Differentiability of stochastic flow of reflected Brownian motions

We prove that a stochastic flow of reflected Brownian motions in a smooth multidimensional domain is differentiable with respect to its initial position. The derivative is a linear map represented by a multiplicative functional for reflected Brownian motion. The method of proof is based on excursion theory and analysis of the deterministic Skorokhod equation.


Introduction
This article contains a result on a stochastic flow X x t of reflected Brownian motions in a smooth bounded domain D ⊂ R n , n ≥ 2. We will prove that for some stopping times σ r defined later in the introduction, the mapping x → X x σr is differentiable a.s., and we will identify the derivative with a mapping already known in the literature.
We start with an informal overview of our research project. We call a pair of reflected Brownian motions X t and Y t in D a synchronous coupling if they are both driven by the same Brownian motion. To make things interesting, we assume that X 0 = Y 0 . The ultimate goal of the research project of which this paper is a part, is to understand the long time behavior of V t := X t − Y t in smooth domains. This project was started in [BCJ], where synchronous couplings in 2-dimensional smooth domains were analyzed. An even earlier paper [BC] was devoted to synchronous couplings in some classes of planar non-smooth domains.
Multidimensional domains present new challenges due to the fact that the curvature of ∂D is not a scalar quantity and it has a significant influence on V t . Eventually, we would like to be able to prove a theorem analogous to the main result of [BCJ], Theorem 1.2. That theorem shows that |V t | goes to 0 exponentially fast as t goes to infinity, provided a certain parameter Λ(D) characterizing the domain D is strictly positive. The exponential rate at which |V t | goes to 0 is equal to Λ(D). The proof of Theorem 1.2 in [BCJ] is extremely long and we expect that an analogous result in higher dimensions will not be easier to prove.
This article and its predecessor [BL] are devoted to results providing technical background for the multidimensional analogue of Theorem 1.2 in [BCJ].
Suppose that |V t | is very small for a very long time. Then we can think about the evolution of V t as the evolution of an infinitesimally small vector, or a differential form, associated to X t . This idea is not new-in fact it appeared in somewhat different but essentially equivalent ways in [A,IK1,IK2,H]. The main theorem of [BL] showed existence of a multiplicative functional governing the evolution of V t , using semi-discrete approximations. The result does not seem to be known in this form, although it is close to theorems in [A,IK1,H]. However, the main point of [BL] was not to give a new proof to a slightly different version of a known result but to develop estimates using excursion techniques that are analogous to those in [BCJ], and that can be applied to study V t .
Suppose that for every x ∈ D we have a reflecting Brownian motion X x t in D starting from X x 0 = x, and all processes X x t , x ∈ D, are driven by the same Brownian motion. For a fixed x 0 ∈ D, let σ r be the first time t when the local time of X x 0 on ∂D reaches the value r.
The main result of the present article, Theorem 3.1, says that for every r > 0, the mapping x → X x σr is differentiable at x = x 0 a.s., and the derivative is a linear mapping defined in Theorem 3.2 of [BL].
The differentiability in the initial data was proved in [DZ] for a stochastic flow of reflected diffusions. The main difference between our result and that in [DZ] is that that paper was concerned with diffusions in (0, ∞) n , and our main goal is to study the effect of the curvature of ∂D. The results in [DZ] have been transferred to SDEs in a convex polyhedron with possibly oblique reflection-see [An]. Differentiability of a stochastic flow of diffusions (without reflection) in the initial condition is a classical topic, see, e.g., [K], Chap. II, Thm. 3.1.
Our main result can be considered a pathwise version of theorems proved in [A,H,IK1] and [IK2], Section V.6 (see also references therein). In a sense, we pass to the limit in a different order than the authors of the cited publications. Hence, our theorem is closer in spirit to the results in [LS, S, DI, DR]. There is a difference, though. The articles [LS, S, DI, DR] are concerned with the transformation of the whole driving path into a reflected path (the "Skorokhod map"). At this level of generality, the Skorokhod map was proved to be Hölder with exponent 1/2 in Theorems 1.1 an 2.2 of [LS] and Lipschitz in Proposition 4.1 in [S]. See [S] for further references and history of the problem. Under some other assumptions, the Skorokhod map was proved to have the Lipschitz property in [DI, DR]. Articles [MM] (Lemma 5.2) and [MR] contain results about directional derivatives of the Skorokhod map in an orthant, without and with oblique reflection, respectively. The first theorems on existence and uniqueness of solutions to the stochastic differential equation representing reflected Brownian motion were given in [T]. Some results on stochastic flows of reflected Brownian motions were proved in an unpublished thesis [W]. Synchronous couplings in convex domains were studied in [CLJ1,CLJ2], where it was proved that under mild assumptions, V t is not 0 at any finite time.
The proof of the main result depends in a crucial way on ideas developed in a joint project with Jack Lee ( [BL]). I am indebted to him for his implicit contributions to this paper. I am grateful to Sebastian Andres, Peter Baxendale, Elton Hsu and Kavita Ramanan for very helpful advice.

