Geography of the Level Sets ofthe Brownian Sheet

Robert C. Dalang1 and John B. Walsh2

Abstract

We describe geometric properties of W > α, where W is a standard real-valued Brownian sheet, in the neighborhood of the first hit P of the level setW > α along a straight line or smooth monotone curve L. In such a neigh-borhood we use a decomposition of the form W (s, t) = α− b(s) + B(t) + x(s, t),where b(s) and B(t) are particular diffusion processes and x(s, t) is comparativelysmall, to show that P is not on the boundary of any connected component ofW > α. Rather, components of this set form clusters near P . An integral testfor thorn-shaped neighborhoods of L with tip at P that do not meet W > α isgiven. We then analyse the position and size of clusters and individual connectedcomponents of W > α near such a thorn, giving upper bounds on their height,width and the space between clusters. This provides a local picture of the levelset. Our calculations are based on estimates of the length of excursions of B andb and an accounting of the error term x.

Running title: Geography of level setsAMS 1991 Subject Classifications: Primary 60G60; secondary 60G15.Key words and phrases. Brownian sheet, level set, Gaussian process.

1Department of Mathematics, Tufts University, Medford, MA 02155, U.S.A. The research of thisauthor was partially supported by NSF grant DMS-9103962.

2Department of Mathematics, University of British Columbia, Vancouver, British ColumbiaV6T1Y4, Canada.

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1 Introduction

Let W (s, t), s ≥ 0, t ≥ 0 be a standard Brownian sheet. For any real number α,consider the level set

L(α) = (s, t) ∈ RI 2+ : W (s, t) = α.

This study originated with some geometrical questions about L(α), not all of which areeasy to make rigorous. For instance, is L(α) a curve—or a countable union of curves—in the classical sense, or is it like the boundary between light and dark in M. Escher’s“Night and Day” [6]? That woodcut pictures a flock of white geese against the nightsky which blends into a flock of black geese flying in the opposite direction in daylight.One side is white on a dark background, the other is black on a light background, andit is difficult to say where the one ends and the other begins.

As posed, this question is probably not answerable—the two alternatives are notmutually exclusive, for example—but Escher’s image gives us a way to think of thelevel sets. They are quite granular in character. L(α) separates the open sets L+(α) =W > α and L−(α) = W < α, and it is easy to see that L+(α) has many smallcomponents, corresponding to local maxima of W which barely exceed α. Each of theseis completely surrounded by the set L−(α). Similarly, there are small components ofL−(α) surrounded by L+(α). Let us call all these small components bubbles. Then wecan say that on one side of L(α) is the set L+(α) containing many bubbles of L−(α)and on the other side, the set L−(α) containing bubbles of L+(α). Because of theirregularity of the sample paths of the Brownian sheet, we expect that each bubble willbe surrounded by smaller bubbles, each in turn surrounded by even smaller bubbles,and so on. The bubbles will tend to cluster, then coalesce, at the boundary, so thatone can visualize the one set of bubbles passing into the other like Escher’s geese.

These considerations are implicit in an article by W.S. Kendall [9]. He showed thatL(α) is totally disconnected at a typical point, but is not totally disconnected at everypoint (indeed, a set dividing the plane into two non-empty disjoint open sets cannot betotally disconnected [8]). Kendall chose his “typical” point as follows. Fix (u, v) ∈ RI 2

+.Let C(u, v) be the component of the set (s, t) : W (s, t) = W (u, v) which contains(u, v). He then showed that C(u, v) almost surely reduces to the singleton (u, v). Inthis case the point is fixed and the level set is chosen randomly, but making free withFubini’s theorem, this implies that for a.e. α the set L(α) is totally disconnected atalmost every one of its points, where the “almost every” is with respect to local timeon L(α) [3, 1]. One can see from Kendall’s proof that this typical point is a.s. noton the boundary of any component of L−(α) or L+(α), but is rather a limit point ofcomponents—bubbles—of each.

We want to study points on the boundary of one of the connected components ofL−(α), so we will choose a different type of “typical” point as follows. Say α = 1 forsimplicity. Consider the connected component C0 of L−(1) ⊂ RI 2

+ which contains the

2

origin (and thus the coordinate axes). Fix t0 > 0, and proceed from the t-axis along thehorizontal line t = t0 until first encountering the boundary of C0 at the point (S, t0),where S = infs ≥ 0 : W (s, t0) ≥ 1. Then the point (S, t0) is in L(1). It is differentfrom the points studied by Kendall, for it is in the boundary of the component C0.

