Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 30, 2005, 353–360

REALIZING STEP FUNCTIONS AS HARMONIC MEASURE DISTRIBUTIONS OF PLANAR DOMAINS

Marie A. Snipes and Lesley A. Ward

University of Michigan, Department of Mathematics Ann Arbor, MI 48109, U.S.A.; masnipes@umich.edu Harvey Mudd College, Department of Mathematics Claremont, CA 91711, U.S.A.; ward@math.hmc.edu

Abstract. The harmonic measure distribution function of a planar domain relates the geometry of the domain to the behaviour of Brownian motion in the domain. The value of the function h(r) specifies the harmonic measure of the part of the boundary of the domain which lies within any given distance r of a fixed basepoint in the domain. A longterm goal is to realize all suitable functions as distribution functions, by explicit construction of appropriate domains. We show here that increasing step functions can be realized as distribution functions of discs with concentric circular arcs deleted from their interiors.

1. Introduction

We consider a function that captures some of the behaviour of Brownian motion in a planar domain. Let D ⊂ C be a bounded or unbounded domain, and fix a basepoint z_{0} in D. For each r ≥ 0, let h(r) = h_{D,z_{0}}(r) denote the probability that the first hit on the boundary of D by a Brownian traveller starting at z_{0} occurs within distance r of z_{0} . We call h: [0, ∞) → [0, 1] the harmonic measure distribution function of D with respect to z_{0} .

By Kakutani’s result [K] on the connection between Brownian motion and harmonic measure, h(r) is equal to the harmonic measure in D at z_{0} of the part of the boundary ∂D within distance r of z_{0} :

(1)  h(r) = ω(z_{0}, ∂D ∩ \overline{B(z_{0}, r)}, D).

Here, given a subset E of the boundary ∂D, the harmonic function u(z) = ω(z, E, D) is the solution to the Dirichlet problem Δu = 0 on D, with boundary values equal to 1 on E and 0 on the rest of the boundary.

2000 Mathematics Subject Classification: Primary 30C85; Secondary 30C20.

Research of both authors was supported in part by the Research Fund of Harvey Mudd College and that of the first-mentioned author by fellowship support from the National Physical Science Consortium.

The current paper is motivated by these questions: Which functions h(r) can arise as the harmonic measure distribution functions of domains? Given such a function, can one construct a domain which generates it?

It follows from the definition that h(r) is a right-continuous function, increasing from 0 towards 1. In [WW1], it is shown that h(r) does not uniquely determine the domain, and a lower bound for h(r) (in the case of simply connected domains) is established using Beurling’s solution to the Milloux problem [A]. Results on the possible asymptotic behaviours of h(r) for r ≈ dist (z_{0}, ∂D) are proved in [WW1] and [WW2]. A result on the effect of moving the basepoint z_{0} is proved in [WW2]. Some of the questions posed in [WW1] are answered in [BS].

In this paper we identify a useful class of functions which do arise as harmonic measure distribution functions. A preliminary version of this work can be found in [S].

Theorem 1. Every increasing step function on [0, ∞) which increases from 0 to 1 by finitely many steps arises as the harmonic measure distribution function of some planar domain.

A more precise version of Theorem 1 is stated as Theorem 2 in Section 2 below.

In [BH, Problem 6.116], Stephenson posed several questions about the related function

(2)  w(r) = w_{D}(r) = ω(0, ∂D_{r} ∩ {|z| = r}, D_{r}).

Here D is a domain in C containing 0, and D_{r} denotes the connected component of D ∩ \overline{B(0, r)} containing 0. For comparison, setting the basepoint for h(r) at 0, we have

(3)  1 − w(r) = ω(0, ∂D_{r} ∩ B(0, r), D_{r}), while
(4)  h(r) = ω(0, ∂D ∩ \overline{B(0, r)}, D).

The main difference is that 1 − w(r) asks about the location of a Brownian traveller’s first exit from the bounded component D_{r}, while h(r) asks about the location of the first exit from the whole domain D. In particular, the values of 1 − w(r) for small r, 0 < r < R, are completely determined by the geometry of the part of the boundary within distance R of 0, while the values of h(r) for 0 < r < R depend on the geometry of the full boundary of D, at all distances from 0.

