1 Various preliminaries Notation... 1

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1 Contents 1 Various preliminaries Notation Brownian motion and harmonic functions Optional sampling theorem Itô s formula calculation Harmonic functions and holomorphic functions Conformal invariance Example: Recurrence A fundamental estimate Green s function 13 5 Riemann mapping theorem 19 6 Analytic boundary points Excursion measure Poisson kernel Toolbox for conformal maps Beurling estimate Koebe distortion theorems Loewner differential equation A class of transformations Compact hulls Boundary behavior Curves Loewner differential equation Loewner chains generated by a curve Non-crossing curves Multiply connected domains 56 A Poisson kernels 61 A.1 Disk A.2 Upper half plane A.3 Half Disk A.4 Infinite Strip Various preliminaries 1.1 Notation We write D = {z C : z < 1} for the open unit disk and H = {x + iy C : y > } for the upper half plane. More generally, we write D r = e r D for the disk of radius e r about the origin and D r (z) = z + D r for the disk of radius e r about z. Let C r = D r, C r (z) = D r (z). We write Ĉ for the Riemann sphere. 1

2 The potential theory of complex Brownian motion makes heavy use of the logarithm function. This affects our choice of notation. There are many times that we will want to write the logarithm of a radius, and by parametrizing our radii exponentially it will make our formulas nicer. Definition A (standard) complex Brownian motion is a process of the form B t = B 1 t + B 2 t i where B 1 t, B 2 t are independent one-dimensional Brownian motions. Equivalently, B t is a standard Brownian motion in R 2 considered as taking values in C. When we say complex Brownian motion, we will always mean standard complex Brownian motion. If the context is clear we will say just Brownian motion. If D C is an open set, then we define τ D = inf{t : B t D}, τ D = inf{t > : B t D}. Notethat τ D = τ D > if B D and τ D = if B C \ D. If B D. Definition We call a boundary point z D regular if P z {τ D = } = 1, that is if Brownian motion starting at z immediately hits the boundary. Here we have introduced a new notation: P w means the probability assuming that B = w. Note that isolated points of D are not regular points. The next lemma shows that Brownian motion starting near a regular point exits the domain quickly with high probability. Lemma 1.1. Suppose D is an open subset of C, Suppose z is a regular boundary point of D. Then for every ɛ >, there exists δ > such that if w z < δ, then P w {diam(b[, τ D ]) ɛ} 1 ɛ. If z is an irregular boundary point of D, then P z {τ D > } = 1. Proof. Without loss of generality, assume that z = and let ξ s = inf{t : B t = e s }. Suppose is a regular point. Since P{τ D = } = 1, we know that P{B(, ξ s ] D} =. Therefore, there exists u such that P{B[ξ u, ξ s ] D} e s. We can find δ > such that the distribution of B(ξ u ) given that B = δ agrees with that assuming B = up to an error of e s. Therefore, for w < δ, and hence P w {B[ξ u, ξ s ] D} 2e s, P w {diam[, τ D ] 2 e s } 2e s. Suppose is an irregular point. Then there exists r with For every ɛ > we can find s r such that As before, we can find u such that if z e u, P{B(, ξ r ] D} = ρ >. P{B[ξ s, ξ ] D} ρ + ɛ. P u {B[ξ s, ξ ] D} ρ + 2ɛ. 2

3 Therefore, ρ = P{B(, ξ ] D} = P{B(, ξ u ] D} P{B(, ξ ] D B(, ξ u ] D} P{B(, ξ u ] D} (ρ + 2ɛ), which implies that Hence, P{B(, ξ u ] D} ρ ρ + 2ɛ. P{τ D > } = lim u P{B(, ξ u] D} = 1. We will call an open set D regular if P z {τ D < } = 1 for every z D. Exercise 1. Suppose B t is a complex Brownian motion starting at the origin. Let θ R, a > and Y t = e iθt B t, Z t = a 1 B a2 t. Then Y t, Z t are (standard) complex Brownian motions. Definition A domain is a connected open subset of C. A domain is simply connected if Ĉ \ D is connected. If D is a domain with D compact, we will also consider D { } as a domain in Ĉ. Definition Let D denote the set of all domains D C such that every z D is regular. Let D denote the set of all domains D C such that there exists z D that is regular. Clearly D D. Giving the exact criterion to be in D is difficult. However, we will now derive a sufficient condition that will suffice for our purposes. We will use the following lemma that makes strong use of the planarity of C. Lemma 1.2. There exists β >, c < such that if B t is a complex Brownian motion starting at z D and τ = τ D = inf{t : B t = 1}, then the probability that the origin lies in the unbounded component of C \ B[, τ] is no more than c z β. If follows that if D is a domain, D, and the connected component of D containing also contains at least one point on D, then P z {B[, τ D ] D} c z β. Proof. Let p(z) be the probability that lies in the unbounded component given B = z, and note that p(z) = p( z ). Let q be the probability that a Brownian motion starting on C 1, the circle of radius e 1 about the origin, disconnects C 1 from C before time τ. By constructing a particular event, it is easy to see that q > and is independent of the angle of the starting point. Using the strong Markov property and the scaling property of Brownian motion we can see that for r 1, p(e r ) q p(e r 1 ). Hence for integer n, p(e n ) e n log q, and more generally p(e u ) p(e u ) e log q e u log q. 3

4 It is significantly harder to show that β = 1/4 is the optimal value in this lemma. We will not need to know the value of β, but only that it is positive. Corollary 1.3. Suppose D C is a domain such that all the connected components of D are all larger than one point. Then D D. In particular, all simply connected D C are in D. This argument establishes a stronger fact. If z K C, we define rad K (z) = sup{ w z : w K}, diamk = sup{ w z : w, z K}. Note that for all z K, diamk 2 rad K (z). Proposition 1.4. Suppose D is a domain and w, w are in the same component of D. Then, for all z, ( ) β z w P z {diam (B[, τ D ]) w w } c w, w where c, β are as in Lemma 1.2. Proposition 1.5. Suppose D is a domain and z is an irregular boundary point of D. Let h be a strictly positive harmonic function on D. Then there exists a sequence z n D with z n z with lim inf n h(z n ) >. Proof. Since is irregular, we can find a compact V D and δ >. such that P z {τ C\V < τ D } δ >. This implies that there exists z n z with P zn {τ C\V < τ D } δ} and hence h(z n ) δ min{h(w) : w V }. Proposition 1.6. If D is a domain and z D, then with probability one, B(τ D ) is a regular point of D. Proof. Let V be a subset of D and let E be the event {B(τ D ) V }. Let M t be the martingale M t = P[E F t τ ]. For t < τ, M t = h(b t ) where h is the positive harmonic function h(w) = P w {B t V }. Assume that V is such that < h < 1 on D. Note that M τ = 1 E and the martingale convergence theorem implies that M τ = M τ with probability one. In particular, with probability one, if ζ := B(τ D ) V, then there exists γ : [, 1) D with γ(1 ) = ζ such that lim t 1 h(γ(t)) =. Since h is a bounded function, one can see using Lemma 1.2 that this last condition implies: if z n ζ, then h(z n ). This is impossible if ζ is an irregular point by the previous proposition. 2 Brownian motion and harmonic functions Although we restrict our discussion here to C = R 2, much of what is done here extends to R d. We will go back and forth between considering a point z = x + iy = (x, y) as a point in C or in R 2. Recall that a function f : D R is harmonic if it satisfies either of the following equivalent conditions. f is C 2 and for all z D. f(z) := xx f(z) + yy f(y) =, f is locally integrable and satisfies the mean value property. That is to say, if z D with dist(z, D) > r, then MV (f; z, r) = f(x). We will not prove these standard results. 4

