Im w − Im ξ}{|w − ξ|^{2}} dμ^{∗}(xi) + ∫_{|ξ|≥Dε} \frac{ε}{ε^{2} + |ξ|^{2}} dμ^{∗}(xi) ≍ 𝒫μ^{∗}(0, ε).

Now recall that Cμ(0) is finite ν -a.e. and apply Claim 4.3.

Next we estimate the “conjugate Poisson part”, Re e^{−α_{z}}Cμ^{∗}(z).

Claim 4.5. We have

Re e^{−α_{z}}C_{ε}μ^{∗}(z) = Re e^{−α_{z}}C_{ε}μ(z)+o(𝒫(0, ε)μ^{∗}(z)) as ε → 0+ for μ_{s}^{∗} -a.e. z ∈ γ.

Proof. Again we can assume that z = 0 ∈ γ, the tangent line to γ at 0 is horisontal and γ lies in {|y| < δ|x|}. Note that

Re C_{ε}μ^{∗}(w) = ∫_{|ξ|>ε} \frac{Re w − Re ξ}{|w − ξ|^{2}} dμ^{∗}(ξ).

To prove the statement we need to compare kernels of Re C_{ε}η_{+}(0) and Re C_{ε}η∗_{+}(0) at the points x + i(f(x) + y) and x + i(f(x) − y) correspondingly. Simple calculations show that, since |f(x)| < δ|x|,

|\frac{x}{|x + if(x) + iy|^{2}} − \frac{x}{|x + if(x) − iy|^{2}} | < C \frac{2y}{x^{2} + 4y^{2}} .

Since the right-hand side is the kernel for the Poisson integral of the “projected” measure P ψ(η∗_{−}) at the point ψ(x + i(f(x) − y)) = x − iy, this inequality and Claim 4.2 imply

| Re C_{ε}μ(0) − Re C_{ε}μ^{∗}(0)| = | Re C_{ε}η_{+}(0) − Re C_{ε}η∗_{+}(0)|

< C Im Cψ(η∗_{−})(x + iy) = o(𝒫μ^{∗}(0, ε)).

Claim 4.6. Suppose that z ∈ γ and Cμ(z) is finite. Then

Re e^{−α_{z}}Cμ^{∗}(z + ie^{α_{z}}ε) = o(𝒫μ^{∗}(z, ε)) as ε → 0+ for μ_{s}^{∗} -a.e. z ∈ γ.

Proof. Again we can assume that z = 0, the tangent line to γ at 0 is horisontal and γ lies in {|y| < δ|x|}.

Denote

h_{α}(z) = \{ \begin{matrix}\frac{Re z}{|z|^{2}} & on \{|z| > α\}, \\ 0 & on \{|z| ≤ α\}. \end{matrix}

Notice that there exists a linear combination of such functions

∑ a_{n}h_{α_{n}}, ∑ a_{n} = 1, α_{n} ≤ 10ε

which approximates the kernel of Re Cμ^{∗}(iε) on B(0, 10ε) ∩ {|y| < δ|x|}:

|\frac{− Re ξ}{|iε − ξ|^{2}} − ∑ a_{n}h_{α_{n}}(ξ)| < C \frac{δ}{ε}

for some absolute constant C. Outside of B(0, 10ε) ∩ {|y| < δ|x|} the condition ∑ a_{n} = 1 will automatically imply

|\frac{− Re ξ}{|iε − ξ|^{2}} − ∑ a_{n}h_{α_{n}}(ξ)| < Cδ \frac{ε}{ε^{2} + |ξ|^{2}}

for any ξ ∈ {|y| < δ|x|} and

|\frac{− Re ξ}{|iε − ξ|^{2}} − ∑ a_{n}h_{α_{n}}(ξ)| < C \frac{ε}{ε^{2} + |ξ|^{2}}

for other ξ ∉ B(0, 10ε). Integrating the last three estimates with respect to μ^{∗} = ν + η^{∗} we obtain

|Re Cμ^{∗}(iε) − ∑ a_{n}C_{α_{n}}μ^{∗}(0)| < C(δ𝒫ν(0, ε) + 𝒫η^{∗}(0, ε)).

Moreover, by the properties of both kernels, for any 0 < s < 1 we will have the estimate

|Re Cμ^{∗}(isε) − ∑ a_{n}C_{sα_{n}}μ^{∗}(0)| < C(δ𝒫ν(0, sε) + 𝒫η^{∗}(0, sε))
= Cδ𝒫ν(0, sε) + o(𝒫ν(0, sε))
= Cδ𝒫μ^{∗}(0, sε) + o(𝒫μ^{∗}(0, sε))

by Lemma 4.8. The statement now follows from the fact that C_{sα_{n}}μ^{∗}(0) = o(𝒫μ^{∗}(0, sε)) by the last claim and that δ can be chosen arbitrarily small near 0.