Preliminaries
2.1. General notation. All constants are assumed to be strictly positive and finite, unless stated otherwise. The open ball in R n with center x and radius r will be denoted B(x, r).
We will use d( · , · ) to denote the distance between a point and a set.
2.2. Differential geometry. We will review some notation and results from [BL]. We will be concerned with a bounded domain D ⊂ R n , n ≥ 2, with a C 2 boundary ∂D. We may consider M := ∂D to be a smooth, properly embedded, orientable hypersurface (i.e., submanifold of codimension 1) in R n , endowed with a smooth unit normal vector field n.
We consider M as a Riemannian manifold with the induced metric. We use the notation · , · for both the Euclidean inner product on R n and its restriction to the tangent space T x M for any x ∈ M, and | · | for the associated norm. For any x ∈ M, let π x : R n → T x M denote the orthogonal projection onto the tangent space T x M, so π x z = z − z, n(x) n(x), (2.1) and let S(x) : T x M → T x M denote the shape operator (also known as the Weingarten map), which is the symmetric linear endomorphism of T x M associated with the second fundamental form. It is characterized by  S(γ(t))γ ′ (t) n(γ(t)) = v ′ (t) + v(t), ∂ t (n • γ)(t) n(γ(t)).
The eigenvalues of S(x) are the principal curvatures of M at x, and its determinant is the Gaussian curvature. We extend S(x) to an endomorphism of R n by defining S(x)n(x) = 0.
It is easy to check that S(x) and π x commute, by evaluating separately on n(x) and on v ∈ T x M.
For any linear map A : R n → R n , we let A denote the operator norm.
We recall two lemmas from [BL].
Lemma 2.1. For any bounded C 2 domain D ⊂ R n and c 1 , there exists c 2 such that the following estimates hold for all x, y ∈ ∂D, 0 ≤ l, r ≤ c 1 , b ≥ 0 and z ∈ R n : (2.4) e lS(x) − e lS(y) ≤ c 2 l |x − y|.
(2.13) 2.3. Probability. Recall that D ⊂ R n , n ≥ 2, is an open connected bounded set with C 2 boundary and n(x) denotes the unit inward normal vector at x ∈ ∂D. Let B be standard d-dimensional Brownian motion and consider the following Skorokhod equation, (2.14) Here x ∈ D and L x is the local time of X x on ∂D. In other words, L x is a non-decreasing continuous process which does not increase when X x is in D, i.e., ∞ 0 1 D (X x t )dL x t = 0, a.s. Equation (2.14) has a unique pathwise solution (X x , L x ) such that X x t ∈ D for all t ≥ 0 (see [LS]). The reflected Brownian motion X x is a strong Markov process. The results in [LS] are deterministic in nature, so with probability 1, for all x ∈ D simultaneously, (2.14) has a unique pathwise solution (X x , L x ). In other words, there exists a stochastic flow (x, t) → X x t , in which all reflected Brownian motions X x are driven by the same Brownian motion B.
We fix a point z 0 ∈ D. We will abbreviate (X z 0 , L z 0 ) by writing (X, L).
We need an extra "cemetery point" ∆ outside R n , so that we can send processes killed at a finite time to ∆. For s ≥ 0 such that X s ∈ ∂D we let ζ(e s ) = inf{t > 0 : X s+t ∈ ∂D}.
Let σ be the inverse of local time L, i.e., σ t = inf{s ≥ 0 : L s ≥ t}, and E r = {e s : s < σ r }. Fix some r, ε > 0 and let {e u 1 , e u 2 , . . . , e um } be the set of all excursions e ∈ E r with |e(0) − e(ζ−)| ≥ ε. We assume that excursions are labeled so that u k < u k+1 for all k and we let ℓ k = L u k for k = 1, . . . , m. We also let u 0 = inf{t ≥ 0 : X t ∈ ∂D}, ℓ 0 = 0, ℓ m+1 = r, and ∆ℓ k = ℓ k+1 − ℓ k . Let x k = e u k (ζ−) be the right endpoint of excursion e u k for k = 1, . . . , m, Recall from Section 2.2 that S denotes the shape operator and π x is the orthogonal pro- ( 2.15) Note that all concepts based on excursions e u k depend implicitly on ε > 0, which is often suppressed in the notation. Let A ε r denote the linear mapping v 0 → v r . We will impose a geometric condition on ∂D. To explain its significance, we consider D such that ∂D contains n non-degenerate (n − 1)-dimensional balls, such that vectors orthogonal to these balls are orthogonal to each other. If the trajectory {X t , 0 ≤ t ≤ r} visits the n balls and no other part of ∂D, then it is easy to see that A ε r = 0. To avoid this uninteresting situation, we impose the following assumption on D.
The following theorem has been proved in [BL].