5 2.1 Optional sampling theorem There is a close relationship between harmonic function and martingales. Before proceeding we will prove a lemma that is one version of the optional sampling or optional stopping theorem for martingales. The assumptions we make are significantly stronger than is needed for the result, but it will suffice for our purposes. Proposition 2.1. Suppose M t, t T is a uniformly bounded continuous martingale, and τ is a stopping time, each with respect to the filtration {F t }. Then M t τ, t T is a continuous martingale with respect to the filtration {F τ t }. We recall that if τ is a stopping time, then F τ is the σ-algebra of all events E such that for all t, E {τ t} F t. One thinks of this as all the events that depend on the process only up to the stopping time τ. Proof. Let Y t = M t τ. It is immediate that Y is a continuous process. Let us first assume that τ takes on only a discrete number of values = s < s 1 < < s k <. If s < t, then M t τ can be written as M t τ = s j<t M sj 1{τ = s j } + M t 1{τ > s j }. Using the definition, it is not hard to show that this is a martingale. To illustrate this, we consider the case s j = s < t, in which case E [M t τ F s ] = k j M sk 1{τ = s k } + E [ M t 1{τ > s j } F sj ]. Since the event {τ > s j } is the complement of the event 1{τ s j } which is F sj -measurable, E [ M t 1{τ > s j } F sj ] = 1{τ > sj } E [ M t F sj ] = 1{τ > sj } M sj, and hence E [ M t τ F sj ] = M sj 1{τ = s j } + M sj 1{τ > s j } k j = M sk 1{τ = s k } + M sj 1{τ s j+1 } = Y sj. k<j 1 For more general τ, we approximate τ by discrete stopping times, τ j = τ (j+1)/n, j n τ < j + 1 n. The random variables M t τ j M t with probability one. Since they are bounded, they also converge in L 1. Let B t be a standard one-dimensional Brownian motion starting at the origin and suppose that a, b >. Let τ = inf{t : B t = b or B t = a}. Then B t τ is a martingale. Therefore, for each t, = E[B ] = E [B t τ ]. With probability one B t τ B τ. Since B t τ is uniformly bounded, we can use the dominated convergence theorem to see that E[B τ ] = lim t E [B t τ ] =. 5

6 But, E[B τ ] = b P{B τ = b} a [1 P{B τ = b}]. Solving, we get P{B τ = b} = a a + b. This relation is often referred to as the gambler s ruin estimate for one-dimensional Brownian motion. From this we can easily see that one-dimensional Brownian motion is recurrent, that is, it keeps returning to the origin. One must be careful in using this proposition. If P{τ < } = 1, then with probability one lim M t τ = M τ. t However, it is not always the case that this limit is in L 1. Indeed, it is possible for E[M τ ] lim t E[M t τ ]. As an example, let M t = B t be a standard one-dimensional Brownian motion starting at the origin and let τ = inf{t : B t = 1}. Recurrence of one-dimensional Brownian motion implies that P{τ < } = 1. However, E[B τ ] E[B ]. 2.2 Itô s formula calculation Suppose D is a domain, h : D R is a harmonic function and B t = B 1 t +ib 2 t is a complex Brownian motion starting at z D. Let τ = τ D = inf{t : B t D}. Then, for t < τ, Itô s formula implies that dh(b t ) = h x (B t ) db 1 t + h y (B t ) db 2 t. Suppose K D is a compact set with z int(k), and let τ = inf{t : B t K}. Then h(b t τ ) is a bounded martingale. It follows that The left-hand side is the same as MV (z; f, K). E z [h(b t τ )] = E z [h(b )] = h(z). Proposition 2.2 (Dirichlet problem). Suppose D D, and h is a bounded continuous function on D. Then there exists a unique bounded continuous function h : D R that extends h and is harmonic in D. In fact. for every z D, h(z) = E z [h(τ D )]. (1) Proof. If h is defined by (1), then h is locally integrable and satisfies the mean value property. Hence, h is harmonic. Conversely if h is harmonic in D and continuous on D, then M t = h t τd is a continuous martingale, and (1) satisfies the mean value property. We need to show that h defined as in (1) is continuous on D, and this uses the fact that every point in D is a regular point. The assumption that h is bounded is necessary for uniqueness. For example if D = (, ) and h() =, there are an infinite number of harmonic extensions to D given by h(x) = cx. As we will see below, if a Brownian motion starts at z D, then, and hence P z {B τd V } = 1 2π ζ =1 1 z 2 ζ z 2 d ζ, E z [F (B τd )] = 1 1 z 2 F (ζ) d ζ. (2) 2π ζ =1 ζ z 2 Using this we can derive the fundamental facts about harmonic functions. 6

7 Proposition 2.3. For every positive integer k, there exists a universal c k < such that if D is a domain, h : D R is harmonic, and z D with dist(z, D) ɛ, then for j k, x j y k j h(z) c k ɛ k sup{ h(w) : w z < ɛ}. Proof. By considering h(w) = h(z + ɛw), we can see that it suffices to prove the result for D = D, z =. In this case we can differentiate under the integral in (2). Proposition 2.4 (Harnack principle). If D is a domain, then every compact K D, there exists c = c(k, D) < such that if h : D (, ) is harmonic in D, then f(z) c f(w), z, w K. Proof. Since D is connected, by choosing K larger if necessary, we may assume that K is connected. If D D and K = { w 1/2}, then the result follows from the explicit form of h in (2). More generally, we can cover K by a finite collection of open balls D r (ζ j ), j = 1,..., N with 2D r (ζ) D. We write w z, if one of these balls contains both w, z. We say that w and z are connected if there exists a sequence z = z, z 1,..., z k = w with z j 1 z j. We claim that all points in D are connected. Indeed, let U 1 be the union of all the disks D r (ζ j ), j = 1,..., N for which z is connected to ζ j and let U 2 be the union of the other disks. If U 2, then U 1, U 2 disconnect K. Hence, for z, w K we can find a sequence z = z, z 1,..., z k = w with z j 1 z j with k N. and f(z) c N f(w). It will be useful to have the following convention. Convention. Suppose D is a domain and h : D R is a harmonic function. We say h : D R is an extension of the harmonic function to the boundary, if h is continuous at all regular points of D. While we have stated the derivative estimates and Harnack principle for harmonic functions in R 2, the analogous results hold for harmonic functions in R d. 2.3 Harmonic functions and holomorphic functions If f(z) = u(z) + iv(z) is a holomorphic function in a domain D, then u, v satisfy the Cauchy-Riemann equations, x u = y v, y u = x v. Using this and equivalence of mixed partials, one can see that u, v are harmonic functions. The next proposition, which is a standard result from a first course in complex variables, gives a partial converse to this fact. Proposition 2.5. Suppose D is a simply connected domain and u is harmonic function on D. Then there is a harmonic function v on D, which is unique up to an additive constant, such that f(z) = u(z) + iv(z) is holomorphic. Sketch of proof. We use the Cauchy-Riemann equations to find v. Let us fix z D and arbitrarily choose v(z ) =. If γ : [, 1] D is a smooth curve with γ() = z, γ(1) = z, then we define v(z) = γ n u dγ := 1 [ y u(γ(t)) x γ(t) x u(γ(t)) y γ(t)] dt. 7