Let us summarize the above claims: We obtained the measure μ^{∗} = ν + η^{∗}, where ν is supported on the C^{1} -graph γ and η^{∗} lies under the graph (in Ω_{−}). We know that for ν_{s} -a.e. ξ ∈ γ, Re e^{α_{z}}Cμ^{∗}(z) = o(Im e^{α_{z}}Cμ^{∗}(z)) as z approaches ξ along the normal line from above. We have to show that then ν_{s} must be discrete.

Following the algorithm of Section 3, we first choose a large ϕ and assume that there exists a set E ⊂ γ such that ν_{s}(E) > 0 and Re e^{α_{z}}Cμ^{∗}(z) = o(Im e^{α_{z}}Cμ^{∗}(z)) as z → ξ, z ∈ Γ_{ϕ}(ξ) for every ξ ∈ E. WLOG all α_{ξ} for ξ ∈ E are smaller than a fixed δ. Then there exists ε_{0} > 0 and E′ ⊂ E, ν_{s}(E′) > 0 such that Re Cμ^{∗}(z) < \frac{1}{2} Im Cμ^{∗}(z) for any z ∈ Γ_{ϕ}(ξ), ξ ∈ E′, |z − ξ| < ε_{0} . Consider a positive sequence {ε_{k}}_{k=1}^{∞} monotonously decreasing to zero, ε_{0} > ε_{1} > ε_{2} > · · ·. Define

Γ_{k} = {z | z ∈ ⋃_{ξ∈E′} Γ_{ϕ}(ξ), ε_{k} < Im e^{α_{ξ}}(z − ξ) < ε_{0}}

and Γ = ∪Γ_{k}. Since Re Cμ^{∗}(z) < \frac{1}{2} Im Cμ^{∗}(z) in Γ_{k},

Re(Cμ^{∗}(z))^{2} = (Re Cμ^{∗}(z))^{2} − (Im Cμ^{∗}(z))^{2}

is a negative harmonic function in Γ_{k}. Therefore it is summable with respect to the harmonic measure on ∂Γ_{k}. Since Re Cμ^{∗}(z) < \frac{1}{2} Im Cμ^{∗}(z), (Im Cμ^{∗}(z))^{2} is summable with respect to the harmonic measure on ∂Γ_{k}. Each Γ_{k} is a Lipschitz domain. Let ζ ∈ Γ_{1} . If ϕ > \frac{1}{4}π + 10δ the Lipschitz constant for the boundary of ∂Γ_{k} is large enough so that the density w_{k} of the harmonic measure on ∂Γ_{k} with respect to ζ satisfies

∫_{∂Γ_{k}} w_{k}^{−1} ds < C < ∞

for all k. (Again, one needs to make the “upper” part of ∂Γ smooth to have this, which can always be done; but on the lower part, which we are mostly interested in, the integral converges as it is.) Then by the Cauchy–Schwarz inequality

∫_{∂Γ_{k}} | Im Cμ^{∗}(z)| ≤ (∫_{∂Γ_{k}} Im Cμ^{∗}(z)w_{k} ds)^{1/2}(∫_{∂Γ_{k}} w_{k}^{−1} ds)^{1/2} < ∞

and therefore Im Cμ^{∗}(z) is summable on ∂Γ_{k} with respect to the arclength. Denote by h_{k}, k = 0, 1, 2, … the summable function on [−1, 1] obtained by “projection” of the values of Im Cμ^{∗} from ∂Γ_{k}:

h_{k}(x) = \{ \begin{matrix}max_{x+iy∈Γ_{k}} | Im Cμ^{∗}(x + iy)| & if x + iy ∈ ∂Γ_{k} for some y , \\ 0 & otherwise. \end{matrix}

Since the domains Γ_{k} “converge” to Γ_{0} , one can show that

∫_{∂Γ_{k}\∂Γ} | Im Cμ^{∗}(z)| ds → 0

as k → ∞. Indeed, since |Im((Cμ^{∗}(z))^{2})| + 1 > |Re((Cμ^{∗}(z))^{2})| the function (Cμ^{∗}(z))^{2} is an H^{1}(w_{k}ds)-function in Γ_{k}. In particular we have

∫_{∂Γ_{k}} Im((Cμ^{∗})^{2})w_{k} ds = Im((Cμ^{∗}(ζ))^{2}).

