Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 21, 1996, 255–270

HARMONIC MEASURE OF SOME CANTOR TYPE SETS

Athanassios Batakis

Université de Paris-Sud XI, Département de Mathématiques F-91405 Orsay Cedex, France; athanassios.batakis@.math.u-psud.fr

Abstract. We show that for some compact sets K ⊂ R^{2} of Cantor type the harmonic measure is supported by a set whose Hausdorff dimension is strictly smaller than the dimension of K.

1. Introduction

In [MV] Makarov and Volberg show that the Hausdorff dimension of the harmonic measure of the complement of a particular kind of Cantor set is strictly smaller than the dimension of the set, i.e. that there exists a subset of the Cantor set of full harmonic measure but of strictly smaller dimension. Furthermore, in [V1], [V2] Volberg has extended this result to cover a large class of Cantor sets on the real axis. In [C], Carleson has shown that techniques and tools of the ergodic theory could be used to study the harmonic measure of “classical” Cantor sets and [MV], [V1], [V2], [Z] contain some of the results that strongly rely on this idea. Carleson proved that the dimension of the harmonic measure of these Cantor sets in the plane is always strictly smaller than 1. A similar but more general result is proved in [JW], under the assumption of the “capacity density condition”.

The purpose of this work is to prove the inequality between the dimension of the harmonic measure and the dimension of the set for a larger class of Cantor sets in the plane (or the space) without using ergodic theory. We have been motivated by an idea of Bourgain which appeared in [B], and we are also making use of some lemmas and results of the works mentioned above. Further information regarding the Hausdorff dimension and measure of Cantor type sets is provided for example in [Be], [F].

This paper is organized in five sections: first we present the main theorem and introduce some notation. In the second section we prove a number of lemmas and in the third we prove the theorem. Two examples are given in the fourth section, and in the final section we make some remarks and investigate the possibilities of the method.

1991 Mathematics Subject Classification: Primary 31A15; Secondary 28A80.

Let {a_{j}} be a sequence of real numbers such that there exist two constants \underline{A}, \overline{A}, 0 < \underline{A} ≤ \overline{A} < \frac{1}{2} with \overline{A} ≥ a_{i} ≥ \underline{A}, for all i ∈ N. We construct a Cantor set K in the following way: we replace the square [0, 1]^{2} with four equal squares of side-length a_{1} situated in the four corners, and each one of them with four new ones of side-length a_{1}a_{2} and so on; see Figure 1. We denote Ĩ_{i_{1}...i_{n}}^{n}, where i_{j} ∈ {1, 2, 3, 4} for 1 ≤ j ≤ n, the 4^{n} squares of the nth generation constructed in this way with the enumeration shown in the figure and the usual condition that Ĩ_{i_{1}...i_{n}}^{n} is the “father” of the sets Ĩ_{i_{1}...i_{n}i_{n+1}}^{n+1} , i_{n+1} ∈ {1, 2, 3, 4}. It is clear that \overline{A} ≥ diam Ĩ_{i_{1}...i_{n}i}^{n+1}/diam Ĩ_{i_{1}...i_{n}}^{n} = a_{n+1} ≥ \underline{A}, i = 1, …, 4. We will denote by I_{i_{1}...i_{n}}^{n} the intersection of Ĩ_{i_{1}...i_{n}}^{n} with K. We will say that a set F ⊂ R^{2} \ K is of full harmonic measure for the domain R^{2} \ K if ω(F) = 1, where ω is the harmonic measure of R^{2} \ K (see Notation 1.1).

Figure 1.

Theorem 1.0. For a Cantor set K as above there exists a subset F of K of full harmonic measure such that dim(F) < dim(K).

Notation 1.1. If there is a constant c independent of the parameters α, β such that α/c ≤ β ≤ cα, we will write α ∼ β (in what follows the symbols c and C will be used to denote the constants). Suppose that Ω is a domain in R^{2} and that F ⊂ ∂Ω. For x ∈ Ω we denote by ω(x, F, Ω) the harmonic measure of F in Ω evaluated at x. We denote by ω(F, Ω) the harmonic measure of F in Ω evaluated at infinity. If Ω = R^{2} \ K, we will write ω(F) instead of ω(F, Ω). For a square F we denote its side-length by l(F) and, finally, h_{ρ} is the ρ-dimensional Hausdorff measure.

2. Preparatory lemmas

The proof of the theorem will be based on a number of lemmas some of which are already well known.

We first remark that there exists a constant c_{0} > 1 depending only on \underline{A}, \overline{A} such that, if c_{0}Ĩ_{i_{1}...i_{n}}^{n} denotes the square of side-length c_{0} -times the side-length of Ĩ_{i_{1}...i_{n}}^{n} with the same center, then c_{0}Ĩ_{i_{1}...i_{n}}^{n} ∩ K = I_{i_{1}...i_{n}}^{n}. For this c_{0} (not depending on i_{1} … i_{n}) we have the following classical lemma:

Lemma 2.1. There exists a δ > 0 depending neither on n nor on the choice of i_{1} … i_{n} such that for all x ∈ \frac{1}{2}(1 + c_{0})Ĩ_{i_{1}...i_{n}}^{n}

(1)  ω(x, I_{i_{1}...i_{n}}^{n}, R^{2} \ K) > δ.

