PATHWISE UNIQUENESS FOR REFLECTING BROWNIAN MOTION IN CERTAIN PLANAR LIPSCHITZ DOMAINS

We give a simple proof that in a Lipschitz domain in two dimensions with Lipschitz constant one, there is pathwise uniqueness for the Skorokhod equation governing reﬂecting Brownian motion. Suppose that D ⊂ R 2 is a Lipschitz domain and let n ( x ) denote the inward-pointing unit normal vector at those points x ∈ ∂D for which such a vector can be uniquely deﬁned (such x form a subset of ∂D of full surface measure). Suppose (Ω , F , P ) is a probability space. Consider the following equation for reﬂecting Brownian motion with normal reﬂection taking values in D , known as the (stochastic) Skorokhod equation: We there a ﬁltration {F t } satisfying the usual conditions, and W = { W t , t ≥ 0 } is a Brownian motion with respect to {F t } . In particular, if s < t , we have W t − W s independent of F s . L = { L t , t ≥ 0 } is the local time of X = { X t , t ≥ 0 on ∂D , is, a continuous nondecreasing process that increases only when X is on the boundary ∂D and such that L does not charge any set of zero surface measure. Moreover we require X to

Suppose that D ⊂ R 2 is a Lipschitz domain and let n(x) denote the inward-pointing unit normal vector at those points x ∈ ∂D for which such a vector can be uniquely defined (such x form a subset of ∂D of full surface measure). Suppose (Ω, F, P) is a probability space. Consider the following equation for reflecting Brownian motion with normal reflection taking values in D, known as the (stochastic) Skorokhod equation: We suppose there is a filtration {F t } satisfying the usual conditions, and on ∂D, that is, a continuous nondecreasing process that increases only when X is on the boundary ∂D and such that L does not charge any set of zero surface measure. Moreover we require X to be adapted to {F t }.
We say that pathwise uniqueness holds for (1) if whenever X and X are two solutions to (1) with possibly two different filtrations {F t } and {F t }, resp., then P(X t = X t for all t ≥ 0) = 1.
In this note we give a short proof of the following theorem.
Theorem 1 Suppose D ⊂ R 2 is a Lipschitz domain whose boundary is represented locally by Lipschitz functions with Lipschitz constant 1. Then we have pathwise uniqueness for the solution of (1).
We remark that there are varying definitions of pathwise uniqueness in the literature. Some references, e.g., [4], allow different filtrations for X and X , while others, e.g., [5], do not. We prove pathwise uniqueness with the definition used by [4], which yields the strongest theorem. Theorem 1 was first proved in [2], with a vastly more complicated proof. Moreover, in that proof, it was required that the Lipschitz constant be strictly less than one. Strong existence was also proved in [2]; it will be apparent from our proof that we also establish strong existence.
In C 1+α domains with α > 0, the assumption that L not charge any sets of zero surface measure is superfluous; see [3], Theorem 4.2. (There is an error in the proof of Theorem 3.5 of that paper, but this does not affect Theorem 4.2.) In a forthcoming paper, the authors plan to prove that pathwise uniqueness holds in C 1+α domains in R d for d ≥ 3 and α > 1/2, but that pathwise uniqueness fails for some C 1+α domains in R 3 with α > 0. We do not have a conjecture as to whether pathwise uniqueness holds in all two-dimensional Lipschitz domains.
Proof. Standard arguments allow us to limit ourselves to domains of the following form where f : R → R satisfies the following conditions: Consider any x 0 ∈ D and processes X and Y taking values in D such that a.s., We will first assume that the filtrations for X and Y are the same, and then remove that assumption at the end of the proof. Here L X is the local time of X on ∂D, that is, a continuous nondecreasing process that increases only when X is on the boundary ∂D and that does not charge any set of zero surface measure. The processes L Y is defined in an analogous way relative to Y . We will write X t = (X 1 t , X 2 t ) and similarly for Y . Let Next we will show that, a.s.,

Electronic Communications in Probability
The following proof of (3) applies to almost all trajectories because it refers to properties that hold a.s. We will define below times t 1 and t 2 . They are random in the sense that they depend on ω in the sample space but we do not make any claims about their measurability. In particular, we do not claim that they are stopping times.
Hence t 2 is strictly less than t 1 . For t ∈ (t 2 , t 1 ), X t − Y t ∈ K, so X t ∈ D, because for any x ∈ ∂D and y ∈ R 2 such that x − y ∈ K, we have y / ∈ D. We see that We have n(x) ∈ K for every x ∈ ∂D where n(x) is well defined. Hence t1 t2 n(X s )dL X s ∈ K. For all x, y, z ∈ R 2 such that x ∈ K, y / ∈ K and −z ∈ K, we have x − y = z. We apply this to x = X t1 − Y t1 , y = X t2 − Y t2 and z = − t1 t2 n(X s )dL X s to obtain a contradiction with (4). This contradiction shows that there does not exist t with X t − Y t ∈ K. By the same argument with X and Y reversed, there does not exist t with Y t − X t ∈ K. Simple geometry shows that if x, y ∈ R 2 , x = (x 1 , x 2 ), y = (y 1 , y 2 ), x 1 = y 1 , x − y / ∈ K and y − x / ∈ K then x = y. We apply this observation to x = X t and y = Y t to conclude that if By the continuity of X and Y , the set I = {t ∈ (0, u 1 ) : Thus it consists of a finite or countable union of disjoint intervals {I n }. For any I n = (s 1 , s 2 ) we have X 1 s1 = Y 1 s1 and, therefore, X s1 = Y s1 . Similarly, X s2 = Y s2 . It follows that In n(X s )dL X s = In n(Y s ) dL Y s .
Suppose without loss of generality that V t0 = Y t0 . Then by (2) By (5), By induction, for any n,