Let us fix k. If l > k is large enough

∫_{∂Γ∩∂Γ_{k}} Im((Cμ^{∗})^{2})w_{l} ds

is close to

∫_{∂Γ∩∂Γ_{k}} Im((Cμ^{∗})^{2})w ds

which, in its turn, for large enough k is close to Im((Cμ^{∗}(ζ))^{2}). This means that

∫_{∂Γ_{l}\∂Γ} Im((Cμ^{∗})^{2})w_{l} ds

is close to 0 for large l. Therefore

∫_{∂Γ_{l}\∂Γ} | Im Cμ^{∗}(z)| ≤ (∫_{∂Γ_{l}\∂Γ} Im Cμ^{∗}(z)w_{l} ds)^{1/2}(∫_{∂Γ_{l}} w_{l}^{−1} ds)^{1/2}

≤ C(∫_{∂Γ_{l}\∂Γ_{0}} Im Cμ^{∗}(z)w_{l} ds)^{1/2} → 0.

Therefore the sequence h_{k} converges in L^{1}[−1, 1]. This means that there exists a subsequence h_{n_{k}} that has a summable majorant H ∈ L^{1}[−1, 1]: |h_{n_{k}}| < H for all k.

Let again ν_{p} = ψ(ν) be the projection of ν on [−1, 1]: ν_{p}(B) = ν({x + iy | x ∈ B}. By Claim 4.4 for ν_{s} -a.e. ξ

Im(e^{α_{z}}Cμ^{∗}(z)) ≍ P ν_{p}(x + e^{−α_{z}}(z − ξ))

as z → ξ, z ∈ Γ_{ϕ}(ξ), where x = ψ(ξ). Using this relation one can show, that P H(z) ≥ CP ν_{p}(z) for z ∈ Γ_{ϕ}(x), Im z = ε_{n_{k}}, k = 1, 2, … for (ν_{p})_{s} -a.e. x ∈ ψ(E′). But this contradicts Lemma 3.2 since H dx ⊥ (ν_{p})_{s}.

Therefore, the set of such ξ ∈ γ for which

(12)  Re e^{−α_{ξ}}Cμ^{∗}(z) = o(Im e^{−α_{ξ}}Cμ^{∗}(z))

as z → ξ, z ∈ Γ_{ϕ}(ξ) for large ϕ has to be a zero-set with respect to ν_{s}.

Using the normal families argument like in Section 3 one can pass from large ϕ to arbitrary sectors and show that there is only a zero set of points ξ with respect to ν_{s} such that for some ϕ = ϕ(ξ) > 0, (12) holds as z → ξ, z ∈ Γ_{ϕ}(ξ).

Now we have to make the last step from sectors to normal lines. Again our argument will be analogous to Section 2.

Definition 4.7. We will call E ⊂ γ porous if for any ξ ∈ E and for any ε > 0 there exists δ < ε such that E ∩ B(ξ, 100δ) \ B(ξ, δ) = ∅.

Now, like in Lemma 2.3, suppose that there is a set E ⊂ γ, ν_{sc}(E) > 0 such that 𝒫(z, ε) → ∞ and

| Re e^{−α_{z}}Cμ^{∗}(z + iεe^{α_{z}})| < h(ε)| Im e^{−α_{z}}Cμ^{∗}(z + iεe^{α_{z}})|

with some uniform function h > 0, h(ε) → 0 as ε → 0+ for every z ∈ E.

We can repeat the proof of Lemma 2.3 almost word by word to show that E is porous. Indeed, based on the fact that (12) cannot hold in a non-zero set of sectors with respect to ν_{s}, for ν_{s} -a.e. z ∈ E and any ε > 0 we can find a ball B centered on the normal line at z + i1000δe^{α_{z}} of the radius 200δ, where 2000δ < ε, such that the directional derivative of Re e^{−α_{z}}Cμ^{∗} in the direction perpendicular to the normal line is large in B in comparison to Im e^{−α_{z}}Cμ^{∗}. Similarly to the proof of Lemma 2.3, this means that those points on γ for which the corresponding normal lines hit B \ \frac{1}{200}B cannot belong to E (note that α_{ξ} → α_{z} as ξ → z since γ is a C^{1} curve). But all normal lines going through the points from (B(z, 100δ − o(δ)) \ B(z, δ + o(δ))) ∩ γ will hit B \ \frac{1}{200}B.

Yet another version of the Lebesgue theorem that we will use is presented in the following statement:

Lemma 4.8. Let μ, ν ∈ M(R^{2}), μ = fν where f ∈ L^{1}(|ν|). Then

lim_{ε→0} \frac{𝒫μ(z, ε)}{𝒫ν(z, ε)} = f(z)

as ε → 0+ for μ-a.e. z. In particular the limit is 0 ν -a.e. if and only if μ ⊥ ν.