Theorem 2.5. Suppose that D satisfies all assumption listed so far in Section 2. Then for every r > 0, a.s., the limit A r := lim ε→0 A ε r exists and it is a linear mapping of rank n − 1. For any v 0 , with probability 1, A ε r v 0 → A r v 0 as ε → 0, uniformly in r on compact sets.
represents the solution to the following ODE, In the 2-dimensional case, and only in the 2-dimensional case, we have an alternative the curvature at x ∈ ∂D, that is, the eigenvalue of S(x). Then The remaining part of this section is a short review of the excursion theory. See, e.g., [M] for the foundations of the excursion theory in the abstract setting and [Bu] for the special case of excursions of Brownian motion. Although [Bu] does not discuss reflected Brownian motion, all results we need from that book readily apply in the present context. An "exit system" for excursions of the reflected Brownian motion X from ∂D is a pair (L * t , H x ) consisting of a positive continuous additive functional L * t and a family of "excursion laws" {H x } x∈∂D . In fact, L * t = L t ; see, e.g., [BCJ]. Recall that ∆ denotes the "cemetery" point outside R n and let C be the space of all functions f : [0, ∞) → R n ∪ {∆} which are continuous and take values in R n on some interval [0, ζ), and are equal to ∆ on [ζ, ∞).
For x ∈ ∂D, the excursion law H x is a σ-finite (positive) measure on C, such that the canonical process is strong Markov on (t 0 , ∞), for every t 0 > 0, with transition probabilities of Brownian motion killed upon hitting ∂D. Moreover, H x gives zero mass to paths which do not start from x. We will be concerned only with "standard" excursion laws; see Definition 3.2 of [Bu]. For every x ∈ ∂D there exists a unique standard excursion law H x in D, up to a multiplicative constant.
Recall that excursions of X from ∂D are denoted e s and σ t = inf{s ≥ 0 : L s ≥ t}. Let I be the set of left endpoints of all connected components of (0, ∞) {t ≥ 0 : X t ∈ ∂D}. The following is a special case of the exit system formula of [M], where W t is a predictable process and f : C → [0, ∞) is a universally measurable function which vanishes on excursions e t identically equal to ∆. Here H x The normalization of the exit system is somewhat arbitrary, for example, if (L t , H x ) is an exit system and c ∈ (0, ∞) is a constant then (cL t , (1/c)H x ) is also an exit system. Let P y D denote the distribution of Brownian motion starting from y and killed upon exiting D. Theorem 7.2 of [Bu] shows how to choose a "canonical" exit system; that theorem is stated for the usual planar Brownian motion but it is easy to check that both the statement and the proof apply to the reflected Brownian motion in R n . According to that result, we can take L t to be the continuous additive functional whose Revuz measure is a constant multiple of the surface area measure on ∂D and H x 's to be standard excursion laws normalized so for any event A in a σ-field generated by the process on an interval [t 0 , ∞), for any t 0 > 0.
The Revuz measure of L is the measure dx/(2|D|) on ∂D, i.e., if the initial distribution of X is the uniform probability measure µ in D then E µ 1 0 1 A (X s )dL s = A dx/(2|D|) for any Borel set A ⊂ ∂D, see Example 5.2.2 of [FOT]. It has been shown in [BCJ] that (L t , H x ) is an exit system for X in D, assuming the above normalization.
3. Differentiability of the stochastic flow in the initial parameter Recall that z 0 ∈ D is a fixed point. Our main result is the following theorem.
Theorem 3.1. Suppose that D satisfies all assumptions of Section 2. Then for every r > 0 Note that in the above theorem, both processes are observed at the same random time σ r , the inverse local time for the process X z 0 . In other words, we do not consider The proof of the theorem will consist of several lemmas. We start by introducing some notation.
We will prove the theorem only for r = 1, and we will suppress r in the notation from now on. It is clear that the same proof applies to any other value of r.
It follows from Lemma 3.2 below that we can find a constant c * and a sequence of stopping times T k such that T k → ∞, a.s., and sup z∈D L z T k ≤ kc * for all k. We fix some integer k * ≥ 1 and let σ * = σ 1 ∧ T k * . The dependence of σ * on k * and c * will be suppressed in the notation.
In much of the paper, we will consider "fixed" starting points z 0 and y. We will write X t = X z 0 t and Y t = X y t , so that X 0 = z 0 and Y 0 = y. Later in this section, we will often take We fix some (small) a 1 , a 2 > 0. We will impose some conditions on the values of a 1 and a 2 later on. Let S 0 = U 0 = inf{t ≥ 0 : X t ∈ ∂D} and for k ≥ 1 define The filtration generated by the driving Brownian motion will be denoted F t . As usual, for a stopping time T , F T will denote the σ-field of events preceding T .
Since D is bounded and ∂D is C 2 , there exists δ 0 > 0 such that if x ∈ D and d(x, ∂D) < δ 0 then there is only one point y ∈ ∂D with |x − y| = d(x, ∂D). We will call this point . For all other points, we let Π x = z * , where z * ∈ ∂D is a fixed reference point.