Tsuji proved an upper estimate [T, Theorem III.67, p. 112] for a function u_{r}(0) almost identical to Stephenson’s w(r).

In a remark after Theorem 2, we indicate how our results answer the analogues for our h(r) of some of Stephenson’s questions.

We thank Bruce Palka and Henry Krieger for reading earlier versions of this paper, Mario Bonk for pointing out Brouwer’s theorem on invariance of domain, and Byron Walden and Jorge Aarão for many helpful discussions.

2. Increasing step functions arise from circle domains

We prove that all step functions that increase from 0 to 1 through finitely many steps can be generated as the harmonic measure distribution functions of certain domains. In fact there are uncountably many essentially different domains for each such step function.

The simplest such step function takes the value 0 for small r, say for 0 < r < r_{1} , and jumps to 1 at r = r_{1} . This step function is generated by the domain which is a disc of radius r_{1} and has its basepoint at the origin. Further, the same step function arises from any domain whose boundary is an arc (or any subset of positive one-dimensional Hausdorff measure) of the circle of radius r_{1} centred at the basepoint. (Brownian motion in the plane is recurrent, and a Brownian traveller will hit such a boundary with probability one, irrespective of whether the domain is bounded.)

This suggests our definition of circle domains as discs with deleted concentric boundary arcs. See Figure 1.

For convenience, by the radius of a circular arc we mean the radius of the circle on which the arc lies.

Definition 1. A circle domain D with n arcs is a bounded domain, containing the origin, whose boundary consists of finitely many concentric circular arcs A_{1}, …, A_{n}, of radii r_{1}, …, r_{n} respectively, centred at the origin, together with a concentric boundary circle A_{n+1} , of radius r_{n+1} , enclosing the arcs. Here 0 < r_{1} < ⋯ < r_{n} < r_{n+1} . Choose the origin as the basepoint of D. For i = 1, …, n, let x_{i} denote the angle subtended by A_{i} at 0; we call x_{i} the length of the arc A_{i}.

We assume that the midpoints of the boundary arcs all lie on the positive real axis, unless otherwise specified. Of course these midpoints could also be set at any preassigned angles from the positive real axis.

The harmonic measure distribution function h_{D}(r) of a circle domain D is a right-continuous increasing step function, whose finitely many discontinuities may occur only at r_{1}, …, r_{n+1} . The step function has height 0 for 0 < r < r_{1} , and height 1 for r ≥ r_{n+1} . We denote by h_{i} the size of the jump in h_{D}(r) at r_{i}, and by y_{i} the height of the step function for r_{i} ≤ r < r_{i+1} , so that for 1 ≤ i ≤ n,

(5)  h_{i} = h_{D}(ri^{+}) − h_{D}(ri^{−}) = h_{D}(r_{i}) − h_{D}(ri^{−}) = ω(0, A_{i}, D),
(6)  y_{i} = ∑_{1≤j≤i} h_{j} = ω(0, A_{1} ∪ ⋯ ∪ A_{i}, D).

Note that 0 ≤ y_{1} ≤ ⋯ ≤ y_{n} ≤ 1.

We parametrize the space 𝒳 of circle domains D with n arcs located at radii r_{1}, …, r_{n+1} by the n-tuple of angles x_{i} subtended at 0 by the arcs.

Figure 1. Circle domains with two arcs at fixed radii: D, D′ with longer inner arc and shorter outer arc than in D, and D_{0} = D ∩ D′.

Definition 2. Fix an integer n ≥ 1 and numbers r_{i} with 0 < r_{1} < ⋯ < r_{n+1} . Let

(7) 𝒳 = 𝒳 (n, r_{1}, …, r_{n+1}) = {(x_{1}, …, x_{n}) : 0 ≤ x_{i} ≤ 2π, 1 ≤ i ≤ n}

denote the space of circle domains with n arcs placed at radii r_{i}, 1 ≤ i ≤ n, and with boundary circle of radius r_{n+1} .

