Existence of principal values of some singular integrals on Cantor sets, and Hausdorff dimension

Consider a standard Cantor set in the plane of Hausdorff dimension 1. If the linear density of the associated measure $\mu$ vanishes, then the set of points where the principal value of the Cauchy singular integral of $\mu$ exists has Hausdorff dimension 1. The result is extended to Cantor sets in $\mathbb{R}^d$ of Hausdorff dimension $\alpha$ and Riesz singular integrals of homogeneity $-\alpha$, 0<$\alpha$<d : the set of points where the principal value of the Riesz singular integral of $\mu$ exists has Hausdorff dimension $\alpha$. A martingale associated with the singular integral is introduced to support the proof.


Introduction
Our main result deals with the Cauchy singular integral on Cantor sets in the plane and the proof extends with minor variations to the Riesz transforms in R d .We first proceed to formulate the result for the Cauchy integral and then we take care of the Riesz transforms.
The appropriate Cantor sets for the Cauchy integral are defined as follows.Let (λ n ) ∞ n=1 a sequence of real numbers satisfying 1 4 ≤ λ n ≤ λ < 1 2 .Let Q 0 := [0, 1]×[0, 1] be the unit square.Take the 4 squares contained in Q 0 with sides of length λ 1 parallel to the coordinate axis having a vertex in common with Q 0 (the 4 "corner squares" of side length λ 1 ).Repeat in each of these 4 squares the same procedure with the dilation factor λ 1 replaced by λ 2 to get 16 squares of side length λ 1 λ 2 .Proceeding inductively we obtain at the n-th step 4 n squares Q n j , 1 ≤ j ≤ 4 n , of side length Let µ be the Borel probability measure on K with µ(Q n j ) = 4 −n and denote by a n the linear density at generation n, that is, Set D n = {Q n j : j = 1, . . ., 4 n } and D = ∪ ∞ n=1 D n . 1 We then have Theorem 1.1.If lim n→∞ a n = 0, then the set of points z ∈ K for which the principal value (1.1) lim ε→0 ˆ|w−z|>ε 1 w − z dµw exists has Hausdorff dimension greater than or equal to 1.
If a n = 1 for all n, then K is the famous Garnett-Ivanov Cantor set, which has positive and finite one-dimensional Hausdorff measure but zero analytic capacity.In this case it was noticed in [CPV] that the principal value does not exist at any point of K.If a n → 0, then the Hausdorff dimension of K is greater than or equal to 1 and it has non-sigma finite one-dimensional Hausdorff measure.If in addition n a 2 n < ∞, then the principal value exists µ almost everywhere.So Theorem 1.1 is relevant only when a n → 0 slowly.That the condition n a 2 n < ∞ implies the almost everywhere existence of principal values can be seen in two ways.First, we introduce a martingale (S n ) ∞ n=0 (see (2.1) below) and show that the increments |S n+1 (x) −S n (x)| are bounded by C a n , with the constant C independent of n and x.In Lemma 2.4 we prove that for any point x the principal value exists at x if and only (S n (x)) ∞ n=0 converges.If n a 2 n < ∞, then S n is an L 2 martingale and consequently it converges almost everywhere.Alternatively, the condition n a 2 n < ∞ implies that the Cauchy singular integral operator is bounded in L 2 (µ).In [MV] it was shown in a very general setting that L 2 boundedness together with zero density of the measure yields the almost everywhere existence of principal values.
The main argument in the proof of Theorem 1.1 deals with case where n a 2 n = ∞.It is a variation of a line of reasoning used in other situations (see [DLN] and the references there).We use a stopping time argument to show that (S n (x)) ∞ n=0 converges to 0 in a set of Hausdorff dimension 1 (indeed, given any complex number z 0 the martingale (S n (x)) ∞ n=0 converges to z 0 in a set of Hausdorff dimension 1).We get the dimension 1 conclusion by applying a lemma of Hungerford [H].For the sake of the reader we present a proof of Hungerford's lemma in our context in section 4.
Our proof extends with only technical modifications to cover the case of other odd kernels, for instance, z m z m+1 , m = 1, 2, . . .But one of the ingredients of our method fails for the odd kernel z+z z 2 and we do not know whether Theorem 1.1 holds in this case.The difficulty is indicated at the fifth line after the statement of Lemma 3.1.