We will study the geography of the sets L+(1) and L−(1) in a neighborhood of(S, t0)—the “lay of the land” on a microscopic scale, so to speak. (Figure 1 shows acomputer simulation of these sets on a macroscopic scale. Earlier such pictures canbe found in R. Adler’s book [2, Chapter 8, Figures 8.0.1 and 8.0.2].) We first showthat L(1) is disconnected at the point (S, t0) and that (S, t0) is not on the boundaryof any single component of L+(1), but is rather a limit point of a sequence of distinctcomponents of that set. We then describe how these components cluster around thepoint (S, t0): we show in Section 3.1 that there is a “thorn-shaped” neighborhood ofthe segment L = ]0, S[×t0 which is contained in L−(1). After this, we concentrateon the distribution and size of the bubbles near (S, t0). As one approaches (S, t0)along the segment L, one passes by infinitely many components of L+(1): if sn is thes-coordinate of the position where we encounter the n-th component, then rn = S− snconverges to 0 at a super-exponential rate (see Theorem 3.7). Bounds on the heightand width of theses components are given in Theorems 3.9 and 3.11.

The key to these results is a local decomposition of the Brownian sheet in theneighborhood of (S, t0) of the form

W (s, t) = 1 − b(s) +B(t) + x(s, t),

where b and B are particular independent diffusion processes and x is small relativeto these two processes. There are four cases to distinguish, depending on the relativepositions of (s, t) and (S, t0) (see (3)–(6)). In the most interesting case, where s < Sand t > t0, b is a Bessel process of dimension 3 and B is a standard Brownian motion.It is natural to expect that components of L+(1) are closely approximated by those ofthe set B > b. We establish this in a companion paper [5] in which we study thestructure of these components.

If we interchange the coordinates s and t in the Brownian sheet, we get anotherBrownian sheet, so that the same results hold when we approach the level set along avertical line as when we approach it on a horizontal line. On the other hand, as Figure1 makes clear, the Brownian sheet is not invariant under rotations: the horizontal andvertical are distinguished directions, so one might expect that things would be differentif we approached the level set along, say, a diagonal line. However, this is not the case.We show in Sections 2.3 and 3.2 that the results are unchanged for lines with positiveslope. This does not follow directly from the results about horizontal lines, and in fact,though the proofs are similar, we know of no single elegant proof which handles bothcases simultaneously.

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Figure 1: A level set: the set W > 0 is in black.

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2 Points of disconnection of level sets

Let (Ω,F , P ) be a complete probability space and let W (s, t), s ≥ 0, t ≥ 0 be aBrownian sheet defined on this space. Recall [12, Chapter 3] that this is a mean zerocontinuous Gaussian process with covariance function

EW (s, t)W (u, v) = min(s, u) min(t, v).

Given a rectangle R = [u1, u2] × [v1, v2], we set

∆RW = W (u2, v2) −W (u2, v1) −W (u1, v2) +W (u1, v1).

Recall that the map R 7→ ∆RW extends to a σ-additive L2-valued measure A 7→W (A)defined on the bounded Borel subsets of RI 2

+, such that W (A) is Gaussian with mean 0and EW (A)W (C) equals the Lebesgue measure of A∩C. In particular, if A∩C = φ,then W (A) and W (B) are independent.

For α > 0, in addition to the level set L(α) defined in the introduction, we set

L−(α) = (s, t) ∈ RI 2+ : W (s, t) < α, L+(α) = (s, t) ∈ RI 2

+ : W (s, t) > α.

Our objective is to “build a map” of these sets in the neighborhood of a particularpoint.

2.1 Approaching the boundary: horizontal lines

Fix t0 > 0 and defineS = infs > 0 : W (s, t0) > 1.

Theorem 2.1 Let ε > 0. With probability one, there exists a closed curve lyingentirely in (s, t) ∈ RI 2

+ : W (s, t) < 1 which contains (S, t0) in its interior and whichis contained in a disc of radius ε centered at (S, t0).