Now suppose that (12) holds as z → ξ along a normal line from Ω_{+} for ν_{sc} - a.e. ξ ∈ γ. Then we can choose a set E, non-zero with respect to ν_{sc}, where that relation holds with a uniform “o” on the right-hand side. As we established above, E has to be porous. By approximating kernels, like in Claim 4.6, we can show that one can replace Re e^{−α_{ξ}}Cμ^{∗}(z), z = ξ + ie^{α_{ξ}}ε with Re e^{−α_{ξ}}C_{ε}μ^{∗}(ξ). The new relation

(13)  | Re e^{−α_{ξ}}C_{ε}μ^{∗}(ξ)| < g(ε)| Im e^{−α_{ξ}}Cμ^{∗}(ξ + iεe^{α_{z}})|

will still hold with some uniform function g > 0, g(ε) → 0 as ε → 0+ for ν_{s} -a.e. ξ. Denote by ν^{E} the restriction of ν on E. Then by Lemma 4.8

(14)  𝒫(ν − ν^{E})(ξ, ε) = o(𝒫ν(ξ, ε))

for ν_{sc} -a.e. ξ ∈ E. Let ξ ∈ E be a point where (13) and (14) hold. WLOG ξ = 0, the tangent line at 0 is horizontal and γ ⊂ {|y| < c|x|} for some small 0 < c < 1.

Since E is porous we can choose a small δ such that E ∩ (B(0, 100δ) \ B(0, δ)) = ∅. Let z_{1} and z_{2} be the points in γ ∩ B(0δ) with the smallest and the biggest real parts correspondingly for which (13) holds. Denote Δ = |z_{1} − z_{2}|. (Note, that we can assume that Δ > 0, i.e. z_{1} ≠ z_{2} . If that was not true, 0 would be an isolated point of E; but ν_{sc} -a.e. point of E is not isolated.)

We can estimate the difference between the kernels of Re e^{−α_{z_{1}}} C_{(2Δ)}μ^{∗}(z_{1}) and Re e^{−α_{z_{2}}} C_{(2Δ)}μ^{∗}(z_{2}) as follows: It is less than −C_{1}Δ/(|z|^{2} + Δ^{2}) on γ \ B(0, 100δ). Its absolute value is bounded by C_{2}Δ/(|z|^{2} + Δ^{2}) on C \ B(0, δ) for some C_{1,2} > 0. Finally, it is 0 on B(0, δ). Therefore

Ree^{−α_{z_{1}}} C_{(2Δ)}μ^{∗}(z_{1}) − Re e^{−α_{z_{2}}} C_{(2Δ)}μ^{∗}(z_{2})

< −C_{1} ∫_{γ\B(0,100δ)} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z) + C_{2} ∫_{γ∩(B(0,100δ)\B(0,δ))} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z)

+ C_{2} ∫_{C\γ} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z).

Note that the left-hand side is small by the absolute value in comparison to 𝒫μ^{∗}(0, Δ) because (13) holds at z_{1} and z_{2} . By Lemma 4.8 𝒫μ^{∗}(ξ, Δ) ∼ 𝒫μ^{∗}|_{γ}(ξ, Δ) at ν_{s} -a.e. ξ. WLOG our point 0 is one of such ξ ’s. Then the third summand in the right-hand side is small in comparison to 𝒫μ^{∗}(0, Δ) as well. The second summand is small in comparison to 𝒫μ^{∗}(0, Δ) because of (14). Therefore

∫_{γ\B(0,100δ)} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z)

is small in comparison to 𝒫μ^{∗}(0, Δ). Then

\begin{aligned}𝒫μ^{∗}(0, Δ) & ∼ ∫_{γ\B(0,100δ)} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z) + ∫_{γ∩(B(0,100δ)\B(0,δ))} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z) \\ & + ∫_{γ∩B(0,100δ)} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z). \end{aligned}

As we have just shown, the first integral on the right-hand side is small. So is the second integral by (14). Therefore

𝒫μ^{∗}(0, Δ) ∼ ∫_{γ∩B(0,100δ)} \frac{Δ}{|z|^{2} + Δ^{2}} dμ^{∗}(z) ≤ C_{3} \frac{μ^{∗}(B(0, δ)}{δ}.