, n(y 1 ) ≤ r 1 /32 + r 1 /64 < r 1 /16. Thus π y 1 X z t 3 − X z have sup s k ≤s,t≤s k+1 |B t − B s | ≤ r 1 /64 for k = 0, 1, . . . , m − 1. We can repeat the above argument on each interval [s k , s k+1 ] to obtain sup z∈D L z s k+1 − L z s k ≤ r 1 /8, and, consequently, sup z∈D L z sm ≤ mr 1 /8. This proves the first assertion of the lemma. By continuity of Brownian motion, for any fixed u, with probability 1, one can find a (random) integer m < ∞ and a sequence 0 = s 0 < s 1 < · · · < s m = u such that sup s k ≤s,t≤s k+1 |B t − B s | ≤ r 1 /64 for k = 0, 1, . . . , m − 1. The second assertion of the lemma follows from this and the first part of the lemma.
Recall σ * defined at the beginning of this section. Lemma 3.3. There exists c 1 such that a.s., for all t ≤ σ * and y, z ∈ D, we have |X y t −X z t | < c 1 |y − z|.
Proof. Fix any y, z ∈ D, let L * t = L y t + L z t , and σ * t = inf{s ≥ 0 : L * s ≥ t}. It follows from (2.10) that x − y, n(x) ≤ c 2 |x − y| 2 for all x ∈ ∂D and y ∈ D. This and (2.14) imply that, Recall from the beginning of this section that sup z∈D L z σ * ≤ k * c * < ∞. This and the definitions of σ * and σ * r imply that σ * ≤ σ * 2k * c * . Hence, |X z t − X y t | < e 4k * c * c 2 |y − z| for all t ≤ σ * .
Recall the notation from the beginning of this section. In particular, ε = |X 0 − Y 0 |.
Lemma 3.5. For some c 1 , Proof. It follows from (3.19) in [BL] that, for any β < 1, some c 2 , and all ε * > 0, The main difference between (3.6) and (3.7) is the presence of the condition X t ∈ ∂D in the supremum. Let Recall that n ≥ 2 is the dimension of the space R n into which D is embedded. Standard estimates show that if T ∂D = inf{t ≥ 0 : X t ∈ ∂D}, x ∈ ∂D, y ∈ ∂B(x, r) ∩ D, r > ρ, and X 0 = y, then We have for every x ∈ ∂D and b > 0, The upper bound in the last estimate follows from (2.17) and Lemma 3.4 (i). The lower bound can be proved in a similar way.
Lemma 3.6. There exists c 1 such that if X 0 ∈ ∂D then, (3.11) It follows from Lemma 3.5 that, for some c 2 , Estimate (3.10) and the exit system formula (2.16) imply that The lemma follows by combining the last estimate with (3.11) and (3.12).
Recall also that ε * is the parameter used in the definition of ξ j and x * j at the beginning of this section.
(ii) The following remark applies to Lemma 3.7 and all other lemmas. Typically, their proofs require that we assume that |X 0 − Y 0 | is bounded above. However, in many cases, the quantity that is being estimated is bounded above by a universal constant, for trivial reasons. Hence, by adjusting the constant appearing in the estimate, we can easily extend the lemmas to all values of |X 0 − Y 0 |.
Let c 6 ∈ (0, 1/12) be a small constant whose value will be chosen later. Let First we will assume that r ≤ ε 1 /2. We will show that We will argue by contradiction. Assume that A 1 holds and )dL t ≥ c 6 r. By (2.12), we may assume that ε 0 > 0 is so small that for r ≤ r 0 < ε 0 and x ∈ B(X 0 , 2c 6 r), we have |π X 0 (n(x))| ≤ 4νc 6 r. This and the estimate (3.13) By (2.9), we can assume that r 0 and ε 0 are so small that if for some y ∈ ∂D we have (3.14) Since d(Y 0 , ∂D) = r, it is easy to see that if r 0 > 0 is sufficiently small then for r ≤ r 0 and , and, therefore, T 4 ]. Since we have assumed that This, (2.14) and (3.13) imply that The last estimate and (3.14) yield holds. This and the We will next show that if A 1 holds then S 1 ≤ T 4 . Assume that A 1 holds and let Recall that c 6 ≤ 1/12 and r ≤ ε 1 /2. It follows that We have from the definition of T 3 that The definition of T 3 and (3.15) imply that for Hence, We have proved that T 5 ≥ T 4 on A 1 , so . We use the definition of x * , (3.18), (3.20) and (2.13) to see that We use the fact that Y T 7 − Y T 4 = X T 7 − X T 4 and apply (2.13), (3.18) and (3.21), to obtain, We combine this estimate with (3.22) to see that

This bound and (3.17) yield
We make c 6 > 0 smaller, if necessary, so that 64c 6 ν ≤ a 2 . Then d( and completes the proof that if A 1 holds then S 1 ≤ T 4 . Assume that A 1 holds and suppose that n(X 0 ), T 4 0 n(X t )dL t ≥ 20r. We will show that these assumptions lead to a contradiction. It follows from (3.15) that for ( 3.24) This implies that In view of the definition of T 5 and (3.14), This contradicts (3.24) so we conclude that if A 1 holds then n(X 0 ), Note that n(X 0 ), n(x) ≥ 1/2 for all x ∈ ∂D ∩ B(X 0 , 2c 6 r), assuming that ε 0 > 0 is small T 4 ] such that X t ∈ ∂D. This and (3.27) imply that, and, therefore, By (3.24) and the fact that This, the definition of T 5 and the fact that T 5 ≥ T 4 on A 1 imply that for t ≤ T 4 , we have |X t − X 0 | ≤ 45r. If we take c 1 = 45 then this and (3.19) show that on A 1 , T 4 ≤ T 1 and, therefore, It is easy to see that P(A 1 ) > p 1 for some p 1 > 0 which depends only on c 6 . This completes the proof of part (i) in the case r ≤ ε 1 /2, with c 1 = 45 and c 2 = 40.