8 In order to show this is well defined, we need to show that we get the same value for v(z) regardless of the curve γ. Equivalently, we need to show that if γ(1) = z, then n u dγ =. γ If D is simply connected, then the region(s) bounded by γ are entirely in D, and this identity follows from Green s theorem and the fact that u is harmonic. By construction, u, v satisfy the Cauchy-Riemann equation and hence f = u + iv is holomorphic. To show uniqueness, suppose that f = u + iṽ is also holomorphic. Then f f is holomorphic and only takes on imaginary values. Hence f f is constant. The function v is often called the complex conjugate. We have used simple connectedness to conclude that there exists an complex conjugate v defined on all of D. This is not alway true if the domain is not simply connected.. For example, if u(z) = log z on D = {z : < z < 1}, then f is harmonic, but there is no holomorphic extension to all of D. However, regardless of the topology of D, we can always find conjugates v defined in a neighborhood of z. When trying to determine if two complex domains are conformally equivalent, it is often the case that one can determine the real or imaginary part (or, perhaps, the radial part which is the real part of the exponential, or something else). This determines the function (up to a constant) locally and then the question becomes whether or not one can extend it to the entire domain D. The term conjugate is overused in complex variables! I have given lectures where I have used conjugate three different ways in the same lecture as the conjugate of a number, as the complex conjugate function above, and also in the algebraic sense of a conjugate function f = g 1 f g. 2.4 Conformal invariance If f is a holomorphic function with f() =, f (), then locally near zero f looks like a dilation by f () and a rotation by arg f(). Brownian motion is invariant under rotation and is also invariant under scaling if one changes the parametrization appropriately. This is the basic reason why the following theorem holds. Theorem 1. Suppose D is a domain in C and B t is a complex Brownian motion starting at z C. Suppose f : D C is a nonconstant holomorphic function. Let and for s < ξ, define σ(s) < τ D by that is ξ = τd σ(t) = σ(s) f (B s ) 2 ds (, ], t ds f (B s ) 2, f (B s ) 2 ds = s. Then Y s = f(b s ), s < ξ is a complex Brownian motion. We will need a lemma that states in some sense that all stochastic integrals are tim changes of standard Brownian motions. Indeed, a stronger fact is true that we will not prove all continuous martingales are time changes of standard Brownian motions. 8

9 Lemma 2.6. Suppose B t is a standard one-dimensional Brownian motion with filtration {F t } and suppose that A t is a continuous, adapted process such that there exists < c 1 < c 2 < with c 1 A t c 2. Let and let that is X t = t A s db s, σ(r) = inf{t : X t = r}, σ(r) X 2 s ds = r. Suppose that for all t, P{σ(t) < } = 1. Then W r := X σ(r) is a standard Brownian motion with respect to F r = F σ(r). Sketch of Proof. To prove this, one shows that conditioned on F s the distribution of W r+s W s is that of a Brownian motion with variance r 2. We will do this in the case s = ; the general case is similar. If λ R, let K t = exp{iλx t }. (If one does not want to use Itô s formula with with complex valued processes one can write this as cos(λx t )+ i sin(λx t ).) Itô s formula shows that [ ] ] dk t = K t i λ db t λ2 2 A2 t dt = K t [i λ db t λ2 2 d X t. If M t = exp{λ 2 X t /2} K t, then M t is a local martingale satisfying, dm t = i λm t db t. Note that M t σ(r) is a bounded martingale, and hence the optional sampling theorem implies that But, M σ(r) = e λ2 r/2 exp{iλ(r)}, and hence E [ M σ(r) ] = E[M ] = 1. E [ e λiwr] = e λ2 r/2. Since the characteristic function determines the distribution, we see that W r N(, r). Proof of Theorem 1. We will give a sketch of the proof relying on some facts from stochastic calculus. Let U D be a subdomain with U compact containing none of the zeros of f, and let τ = τ U < τ. Let us write B t = Bt 1 + ibt 2 and let X t = u(bt 1, Bt 2 ), Y t = v(bt 1, Bt 2 ). Using the fact that u, v are harmonic functions, Itô s formula and the Cauchy-Riemann equations give dx t = u x (B t ) db 1 t + u y (B t ) db 2 t, dy t = v x (B t ) dbt 1 + v y (B t ) dbt 2 = u y (B t ) dbt 1 + u x (B t ) dbt 2. Note that t σ(t) = f (B t ) 2 = u(b t ) 2. If we let ˆX t = X σ(t), Ŷt = Y σ (t), then d ˆX t = u x( ˆB t ) u( ˆB t ) dw 1 t + u y( ˆB t ) u( ˆB t ) dw 2 t, 9

10 dŷt = u y( ˆB t ) u( ˆB t ) dw 1 t + u x( ˆB t ) u( ˆB t ) dw 2 t, where W 1 t, W 2 t are independent, standard Brownian motions. This means that ( ˆX t, Ŷt) are independent standard Brownian motions, that is, ˆB t = ˆX t + i Ŷt, is a standard complex Brownian motion at least for t τ. Since this holds for every U, and with probability one the Brownian motion avoids the singular points of U, we can conclude that it holds for t < τ D. Exercise 2. Suppose f is a nonconstant holomorphic function on a domain D. Let τ < τ D be a bounded stopping time. Show that [ τ ] ds E f (B s ) 2 <. We allow the case where f (B ) =. The statement of Theorem 1 is a little nicer if f is a conformal transformation. We say that f : D D is a conformal transformation if f is holomorphic, one-to-one, and onto. Theorem 2. Suppose D is a domain in C and f : D f(d) is a conformal transformation. Suppose B t is a complex Brownian motion starting at z C. Let and for s < ξ, define σ(s) < τ D by ξ = τd σ(s) f (B s ) 2 ds (, ], f (B u ) 2 du = s. Then Y s = f(b σ(s) ), s < ξ is a complex Brownian motion, and ξ = τ f(d) = inf{t : Y t f(d)}. Proof. Note that if ξ <, we can extend Y s, s ξ by continuity. If ξ <, we claim that Y ξ f(d). Indeed, if Y ξ = w f(d), then B τd = f 1 (w) D Example: Recurrence Here we show that two-dimensional Brownian motion is neighborhood recurrent. To be more precise, with probability one, for all z C, ɛ >, T <, there exists t > T with B t z < ɛ. It suffices to prove this for z with rational coordinates, and the proof is essentially the same for all of them, so let us consider z =. Let B t be a complex Brownian motion and let f(z) = e z, Y t = f(b t ) = e Bt = e B1 t e ib 2 t. Then Yt is a time change of a Brownian motion. Since Y t = e B1 t, we see from the recurrence of the one-dimensional Brownian motion B 1 that lim inf Y t =. t We can also see from this that the Brownian motion is not pointwise recurrent. Indeed with probability one, a Brownian motion never visits the origin after time zero. This is obvious for Y t since is not in the range of the exponential function. An important corollary of the neighborhood recurrence of Brownian motion is the following: if D is a domain with at least one regular boundary point, then for all z D, P z {τ D < } = 1. 1