Remark 2.2. If δ_{n} is the side-length of the square I_{i_{1}...i_{n}}^{n}, the Green’s function G of the square c_{0}Ĩ_{i_{1}...i_{n}}^{n} satisfies

\frac{1}{2π} log \frac{C^{−1}δ_{n}}{|x − y|} ≤ G(x, y) ≤ \frac{1}{2π} log \frac{Cδ_{n}}{|x − y|}

for x, y ∈ \frac{1}{2}(1+c_{0})Ĩ_{i_{1}...i_{n}}^{n}, where the constant C depends only on c_{0} . Furthermore, we have

ω(x, I_{i_{1}...i_{n}}^{n}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ K) = ^{c_{0}Ĩ_{i_{1}...in}^{n}} R_{1}^{I_{i_{1}...in}^{n}} (x) = Gμ(x)

where ^{Ω}R_{1}^{F} is the capacitary potential of the set F in the domain Ω and μ is the capacitary measure of I_{i_{1}...i_{n}}^{n} in c_{0}Ĩ_{i_{1}...i_{n}}^{n}, ∥μ∥ = cap_{c_{0}Ĩ_{i_{1}...in}^{n}} (I_{i_{1}...i_{n}}^{n}).

Proof of Lemma 2.1. We first show that there exists a constant c_{1} > 0 such that for x ∈ \frac{1}{2}(1 + c_{0})Ĩ_{i_{1}...i_{n}}^{n}

(2)  ω(x, I_{i_{1}...i_{n}}^{n}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ K) > c_{1}.

If μ is the probability measure on K charging every square of the nth generation with mass 4^{−n}, let μ_{n} = 4^{n}μ_{|_{I{n \atop i_{1}...in}}} be the restriction of the renormalized measure

μ on the square I_{i_{1}...i_{n}}^{n}.

Let us calculate the potential of μ_{n} for y ∈ I_{i_{1}...i_{n}}^{n}:

Gμ_{n}(y) ≤ 4^{n} ∑_{κ>n} 3 · 4^{−κ} log(\frac{C ∏_{i=1}^{n}a_{i}}{∏_{i=1}^{κ}a_{i}} ) = 3 · 4^{n} ∑_{κ>n} 4^{−κ} log(C ∏_{i=n+1}^{κ} a_{i}^{−1})

= 3 ∑_{κ=1}^{∞} 4^{−κ} log(C ∏_{i=1}^{κ} a_{i+n}^{−1}) ≤ C̃(\underline{A}, \overline{A}) < ∞.

The same reasoning provides a constant c_{2} > 0 such that

\frac{1}{c_{2}} ≤ Gμ_{n}(y) ≤ c_{2} for all y ∈ I_{i_{1}...i_{n}}^{n}.

By the maximum principle we get

(3)  ω(x, I_{i_{1}...i_{n}}^{n}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ K) ∼ Gμ_{n}(x)  for  x ∈ c_{0}Ĩ_{i_{1}...i_{n}}^{n}.

We can easily see that

Gμ_{n}(x) ≥ c_{3}  for  x ∈ ∂\{\frac{1}{2}(1 + c_{0})Ĩ_{i_{1}...i_{n}}^{n}\}.

On the other hand, the harmonic measure is non-decreasing as a function of the domain; hence

ω(x, I_{i_{1}...i_{n}}^{n}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ K) ≤ ω(x, I_{i_{1}...i_{n}}^{n}, R^{2} \ K)

and the lemma is proved.

Lemma 2.3. There exists a δ > 0 not depending on n such that for the squares of the nth generation I_{1...11}^{n} and I_{1...14}^{n} we have

(4)  ω(I_{1...11}^{n}) > (1 + δ)ω(I_{1...14}^{n}).

Remark 2.4. This lemma has been proved in [MV] in the case of standard planar Cantor sets. The proof given below is similar.

We will make repeated use of the following well-known formula (see for instance [Br]): If Ω ⊂ Ω̃ are two domains and if F ⊂ ∂Ω ∩ ∂Ω̃, the harmonic measures of the domains, ω and ω̃, are associated in the following way:

ω(x, F) = ω̃(x, F) − ∫_{∂Ω∩_{Ω}} ω̃(y) ω(x, dy).

Proof of Lemma 2.3. To begin with, let us point out that the symmetry of the set implies

(5)  ω(I_{1...11}^{n}, R^{2} \ I_{1...1}^{n−1}) = ω(I_{1...14}^{n}, R^{2} \ I_{1...1}^{n−1}).

For the same reason, if x lies on the I_{1...1}^{n−1} square’s diagonal separating I_{1...11}^{n} and I_{1...14}^{n}, we have

(6)  ω(x, I_{1...11}^{n}, R^{2} \ I_{1...1}^{n−1}) = ω(x, I_{1...14}^{n}, R^{2} \ I_{1...1}^{n−1}).