Next we will prove that Recall that T 7 = sup{t ≤ T 4 : X t ∈ ∂D} and note that we can use (3.31) because This contradicts (3.29) because T 7 ≤ T 4 ≤ T 11 . This proves that if A 2 holds then Recall the definition of T 11 and the fact that This completes the proof that A 2 ⊂ A 3 .
It is easy to see that P(A 2 ) > p 2 , for some p 2 > 0. It follows that P(A 3 ) > p 2 .
We may now apply the strong Markov property at the stopping time T 9 and repeat the argument given in the first part of the proof, which was devoted to the case r ≤ ε 1 /2. It is straightforward to complete the proof of part (i), adjusting the values of c 1 , c 2 , ε 0 , r 0 and p 0 , if necessary.
(ii) We will restart numbering of constants, i.e., we will use c 6 , c 7 , . . . , for constants unrelated to those with the same index in the earlier part of the proof.
Let c 1 , c 2 , ε 0 and r 0 be as in part (i) of the lemma, ε 2 = ε 0 ∧ r 0 , and ε 1 ≤ ε 2 . Recall that We will estimate Ed(Y T k 5 , ∂D). By Lemma 3.4 (i) and the definition of T k 1 , on the event s. Note that between times T k−1 5 and T k 4 , the process Y t does not hit the boundary of D. Between times T k 4 and T k 5 , the process X t does not hit ∂D. If Y t does not hit the boundary on the same interval, it is elementary to see that d( , and assume that t * is the largest time with this property. Since both Y t * and X T k 5 belong to ∂D, easy geometry shows that in this case d(Y T k 5 , ∂D) ≤ c 10 ε 2 2 −j . This completes the proof that d(Y T k 5 , ∂D) ≤ R + c 8 ε 2 2 −j , a.s. Let j 0 be the smallest integer such that 2 −j 0 ≥ diam(D) and let j 1 be the largest integer . By the strong Markov property applied at T k−1 5 and part (i) of the lemma, on the event By the strong Markov property and induction, This, (3.40) and (3.41) imply, We assume without loss of generality that p 0 > 0 is so small that (1 − p 0 )p −1 0 > 1. We obtain by induction, 2 ). Note that, by (3.34) and (3.37), Hence, (3.44) It follows from this and (3.43) that If we assume that ε 2 > 0 is sufficiently small, this is bounded by c 19 (r + ε 3 2 ).
(iii) We will restart numbering of constants, i.e., we will use c 6 , c 7 , . . . , for constants unrelated to those with the same index in the earlier part of the proof.
If we assume that ε 2 > 0 is sufficiently small, this is bounded by c 11 | log r|(r + ε 3 2 ). (iv) Once again, we will restart numbering of constants, i.e., we will use c 6 , c 7 , . . . , for constants unrelated to those with the same index in the earlier part of the proof.
Recall that j 0 is the smallest integer such that 2 −j 0 ≥ diam(D). Let j 3 be the smallest j with the property that 2 −j ≤ d(Y T k 5 , ∂D). It follows from (3.38) that for any β 2 < 1, on the . It follows from the definition of S 1 that |Π(X S * 1 ) − X S * 1 | ≤ c 11 ε 2 2 if S 1 < σ * ∧ τ + (ε 2 ). In the case when S * 1 = σ * ∧ τ + (ε 2 ), the distance between X and Y is increasing at this instance, so it is easy to see that the vector X S * 1 − Y S * 1 must also have a position such that |Π(X S * 1 ) − X S * 1 | ≤ c 11 ε 2 2 . (3.46) Recall that we assume that X 0 ∈ ∂D, |X 0 − Y 0 | = ε 1 , d(Y 0 , ∂D) = r. Recall also that ε * is the parameter used in the definition of ξ j and x * j at the beginning of this section. It follows from (3.33)-(3.37) that if ε * ≥ c 1 ε 2 then at most one ξ i may belong to any given interval (T k If we assume that ε 2 > 0 is sufficiently small, this is bounded by c 13 ε β 2 2 (r + ε 3 2 ). Recall definitions of σ * and S 1 , and Lemma 3.3. There exists c 14 such that if ε 1 ≤ c 14 ε 2 then σ * < τ + (ε 2 ). Hence, if ε 1 ≤ c 14 ε 2 then The following estimate can be proved just like We use this estimate, (3.47), the strong Markov property at S k , and the definition of S k to see that Lemma 3.9. There exist c 1 and a 0 > 0 such that for a 1 , a 2 < a 0 , if |X 0 − Y 0 | = ε then a.s., for every k ≥ 1, on the event U k < σ * , Proof. It is elementary to see that one can choose c 1 , a 0 > 0 and ε 0 > 0 so that for a 1 < a 0 , ε ≤ ε 0 , x ∈ ∂D, y ∈ D, |x − y| ≤ ε, z ∈ ∂D, |x − z| ≤ 2a 1 ε and |y − z| ≤ 2a 1 ε, then

by Lemma 3.3. It follows easily from
(3.1) that we can adjust the values of c 1 and ε 0 and choose a 2 > 0 so that if |X 0 −Y 0 | = ε ≤ ε 0 then on the event S k < σ * , We will show that A = ∅. Suppose otherwise and let T 1 = inf A. Then We must have either X T 1 ∈ ∂D or Y T 1 ∈ ∂D. It follows from (3.48) that either X T 1 / ∈ ∂D or Y T 1 / ∈ ∂D. Suppose without loss of generality that X T 1 ∈ ∂D and Y T 1 / ∈ ∂D. Then by (3.48), By the definition of T 1 , for every δ > 0, L t must increase on the interval [T 1 , T 1 + δ]. It is easy to see that this implies that the function is decreasing on the interval [T 1 , T 1 + δ 1 ], for some δ 1 > 0. This contradicts the definition of T 1 . Hence, for all t ∈ [S k , U k ], In particular, The lemma follows from the above estimate and (3.49).