11 Indeed, if w is a regular point, then there exists a δ such that if the Brownian motion is within distance δ of w, then with probability 1/2 it goes leaves the domain before it goes distance one from z. Since we keep returning to the δ neighborhood of z we get infinitely many chances to escape D near w and we will eventually succeed. This fact is used implicitly in the following definition. Definition If D D and z D, then the harmonic measure hm D (z, ) is defined to be the distribution of B(τ D ) assuming B = z. In other words, the probability that a Brownian motion starting at z exits D at V is hm D (z, V ). More generally if f is a function defined on D, we let hm D (z, f) = E z [f(b τ )] = f(w) dhm D (z, w). In particular, the harmonic measure hm D (z, ) is well defined. If f : D f(d) is a conformal transformation, that extends to a homemomorphism of D, then then hm f(d) (f(z), f(v )) = hm D (z, V ). (3) There is a similar formula that holds if f does not necessarily extend to a homemorphism, but to explain it requires a discussion of prime ends. Suppose for ease that D, f(d) are bounded domains. Consider the set of curves γ : [, t ) D with γ(t ) D and such that the limit D f(γ(t )) = lim t t f(γ(t)) exists. We say that γ 1 is equivalent to γ 2 if γ 1 (t 1 ) = γ 2 (t 2 ) and f(γ 1 (t 1 )) = f(γ 2 (t 2 )). Then (3) holds for these equivalence classes. For example, if D = D H, then f(z) = z 2 is a conformal transformation of D onto f(d) = D \ [, 1). The boundary point 1/4 f(d) corresponds to two equivalences classes, one for curves approaching 1/4 from above and the other for curves approaching 1/4 from below. These correspond to curves in D that leave D at 1/2 and 1/2 respectively. Examples like this were there are two-sided points are easy to handle, even if we have to be careful about our notation. If D, f(d) are bounded domains and B t is a Brownian motion in D, then γ := B[, τ D ] corresponds to an equivalence class in D and f γ an equivalence class in D. The equation (3) is interpreted in terms of these classes. We will use (3) for general domains D implicitly meaning this interpretation of the equation. The definition of harmonic measure does not require any smoothness of the boundary. However, if the boundary is nice, then one can write harmonic measure as an integral of a kernel called the Poisson kernel. Rotational invariance of Brownian motion shows that harmonic measure on D centered at zero is the uniform distribution. We define H D (z, w) to be the Poisson kernel normalized so that In other words, if V is sufficiently smooth, hm D (z, V ) = 1 π H D (, e iθ ) = 1 2. V H D (z, w) dw. Here dw represents integration with respect to arc length, that is, a traditional line integral from vector calculus rather than a complex integral along a curve. The term sufficiently smooth is a little vague. We will only use the Poisson kernel at places where D is locally an analytic curve. If f : D f(d) is a conformal transformation, D is locally analytic at w,. and f(d) is locally analytic at f(w), then the Poisson kernel satsifes the conformal covariance relation H D (z, w) = f (w) H f(d) (f(z), f(w)). 11

12 We write MV(z; f, r) to be the mean value of the function on the circle of radius e r about z, MV(z; f, r) = 1 2π 2π f(z + e r+iθ ) dθ = E z [ f(b τdr ) ]. More generally, if z D and f is a function defined on D, we define the mean value of f on D with respect to z by MV(z; f, D) = E z [ f(b τdr ) ]. Suppose f : D D is a conformal transformation. Then boundary D can be very rough. For example, there can be points w D such that there is no continuous path η : [, 1) D with η(1 ) = w. However, the harmonic measure of such points has to be zero. This is immediate from the definition of harmonic measure in terms of Brownian motion A fundamental estimate Here we prove and estimate that we will use continually. Recall that D r = e r D and define A r to be the annulus A r = D \ D r = {z : e r < z < 1}. Proposition 2.7. If z A r and τ = τ Ar, then P z { B τ = e r } = log z. (4) r Proof. Let us give two similar proofs. First, note that φ(z) = log z is a bounded harmonic function in A r that is continuous on A r. This can be checked by differentiation or by noting that it is the real part of the (locally) analytic function log z. Therefore, by Proposition 2.2, φ(z) = φ(b ) = E z [φ(b τ )] = r P z { φ(b τ ) = e r}. Alternatively we can let Y t = exp{b t }. Then the probability is the same as the probability that the one-dimensional Brownian motion Bt 1 starting at log z reaches level r before reaching level which by the gambler s ruin estimate is log z /r. Proposition 2.8. Suppose φ is a harmonic function on the annulus A r = {e r < z < 1}. Let M s = 1 2π 2π φ(e s+iθ ) dθ, be the average value of φ on the circle of radius e s. Then there exist a, b such that M s = as + b, < s < r. Moreover, for each s, where n is the outward unit normal. C s n φ(w) dw = 2πa, 12