We also consider the space 𝒴 of all right-continuous step functions, increasing from 0 to 1, with at most n + 1 discontinuities, occurring only at the points r_{1}, …, r_{n+1} . We parametrize 𝒴 by the n-tuple of values y_{i} taken by the step function on the intervals [r_{i}, r_{i+1}), for 1 ≤ i ≤ n.

Definition 3. Fix an integer n ≥ 1 and numbers r_{i} with 0 < r_{1} < ⋯ < r_{n+1} . Let

(8)  𝒴 = 𝒴 (n, r_{1}, …, r_{n+1}) = {(y_{1}, …, y_{n}) : 0 ≤ y_{1} ≤ ⋯ ≤ y_{n} ≤ 1}

denote the space of step functions associated with the radii r_{1}, …, r_{n+1} .

The boundary ∂𝒳 consists of those elements x⃗ ∈ 𝒳 such that x_{i} = 0 or x_{i} = 2π for at least one i ∈ {1, …, n}. In other words, such an x⃗ represents a circle domain whose i^{th} arc A_{i} has length zero (so A_{i} is either the empty set or a single point) or is a full circle. The boundary ∂𝒴 consists of those elements y⃗ ∈ 𝒴 such that at least one of the following holds: y_{1} = 0, or y_{n} = 1, or y_{i} = y_{i+1} for some i ∈ {1, …, n − 1}. In other words, such a y⃗ represents a step function in which at least two adjacent steps have the same height.

As noted above, the step function generated as the harmonic measure distribution function of a circle domain in 𝒳 is an element of 𝒴 . We now prove that all suitable step functions (in other words, all step functions in 𝒴 ) can be generated as harmonic measure distribution functions of circle domains. Here is a more precise version of Theorem 1.

Theorem 2. Let f(r) be a right-continuous step function, increasing from 0 to 1, with its discontinuities at r_{1}, …, r_{n+1} , where 0 < r_{1} < ⋯ < r_{n+1} . Then there exists a circle domain D with n arcs whose harmonic measure distribution function h_{D}(r) is equal to f(r). The radii of the n arcs and of the boundary circle in D are given by r_{1}, …, r_{n} and by r_{n+1} respectively.

Proof. Fix the number n of concentric circular arcs, and the radii r_{1} , …, r_{n+1} . Let F: 𝒳 → 𝒴 be the function which takes the point (x_{1}, …, x_{n}) ∈ 𝒳 representing the circle domain D with arc lengths x_{1}, …, x_{n} to the point (y_{1}, …, y_{n}) ∈ 𝒴 representing the harmonic measure distribution function h_{D}(r) of D. Here h_{D}(r) is an increasing step function with step heights 0, y_{1}, …, y_{n}, 1.

(Numerical evidence [C] suggests that F is nonlinear. Specifically, it suggests that in the case of circle domains with two arcs at fixed radii, F is not the restriction to 𝒳 of a linear transformation of R^{2} .)

The map F is not one-to-one on the boundary ∂𝒳 at points where any of the lengths x_{i} are equal to 2π, for i < n + 1. For once the i^{th} arc A_{i} becomes a full circle, the outer arcs are no longer accessible to a Brownian traveller from the origin. Therefore changing the lengths of any arcs with radii larger than r_{i} does not change h_{D}(r).

We observe that F maps the boundary of 𝒳 into the boundary of 𝒴 , and the interior Int 𝒳 into the interior Int 𝒴 . First, if x⃗ ∈ ∂𝒳 and x_{i} = 0, then the harmonic measure h_{i} of A_{i} is zero, and so y_{i−1} = y_{i} and y⃗ = F (x⃗) is in ∂𝒴 . If x⃗ ∈ ∂𝒳 and x_{i} = 2π, then 1 = ω(0, A_{1} ∪⋯∪A_{i}, D) = y_{i} = y_{i+1} = ⋯ = y_{n}, and so y⃗ = F (x⃗) is in ∂𝒴 . Second, if x⃗ ∈ Int 𝒳 , then each arc A_{j} has length strictly between 0 and 2π, so each arc has a positive (but less than one) probability of being hit by a Brownian traveller from 0, so the jumps h_{j} in the step function y⃗ = F (x⃗) are all greater than zero. Therefore y⃗ ∉ ∂𝒴 , and so F (Int 𝒳 ) ⊂ Int 𝒴 .