Let H^{−} be the half-plane limited by the line containing this diagonal, such that the square I_{1...14}^{n} is contained in H^{−}. Using (2), the monotony of the harmonic measure, and Harnack’s inequalities, one can verify the existence of a constant c_{4} > 0 such that

(7)  ω(x, I_{1...14}^{n}, H^{−} \ I_{1...1}^{n−1}) ≥ c_{4}  for all x ∈ I_{1...4}^{n−1}.

By the maximum principle and (6) we obtain

ω(x, I_{1...14}^{n}, H^{−} \ I_{1...1}^{n−1}) = ω(x, I_{1...14}^{n}, R^{2} \ I_{1...1}^{n−1}) − ω(x, I_{1...11}^{n}, R^{2} \ I_{1...1}^{n−1})

for all x ∈ H^{−}. Combining with (7),

(8) ω(x, I_{1...14}^{n}, R^{2} \ I_{1...1}^{n−1}) − ω(x, I_{1...11}^{n}, R^{2} \ I_{1...1}^{n−1}) ≥ c_{4}  for all x ∈ I_{1...4}^{n−1},

and (5), (8) imply that

(9)

ω(I_{1...11}^{n}, R^{2} \ K) − ω(I_{1...14}^{n}, R^{2} \ K) =
= ∫_{K\I_{1...1}^{n−1}}(ω(y, I_{1...14}^{n}, R^{2} \ I_{1...1}^{n−1})
− ω(y, I_{1...11}^{n}, R^{2} \ I_{1...1}^{n−1}))ω(dy, R^{2} \ K)
≥ c_{4}ω(I_{1...4}^{n−1}, R^{2} \ K).

Finally, using Harnack’s principle and (7) we obtain a constant c_{5} > 0 not depending on n, verifying

ω(I_{1...4}^{n−1}) ≥ c_{5}ω(I_{1...14}^{n}).

Hence, (9) turns into

ω(I_{1...11}^{n}) − ω(I_{1...14}^{n}) ≥ c_{4}c_{5}ω(I_{1...14}^{n})

and the lemma is proved.

Lemma 2.5 ([C], [MV]). Let Ω be a domain containing ∞ and let A_{1} ⊂ B_{1} ⊂ A_{2} ⊂ B_{2} ⊂ ⋯ ⊂ A_{n} ⊂ B_{n} ⊂ Ω be conformal discs such that the annuli B_{i} \ A_{i} are contained in Ω, for 1 ≤ i ≤ n. If the modules of the annuli are uniformly bounded away from zero and if ∞ ∈ Ω \ B_{n}, then, for all pairs of positive harmonic functions u, v vanishing on ∂Ω \ A_{1} and for all x ∈ Ω \ B_{n}, we have

(10)  |\frac{u(x)}{v(x)} : \frac{u(∞)}{v(∞)} − 1| ≤ Kq^{n},

where q < 1 and K are two constants that depend only on the lower bound of the modules of the annuli.

260 Athanassios Batakis

Lemma 2.6. There exists an N_{0} = N_{0}(δ, \underline{A}, \overline{A}) large enough such that for all n ∈ N and all squares I_{i_{1}...i_{n}}^{n}

(11)  ω(I_{i_{1}...i_{n}11...1}^{n+N_{0}}) > (1 + \frac{1}{2}δ)ω(I_{i_{1}...i_{n}11...4}^{n+N_{0}}),

where δ is the positive constant defined in Lemma 2.3.

Proof of Lemma 2.6. Let us first show the following estimate for x ∈ \frac{1}{2}(1 + c_{0})Ĩ_{i_{1}...i_{n}}^{n}:

(12a)  ω(x, I_{i_{1}...i_{n}11...1}^{n+N_{0}}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n}) ∼ ω(x, I_{i_{1}...i_{n}11...1}^{n+N_{0}}, R^{2} \ K).

For N ∈ N we choose x such that ω(x, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2}\K) = sup\{ω(y, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2}\K) : y ∈ ∂\{\frac{1}{2}(1+c_{0})Ĩ_{i_{1}...i_{n}}^{n}\}\}. Then,

ω(x,I_{i_{1}...i_{n}11...1}^{n+N}, R^{2} \ K) ≥ ω(x, I_{i_{1}...i_{n}11...1}^{n+N}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n})
≥ ω(x, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2} \ K)
− ∫_{∂{c_{0}Ĩ_{i_{1}...in}^{n} }} ω(y, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2} \ K)ω(x, dy, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n})
≥ ω(x, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2} \ K)
− (1 − ω(x, I_{i_{1}...i_{n}}^{n}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n}))ω(x, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2} \ K)
≥ c_{1}ω(x, I_{i_{1}...i_{n}11...1}^{n+N}, R^{2} \ K)

because of (2).

Then (12a) follows on our using again Harnack’s inequalities.

We have, of course, the same estimate for I_{i_{1}...i_{n}11...4}^{n+N_{0}} :

(12b)  ω(x, I_{i_{1}...i_{n}11...4}^{n+N_{0}}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n}) ∼ ω(x, I_{i_{1}...i_{n}11...4}^{n+N_{0}}, R^{2} \ K).