In R d our proof works for the Riesz transforms of any homogeneity −α, 0 < α < d.These are the vector valued singular integrals with kernel The appropriate Cantor sets for the α-Riesz transform are those of Hausdorff dimension α.They are constructed by the procedure outlined before in the planar case with dilation factors that satisfy 2 The Cantor set is defined by In section 5 we give some indications on how to adapt the proof for the Cauchy kernel to the Riesz transforms in higher dimensions.
We let diam(A) denote the diameter and dim A the Hausdorff dimension of a set A. We use the notation a b to mean that a ≤ C b for some constant C which may depend on λ and d, and a ∼ b for a b and b a.

Martingales
Let C be the Cauchy kernel, C(x) = 1/x for x ∈ C, x = 0.For each x ∈ K let Q n (x) be the square in D n containing x. Define the truncated Cauchy Integral at generation n as and a martingale (S n (x)) ∞ n=0 by (2.1) T n dµ, x ∈ K.
Remark 2.1.That S n is a martingale is easily checked.The reader will realise that the martingale condition also holds for kernels K(x, y) satisfying the antisymmetry condition K(x, y) = −K(y, x).

We shall prove
Theorem 2.2.If lim n→∞ a n = 0, then the set of points x ∈ K for which (S n (x)) ∞ n=0 converges has Hausdorff dimension greater than or equal to 1.
We first show that the martingale (2.1) has uniformly bounded increments.
Lemma 2.3.There exists a positive constant C = C(λ) such that Thus if n a n converges, (S n (x)) ∞ n=0 converges for all x ∈ K.As mentioned in the introduction, even the weaker condition n a 2 n < ∞ implies that (S n (x)) ∞ n=0 converges for µ almost all x ∈ K. Hence we shall assume that n a 2 n = ∞.Under this assumption one proves in [CPV] that the set where the principal values fail to exist has full µ measure.In Lemma 2.4 below we show that principal values exist if and only if the martingale converges.Hence (S n (x)) ∞ n=0 is not convergent for µ almost all x ∈ K.By a standard result in martingale theory (see, for example, [S, Corollary 6, p.561]) we get The last double integral is a n , where the implicit constant depends on λ here and for the rest of the proof.
To estimate the first summand above we remark that for each By the following lemma Theorem 2.2 is equivalent with Theorem 1.1.
Lemma 2.4.If lim n→∞ a n = 0, then for each x ∈ K the principal value (1.1) exists if and only if the sequence where the constant depends on λ.Compare now a given truncation ´K\B(x,ε) We proceed now to discuss relative martingales.
Then for some constant C, Indeed, we have The first summand above is bounded in absolute value by a constant times a m by the same argument as in the proof of (2.2).
As for (2.2) we have for

The stopping time argument
The proof of Theorem 2.2 is based on a stopping time argument for which we need some preliminary facts.
Given a non-zero complex number z consider the sector σ(z, θ), 0 < θ < π, with vertex at z and aperture θ whose axis is the semi-line emanating from z and passing through 0. That is, w ∈ σ(z, θ) if and only if where •, • denotes the scalar product in the plane.
The octants with vertex 0 are the eight sectors These are the sectors with vertex the origin of amplitude 45 • degrees and having an edge over a coordinate axis.It will be convenient to expand these octants so that they have the same axis and amplitude of 75 • .In other words, we are adding 15 • in each direction.Denote the expanded sectors by σj .The octants with vertex z are the sectors σ j (z) = z + σ j , 1 ≤ j ≤ 8, and the expanded octants σj (z) = z + σj .
We have the following obvious lemma.
Lemma 3.1.Given any sector σ of vertex z and amplitude 120 • there exists an octant with vertex z, say σ j (z) for some index j between 1 and 8, such that σj (z) ⊂ σ.
Consider the symmetries with respect to the coordinate axis and the main diagonal.That is, f 1 (x + iy) = −x + iy, f 2 (x + iy) = x − iy and f 3 (x + iy) = y + ix for x + iy ∈ C. For any j, k = 1, . . ., 8, by composing two such symmetries we obtain a linear mapping f j,k that maps the octant σ j onto the octant σ k .Observe that C(f j (z)) = f j (C(z)) for j = 1, 2, and C(f 3 (z)) = −f 3 (C(z)).It is precisely this last identity that fails for the kernel (z + z)/z 2 .
Let Q ∈ D and let c Q be its center.Define where ε j,k = ±1.We check (3.1) by the general formula for the image (push-forward) ν ♯,f of a measure ν under a Borel map f (see, for example, [M, Theorem 1.19]) from which (3.1) follows.