At the same time, looking at kernels of Re e^{−α_{z_{1}}} C_{Δ}μ^{∗}(z_{1}) and Re e^{−α_{z_{2}}} C_{Δ}μ^{∗}(z_{2}), we get

Ree^{−α_{z_{1}}} C_{Δ}μ^{∗}(z_{1}) − Re e^{−α_{z_{2}}} C_{Δ}μ^{∗}(z_{2})

≥ Re e^{−α_{z_{1}}} C_{(2Δ)}μ^{∗}(z_{1}) − Re e^{−α_{z_{2}}} C_{(2Δ)}μ^{∗}(z_{2}) + C_{4} \frac{μ^{∗}(B(0, δ)}{δ} + o(𝒫μ^{∗}(0, Δ))

≥ C_{5}𝒫μ^{∗}(0, Δ).

This contradicts the fact that each Re e^{−α_{z_{k}}} C_{Δ}μ^{∗}(z_{k}), k = 1, 2 is small in comparison to Im e^{−α_{z_{1}}} C_{Δ}μ^{∗}(z_{1}), which in its turn is larger than C_{6}𝒫μ^{∗}(0, Δ) by Claim 4.4.

This finishes the proof of Theorem 1.4.

Note that instead of the existence of Cμ a.e. on γ we only used the fact that the relation C_{ε}μ(ξ) = o(𝒫μ(ξ, ε)) holds μ_{s} -a.e. on γ. This gives a slightly stronger version of Theorem 1.4 as mentioned in the introduction. The requirement that μ is positive is not crucial as well.

4.2. Radial inner functions. If μ ∈ M(T) we denote by P μ(z) and Qμ(z) its Poisson and conjugate Poisson integrals in the unit disk:

P μ(z) = ∫_{T} \frac{1 − |z|^{2}}{|z − ξ|^{2}} dμ(ξ)  and  Qμ(z) = ∫_{T} \frac{2 Im zξ̄}{|z − ξ|^{2}} dμ(ξ).

Proof of Theorem 1.6. If θ is an inner function in the unit disk, consider the family of Clark measures, i.e. positive singular measures {μ_{α}}_{α∈T} on T uniquely defined by the equation

P μ_{α}(z) = Re \frac{α + θ}{α − θ}

120 P. W. Jones and A. G. Poltoratski

(see, for instance, [2], [8] or [6] for more information on such families of measures). One of the basic properties of μ_{α} is the formula

(15)  ∫ μ_{α}(E) dm(α) = m(E),

where E is any Borel subset of T and m is the normalized Lebesgue measure on T, see [1]. In particular, a set of Lebesgue measure zero on T will have measure zero with respect to μ_{α} for a.e. α ∈ T. Hence if θ is radial near Σ ⊂ T, then for a.e. α ∈ T, μ_{α} -a.e. radius that ends in Σ is mapped by θ into a curve, tangent to a radius. The definition of μ_{α} together with elementary calculations show that this property can be translated into the relation

Qμ_{α}(rξ) = o(P μ_{α}(rξ))

as r → 1− for μ_{α} -a.e. ξ ∈ Σ. Hence, by Theorem 1.1 almost all μ_{α} are pure point on Σ. Another well-known property of the Clark measures says that μ_{α}({ξ}) > 0 if and only if θ(ξ) = α and θ has a non-tangential derivative at ξ (see, for instance, [8]). Utilizing again (15), one can see that θ has angular derivatives a.e. on T.

References

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[2] Aleksandrov, A. B.: Invariant subspaces of the backward shift operator in the space H^{p} (p ∈ (0, 1)). - Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 92, 1979 (Russian).

[3] Mattila, P.: Cauchy singular integrals and rectifiability of measures in the plane. - Adv. in Math. 115, 1995, 1–34.

[4] Mattila, P., and D. Preiss: Rectifiable measures in R^{n} and existence of principal values for singular integrals. - J. London Math. Soc. (2) 52, 1995, 482–496.

[5] Melnikov, M., and J. Verdera: A geometric proof of the L^{2} boundedness of the Cauchy integral on Lipschitz graphs. - Internat. Math. Res. Notices 7, 1995, 325–331.

[6] Poltoratski, A.: On the boundary behavior of pseudocontinuable functions. - St. Petersburg Math. J. 5, 1994, 389–406.

[7] Poltoratski, A.: Images of non-tangential sectors under Cauchy transforms. - J. Anal. Math. (to appear).

[8] Sarason, D.: Sub-Hardy Hilbert Spaces in the Unit Disk. - The University of Arkansas Lecture Notes in the Mathematical Sciences 10, Wiley & Sons, New York, 1994.

[9] Tolsa, X.: Principal values for the Cauchy integral and rectifiability. - Proc. Amer. Math. Soc. 128, 2000, 2111–2119.

Received 17 February 2003