Remark 2.2 The point (S, t0) is of course on the boundary of the component ofL−(1) which contains the origin. By Theorem 2.1, there is a sequence of curves inL−(1) decreasing to (S, t0); these evidently disconnect the point from L(1), and alsofrom L+(1). Now the latter set, being open, is the union of a countable number ofconnected components. The point (S, t0) is not in the boundary of any single oneof them, for otherwise this component would intersect one of the curves. Instead,(S, t0) must be a limit point of different components of L+(1), and we can think ofL+(1)—locally at least—as a collection of bubbles clustering about the point (S, t0).The nature of this clustering is described in Section 3 (see also [5]).

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We will need some definitions and lemmas before we prove Theorem 2.1, but wecan explain the idea right now (this idea was the key to Kendall’s analysis). Near(S, t0), W (s, t) is very nearly a sum of two functions of one variable, say W (s, t) ∼1 + f(s) + g(t). We then choose s1 < S < s2 and t1 < t0 < t2 such that f(si) and g(ti)are negative, and then let the curve be the boundary of the rectangle [s1, s2] × [t1, t2].The sum f(s) + g(t) will be negative at the corners; with some care in choosing thepoints, we can make it negative on the entire boundary. For ease of reference, this ideais formalized in the following elementary lemma, whose proof is left to the reader. Fora real-valued function f on RI +, we define

f ∗(h) = sup0≤u≤h

f(u).

Lemma 2.3 Let h1, h2 > 0, and consider functions f, g : [0, 1] → RI and γ : [0, 1]2 →RI such that

max(f(h1), g(h2)) + max(f ∗(h1), g∗(h2)) + sup

0≤u≤h1, 0≤v≤h2

γ(u, v) < 0.(1)

Then the function (u, v) 7→ f(u)+g(v)+γ(u, v) is strictly negative on the two segments[0, h1] × h2 and h1 × [0, h2].

By Brownian scaling, we can suppose that t0 = 1, and we will assume this fromnow on. Start at the point (S, 1). Move along any one of the four directions parallelto the axes. The processes we see are Brownian motions in two of the directions,a Brownian bridge in the third, and a Bessel process of “dimension 3” (or Bessel(3)process, for short) in the fourth; all four will be independent. To be more precise, notethat W (S, 1) = 1 and define, for u ≥ 0 and v ≥ 0,

b(u) = 1 −W (S − u, 1) ; c(u) = W (S + u, 1) − 1 ;

B(v) = W (S, 1 + v/S) − 1 ; C(v) = 1 − W (S, 1 + v/S) ,

where W (s, t) = tW (s, 1/t). Notice that by the time inversion property of the Browniansheet [12, Chapter 3], W is again a Brownian sheet. Let us also define four two-parameter processes x, x, X and X by

x(u, v) = ∆R1(u,v)W, x(u, v) = ∆R1(u,v)W ,

X(u, v) = ∆R2(u,v)W, X(u, v) = ∆R2(u,v)W ,

where u, v ≥ 0 and

R1(u, v) = [S − u, S] × [1, 1 + v], R2(u, v) = [S, S + u] × [1, 1 + v].

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Lemma 2.4 (i) The processes B(v), v ≥ 0, C(v), v ≥ 0, c(u), u ≥ 0 andb(u), 0 ≤ u ≤ S are independent. The first three are Brownian motions independentof S and the fourth is equivalent in distribution to a Bessel(3) process b restricted tothe interval [0, U ], where U = supu : b(u) = 1.

(ii) X and X are independent Brownian sheets, and, conditioned on S = s, x andx are independent Brownian sheets restricted to [0, s] × RI +.

Proof. Set F(s, t) = σW (u, v) : 0 ≤ u ≤ s, 0 ≤ v ≤ t. Then S is a stopping timerelative to F(s,∞), s ≥ 0, the process B is measurable with respect to F(S,∞),and c is independent of this σ-field by the strong Markov property applied to S.

Now S is also a stopping point relative to F(s, 1), s ≥ 0, b and C are measurablewith respect to F(S, 1), and B and c are independent of this σ-field. The remainingindependence follows by inverting time in the second variable: W equals W on the linet = 1, so S is also a stopping time relative to F(s, 1), s ≥ 0. The above argumentshows that c and C are independent of F(S, 1), hence of b and B.

It follows that all four processes are independent. The definition of the Brown-ian sheet implies that the first three are Brownian motions, and the fact that b is aBessel(3) process follows from D. Williams’ decomposition of the Brownian excursion[11, Chap.XII (4.4)]. We get (ii) from similar considerations. ♣

Define

M(h, k) = sup|x(u, v)|, |X(u, v)|, |x(u, v)|, |X(u, v)| : u ≤ h, v ≤ k.