(We show below, in part (iii) of the proof, that in fact F maps ∂𝒳 onto ∂𝒴 and Int 𝒳 onto Int 𝒴 .)

We show that F is a homeomorphism between the interiors of 𝒳 and 𝒴 , and that F is a continuous map of 𝒳 onto 𝒴 . The main part of the proof is to show that F is onto.

(i) F: 𝒳 → 𝒴 is continuous. One can easily see that, since harmonic measure is continuous up to the boundary, F is continuous up to and including the boundary.

(ii) F is one-to-one on the interior of 𝒳 . Consider two distinct circle domains D and D′ in 𝒳 . Let S be the collection of arcs in ∂D that are strictly longer than their counterparts (at the same radii) in the collection S′ ⊂ ∂D′. We may assume S is non-empty, exchanging D and D′ if necessary. Let T be the collection of those arcs (if any) in ∂D that are strictly shorter than their counterparts in T′ ⊂ ∂D′. Construct a new domain D^{∗} ∈ 𝒳 using at each radius r_{i} the longer of the boundary arcs A_{i} and A′_{i}. In particular S^{∗} = S and T^{∗} = T′. Then by the monotonicity of harmonic measure,

(9)  ω(0, S, D) ≥ ω(0, S^{∗}, D^{∗}) ≥ ω(0, S′, D′).

Since S is non-empty, the second inequality is actually strict. But then D and D′ cannot have the same step function h(r), since for instance ω(0, S, D) is the sum of the jump heights in h(r) at the values r_{i} corresponding to the arcs in S, and this is strictly greater than the analogous sum for D′.

(iii) F: 𝒳 → 𝒴 is onto. We use induction on the dimension n of 𝒳 (and 𝒴 ).

When D has only one boundary arc, then 𝒳 = [0, 2π], 𝒴 = [0, 1], F (0) = 0, and F (2π) = 1, and so by continuity F is onto.

Suppose that F maps 𝒳 onto 𝒴 when there are n arcs. Consider the case of n + 1 arcs, so 𝒳 = 𝒳 (n + 1, r_{1}, …, r_{n+2}) and 𝒴 = 𝒴 (n + 1, r_{1}, …, r_{n+2}). We show first that F maps the interior Int 𝒳 onto Int 𝒴 , and then that F maps the boundary ∂𝒳 onto the boundary ∂𝒴 .

We have shown that F: 𝒳 → 𝒴 is continuous and injective on the open subset Int 𝒳 of R^{n+1} . Therefore the image F (Int 𝒳 ) is an open subset of R^{n+1} , by Brouwer’s theorem on invariance of domain. (See for example [M, p. 207].) Since F (Int 𝒳 ) ⊂ Int 𝒴 , we have that F (Int 𝒳 ) is an open subset of Int 𝒴 .

Suppose that F (Int 𝒳 ) is not the whole of Int 𝒴 . We claim there is a point y⃗ ∈ ∂F (𝒳 )∩Int 𝒴 . For by assumption, there is a point y⃗_{1} ∈ Int 𝒴 \F (𝒳 ). Take any point y⃗_{2} in F (Int 𝒳 ) ⊂ Int 𝒴 . Since 𝒴 is convex, the line segment γ joining y⃗_{1} to y⃗_{2} lies entirely in Int 𝒴 . Since y⃗_{2} is in F (Int 𝒳 ) ⊂ F (𝒳 ) and y⃗_{1} is not in F (𝒳 ), there is some point y⃗ on γ such that y⃗ ∈ ∂F (𝒳 ). So y⃗ ∈ ∂F (𝒳 )∩Int 𝒴 .