To simplify the notation in what follows we will write ω_{1}(x)(ω̃_{1}(x)) and ω_{4}(x)(ω̃_{4}(x)) instead of

ω(x, I_{i_{1}...i_{n}11...1}^{n+N_{0}}, R^{2} \ K)(ω(x, I_{i_{1}...i_{n}11...1}^{n+N_{0}}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n}))

and

ω(x, I_{i_{1}...i_{n}11...4}^{n+N_{0}}, R^{2} \ K)(ω(x, I_{i_{1}...i_{n}11...4}^{n+N_{0}}, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n})),

respectively.

By the relation (10) in Lemma 2.5, for z ∈ ∂\{\frac{1}{2}(1 + c_{0})Ĩ_{i_{1}...i_{n}}^{n}\}
(13)  \frac{ω_{1}(∞)}{ω_{4}(∞)} ∼_{q^{N_{0}}} \frac{ω_{1}(z)}{ω_{4}(z)} ∼_{q^{N_{0}}} \frac{ω_{1}(y)}{ω_{4}(y)}  for all y ∉ K ∪ c_{0}Ĩ_{i_{1}...i_{n}}^{n}.

From (13) it follows that

(14) \begin{aligned}| & \frac{ω̃_{1}(z)}{ω̃_{4}(z)} : \frac{ω_{1}(∞)}{ω_{4}(∞)} − 1| ∼ |\frac{ω̃_{1}(z)}{ω̃_{4}(z)} : \frac{ω_{1}(z)}{ω_{4}(z)} − 1| = \frac{ω_{4}(z)}{ω̃_{4}(z)}|\frac{ω̃_{1}(z)}{ω_{1}(z)} − \frac{ω̃_{4}(z)}{ω_{4}(z)}| \\ & = \frac{ω_{4}(z)}{ω̃_{4}(z)} ∫_{∂\{c_{0}Ĩ_{i_{1}...in}^{n} \}}|\frac{ω_{4}(y)}{ω_{4}(z)} − \frac{ω_{1}(y)}{ω_{1}(z)}|ω(z, dy, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n}) \\ & ≤ \frac{1}{c_{1}} ∫_{∂\{c_{0}Ĩ_{i_{1}...in}^{n} \}} \frac{ω_{4}(y)}{ω_{4}(z)}|\frac{ω_{1}(y)}{ω_{1}(z)} : \frac{ω_{4}(y)}{ω_{4}(z)} − 1|ω(z, dy, c_{0}Ĩ_{i_{1}...i_{n}}^{n} \ I_{i_{1}...i_{n}}^{n}) \\ & ≤ Cq^{N_{0}}. \end{aligned}

If we take i_{1} = ⋯ = i_{n} = 1, Lemma 2.3 implies ω_{1}(∞)/ω_{4}(∞) > 1+δ. Then (14) shows that there exists an N_{0} large enough such that ω̃_{1}(z)/ω̃_{4}(z) > 1 + \frac{3}{4}δ. On the other hand, ω̃_{1}(z)/ω̃_{4}(z) does not depend on the choice of i_{1}, …, i_{n}. It follows that ω_{1}(∞)/ω_{4}(∞) > 1 + \frac{1}{2}δ for all the possible choices of i_{1}, …, i_{n}.

Lemma 2.7. There exists an N_{1} ∈ N independent of n and of i_{1} … i_{n} such that for all the squares I_{i_{1}...i_{n}}^{n} there is a square J_{m} = I_{i_{1}...i_{n}...i_{n+N_{1}}}^{n+N_{1}} ⊂ I_{i_{1}...i_{n}}^{n} of the (n + N_{1})th generation such that

ω(J_{m}) < \frac{1}{4} \frac{ω(I_{i_{1}...i_{n}}^{n})}{4^{N_{1}}}.

(In fact, \frac{1}{4} could be replaced with any constant ε > 0).

Proof of Lemma 2.7. Choose a square I_{i_{1}...i_{n}}^{n}. According to the preceding lemma, there exists an α < 1 independent of the choice of i_{1}, …, i_{n} and a J_{1} = I_{i_{1}...i_{n}...i_{n+N_{0}}}^{n+N_{0}} such that

ω(J_{1}) < α\frac{ω(I_{i_{1}...i_{n}}^{n})}{4^{N_{0}}}.

Similarly there exists a J_{2} = I_{i_{1}...i_{n}...i_{n+2N_{0}}}^{n+2N_{0}} ⊂ J_{1} with

ω(J_{2}) < α\frac{ω(J_{1})}{4^{N_{0}}} < α^{2} \frac{ω(I_{i_{1}...i_{n}}^{n})}{4^{2N_{0}}},

and after k steps we obtain a square J_{k} verifying

ω(J_{k}) < α^{k} \frac{ω(I_{i_{1}...i_{n}}^{n})}{4^{kN_{0}}}.

To finish the proof, take k = m such that α^{k} < \frac{1}{4} and let N_{1} = kN_{0} .

3. Proof of Theorem 1.0

The theory of martingales provides a well-known technique for proving the inequality between the dimensions of the two measures by using Lemma 2.7. However, we propose here a different path which does not involve probabilistic tools and is inspired by [B].

We introduce some more notation. For n ∈ N we will denote by ℰ_{n} the collection of squares {I_{i_{1}...i_{n}}^{n} : i_{j} = 1, …, 4, j = 1, …, n}, and for I ∈ ℰ_{n}, ℰ_{n+s}(I) will represent those squares J ∈ ℰ_{n+s} that are contained in I.