Lemma 3.10. There exists c 1 such that if |X 0 − Y 0 | ≤ ε then for every k, Proof. We use the strong Markov property at the hitting time of ∂D by X and Lemma 3.7 (ii) to see that (3.50) We will estimate (L S k+1 −L U k )1 {U k <τ + (ε)} for k ≥ 1. Fix some k ≥ 1 and assume that U k < τ + (ε). Note that d( Let j 0 be the greatest integer such that 2 −j 0 is greater than the diameter of D and let j 1 be the least integer such that 2 −j 1 ≤ |X U k − Y U k |. By Lemma 3.4, for j 0 ≤ j ≤ j 1 , (3.51) Next we will estimate d(Y T 1 , ∂D). Between times U k and T 1 , the process X t does not hit ∂D. If Y t does not hit the boundary on the same interval, it is elementary to see from Lemma 3.9 that for j 0 ≤ j ≤ j 1 , Suppose that for some t * ∈ [U k , T 1 ] we have Y t * ∈ ∂D, and assume that t * is the largest time with this property. If t * = T 1 then d(Y T 1 , ∂D) = 0. Otherwise we must have τ + (ε) > t * , Since both Y t * and X T 1 belong to ∂D, easy geometry s. By Lemma 3.7 (ii) and the strong Markov property applied at U k , Hence, using (3.51), It is elementary to check that and the conditional distribution of L U k − L S k given {S k < τ + (ε)} is stochastically bounded by an exponential random variable with mean c 14 It follows that is a submartingale with respect to the filtration F * m = F X,Y S m+1 . If We let i → ∞ and obtain by the monotone convergence Hence, This and (3.50) imply the lemma.
Recall parameters a 1 and a 2 and operator G k defined in (3.2).
Estimate (3.52) and C 2 -smoothness of ∂D can be used to show that for any c 2 one can choose small a 1 , a 2 > 0 and ε 0 > 0 so that for every k ≥ 1 and all t ∈ [S k , U k ] such that X t ∈ ∂D, We obtain from (2.14) and the triangle inequality, The expression on the last line is equal to zero for elementary reasons, so The first term on the right hand side is equal to 0 by the definition of Θ. It is easy to see that this claim holds even if the definition of Θ involves the division by 0. We have obtained It follows from the definitions of S k , U k and Π x that for sufficiently small a 1 and a 2 , we and a similar formula holds for X in place of Y on the left hand side. Hence, by (2.7), for some c 3 , This shows that if we take a 1 sufficiently small then |Θ| ≤ c 1 .
We use (3.53) to derive the following estimate, We combine (3.54)-(3.55) to see that For any c 2 , we can choose small ε 0 so that This and (3.56) imply that We obtain the lemma by choosing sufficiently small c 2 .
Lemma 3.12. If a 1 is sufficiently small then for some c 1 , ε 0 > 0 and all ε < ε 0 , if |X 0 −Y 0 | = ε then a.s., for all k ≥ 1, Proof. Let w = n(Π(X S k )). It follows from the definition of U k that for t ∈ [S k , U k ]. This and (2.8) imply that for some c 3 and t ∈ [S k , U k ], (3.58) We appeal to (2.13) to see that if a 1 is sufficiently small and y ∈ ∂D and z ∈ D are such (3.60) We claim that I = ∅. Suppose otherwise and let t 1 = inf I and t 2 = sup{t ∈ [S k , t 1 ] : Y t ∈ ∂D}, with the convention that sup ∅ = S k . By (3.57), (3.59) and (3.60), This contradicts the definition of t 1 , so we see that I = ∅. Similarly, one can prove that This and (3.57)-(3.58) yield, By the definition of σ * , L y U k − L y S k ≤ c 5 , so the above estimate implies The lemma follows from the last two estimates.
