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Harmonic Measure Chapter 6 Harmonic Measure Prologue: It is unfortunate that most graduate com- plex analysis courses do not treat harmonic measure. Indeed, most complex analysis texts do not cover the topic. The term “harmonic measure” was introduced by R. Nevanlinna in the 1920s. But the ideas were antici- pated in work of the Riesz brothers and others. It is still studied intensively today. Harmonic measure is a natural outgrowth of the study of the Dirichlet problem. It is a decisive tool for mea- suring the growth of functions, and to seeing how they bend. It has been used in such diverse applications as the corona problem and in mapping problems. Harmonic measure has particularly interesting applications in prob- ability theory—especially in the consideration of diffus- tions and Brownian motion. It requires a little sophistication to appreciate har- monic measure. And calculating the measure—working through the examples—requires some elbow grease. But the effort is worth it and the results are substantial. This chapter gives a basic introduction to the idea of harmonic measure. 169 79344-3_Printer Proof.pdf 183 “CH06” — 2015/10/26 — 17:20 — page 169 — #1 11/6/2015 1:52:49 PM 170 CHAPTER 6. HARMONIC MEASURE 6.1 The Idea of Harmonic Measure Capsule: Harmonic measure is a way of manipulating the data that comes from the Dirichlet problem. It is a way of measuring the geom- etry of a domain, and can be used to control the growth of harmonic functions. Let Ω ⊆ C be a domain with boundary consisting of finitely many Jordan curves (we call such a domain a Jordan domain). Let E be a finite union of arcs in ∂Ω (we allow the possibility of an entire connected boundary component—a simple, closed curve—to be one of these arcs). Then the harmonic measure of E at the point z ∈ Ω with respect to Ω is the value at z of the bounded harmonic function ω on Ω with boundary limit 1 at points of E and bound- ary limit 0 at points of ∂Ω \ E (except possibly at the endpoints of the arcs that make up E). We denote the harmonic measure by ω(z,Ω,E). The first question to ask about harmonic measure is that of existence and uniqueness: does harmonic measure always exist?Ifit does exist, is it unique? The answer to both these queries is “yes.” Let us see why. Fix a Jordan domain Ω. It is a classical result (see [AHL2], as well as our Section 1.5) that there is a conformal mapping Φ : Ω → U of Ω to a domain U bounded by finitely many circles. If F ⊆ ∂Ω consists of finitely many circular arcs, then it is obvious that the harmonic measure of F exists. It would just be the Poisson integral of the characteristic function of F , and that Poisson integral can be shown to exist by elementary conformal mapping arguments (i.e., mapping to the disc). Now a classic result of Carath´eodory guar- antees that Φ and its inverse extend continuously to the respective boundaries, so that the extended function is a homeomorphism of the closures. Thus we may pull the harmonic measure back from U to Ω via Φ. As for uniqueness: if Ω is bounded, then uniqueness follows from the standard maximum principle. For the general case, we need an extended maximum principle that is due to Lindel¨of: Proposition 6.1 Let Ω be a domain whose boundary is not a finite set. Let u be a real-valued harmonic function on Ω, and assume that there is a real constant M>0 such that u(z) ≤ M, for z ∈ Ω. 79344-3_Printer Proof.pdf 184 “CH06” — 2015/10/26 — 17:20 — page 170 — #2 11/6/2015 1:52:49 PM 6.1. THE IDEA OF HARMONIC MEASURE 171 Suppose that m is a real constant and that lim sup u(z) ≤ m (6.1.1) z→ζ for all except possibly finitely many points ζ ∈ ∂Ω. Then u(z) ≤ m for all z ∈ Ω. Remark 6.2 Observe that any Jordan domain certainly satisfies the hypotheses of the proposition. Furthermore, by classical results com- ing from potential theory (see [AHL2]), any domain on which the Dirichlet problem is solvable will satisfy the hypotheses of the theo- rem. Note, in particular, that Ω certainly need not be bounded. We conclude by noting that this result is a variant of the famous Phragmen-Lindel¨oftheorem (see [RUD2]). Proof of Proposition 6.1: First assume that