It can be shown (see for instance Lemma 2 of [Be]) that if ρ is the Hausdorff dimension of K, then

ρ = sup{s > 0 : lim inf_{n→∞} 4^{n} ∏_{i=1}^{n} a_{i}^{s} = ∞} = inf{s > 0 : lim inf_{n→∞} 4^{n} ∏_{i=1}^{n} a_{i}^{s} = 0}

simply because in order to obtain the Hausdorff dimension of the Cantor set K it suffices to consider coverings of K with the squares of construction Ĩ_{i_{1}...i_{n}}^{n}. However, the ρ-Hausdorff measure of the Cantor sets considered here could be infinite.

It easily follows that for ε > 0 there exists a strictly increasing sequence of integers {n_{j}}_{j=1}^{∞} such that

(15)  4^{n_{j}} ∏_{i=1}^{n_{j}} a_{i}^{ρ+ε} > 4^{n_{j+1}} ∏_{i=1}^{n_{j+1}} a_{i}^{ρ+ε}.

We will also assume that n_{j+1} − n_{j} > 2N_{1} .

Lemma 2.8. There exists a β < 1 such that the following inequality holds for ε > 0 and I ∈ ℰ_{n_{j}}:

(16)  ∑_{J∈ℰ_{n_{j+1}} (I)} ω(J)^{1/2}l(J)^{(ρ+ε)/2} ≤ β^{n_{j+1}−n_{j}}ω(I)^{1/2}l(I)^{(ρ+ε)/2};

where n_{j} is the sequence corresponding to ε given by (15).

Proof of Lemma 2.8. Let us start by showing that there is a β̃ such that for I ∈ ℰ_{n}

(17)  ∑_{J∈ℰ_{n+N_{1}} (I)} ω(J)^{1/2}(\frac{1}{4})^{(n+N_{1})/2} ≤ β̃ω(I)^{1/2}(\frac{1}{4})^{n/2}.

Take J_{m} ∈ ℰ_{n+N_{1}}(I) to be the square provided by Lemma 2.7, i.e. a square such that ω(J_{m}) < \frac{1}{4}ω(I) · 4^{N_{1}} . We have

ω(J_{m})^{1/2}(\frac{1}{4})^{(n+N_{1})/2} ≤ \frac{1}{2} \frac{1}{4^{N_{1}}} (\frac{1}{4})^{n/2}ω(I)^{1/2}

∑_{J∈ℰ_{n+N_{1}} (I), J≠J_{m}} ω(J)^{1/2}(\frac{1}{4})^{(n+N_{1})/2} ≤ ω(I)^{1/2}(4^{N_{1}} − 1)^{1/2}(\frac{1}{4})^{(n+N_{1})/2}

by the Cauchy–Schwarz inequalities. Summing up we get

(18) ∑_{J∈ℰ_{n+N_{1}} (I)} ω(J)^{1/2}(\frac{1}{4})^{(n+N_{1})/2} ≤ ω(I)^{1/2}(\frac{1}{4})^{n/2}(\frac{1}{2} \frac{1}{4^{N_{1}}} + (\frac{4^{N_{1}} − 1}{4^{N_{1}}})^{\frac{1}{2}} )

and we may let

β̃ = \frac{1}{2} \frac{1}{4^{N_{1}}} + (\frac{4^{N_{1}} − 1}{4^{N_{1}}})^{1/2} < 1.

Choose ε > 0 and let {n_{j}} be a corresponding sequence given by (15). Then by (17)

\begin{aligned}∑_{J∈ℰ_{n_{j+1}} (I)} ω(J)^{1/2}l(J)^{(ρ+ε)/2} & ≤ 4^{n_{j}/2}β̃ (∏_{i=1}^{n_{j}} a_{i}^{(ρ+ε)/2})× \\ & × ∑_{J∈ℰ_{n_{j+1}−N_{1}} (I)} ω(J)^{1/2}(\frac{1}{4^{n_{j+1}−N_{1}}} )^{1/2}. \end{aligned}

We repeat the procedure and apply the Cauchy–Schwarz inequalities. We then get

∑_{J∈ℰ_{n_{j+1}} (I)} ω(J)^{1/2}l(J)^{(ρ+ε)/2} ≤ 4^{n_{j}/2}β̃^{(n_{j+1}−n_{j})/2N_{1}}(∏_{i=1}^{n_{j}} a_{i}^{(ρ+ε)/2})(\frac{ω(I)}{4^{n_{j}}})^{1/2}.

The existence of β is now obvious. For instance, one may take β = β̃^{1/2N_{1}} .

Proof of Theorem 1.0. Let ℒ_{j} = {J ∈ ℰ_{n_{j}} | ω(J) > l(J)^{ρ−ε}} and ℒ′ _{j} = ℰ_{n_{j}} \ ℒ_{j}, where ε > 0 is to be chosen later, and let {n_{j}} be a sequence corresponding to ε as above. It is clear that

(19)  ∑_{J∈ℒ_{j}} l(J)^{ρ−ε} < ∑_{J∈ℒ_{j}} ω(J) ≤ 1.