Lemma 3.13. For some c 1 there exist a 0 , ε 0 > 0 such that if a 1 , a 2 ∈ (0, a 0 ), ε ≤ ε 0 and |X 0 − Y 0 | = ε then for all k ≥ 1, Let T 1 = inf{t ≥ U k : X t ∈ ∂D}. It follows from Lemma 3.3 and definition of U k that d(X U k , ∂D) ≤ c 3 ε. Let j 0 be the smallest integer such that ε2 j 0 is greater than the diameter of D. Lemma 3.4 (i) shows that for some c 4 and all j = 1, 2, . . . , j 0 , By Lemma 3.7 (iii), the strong Markov property applied at T 1 , and Chebyshev's inequality, The fact that |X S k − X U k | ≤ c 6 ε and the last two estimates show that Lemma 3.14. For some c 1 there exist a 0 , ε 0 > 0 such that if a 1 , a 2 ∈ (0, a 0 ), ε ≤ ε 0 and |X 0 − Y 0 | = ε then for all k ≥ 1, Proof. Fix some k and let We will assume from now on that X T 1 ∈ ∂D. The rest of the argument It follows from Lemma 3.3 and definition of U k that d(X U k , ∂D) ≤ c 2 ε 1 . Let j 0 be the smallest integer such that ε 1 2 j 0 is greater than the diameter of D. Lemma 3.4 shows that for some c 3 and all j = 1, 2, . . . , j 0 , (3.61) By (2.9), we can choose c 4 so small that for x ∈ ∂D ∩ B(X T 1 , 5c 4 ε 1 ), By the definition of σ * , |Y t − X t | ≤ c 5 ε 1 for t ≤ σ * . We make c 4 smaller, if necessary, so that, in view of (2.11), | y − x, n(z) | ≤ a 2 ε 2 1 /400, (3.63) assuming that x, y, z ∈ ∂D, |y − z| ≤ (c 5 + 5c 4 )ε 1 and |x − y| ≤ 10c 4 ε 1 .
The following definitions contain a parameter c 6 , the value of which will be chosen later. Let Note that neither X nor Y touches the boundary of D between times U k and T 1 , so Hence, by Lemma 3.9 and the strong Markov property applied at The angle between n(Π(X U k )) and n(X T 1 ) is bounded by c 8 ε 1 2 J because ∂D is C 2 . This and (3.64) imply that Let k 1 be such that c 9 ε 1 2 J ≤ 1/10 if J ≤ k 1 , and let F 1 = {J ≤ k 1 }. If F 1 holds then (3.65) implies that, Case(i). This case is devoted to an estimate of the random variable in the statement of the lemma assuming that Let c 11 = 5c 4 and We will show that T 4 = T 2 ∧ T 3 , if ε (and, therefore, ε 1 ) is sufficiently small. By (2.11), x − y, n(X T 1 ) ≤ c 12 ε 2 1 (3.68) for all x, y ∈ B(X T 1 , c 11 ε 1 ) such that x ∈ ∂D and y ∈ D. Since T 5 ≤ T 3 , we have This and (3.68) imply that For x ∈ ∂D ∩ B(X T 1 , c 11 ε 1 ) we have by (2.8), for small ε 1 , n(x), n(X T 1 ) ≥ 1 − c 14 ε 2 1 ≥ 1/2.
so, in particular, This and (3.67) imply that ≤ c 28 ε 2 1 + c 10 ε 2 1 2 J ≤ c 29 ε 2 1 2 J . Recall that we assume that A 1 holds so that T 3 ≤ T 2 . By (2.10), for x ∈ D ∩ B(X T 1 , c 19 ε 1 ), x − X T 1 , n(X T 1 ) ≥ −c 30 ε 2 1 , so, in view of (3.79), We now choose the parameter c 6 in the definition of T 3 so that −c 6 + c 29 ≤ −2c 30 . We will show that given this choice of c 6 , we must have for the same range of t's. It follows from (3.87) and from the definition of T 3 that ≤ −c 6 ε 2 1 2 J + c 29 ε 2 1 2 J ≤ −2c 30 ε 2 1 . This contradicts (3.88), so we conclude that Y must cross ∂D between times T 1 and T 3 .
Hence, T 7 is well defined. Since we are assuming that A 1 holds, T 7 ≤ T 3 = T 6 . Therefore, (3.89) By (3.79), |Y T 7 −X T 1 | ≤ (c 5 + 5c 4 )ε 1 . This and (3.89) imply that the following can be derived as a special case of (3.63), for x ∈ ∂D ∩ B(Y T 7 , 2c 20 ε 1 ). By the definition of T 3 , This and the fact that We use this estimate and (3.90) to conclude that d(Y T 3 , ∂D) ≤ a 2 ε 2 1 /400. Recall that we are assuming that F 1 holds. It follows from (3.66) that and, therefore,

Let
It is routine to check that (3.68)-(3.73) hold with X T 1 replaced with Π(X S k+1 ), and T 5 replaced with T 8 (the values of the constants may have to be adjusted). Hence, we obtain as in (3.74) that Similarly, an argument analogous to that in (3.80)-(3.83) yields This and (3.92) imply that n(Y s )dL y s ≤ c 33 ε 3 1 2 J . We obtain from this and (3.61), Case (ii). We will now analyze the case when A 1 does not occur. The rest of the proof is an outline only. Most steps are very similar to those in Case (i), so we omit details to save space.