Assume that we have fixed an octant σ j and that for some square R ∈ D contained in Q we have S Q,R ∈ σ k with k = j.We claim that we can find a square If ε j,k = 1 then the value of the relative martingale at the square To treat the case ε j,k = −1 let us introduce the mapping γ : We shall need the following elementary lemma.
Proof.Let R = |z|, r = |w − z| and let v be the third vertex, in addition to 0 and z, of the equilateral triangle containing w.Under the assumptions of the lemma |w| is maximized when w lies on the side connecting z and v. Assuming that w is on that side, project w on the side connecting 0 and z and apply Pythagoras to obtain because of the assumption r < R/2.
Proof of Theorem 2.2.We assume, as we may, that n a 2 n = ∞.Then for µ almost all x the sequence (S n (x)) ∞ n=0 diverges and (2.3) holds.Let M be a big positive integer to be chosen later.We replace (a n ) ∞ n=0 by the non-increasing sequence b n = C max m≥n a m , where C is as in inequalities (2.2), (2.4) and (2.5), which now read We plan to define a sequence of stopping time conditions.At each step a family of stopping time squares will arise, which is going to be the family F n in Lemma 4.1 (Hungerford's lemma).The first stopping time is special and its goal is to have a family of squares with relatively large |S Q | for each square Q in the family.
The first stopping time condition is We call F 1 the set of stopping time squares of first generation.One may think at this as a process as follows.One takes a point x ∈ K and looks at the squares in D containing x.One examines all those squares, starting at Q 0 and checks whether condition (3.5) is satisfied.If it is not, then one proceeds to the square containing x in the next generation.The process stops when one finds a square Q containing x for which (3.5) holds.Note that the set of x for which the process never stops has vanishing µ measure by (2.3).Hence 2) that it is necessary to descend at least M + 1 generations to find the first stopping time square.
The second stopping time condition is slightly different.Let Q ∈ F 1 .The second stopping time is performed on the relative martingale associated with Q and its condition is By (2.3) and (3.3) the stopping time squares of second generation cover almost all Q.Again, by (3.4) and the fact that S Q,Q = 0 one has to descend through at least M + 1 generations to find a stopping time square of second generation.Hence if R is a stopping time square of second generation and R ∈ D n then n ≥ 2(M + 1).We do not put all stopping time squares of second generation in F 2 (Q).We put a stopping time square of second generation R in That there are many such stopping time squares can be shown as follows.
Let R be a stopping time square of second generation.Let α denote the angle between the vectors S R − S Q and S Q,R .Then by (3.3), But it may occur that S Q,R ∈ σ k , k = j.Applying two symmetries f Q,j,k of Q, or a symmetry of the form γ • f Q,j,k in the worst case, as we discussed before Lemma 3.2, we obtain a stopping time square R ′ of second generation and of the same size as R such that S Q,R ′ ∈ σ j and so S R ′ ∈ σj (S(Q)) ⊂ σ(S Q , 120 • ), as desired.
Therefore, subdividing the stopping time squares of second generation in eight classes, according to the octant to which S Q,R belongs, we get Let us obtain some properties of stopping time squares R in F 2 (Q).Let R be the father of R. Then S Q, R ≤ M b M and so Therefore at least one of the following two inequalities holds: either We can proceed to define inductively F n for n ≥ 3, in a way analogous to what we did to define F 2 from F 1 .Assume that we have defined Given Q ∈ F n−1 we set the n generation stopping time in the relative martingale associated with Q as Note that if R is a stopping time square of generation n, we can take advantage of the symmetries of Q, as before, to find another one, say R ′ , of the same size with the additional property that Given R ∈ F n (Q), we have as before that at least one of the following two inequalities holds: either To conclude that lim m→∞ S m (x) = 0 it is enough to show that lim n→∞ S mn (x) = 0.