But, we can also estimate

\begin{aligned}∑_{J∉ℒ_{j}} ω(J) & = ∑_{J∉ℒ_{j}} ω(J)^{1/2}ω(J)^{1/2} ≤ ∑_{J∈ℰ_{n_{j}}} ω(J)^{1/2}l(J)^{((ρ+ε)/2)−ε} \\ & ≤ ∏_{i=1}^{n_{j}} a_{i}^{−ε}β^{n_{j}−n_{j−1}} ∑_{J∈ℰ_{n_{j−1}}} ω(J)^{1/2}l(J)^{(ρ+ε)/2} \end{aligned}

because of (16). By iterating the procedure we get

∑_{J ∉ ℒ_{j}} ω(J) ≤ β^{n_{j}} ∏_{i=1}^{n_{j}} a_{i}^{−ε} ≤ β^{n_{j}} \underline{A}^{−εn_{j}}.

Let ε > 0 be such that β < \underline{A}^{ε}. It is then immediate from the above that

(20)  lim_{j→∞} ∑_{J ∉ ℒ_{j}} ω(J) = 0.

Clearly, (19) and (20) allow us to construct a subset of K of Hausdorff dimension < ρ but of full harmonic measure, and the proof is completed.

4. A counterexample

We state the following simple result:

Proposition 4.0. For a Cantor set K as described in the introduction, the harmonic measure ω of its complement is “monodimensional”, i.e. there is a dimension σ (the dimension of the harmonic measure) such that there exists a subset F ⊂ K of Hausdorff dimension σ with ω(F) = 1, and ω(F′) = 0 for every set F′ ⊂ K of dimension smaller than σ.

The proof given below applies to all self-similar Cantor sets and therefore the proposition remains valid even for “general” Cantor sets.

Proof of Proposition 4.0. Suppose that the proposition is false. Then there is a dimension σ and a real number 0 < α < 1 such that

sup{ω(F) : F ⊂ K, dim(F) ≤ σ} = α

or, equivalently, there exist a dimension σ and a γ > 0 such that

sup\{ inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, F, c_{0}[0, 1]^{2} \ K) : F compact, F ⊂ K, dim(F) ≤ σ\} = γ

and

γ < inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, K, c_{0}[0, 1]^{2} \ K),

where c_{0} is the constant defined in Section 2.

For every real number τ, 0 < τ < 1 there is a compact set F ⊂ K of Hausdorff dimension σ with

τγ < inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, F, c_{0}[0, 1]^{2} \ K) < \frac{1}{τ} γ.

Moreover, we can find a covering ℱ = {I_{j}}_{j∈J} of F with squares I_{j} of the same generation of the construction of K, satisfying

τγ < inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, ⋃_{I∈ℱ} I, c_{0}[0, 1]^{2} \ K) < \frac{1}{τ} γ.

There exists at least one I_{j} ∈ ℱ with the following property:

“There is a compact set F_{j} ⊂ I_{j} ∩ K of Hausdorff dimension σ with

inf_{x∈ \frac{1}{2} (1+c_{0})I_{j}} ω(x, F_{j}, c_{0}I_{j} \ K) > cτγ inf_{x∈ \frac{1}{2} (1+c_{0})I_{j}} ω(x, K, c_{0}I_{j} \ K),

where c is a Harnack constant depending only on K.”

We then say that F_{j} is a γ -subset of I_{j}.

To prove this claim, we first show the existence of at least one I_{j} satisfying

inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, F_{j}, c_{0}[0, 1]^{2} \ K) > τγ inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, K ∩ I_{j}, c_{0}[0, 1]^{2} \ K)

and then proceed with standard arguments, using the Brelot formula.

Recall that all squares of the same generation of the construction of K are identical, and therefore the preceding property is valid for any square of the generation of I_{j}, i.e. every such square has a γ -subset. Let ℱ̃ be the collection of all squares of the same generation with I_{j} that do not belong to ℱ, and let S be the union of F with the γ -subsets of the squares in ℱ̃. Thus S is a subset of K of Hausdorff dimension σ. By the above it is clear that

inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, S, c_{0}[0, 1]^{2} \ K)
> τγ + cτγ(  inf_{x∈ \frac{1}{2} (1+c_{0})[0,1]^{2}} ω(x, K, c_{0}[0, 1]^{2} \ K) − \frac{1}{τ} γ),

which is greater than γ if τ is close enough to one; since γ is taken to be the maximal value of harmonic measure for subsets of K of Hausdorff dimension equal to σ, we have reached a contradiction. The proof is now complete.