Next, F (𝒳 ) is compact, because 𝒳 is compact and F is continuous. Therefore y⃗ ∈ F (𝒳 ). So there is some x⃗ ∈ 𝒳 such that y⃗ = F (x⃗). Since F (∂𝒳 ) ⊂ ∂𝒴 and y⃗ ∈ Int 𝒴 , x⃗ must lie in Int 𝒳 , and so y⃗ ∈ F (Int 𝒳 ).

As noted above, F (Int 𝒳 ) is an open subset of 𝒴 . So there is a neighbourhood U of y⃗ such that y⃗ ∈ U ⊂ F (Int 𝒳 ). This contradicts y⃗ ∈ ∂F (𝒳 ), since U does not intersect the complement of F (𝒳 ).

Therefore F maps Int 𝒳 onto Int 𝒴 , as required.

Next we consider the boundary. Let y⃗ be a step function in the boundary ∂𝒴 (n + 1, r_{1}, …, r_{n+2}). Denote by h_{j} the height of the j^{th} step, i.e., h_{1} = y_{1} , h_{j} = y_{j} − y_{j−1} for 2 ≤ j ≤ n + 1, and h_{n+2} = 1 − y_{n+1} . Then for some j ∈ {1, …, n + 2}, the jump h_{j} = 0 in y⃗. Therefore, y⃗ is also an element of the new space of step functions 𝒴 ′(n, r′_{1}, …, r′_{n+1}), where {r′_{1}, …, r′_{n+1}} = {r_{1}, …, r_{n+2}} \ {r_{j}}. By the induction hypothesis, there exists a circle domain x⃗ with n arcs in the space 𝒳 ′(n, r′_{1}, …, r′_{n+1}) such that F (x⃗) = y⃗. The circle domain x⃗ is also an element of the boundary ∂𝒳 (n+1, r_{1}, …, r_{n+2}), with x_{j} = 0, where x_{j} is the length of the arc A_{j} at radius r_{j}. Thus F maps ∂𝒳 onto ∂𝒴 .

(iv) F^{−1}: Int 𝒴 → Int 𝒳 is continuous. Since F is a continuous function from the compact set 𝒳 to the compact subset 𝒴 of R^{n}, it maps closed subsets of 𝒳 to closed subsets of 𝒴 .

We have shown that F: Int 𝒳 → Int 𝒴 is a homeomorphism, and that F maps 𝒳 onto 𝒴 . This completes the proof of Theorem 2.

Remark 1. Non-uniqueness: In fact, there are uncountably many circle domains that generate the given step function. The proof above establishes the existence of such a domain with the midpoints of the arcs located on the positive real axis. The same arguments show that there is such a domain with the arc midpoints set at any preassigned angles. One could also construct examples in which several boundary arcs lie on the same circle.

Remark 2. Analogues of Stephenson’s questions: First, the circle domains discussed in Remark 1 give examples, different from those in [WW1], which answer negatively the analogue for h(r) of Stephenson’s question (b) on uniqueness. Namely, there are essentially different domains (differing on sets of positive capacity, in particular) whose harmonic measure distribution functions are all the same step function. So h_{D}, or h_{D} restricted to some interval R_{1} < r < R_{2} , do not uniquely determine the domain.

Next, Theorem 2 is related to the analogue of question (c): Given a function h_{D}, can one ‘reconstruct’ D? However, our theorem refers only to the special case of suitable step functions, and the reconstruction is not completely explicit, in that the lengths of the boundary arcs are not specified.

Finally, one cannot infer the connectivity of D (analogue of question (d)) from h_{D}, at least within the class of step functions. For example, the step function with a single jump from 0 to 1 at r_{1} arises as the harmonic measure distribution function of every domain D whose boundary lies entirely on the circle |z| = r_{1} and has positive harmonic measure (or, equivalently, ∂D ⊂ {|z| = r_{1}} and ∂D has positive length). Here ∂D can have any number of components.

360 Marie A. Snipes and Lesley A. Ward

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Received 9 August 2004