13 Proof. Suppose that < p < s < q < r. Let B t be a Brownian motion starting uniformly on the circle of radius e s and let T be the first time t that B t = e p or B t = e q. Then P{ B T = e p } = q s q p. By rotational symmetry, given that B T = e p (or given B T = e q, the distribution of the angular part is uniform. Therefore, M s = q s q p M p + s p q p M q = M q M p q p s + qm p pm q. q p For the final assertion note that 2π 2πa = 2π s M s = s φ(e s+iθ ) dθ = s φ(w) e C s dw = n φ(w) dw. s C s 3 Green s function The Green s function G D (z, w) is the normalized probability that a Brownian motion starting at z visits w before leaving D. As stated this does not make sense since the probability that the Brownian motion visits w is zero. However, we can make sense of it as a limit, G D (z, w) = lim ɛ log[1/ɛ] P z {dist(w, B[, τ D ]) ɛ}. We will show this limit exists in this section and derive some properties. We will first consider the case w = and write just G D (z) for G D (z, ). Throughout this section, we let σ s = inf{t : B t = e s }. We will make no topological assumptions about the domain D. We only require that the boundary contain a regular point, that is, that P z {τ D < } = 1. Definition Let U r (resp., U s r) denote the set of domains (resp., simply connected domains) containing the origin with at least one regular boundary point and dist(, D) r. If r = 1 we write just U, U s. There are natural bijections U U r and U s U s r given by D rd. Proposition 3.1. Suppose D U r and z D \ {}. Then the limit G D (z) = lim s s Pz {σ s < τ D }, exists and lies in (, ). Moreover, G rd (rz) = G D (z). We note that (4) establishes the result for D = D for which G D (z) = log z. Proof. We write τ = τ D. It suffices to prove the result for D U, after which we can use conformal invariance of Brownian motion to see that G eu D(e u z) = lim s s Peuz {σ s < τ eu D} = lim s s Pz {σ s u < τ} = G D (z). Since D D and D contains a regular point, the Harnack principle implies that inf ζ =1 Pζ {τ < σ 1 } = ρ = ρ D >. (5) 13

14 For s > 1, let q s = sup P ζ {σ s < τ}. ζ =1 By the strong Markov property it is the same as the supremum over ζ 1. We claim that To see this, note that if ζ = 1 and s > 1, then P ζ {σ s < τ} P ζ {σ 1 < τ} If ζ = 1, then using (4), we get q s 1 s ρ, (6) sup P ζ {σ s < τ} (1 ρ) sup P ζ {σ s < τ}. ζ =1/e ζ =1/e P ζ {σ s < τ} = P ζ {σ s < σ } + P ζ {σ s > σ } P ζ {σ s < τ σ s > σ } 1 s + s 1 q s. s By taking the supremum over ζ = 1, we see that [ 1 q s (1 ρ) s + s 1 ] q s 1 s s + (1 ρ) q s. which gives (6). We now fix z and let f(s) = P z {σ s < τ}. Then, if z > e s, If ζ = e s, then (4) and (6) imply that f(s + 1) = P z {σ s+1 < τ} = P z {σ s < τ} P z {σ s+1 < τ σ s < τ} = f(s) P z {σ s+1 < τ σ s < τ}. P ζ {σ s+1 < τ} = P ζ {σ s+1 < σ } + P ζ {σ < σ s+1 } P ζ {σ s+1 < τ σ < σ s+1 } s s s + 1 ρ (s + 1) = s s O(s 2 ), where here and throughout the remainder of the proof, the O( ) terms can depend on D. Therefore, [ f(s + 1) = f(s) 1 1 ] s O(s 2 ), log f(s + 1) = log f(s) 1 s O(s 2 ). This equation implies the existence of a constant which we call G D (z) such that for integer s, If u 1, the same argument shows that P z {σ s < τ} = f(s) = G D(z) s [ 1 + O(s 1 ) ]. (7) and hence (7) holds for all s. f(s + u) = f(s) s s + u [ 1 + O(s 1 ) ] = G D(z) s + u [ 1 + O(s 1 ) ], 14

15 We extend G D (z) to be a function on C \ {} by setting G D (z) =, z D. If D is open but not connected and D is the connected component of D containing the origin, we define G D (z) = G D (z). We state the next proposition for D U but it extends immediately to a result about D U r by scaling. Proposition 3.2. There exists c < such that if D U, the following holds. 1. G D is a positive harmonic function on D \ {} that vanishes on D. 2. There exists c D < such that for all z, 3. G D is continuous at every regular point of D. 4. If then h D is a harmonic function on D. G D (z) log + (1/ z ) + c D. h D (z) = G D (z) + log z, z D \ {}, h D () = 1 2π 2π G D (e r+iθ ) dθ, 5. If z < e 1, h D (z) h D () c h D () z. (8) 6. If z D, h D (z) = E z [log B τ ] lim r r Pz {σ r < τ}, where τ = τ D = inf{t > : B t D}. In particular, if D is bounded, then 7. If D D, then G D (z) G D (z). h D (z) = E z [log B τ ]. 8. If D 1 D 2 and D = n=1 D n, then for z D, G D (z) = lim n G D n (z). (9) 9. If f : D f(d) is a conformal transformation with f() =, then G f(d) (f(z)) = G D (z). Proof. 1. We have shown that G D (z) > if z D and it is defined to be zero on D. Suppose U D is a disk with U D \ {}. Then by the definition of G D we can see that G D (z) = E z [G D (B τu )] = MV(z; G D, U) from which we see that G D is harmonic on D \ {}. 2. Note that (6) implies that G D (z) 1/ρ for z = 1. If z = e r with r < s, then P z {σ s < τ D } P z {σ s < σ } + P z {σ < σ s < τ D } r s + s r s 2 ρ. Letting s, we see that G D (z) r + (1/ρ). 15

16 3. Let z be a regular boundary point and let ξ = inf{t : B t z = z /2}. Then for w z < z /2, G D (w) ρ P w {ξ < τ D } where ρ = sup{g D (ζ) : z ζ = z /2} <. For every ɛ >, we can find δ > such that w z < δ implies that P w {ξ < τ D } < δ and hence G D (w) δ ρ. 4. For z D \ {}, let Suppose z < 1. Then, h D (z) = G D (z) + log z. P z {σ s < τ} = P z {σ s < σ } + P z {σ < σ s < τ} log z = + P w {σ s < τ} dhm As (z, w). s C If we multiply both sides by s and take the limit as s, we get In other words, G D (z) = log z + G D (w) dhm D (z, w) = log z + 1 C π h D (z) = 1 π 2π 2π G D (e iθ ) H D (z, e iθ ) dθ, < z < 1, which can be extended to by setting z = on the right-hand side. 5. Using the exact form of the Poisson kernel, we can see that H D (z, e iθ ) = O( z ), G D (e iθ ) H D (z, e iθ ) dθ. and hence G D (z) + log z h D () c z h D (). 6. Let We claim that θ = lim inf r r P {τ D > σ r }. θ = lim r r P {τ D > σ r }. To see this we first note that since D is regular, if Since D is regular, q(r) as r. Hence, if then for ζ = e r, and s > r, q(r) = sup P ζ {σ r < τ D }, ζ =1 p(r, s) = sup ζ =e r {σ s < τ D }, P ζ {σ s < τ D } P ζ {σ s < σ } + P ζ {σ < σ s } P ζ {σ s < τ D σ < σ s } r s + q(r) p(r, s) r + q(r) p(r, s), s 16