We will now construct a Cantor set K′ as in the introduction, except that here we replace a square J of the k th generation, k ≥ 1, with four equal squares, J_{1}, …, J_{4} , whose size depends not only on the generation k but also on the square J; we still require \underline{A} ≤ l(J_{i})/l(J) ≤ \overline{A} with 0 < \underline{A} ≤ \overline{A} < 1/2. We will show that for an appropriate choice of the sizes of the squares the Hausdorff dimension of K′ will be equal to the dimension of its harmonic measure. The idea of the construction was suggested to us by a remark of A. Ancona. Let us begin with the standard planar Cantor set K_{1/4} of dimension 1, i.e. a Cantor set as defined in the introduction with \underline{A} = \overline{A} = 1/4. Let D be the dimension of its harmonic measure; if F is a compact subset of K_{1/4} such that ω(F) > 1/2, it follows from Proposition 4.0 that its dimension will be at least D. We may therefore find such a subset F of K_{1/4} of Hausdorff dimension D. We then construct the desired Cantor set in the following way: In each generation we replace every square J that does not intersect F with squares of size 4^{−M} times the size of J, where M is a fixed integer with M > 1/D, and every square J′ that intersects F is replaced with four squares of size 1/4 times the size of J′. Let K′ be the Cantor set constructed in this way. Observe that K′ ⊂ K_{1/4} by construction and that dim K′ = D because of the choice of M. It is clear (by the monotonicity of the harmonic measure as a function of the domain) that the dimension of the harmonic measure of K′ is also D, and the construction is complete.

We should remark here that the preceding process gives us Cantor sets whose Hausdorff dimension is equal to their harmonic measure dimension for every possible value of the dimension of the harmonic measure of a Cantor set as described in the introduction. Also, a result of [JW] implies that we cannot have dim ω = 1 for Cantor sets of this type. It is therefore natural to ask if we can have dimensions arbitrarily close to one. The following proposition answers the question.

Proposition 4.1. For the self-similar Cantor set K_{δ}, 0 < δ ≤ \frac{1}{4} , as defined in the introduction with \frac{1}{2} − δ = \underline{A} and \underline{A} = \overline{A}, the dimension of the harmonic measure dim ω is greater than 1 − Cδ for some constant C > 0.

This proposition as well as the proof given below is due to Professor A. Ancona (compare with [MV, pp. 15–22, 28]).

Proof of Proposition 4.1. We will need some more notation. Let K_{n} be the nth approximation of K_{δ} by squares of the nth generation, let g_{n} be the Green function of the complement of K_{n} and 𝒞_{n} its critical points. We shall rely on the formula

dim ω = 1 − \frac{lim_{n→∞} n^{−1} ∑_{𝒞_{n}} g_{n}(c)}{χ_{μ}},

where χ_{μ} = log(2/(1 − 2δ)) and the critical points in the sum are counted with their multiplicity.

This formula is a simple variant of the Carleson formula given in [MV, p. 15] (see also [C]); here we consider the sum over the critical points of g_{n} instead of those of the Green function of the complement of K_{δ}.

It remains to be proved that the limit in the previous formula is O(δ) as δ tends to 0.

We extend g_{n} on K_{n} by the value 0 and consider the critical domains of g_{n}, i.e. any region U which is a connected component of {g_{n} < β} for some β > 0 and with a critical point c ∈ ∂U. Let 𝒰 be the collection of all critical domains.

Note that if U ∈ 𝒰 and if τ = max{g_{n}(z) : z ∈ 𝒞_{n} ∩ U}, the number of critical domains U′ ⊂ U associated with τ is exactly equal to the number of critical points z ∈ U with g_{n}(z) = τ (counted with their multiplicity) plus one.

To each U ∈ 𝒰 we attach a square I = I_{U} of some stage k of the construction of the Cantor set, k ≤ n, with the following property:

(PI) We have I ∩K_{n} ⊂ U, and if Ĩ denotes the “father” of I, there exists a square I′ ⊂ Ĩ of the k th generation such that I′ ∩ K_{n} ∩ U = ∅ and I and I′ lie on the same side of Ĩ.

The existence of I_{U} is easily checked. For instance, one may take for Ĩ a minimal square such that Ĩ∩ K_{n} ∩ U^{c} ≠ ∅ and Ĩ∩ K_{n} ∩ U ≠ ∅, and then easily verify the existence of I ⊂ Ĩ with the property (PI).

We now proceed with the following simple algorithm which leads to the construction of a subcollection 𝒰_{0} ⊂ 𝒰 (the “nice” domains) and to the choice of some square c_{U} ⊂ I_{U} of the nth generation (the last generation for K_{n}), for every U ∈ 𝒰_{0} .

Each domain U ∈ 𝒰 which is maximal is “nice”, and we choose the square c_{U} arbitrarily in I_{U}. For U ∈ 𝒰 , if the construction has been achieved for all U′ ⊃ U, U ≠ U′, we decide that U ∉ 𝒰_{0} if there exists c_{U′} ⊂ U for some U′ ⊃ U, U ≠ U′. Otherwise we say that U ∈ 𝒰_{0} and we associate with it some square c_{U} ⊂ I_{U} of the nth generation.

At the end of the procedure every critical domain U ∈ 𝒰 contains exactly one c_{U′} for some U′, U ⊂ U′, and for U, U′ ∈ 𝒰_{0} we have I_{U} = I_{U′} if and only if U = U′.

Hence we have

\frac{1}{n} ∑_{𝒞_{n}}g_{n}(c) = \frac{1}{n}( ∑_{U∈𝒰_{0}} g_{U} − g_{max}),

where g_{U} is the value of g_{n} on ∂U and g_{max} is the maximal critical value of g_{n}.