An upper bound on M is given by the law of the iterated logarithm for the Browniansheet [12, Theorem 3.5.A]:

a < 1/2 ⇒ limε→0

sup0≤h≤ε, 0≤k≤ε

(hk)−aM(h, k) = 0.(2)

Proof of Theorem 2.1. Recall that we have set t0 = 1. For h > 0, consider therectangle

Rh = [S − h, S + h] × [1/(1 + h/S), 1 + h/S]

and the events Γh = W < 1 on ∂Rh and

Λh = max(−b, B,−C, c)(h) < −5√h, max(|B|∗, |C|∗, |c|∗)(h) <

√h.

Note that p ∼ P (Λh) is strictly positive and independent of h by Brownian scaling. Let(hn) be any sequence which decreases to 0. By Fatou’s Lemma, P (lim sup Λhn

) ≥ p > 0.Then Blumenthal’s zero-one law for the four dimensional diffusion (b, B, c, C) impliesthat P (lim sup Λhn

) = 1.

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We claim that for a.e. ω ∈ Ω, there exists η(ω) > 0 such that if 0 < h < η(ω),then ω ∈ Λh implies ω ∈ Γh. This will complete the proof, since then P (lim sup Γhn

) ≥P (lim sup Λhn

) = 1. In order to verify the claim, for ω ∈ Ω, let η(ω) be small enoughso that if max(h, k) ≤ η(ω), then M(h, k) < (hk)1/4 (such an η(ω) exists by (2)).

Fix ω ∈ lim sup Λhnand choose n ∈ NI be such that hn < η(ω) and ω ∈ Λhn

. Let usshow that ω ∈ Γhn

. We shall omit the subscript n below. For u, v > 0, it is immediateto check that

W (S − u, 1 + v/S) = 1 − b(u) +B(v) − x(u, v/S),(3)

W (S + u, 1 + v/S) = 1 + c(u) +B(v) +X(u, v/S).(4)

It is also straightforward, but a bit lengthier, to check that

W (S + u, 1/(1 + v/S)) = 1 + c(u) − C(v) + X(u, v/S)(5)

−(v/S)W (S + u, 1/(1 + v/S)),

W (S − u, 1/(1 + v/S)) = 1 − b(u) − C(v) − x(u, v/S)(6)

−(v/S)W (S − u, 1/(1 + v/S)).

Let us consider the restriction of W to the two segments

∂1Rh = ∂Rh ∩ (s, t) : 0 ≤ s ≤ S, t ≥ 1.

We apply Lemma 2.3 with h1 = h2 = h, f(u) = −b(u;ω), g(u) = B(u;ω) and γ(u, v) =−x(u, v/S;ω). Given our choice of h and since ω ∈ Λh, condition (1) is satisfied sincethe left-hand side of the inequality is less than −5

√h + max(0,

√h) +

√h < 0. It

follows from Lemma 2.3 that W (· ;ω) < 1 on ∂1Rh.The argument in the other three quadrants is similar. For instance, on

∂3Rh = ∂Rh ∩ (s, t) : s ≥ S, 0 ≤ t ≤ 1,

we apply Lemma 2.3 with h1 = h2 = h, f(u) = c(u;ω), g(u) = −C(u;ω) and

γ(u, v) = X(u, v) − (v/S)W (S + u, 1/(1 + v/S)).

Condition (1) is satisfied since the left-hand side of the inequality is less than −5√h+√

h+√h < 0. It follows again from Lemma 2.3 that W (· ;ω) < 1 on ∂3Rh. The proof

of the remaining two cases is similar and is left to the reader. ♣

2.2 Surrounding (S, 1)

In [9], Wilf Kendall showed that a “typical” point in a level set is disconnected fromthe rest of the level set in a strong way: there exist two sets of closed curves containing

8

the point in their interior and decreasing to this point, such that W > 1 on one setof curves and W < 1 on the other. The point (S, 1) is not of this type: Theorem 2.1shows that there are indeed curves around it on which W < 1, but it cannot be in theinterior of a closed curve on which W > 1, since the segment [0, S[×1 is containedin L−(1). There are two natural questions to ask about the contrast between these twotypes of points. First, to what extent does the set L+(1) “surround” the point (S, 1)?Second, how close does the set L+(1) come to the line segment [0, S[×1? We willaddress the first question here, and leave the second to Section 3.1.