17 which implies that Hence, Therefore, p(r, s) r s (1 q(r)). P z {σ s < τ D } = P z {σ r < τ D } P z {σ s < τ D σ r < τ D } P z r {σ r < τ D } s (1 q(r)). lim sup s P z r {σ s < τ D } lim inf s r 1 q(r) Pz {σ r < τ D } = θ. 7. Monotonicity in D follows immediately from the definition. 8. Assume D U. If D 1 D 2 and D = D n, then monotonicity implies that the limit L = lim n G D n (z) exists and is bounded above by G D (z). Assume that D s D n. Then for z < e s, Therefore, for m n and u s, Letting m, we see that L (u s) Letting u, we get L G D (z). G Dn (z) G Ds (z) = G D (e s z) = log z s. G Dm (z) (u s) P z {σ u < τ Dm }. lim m Pz {σ u < τ Dm } = (u s) P z {σ u < τ D }. 9. Note that it follows from the definition, that for any u >, lim s s Pz {σ s+u < τ} = G(z). (1) Let C s = {ζ : ζ = e s } and ˆσ s = inf{t : B t f(c s )}. Conformal invariance of Brownian motion implies that for z D, z > e s, P z {σ s < τ D } = P f(z) {ˆσ s < τ f(d) }. Let θ = log f (). If ɛ >, then for all s sufficiently large, Hence, using (1), σ s+θ+ɛ ˆσ s σ s+θ ɛ. G f(d) (f(z)) = lim s s Pf(z) {σ s < τ f(d) } = G D (z). Definition If D D, then the Green s function G D (z, w) is defined for distinct z, w D by G D (z, w) = G D z (w z). In other words, if then σ s (w) = inf{t : B t w e s }, G D (z, w) = lim s s Pz {σ s (w) < τ D }. 17

18 Proposition 3.3. If D D and z, w D, G D (z, w) = G D (w, z) Proof. Consider a Brownian motion B t with B = z, and let W t = z B t + w. Then W t is a Brownian motion starting at w such that W t = z when B t = w. Roughly speaking, the probability that B visits w before leaving D is the same as the probability that W visits z before leaving D. To make this precise, we need to replace visits with gets close to. Let D ɛ = {ζ D : dist(ζ, D) > ɛ}. Let σ s = σ s (w) = inf{t : B t w e s }, σ s = σ s(z) = inf{t : W t z = e s }. Note that if z w > e s, then σ s = σ s. Note that if B[, σ s ] D ɛ, then W [, σ s] D. From this we see that for every ɛ, G D (w, z) G Dɛ (z, w). Letting ɛ and using (9), we see that G D (w, z) G D (z, w). Lemma 3.4. If τ = τ D, and z D, E z [τ D ] = 1 2 [1 z 2 ]. Proof. Itô s formula shows that M t = B t 2 2t is a martingale. Since E z [M τ ] = E[M ] = z 2, 2 E z [τ] = E z [Bτ 2 D ] z 2 = 1 z 2. Proposition 3.5. Suppose U D is open, and let τ = τ D. Let Then, if z D, Y U = E z [Y U ] 1 π U τ 1{B t U} dt. G(z, w) da(w) E z [Y U ], where da represents integral with respect to area. In particular, if area( U) = we have equalities. Proof. We will show that there is a universal constant c such that for all z, U, D, E z [Y U ] = c G(z, w) da(w). U One we have this we can determine that c = 1/π, by noting that if z =, U = D = D, then E[Y D ] = E[τ] = 1/2 and 2π 1 1 G(, w) da(w) = log w da(w) = (r log r)dr dθ = π r dr = π 2. D D Let us now consider the case, r < s, D = D r, U = D s, z = e s. We claim that there exists a universal constant c such that E z [Y U ] = c (s r) e 2s. By scaling, it suffices to prove this when s =. By rotational symmetry, the left-hand side does not depend on the argument of z. Let τ r = σ r = inf{t : B t = e r }. Let z = 1, and Y r = τr 1{ B t 1} ds. 18

19 If r < u, and ξ = ξ r = inf{t > τ r : B t = 1}, and we write E for E z, E[Y u ] = E[Y r ] + E[Y u Y r ] = E[Y r ] + P{ξ < τ u } E[Y u Y r ξ < τ u ] = E[Y r ] + u r u E[Y u]. Therefore E[Y u ] = (u/r) E[Y r ] and E[Y s ] = s E[Y 1 ]. More generally, if D U and U = D s, then for z = e s, we can write τd τd τd 1{ B t e s } dt = 1{ B t e s } dt + 1{ B t e s } dt, τ D and note that and From this we see that More generally, if z D, z e r, E z [ τd E z [ τd τ D ] 1{ B t e s } dt = c s e 2s, ] 1{ B t e s } dt = O(e 2s ). E z [Y Ds ] = c s e 2s [1 + O(s 1 )]. E z [Y Ds ] = c s e 2s P z {τ s < } [1 + O(s 1 )], (11) where in this case the error term may depend on D and on r. For ease, let us assume that U is bounded and that dist(u, D) = e j >. We fix z, and let Then for s > j, Y (w, s) = e2s π Y Us U τ 1{ B t w e s } dt. Y (w, s) da(w) Y U s, where U s = {ζ U : dist(ζ, U) > e s }, U s = {ζ D : dist(ζ, U) < e s }. Let σ s (w) = inf{t : B t w = e s }. Using (11), we see that E z [Y (w, s)] = c s e 2s P z {σ s (w) < } [1 + O U (s 1 )]. where the O U term can depend on U, D, j. Letting s, we get the result. 4 Riemann mapping theorem We will prove one of the most important theorems in conformal mapping, the Riemann mapping theorem. Our proof will not be the shortest, but we hope it is illuminating. Theorem 3 (Riemann mapping theorem). Suppose D U s r. Then there exists a unique conformal transformation f : D D with f() =, f () >. 19