If U ∈ 𝒰_{0} , let I = I_{U} be the square attached to it, Ĩ its “father”, and I′ as in (PI) (see Figure 2). There are at least s = [1/4δ] parallel segments, l_{1}, …, l_{s}, joining points of K_{n} ∩ I with points of K_{n} ∩ I′, the distance between any two segments being ≥ δ l(I). Necessarily, ∂U cuts through all these segments and therefore sup{g_{n}(t) : t ∈ l_{i}} ≥ g_{U} , 1 ≤ i ≤ s.

Figure 2.

For every l_{i}, i = 1 …, s, let z_{i}^{1}, z_{i}^{2} be the endpoints of l_{i}, z_{i}^{1} ∈ K_{n} ∩ I, z_{i}^{2} ∈ K_{n} ∩I′. It is clear that the set B(z_{i}^{1}, δl(I)/2)∩K_{n} ∩I has capacity ≥ C_{0} > 0 in the domain B(z^{1}, δl(I)), with C_{0} independent of δ ∈ [\frac{1}{4}, \frac{1}{2}). By standard arguments it follows that

g_{n}(t) ≤ Cω(B(z_{i}^{1}, δl(I)/2), R^{2} \ K_{n})

on the segment l_{i} with a constant C indepent of δ.

The above finally yields

\frac{1}{4δ} g_{U} ≤ ∑_{i} sup\{g_{n}(t) : t ∈ l_{i}\} ≤ C ω(I, R^{2} \ K_{n}).

Summing up we find

\frac{1}{n} ∑_{𝒞_{n}}g_{n}(c) ≤ \frac{1}{n} ∑_{U∈𝒰_{0}} g_{U} ≤ Cδ \frac{1}{n} ∑_{I∈ℱ_{n}} ω(I, R^{2} \ K_{n}) ≤ Cδ,

where ℱ_{n} is the collection of all squares of some stage k, k ≤ n, of the construction of K_{δ}. The proof of the proposition is complete.

5. Conclusion – Further remarks

It is clear that the method we developed in Sections 2 and 3 applies not only to the Cantor sets described above but also to other Cantor sets, for example those indicated by Figure 3. The proof can also be applied to some Cantor sets in higher dimensions.

For a general Cantor set K ⊂ R^{d}, a sufficient condition to conclude that dim(ω) < ρ = dim(K) is the following: if I_{i_{1}...i_{n}}^{n} is a square of the nth generation and if I_{i_{1}...1}^{n+1}, …, I_{i_{1}...s}^{n+1} are the squares of the next generation contained in I_{i_{1}...i_{n}}^{n}, there exist 0 < α < 1, 1 ≤ τ ≤ s, and constants a_{j}^{n} > 0 such that

(∗)  ω(I_{i_{1}...i_{n}τ}^{n+1} ) < α\frac{diam(I_{i_{1}...i_{n}τ}^{n+1} )^{ρ}}{∑_{j=1}^{s} diam(I_{i_{1}...i_{n}j}^{n+1})^{ρ}} ω(I_{i_{1}...i_{n}}^{n}),

and

diam(I_{i_{1}...i_{n}j}^{n+1}) = a_{j}^{n}diam(I_{i_{1}...i_{n}}^{n}), 2^{−d} < \underline{A} ≤ a_{j}^{n} ≤ \overline{A} < 1 for all j ∈ \{1, …, s\},

where a_{j}^{n} depends only on j, n but not on the square I_{i_{1}...i_{n}}^{n} and \underline{A}, \overline{A} are two constants not depending on n. Lemmas 2.7 and 2.8 can both be applied to prove a formula similar to (16), and the proof of the theorem may be completed in the same way.

Figures 3a and 3b

In general (∗) seems hard to check; however, under certain assumptions of symmetry on the Cantor set K one may verify it by proving some lemmas similar to those presented above. Even though the method presented here seems to be rather general, we have not been able to get rid of these assumptions of symmetry, and the proof of Lemma 2.3 strongly depends on them.

270 Athanassios Batakis

Added in proof. It is perhaps interesting to point out that the result of Theorem 1.0 can also be proved under some weaker assumptions on the size of the squares:

For a sequence {a_{n}}_{n∈N} as in the introduction we construct a Cantor set in a similar way. We allow the squares of the nth generation I_{i_{1}···i_{n}}^{n} to have sidelengths l_{i_{1}···i_{n}} not necessarily equal but require that a_{n}(1 − ε)l_{i_{1}···i_{n−1}} ≤ l_{i_{1}···i_{n}} ≤ a_{n}(1 + ε)l_{i_{1}···i_{n−1}} , where ε > 0 and l_{i_{1}···i_{n−1}} is the sidelength of the “father” of I_{i_{1}···i_{n}}^{n}. For such a Cantor set the dimension of the harmonic measure is smaller than the dimension of the set, provided ε is small enough (the proof is slightly different).

Acknowledgement. The author would like to express his deep gratitude to Professor Alano Ancona for all the help and the encouragement he kindly provided. He is also very grateful to the referee for requesting the counterexample given in Section 4. Thanks are also due to Y. Heurteaux for his useful criticism.

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Received 28 February 1995