Standard estimates show that
Recall that we have assumed that X T 1 ∈ ∂D. Let For some c 37 and c 38 , we let Let T 10 = sup{t ≤ T 9 : X t ∈ ∂D} and note that X T 9 − Y T 9 = X T 10 − Y T 10 . Using the fact that X T 1 ∈ ∂D and definitions of T 1 , T 2 and K, one can show that This implies that d(X T 9 , ∂D) ≤ c 40 ε 2 1 2 K . We can repeat the argument proving (3.94), with the roles of X and Y interchanged and T 1 replaced by T 9 , to see that if A 2 holds then S k+1 ≤ T 9 and The angle between n(Y T 9 ) and n(Π(X S k+1 )) is less than c 42 ε 1 . We know from (3.67) that d(Y T 1 , ∂D) ≤ c 43 ε 2 1 2 J . These facts and (3.97) imply that n(Π(X S k+1 )), Let k 2 be the largest integer such that if K ≤ k 2 then for x ∈ ∂D ∩ B(Y T 2 , 2ε 1 2 K ) we have n(x), n(Π(X S k+1 )) ≥ 1/2. Assume that F 2 := {K ≤ k 2 } holds. It follows that We also have L T 2 − L T 1 ≤ c 46 ε 2 1 2 J by (3.72). Hence, T 1 n(X s )dL s ≤ c 49 ε 3 1 2 (J∨K)+K .
Combining the last two estimates with (3.98), we obtain, This implies that By (3.67) and an estimate similar to that in Lemma 3.4 (i), This,(3.61) an the strong Markov property applied at T 1 yield, (3.101) For K ≥ J we have 2 (J∨K)+K = 2 2K so the the right hand side of (3.99) is bounded by c 55 ε 3 1 2 2K . This and (3.101) imply that the corresponding contribution to the expectation in (3.100) is bounded by j 0 j=1 j 0 k=j c 54 ε 1 2 −k c 55 ε 3 1 2 2k ≤ c 56 ε 3 1 | log ε 1 |. (3.102) For K < J we have 2 (J∨K)+K = 2 J+K so the corresponding contribution to the expectation in (3.100) is bounded by Combining this with (3.102) yields 103) The probability that A 2 does not occur, conditional on J and K, is bounded above by c 59 ε 2 1 2 K /ε 1 = c 59 ε 1 2 K . If A c 1 ∩ A c 2 holds, we use the following crude estimate, Therefore, using (3.101), It remains to address the cases when F 1 or F 2 fail. The probability of F c 1 ∩ F c 2 is bounded by c 61 ε 2 1 . Hence, (3.105) If F 1 fails but F 2 does not. we can repeat the analysis presented in Case (ii). Hence, (3.103) holds with 1 A c 1 ∩A 2 ∩F 2 replaced with 1 F c 1 ∩A 2 ∩F 2 . The lemma follows from these remarks, (3.95), (3.103), (3.104) and (3.105).
Lemma 3.15. We have for some c 1 , Proof. We will use modified versions of stopping times S k and U k by dropping σ * from the By Brownian scaling and the strong Markov property, P(A | F * k ) ≥ p 1 on {S * k ≤ σ * }, for some p 1 > 0 that does not depend on ε or k. An argument similar to that in the proof of Lemma 3.7 (i) can be used to show that if ε, a 1 and a 2 are small and A holds then We use this estimate to see that It is elementary to check that for all j, Hence, σ * ≤ σ 1 is stochastically majorized by a geometric random variable with mean depending only on D, so Eσ * < c 4 < ∞. (3.107) We combine this, (3.106) and (3.107) to complete the proof.
Recall operator H k defined in (3.2).
Lemma 3.20. For any c 1 , ε 0 > 0 there exists a 0 > 0 such that if a 1 , a 2 < a 0 and |X 0 −Y 0 | = ε then, Proof. We have This, (2.3) and (3.126) imply that Recall notation from the beginning of this section.
Lemma 3.21. We have for any β 1 < 1 and some c 0 and c 1 , assuming that |X 0 − Y 0 | = ε and ε * ≥ c 0 ε, Proof. By Lemma 3.7 (iv), for every k, This and Lemma 3.15 imply that For the notation used in the following lemma and its proof, see the beginning of this section.
By (2.3), the compositions of operators before and after the parentheses in the summation above are uniformly bounded in operator norm by a constant. Therefore, Using the fact that S i and π i commute, as do S i and π i , we obtain, We will deal with each of these terms separately.
We have assumed that x i = x i so one of the following four events holds, (3.138) If F 1 i holds then, (3.149) We continue our discussion of the right hand side of (3.134). We now consider the terms of the form |x i+1 − x i+1 | | x i+1 − x i |. The overall structure of our argument is similar to that used to analyze the terms of the form |x i+1 − x i | |x i − x i |.