20 Proof. It suffices to prove the result for D U s. Uniqueness in the case D = D follows as a consequence of the Schwarz lemma. More generally, if f : D D, g : D D are two such transformation then f g 1 is a conformal transformation of the D onto itself, and hence f g 1 is the identity. We will construct f using the Green s function. Recall that we can write G D (z) = log z + u(z), where u(z) = u D (z) is a harmonic function in D that is bounded in the unit disk. Since D is simply connected, there is a holomorphic function h : D D such that Reh = u and Imh() =. Let f(z) = z e h(z). Then f is holomorphic on D \ {} with f(z) = e G D(z). Since f is bounded in a punctured neighborhood of the origin, we can extend f to be holomorphic on D. Note that f() =, f () = e h() = e u() >. This will be the map f. We need to show that f is one-to-one and onto. Since f () >, f is one-to-one in a neighborhood of the origin. For r >, let Vr = {z Ĉ : G D(z) < r}, V r = {z : G D (z) = r}, V r + = {z D : G D (z) > r}, and note that C \ D Vr, f(vr D) A r, f(v r ) C r, f(v r + ) D r. We first claim that V r + is connected. This does not require D to be simply connected. To see this, suppose z V r + and U is the connected component of V r + containing z. If U, G D ( ) is a bounded harmonic function on U and hence by the optional sampling thoerem (or the maximal principle), G D (z) max G D(ζ) max G D(ζ) r. ζ U ζ V r D Therefore, there is a single connected component of V + r. We also claim that Vr disjoint open sets, say U 1, U 2. Since is connected. To see this, suppose we could write Vr as the union of two Ĉ \ D is connected, it must be contained entirely in one of the sets U 1. This implies that the boundary of U 2 consists entirely of points with G D (z) >, and hence the boundary must be contained in V r. Using the optional sampling theorem (or the maximal principle), we see that G D (z) = r for z U 2. We claim that V + r is bounded. Since Ĉ \ C, and Ĉ \ C is connected, we can see that G D(z) as z. We next claim that f is onto by showing that for each r >, f(v r + ) = D r. To see this, we first note that f(v r + ) is an open subset of D r containing the origin. Suppose z f(v r + ). Then there exist w 1, w 2,... V r +, such that f(w j ) z. Since V r + is bounded, there is a subsequence that we will write as w 1, w 2,... that converges to a point w V r + V r. If w V r +, then f(w) = z V r +. If w V r +, then w V r and z = f(w) C r. We have shown that f(v r + ) is an open subset of D r containing the origin whose boundary is C r. Hence f(v r + ) = D r. We will now show that f (z) for every z. We have already shown that f () >. Suppose z V r with r >. Let L(z) = log f(z) which can be defined in a neighborhood of z. Using the power series representation of f around z, we can see that if f (z) =, then L (z) =, and we can find ɛ > and θ 1 < θ 2 < θ 3 < θ 4 < θ 1 + 2π such that for < δ < ɛ, z + δ e iθj V + r, j = 1 or 3, z + δ e iθj V r, j = 2 or 4. With a little topological thinking, one can see that this contradicts the fact that Vr, V r + are connected sets. 2

21 We now finish the proof by showing that f is globally one-to-one. Assume otherwise. Since f (z) for each z, it is locally one-to-one at each point. Let s be the supremum of all r such that there exists distinct z, w V r with f(z) = f(w). Since f is locally one-to-one at the origin, we see that s <. We can find r n s and z n w n in V rn with f(z n ) f(w n ). By taking a subsequence if necessary, we can assume that z n z, w n w for some z, w which must be in V s. Since f is locally one-to-one around z, we can see that z w. We can find disjoint neighborhoods U z, U w about z, w such that f maps U z, U w onto neighborhoods of f(z) = f(w). In particular, there exists ɛ > and ζ C s+ɛ such that f 1 ({ζ}) contains points in both U z and U w. This contradicts the defintion of s. The basic idea of this proof will be used for proving conformal equivalence of other domains. The idea is to assume that a transformation exists and try to construct the function. After a candidate is found, we then try to see if the candidate works. The final part of the argument can be considered a special case of what is known as the argument principle which is related to Rouché s theorem. 5 Analytic boundary points Definition Let K denote the set of domains D H such that dist(, H \ D) >. Let K denote the set of domains D K with dist(, H \ D) > 1. Suppose D K. Let f : D D is a conformal transformation. We say that f() is an analytic boundary point of D, if f extends to an analytic function in a neighborhood of the origin. More precisely, we say that the pair (f, ) gives a (locally) analytic prime end. As an example, suppose that D = C \ [ 1, 1]. Then D is a simply connected domain of the Riemann sphere C, and we can find a conformal transformation f : H D that sends to what we will call +, the positive-y side of in D. This map can be extended to an analytic map. Hence + is an analytic boundary point (prime end). Note that is also an analytic boundary point, but it is considered as a different point. What we define as analytic boundary point should probably be called analytic prime end. In this case we could define a point to be an analytic boundary point if there exists an f as above such that there is a neighborhood of such that f(d \ H) D =. While a point z D might correspond to many prime ends, it can correspond to at most two analytic prime ends. If z is an analytic boundary point, then there is a well-defined inward unit normal derivative n = n(z, D) pointing into D. (If z is a two-sided point, then each prime end has a normal derivative. Hence we consider n(z, D) as a function of the prime end z.) If f : D D is a map as above, then we write f(iy) = z + y f () n + O( y 2 ), y. If φ is a harmonic function on D with boundary value in a neighborhood of z, then we define φ on D = f 1 (D) by φ(w) = φ(f 1 (w)). Note that φ is harmonic with boundary value in an interval [ δ, δ] and hence y φ() = lim y y 1 φ(iy), is well defined. We define n φ(z) = lim y y 1 φ(z + yn) = lim y y 1 φ( f () 1 yi + O(y 2 )) = f () 1 y φ(z). 21

22 Proposition 5.1. Suppose D K. If z, w D with z, w 1/2, H D (z, ) = H H (z, ) [1 + O( z )]. G D (z, w) = G H (z, w) [1 + O( z )]. Proof. Note that H D+ (z, ) H D (z, ) H H (z, ) G D+ (z, w) G D (z, w) G H (z, w). Using conformal transformation, we can show the estimates for D = D +. H D+ (z, ) = H H (z, ) [1 + O( z )]. G D+ (z, w) = H H (z, w) [1 + O( z )]. Proposition 5.2. Suppose D K and h : D R is harmonic with h on [ x, x] for some x > 1. Then y h() = 2 π π h(e iθ ) sin θ dθ. Proof. By Schwarz reflection, we can extend h to a harmonic function on D {z : z < x} and from this we see that h is bounded and continuous on {z : z 1}. The optional sampling theorem implies that if z D +, then [ ] h(z) = E z h(b τd+ ) = h(w) hm D+ (z, dw) = 1 π h(e iθ ) H D+ (z, e iθ ) dθ. D + π Using conformal invariance (see Appendix) we can see that H D+ (iy, e iθ ) = 2y sin θ [1 + O(y)], y. Proposition 5.3. Suppose D K. If < ɛ 1/2, let D ɛ = D { z > ɛ} and τ ɛ = τ Dɛ. Then for z D with z 1, By definition, H Dɛ (z, ɛe iθ ) = π 2 Pz { B τɛ = ɛ} ɛ 1 sin θ [1 + O(ɛ)]. P z { B τɛ = ɛ} = 1 π ζ =ɛ H Dɛ (z, ζ) dζ = ɛ π Informally, we can write the conclusion of this proposition as d θ P { B τɛ = ɛe iθ B τɛ = ɛ } = sin θ dθ 2 π H Dɛ (z, ɛe iθ ) dθ. [1 + O(ɛ)]. More precisely, if < θ 1 < θ 2 < π, and V ɛ = V ɛ (θ 1, θ 2 ) = {ɛe iθ : θ 1 θ θ 2 }, then P z {B τɛ V ɛ B τɛ = ɛ} = [1 + O(ɛ)] θ2 θ 1 sin θ 2 dθ = [1 + O(ɛ)] cos θ 1 cos θ 2. 2 The statement of the proposition implies a uniformity in the estimate. We could restate the conclusion as follows: there exists c < such that for all D K, z 1, < ɛ 1/2, Pz { B τɛ = ɛ} 2 ɛ H D ɛ (z, ɛe iθ ) π sin θ c ɛ. 22