We can rephrase the first question as follows. What kind of curves in L+(1), con-nected but not necessarily closed, lie near (S, 1)? We will show that the point can be“surrounded” on its right hand side by such curves.

More exactly, for each h > 0 let Γh be the square of side 2h centered at the point(S, 1) and let γh be the curve consisting of the top, right hand, and bottom sides ofΓh (so γh consists of three sides of the square, open on the left). Then we have thefollowing result.

Theorem 2.5 With probability one, there exists a sequence of an ↓ 0 such that W isstrictly greater than 1 on each γan

.

Proof. For a process Y (t), let Y∗(t) = infY (v) : 0 ≤ v ≤ t. Set Γh = W >1 on γh and

Λh = min(B, c,−C)(h) > 5√h, min(−b∗, B∗, c∗, (−C)∗)(h) > −

√h.

As in the proof of Theorem 2.1, note that P (Λh) is strictly positive and independent ofh, and thus P (lim sup Λhn

) = 1 if hn ↓ 0. And as in that proof, we only need to showthat for a.e. ω ∈ Ω, there is η(ω) > 0 such that if 0 < h < η(ω), then ω ∈ Λh impliesω ∈ Γh. Using (2), we choose η(ω) so that max(h, k) < η(ω) implies M(h, k) < (hk)1/4.

Fix ω ∈ lim sup Γhnand let n ∈ NI be such that hn < η(ω) and ω ∈ Λhn

. We showthat ω ∈ Γhn

. We will omit the subscript n below. Looking at the decompositions(4) and (5) of W near (S, 1), it is clear that a slight variation on Lemma 2.3 impliesthat W (· ;ω) > 1 on Γh ∩ (s, t) : s ≥ S, t ≥ 0. In order to handle the segment[S − h, S] × 1 + h, we use (3). The lower bound on B(h), together with the lowerbound on −b∗(h) and −x(u, v/S) and another slight variation on Lemma 2.3 show thatW (· ;ω) < 1 on this segment. The last segment is handled in a similar fashion using(6). ♣

2.3 Approaching the boundary: other lines

By symmetry, the result of Theorem 2.1 will hold if we approach the boundary of C0

along a vertical rather than a horizontal line. Since most properties of the Brownian

9

sheet depend strongly upon orientation, it is not obvious that the conclusion of The-orem 2.1 remains valid if we approach the boundary of C0 along lines with positiveslope. We will show here how to modify the arguments of Section 2.1 to handle thissituation.

We can reduce the problem to one special case. If a > 0, and if W is definedby W (s, t) = W (s

√a, t/

√a), then W is a Brownian sheet. Thus to understand the

behavior of W along the line t = a s, it is sufficient to study the behavior of W alongthe diagonal t = s. We will limit ourselves to this case. Let S = infs : W (s, s) ≥ 1.

Theorem 2.6 Let ε > 0. With probability one, there exists a closed curve lyingentirely in (s, t) ∈ RI 2

+ : W (s, t) < 1 which contains (S, S) in its interior and whichis contained in a disc of radius ε centered at (S, S).

The key to Theorem 2.1 was the approximation ofW by a sum of two one-parameterprocesses. We will use the same idea but need different processes here. Set ϕ(u) = u

1

2

andR1u = (s, t) : 0 ≤ t ≤ s ≤ u, R2

u = (s, t) : 0 ≤ s ≤ t ≤ u.Consider the stochastic processes

B1(u) = W (R1ϕ(u)), B2(u) = W (R2

ϕ(u)).

Lemma 2.7 Set X(u) = B1(u) + B2(u) and Y (u) = B1(u) − B2(u). Then X =X(u), u ≥ 0 and Y = Y (u), u ≥ 0 are independent Brownian motions.

Proof. Clearly both X and Y are continuous Gaussian processes of independent in-crements and therefore are continuous martingales. Standard properties of the Brow-nian sheet show that EX(u)2 = EY (u)2 = u, and a simple calculation showsthat EX(u)Y (v) = 0, so these two Gaussian processes are independent Brownianmotions. ♣

Note that along the diagonal, W (t, t) = X(t1

2 ). Set σ = infu ≥ 0 : X(u) = 1 andfor u ≥ 0 define

b(u) = X(σ) −X(σ − u) , c(u) = X(σ + u) −X(σ) ,

B(u) = σ− 1

2 (Y (σ) − Y (σ − σu)) , C(u) = σ− 1

2 (Y (σ + σu) − Y (σ)).