Ω is bounded. We will remove this extra hypothesis later. Let the diameter of Ω be d. Let the exceptional boundary points (at which (6.1.1) does not hold) be called ζ1,...,ζk. Let >0. We then may apply the ordinary textbook maximum principle (see [GRK1]) to the auxiliary function k |z − ζ | h(z)=u(z)+ log j . d j=1 Notice that h (instead of u) satisfies the hypotheses of the proposition at every boundary point.Soh ≤ m on all of Ω. Then we let → 0 and the result follows. Now consider the case of Ω unbounded. If Ω has an exterior point (i.e., a point in the interior of the complement of the closure of Ω), then we may apply an inversion and reduce to the case in the preceding paragraph. Finally, if Ω has no exterior point, then let R>|ζj| for all j = 1,...,k. Let Ω1 = {z ∈ Ω:|z| <R} and Ω2 = {z ∈ Ω:|z| >R}. These may not be domains (i.e., they could be disconnected), but they are certainly open sets. Let S = {z ∈ Ω:|z| = R}. If u ≤ m on S, then we apply the result of the first paragraph on Ω1 and the result of the second paragraph on Ω2. 79344-3_Printer Proof.pdf 185 “CH06” — 2015/10/26 — 17:20 — page 171 — #3 11/6/2015 1:52:49 PM 172 CHAPTER 6. HARMONIC MEASURE If it is not the case that u ≤ m on S, then u will have a max- imum N on S with N>m. But this will be a maximum for u on all of Ω. Since u ≤ m at the ends of the arcs of S, it follows that u actually achieves the maximum N on S. Therefore, by the usual maximum principle, u is identically equal to the constant N. But then the boundary condition (6.1.1) cannot obtain. That is a contradiction. It is important to realize that, if Ω has reasonably nice boundary, then ω(z,Ω,E) is nothing other than the Poisson integral of the characteristic function of E. (We used this fact in our discussion of the existence of harmonic measure.) In any event, by the maximum principle the function ω takes real values between 0 and 1. Therefore 0 <ω(z,Ω,E) < 1. One of the reasons that harmonic measure is important is that it is a conformal invariant. Essential to this fact is Carath´eodory’s the- orem, that a conformal map of Jordan domains will extend continu- ously to the closures. We now formulate the invariance idea precisely. Proposition 6.3 Let Ω1, Ω2 be domains with boundaries consisting of Jordan curves in C and ϕ :Ω1 → Ω2 a conformal map. If E1 ⊆ ∂Ω1 is an arc and z ∈ Ω1 then ω(z,Ω1,E1)=ω(ϕ(z), Ω2,ϕ(E1)). Here ϕ(E1) is well defined by Carath´eodory’s theorem. Proof: Let h denote the harmonic function ω(ϕ(z), Ω2,ϕ(E1)). Then h ◦ ϕ equals harmonic measure for Ω1 (by Carath´eodory’s theorem). Remark 6.4 It is worth recording here an important result of F. and M. Riesz (see [KOO, page 72]). Let Ω ⊆ C be a Jordan domain bounded by a rectifiable boundary curve. Let ϕ : D → Ω be a con- formal mapping. If E ⊆ ∂D has Lebesgue linear measure zero, then ϕ(E) also has Lebesgue linear measure zero. 6.2 Some Examples Capsule: An idea like harmonic measure is a bit hollow without some concrete examples. In this section, we produce several very concrete instances. 79344-3_Printer Proof.pdf 186 “CH06” — 2015/10/26 — 17:20 — page 172 — #4 11/6/2015 1:52:49 PM 6.2. SOME EXAMPLES 173 Harmonic measure helps us to understand the growth and value distribution of harmonic and holomorphic functions on Ω. It has become a powerful analytic tool. We begin to understand harmonic measure by first calculating some examples. Example 6.5 Let U be the upper-half-plane. Let E be an interval [−T,T] on the real axis, centered at 0. Let us calculate ω(z,U,E) for ∼ z = x + iy = (x, y) ∈ Ω. It is a standard fact (see [GRK1]) that the Poisson kernel for U is 1 y P (x, y)= · . π x2 + y2 Here we use the traditional real notation for the kernel. In other words, the harmonic function Ω that we seek is given by 1 ∞ y ω(x + iy)= χ[−T,T](t) · dt π −∞ (x − t)2 + y2 1 T y = dt. − 2 2 π −T (x t) + y Now it is a simple matter to rewrite the integral as 1 1 T 1 ω(x + iy, Ω,E)= 2 dt π y −T x−t y +1 − T 1 −1 x t = − Tan π y −T 1 x + T 1 x − T = Tan−1 − Tan−1 . π y π y The reader may check that this function is harmonic, tends to 1 as (x, y) approaches E ⊆ ∂U, and tends to 0 as (x, y) approaches ∂U\E.
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