23 Proof. We fix < θ 1 < θ 2 < π, let V ɛ be as above, and let p = (cos θ 1 cos θ 2 )/2. Let U ɛ = {w H : ɛ < w < 1} and let η ɛ = τ Uɛ. Using conformal invariance, we can see that if ζ U ɛ with ζ = 3/4, then Let q be the supremum of P ζ {B ηɛ V ɛ B ηɛ = ɛ} = p [1 + O(ɛ)]. P ζ {B ηɛ V ɛ B ηɛ = ɛ}. Then q = p [1 + O(ɛ)]. where the supremum is over ζ = 3/4. Then if D K, we can see that P ζ {B τɛ V ɛ B τɛ = ɛ} q. The same argument works with q being the infimum rather than the supremum. Proposition 5.4. If D is a domain containing the origin with analytic boundary point z, H D (, z) = 1 2 ng(z). To check that the constant is correct, recall that we have normalized our quantities so that and hence n G D (, 1) = 1. Proof. Suppose that D K and z 1. Then Similarly, if y < ɛ/2, In particular, if y = ɛ 2, h D (, 1) = 1 2, G D(, x) = log x, H D (z, ) = E z [H D (B τɛ, ); B τɛ = ɛ] = [1 + O(ɛ)] E z [H H (B τɛ, ); B τɛ = ɛ] ɛ [1 + O(ɛ)] π = H Dɛ (z, ɛe iθ ) H H (ɛ e iθ, ) dθ π π = P z sin θ { B τɛ = ɛ} [1 + O(ɛ)] 2 H H(ɛ e iθ, ) dθ = ɛ 1 P z { B τɛ = ɛ} [1 + O(ɛ)] G D (z, iy) = E z [G D (B τɛ, iy); B τɛ = ɛ] π sin 2 θ 2 = [1 + O(ɛ)] E z [G H (B τɛ, iy); B τɛ = ɛ] ɛ [1 + O(ɛ)] π = H Dɛ (z, ɛe iθ ) G H (ɛe iθ, iy) dθ π π = P z sin θ { B τɛ = ɛ} [1 + O(ɛ)] 2 G H(ɛe iθ, iy) dθ = (y/ɛ) P z { B τɛ = ɛ} [1 + O(ɛ) + O(ɛ/y)] G D (z, iy) y Therefore, y G D (z, w) w= = 2 H D (z, ). = 2 H D (z, ) [1 + O(y 1/2 )]. dθ π sin 2 θ dθ 23

24 Corollary 5.5. If D K and z > 1, then 5.1 Excursion measure H D (z, ) = 1 π π G D (e iθ, z) sin θ dθ. If D is a domain and z is an analytic boundary point, we define the (point-to-set) excursion measure E D (z, ) to be the derivative of the harmonic measure, E D (z, V ) = n hm D (z, V ). If D is an open set, non necessarily connected, we define E D (z, V ) to be E D (z, V ) where D is the connected component containing z on the boundary. The measure E D (x, ) is an infinite measure on D, but if dist(z, V ) >, then E D (z, V ) <. If V is an analytic arc, we can write n hm D (z, V ) = 1 n H D (z, w) dw = 1 H D (z, w) dw. π π V Note that if f : D f(d) is a conformal transformation that is analytic in a neighborhood of z, E D (z, V ) = f (z) E f(d) (f(z), f(v )). If V 1, V 2 are two analytic arcs, we define the (set-to-set) excursion measure E D (V 1, V 2 ) = E D (z, V 1 ) dz = H D (z, w) dw dz. V 1 V 2 The important fact is that the set-to-set excursion measure is a conformal invariant. Proposition 5.6. If f : D f(d) is a conformal transformation that is analytic on the arcs V 1, V 2 D, then E D (V 1, V 2 ) = E f(d) (f(v 1 ), f(v 2 )). Since the set-to-set excursion measure is a conformal it is well defined even if the boundary is not analytic. For example, if V 1, V 2 are on the same connected component of the boundary, we can first map D to H mapping this component to the real line. If they are in different components K 1, K 2, then we can first map Ĉ \ (K 1 K 2 ) to an annulus. Proposition 5.7. Suppose D is a domain and z is a locally analytic point. Suppose D D and D, D 1 agree in a neighborhood of z. Let w D \ D. Then H D (w, z) = 1 G D (ζ, w) E D (z, dζ). 2 D 1 If D D is analytic, we can write H D (w, z) = 1 2π V 1 D 1 G D (ζ, w) H D1 (z, ζ) dζ. As an example (and a check on the constants), suppose that D = D, D 1 = A r = {e r < z < 1}, z = 1, w =. Then H D (, 1) = 1/2, and E Ar (1, C r ) = 1 r, G D(e r+iθ, ) = r, H Ar (1, e r+iθ ) = er [1 + o(1)]. 2r V 24

25 Proof. We first consider the case with z =, D, D K. The function h(ζ) = G(ζ, w) is a bounded harmonic function on D. Therefore, for y < 1, h(iy) = E iy [ h(b τd ) ] = G(ζ, w) hm D (iy, dζ). D Letting y, we get 2H D (, w) = n G(, w) = E iy [ h(b τd ) ] = G(ζ, w) E D (, dζ). D Examples If A r = {e r < z < 1} is the annulus with boundaries C, C r, then we have see that E Ar (e iθ, C r ) = 1 r, and hence E Ar (C, C r ) = 2π E Ar (e iθ, C r ) dθ = 2π r. In particular, we see that if r s, then E Ar (C, C r ) E As (C, C s ). From this we can see that A r and A s are not conformally equivalent. Let R L = {x + iy : < x < L, < y < π} and let 1 = [, iπ], 2 = 2,L = [L, L + iπi] be the vertical boundaries. Let h be the harmonic function on R L with boundary value 1 on 2 and on R L \ 2. This can be found by separation of variables, x h(iy) = 4 π h(x + iy) = 4 π cosh(nx) sin(ny) sinh(nl) n=1 E RL ( 1, 2 ) = π n=1 sinh(nx) sin(ny), n sinh(nl) = 4 sin y π e L [ 1 + O(e L ) ], L. x h(iy) dy = 8 π e L [ 1 + O(e L ) ], L. Although it is not immediately obvious from the last expression, one can use the definition to see that the function L E RL ( 1, 2 ) is strictly decreasing. Let D = D(a, b) = {a < Im(z) < b} with boundaries I a = {Im(z) = a}, I b = {Im(z) = b}. Then, the gambler s ruin estimate implies that if x+ia I a, then E D (x+ia, I b ) = 1/(b a), and hence if V I a, E D (V, I b ) = E D (I b, V ) = 1 l(v ), b a where l denotes Lebesgue measure. More generally, suppose that D H is a domain with H {Im(z) a} = D {Im(z) a}. The strong Markov property, implies that if b > a, then Then E D (I b, V ) = 1 b a hm D {Im(z)<b} (x + ia, V ) dx lim b E D(I b, V ) = hm D (x + ia, V ) dx (12) b 25