Then for Iterating this we have In particular, each ν n is a probability measure and some subsequence of (ν n ) converges weakly to a probability measure ν such that ν It remains to verify (4.1).First of all we have by condition (c) for and by induction, Take n such that ε n+1 ≤ µ(Q) < ε n .We may assume that ν(Q) > 0. Then Q intersects a square R in the family F n+1 .Since by (b) µ(R) ≤ ε n+1 ≤ µ(Q), one has R ⊂ Q.We have, by (4.3) and (4.4), To finish, let x ∈ E and 0 < r ≤ 1.For some n, x belongs to a square We first slightly modify the argument in [CPV] to show that ∞ n=1 a 2 n = ∞ yields divergence a.e. of the martingale.If the martingale converges in a set of positive measure, then also the principal values of the Riesz transform exist in a set E of positive measure, by the analog of Lemma 2.4.By a result of Tolsa [T, Theorem 8.13] we find a set F ⊂ E of positive measure on which the singular Riesz transform operator is bounded on L 2 (µ |F ).In particular, the capacity of F associated with the Riesz kernel is positive and so also that of the Cantor set.The main result of [MT] (see Theorem 1.2, p. 678 and its extension in the last formula in p. 696) states that the α-Riesz capacity of the Cantor set is comparable to ( ∞ n=1 a 2 n ) − 1 2 , so that positive capacity yields a convergent series.We remark that the previous argument uses very strong results, in particular the non-homogeneous T (1)-Theorem of Nazarov, Treil and Volberg, to extract the subset F on which the singular Riesz transform is L 2 (µ |F ) bounded.In [CPV] one resorts to Menger curvature, which is not available for kernels of homogeneity −α with 1 < α < d, and the proof is slightly simpler.It would be desirable to have a direct argument relating the series to the convergence of the martingale.
The part of the stopping time argument of section 3 that does not obviously extend to higher dimensions is related to the sector σ(z, 120 • ).In particular, one should replace the 45 • degrees sectors centred at the origin with one edge on a coordinate axis with other regions.We proceed as follows.Divide R d into 2 d regions (which in R 3 are the usual octants) by requiring that each coordinate has a definite sign.For example, O = {x ∈ R d : x 1 ≥ 0, x 2 ≥ 0, . . .Note that the maximal angle between two vectors lying in a subregion O σ is precisely arccos(d −1/2 ), which approaches 90 • as d → ∞.Given a cone Γ with vertex at the origin and aperture θ, we would like to find a region O σ contained in the cone Γ.This can be done as follows.The axis of the cone is a ray emanating from the origin contained in O σ for some σ.Taking θ = θ(d) < π close enough to π one can achieve O σ ⊂ Γ.Indeed, something stronger can be obtained: there exists a sufficiently small angle γ = γ(d) such that expanding O σ in all directions by at most γ degrees one still remains in the cone Γ.
The planar argument now works with θ in place of 120 • .
One also needs to have enough linear isometries to transport one region O σ into another O σ ′ .Consider the following kinds of linear isometries.Fix a variable x i and take the mapping that leaves the other variables invariant and changes the sign to the x i variable.Given two variables x i and x j with i = j consider the mapping that leaves the other variables invariant and interchanges x i and x j .Finally take the mapping x → −x.Let S the set of such linear isometries.One can easily check that given two regions O σ and O σ ′ one can map one into the other by composing finitely many isometries in S.
All these elements lead to a stopping time that proves Theorem 1.2.

n=1
Q∈Fn Q.To complete the proof we shall show that the hypotheses of Hungerford's Lemma 4.1 are fulfilled and that(3.14)lim m→∞ S m (x) = 0, x ∈ F.For (b) in Hungerford's Lemma 4.1 recall that each stopping time square has descended at least M + 1 generations from the generating square in the previous family.Then one has (b) with ε replaced by 1 4 M and taking M big enough one has1 4 M < ε.Condition (c) with c = 1 8 is (3.11).To prove (3.14), take x ∈ F .For every n = 1, 2, . . ., there is a unique Q n ∈ F n such that x ∈ Q n .Let m n be the unique positive integer satisfying Q n ∈ D mn .Clearly the sequence m n is increasing and m n > M n.Since S Qn = S mn (x) we have by (3.12) and (3.13) that either(3.15)|S mn (x)| ≤ 3(M + 2)b (n−1)M or (3.16) |S mn (x)| ≤ |S m n−1 (x)| − b (n−1)M , n = 1, 2, . ..For m n−1 < m < m n we have by (3.10) (3.17) S m (x) − S m n−1 (x) ≤ (M + 1)b (n−1)M .
one gets the self similar Cantor set of dimension α.If a n → 0 then our Cantor set has Hausdorff dimension ≥ α and non σ finite Hausdorff α-dimensional measure.One has Theorem 1.2.If lim n→∞ a n = 0, then the set of points x ∈ K for which the principal value