Lemma 2.8 The processes b(u), 0 ≤ u ≤ σ, c(u), u ≥ 0, B(u), 0 ≤ u ≤ 1 andC(u), 0 ≤ u ≤ 1 are independent. The first is a Bessel(3) process run until it lasthits one, and the other three are Brownian motions.

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Proof. The fact that b is a Bessel(3) process follows directly from [11, Chap XII (4,4)],and the fact that c and C are Brownian motions follows from Lemma 2.7 and the strongMarkov property. Independence of these two processes is checked straightforwardlyby using the definition of independence and conditioning on σ. Observe that theconditional law of B given σ = s is the same as that of s−

1

2 (Y (s) − Y (s− su)), whichis that of Brownian motion. Since this conditional distribution does not depend on s,B itself is a Brownian motion.

To see the independence, notice that c and C are independent of the pre-σ σ-fieldof X and Y , and hence of b and B, which are measurable with respect to this σ-field.On the other hand, for all choices of n, times u0 < · · · < un and Borel sets A0, . . . , An,we have

PB(u0) ∈ A0, B(uk) − B(uk−1) ∈ Ak, k = 1, . . . , n | σ(X)= Pσ− 1

2 (Y (σ) − Y (σ − σu0) ∈ A0, σ− 1

2 (Y (σ − σuk−1) − Y (σ − σuk)) ∈ Ak,

k = 1, . . . , n | X(u), u ≥ 0= g(σ),

where

g(s) = Ps− 1

2 (Y (s) − Y (s− su0)) ∈ A0, s− 1

2 (Y (s− suk−1) − Y (s− suk)) ∈ Ak,

k = 1, . . . , n= PY (u0) ∈ A0, Y (uk) − Y (uk−1) ∈ Ak, k = 1, . . . , n.

Therefore g(s) does not depend on s. It follows that the conditional probability aboveis constant, which proves that B and σ(X) are independent. Thus all four processesare independent. ♣

For (s, t) ∈ RI 2+, let T (s, t) be the right triangle whose apex is at (s, t) and whose

hypotenuse is on the main diagonal (see Figure 2). Let z(s, t) = W (T (s, t)) and notethat

W (s, t) = B1(s2) +B2(t2) − z(s, t)(7)

=1

2(X(s2) + Y (s2) +X(t2) − Y (t2)) − z(s, t).

The processes X and Y can be expressed in terms of b, B, c, and C, but theexpression depends on the values of s and t. If s < S, then

X(s2) = X(σ) − b(σ − s2) and Y (s2) = Y (σ) − σ1

2B(1 − s2/σ) ,(8)

while if s > S, then

X(s2) = X(σ) + c(s2 − σ) and Y (s2) = Y (σ) + σ1

2C(s2/σ − 1) .(9)

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Figure 2: Decomposition of W and the square Qh.

Finally, the following supplies a global bound on the error term z(s, t), which willplay the role of (2) in Section 2.1. For each 1 > δ > 0 and K > 0,

limε−→0

ε−(1−δ) sup|z(s, t)| : s+ t ≤ 2K, |s− t| < ε = 0 a.s.(10)

This is a consequence of [7, Theorem 2], which implies that in the region where |s−t| <η and s+ t ≤ 2K, the modulus of continuity ω(ε) of z is bounded by

√ηε

1

2−δ, for any

δ > 0. Thus, taking η = ε, (10) follows.

Proof of Theorem 2.6. The curves around (S, S) on which W < 1 will againbe rectangles, with sides parallel to the axes rather than at 45o angles. Set Qh =[√σ − h,

√σ + h]2. Let Γh = W < 1 on ∂Qh and

Λh = max(−b, c)(h) < −5√h, max(|c|∗(h), σ 1

2 |B|∗(h/σ), σ1

2 |C|∗(h/σ)) <√h.

As in the proof of Theorem 2.1, note that p ≡ PΛh is strictly positive and indepen-dent of h, and this implies P (lim sup Λhn

) = 1 if hn ↓ 0.The remainder of the proof is also similar to that of Theorem 2.1. This will be clear

once we write out the analogues of (3), (4), (5) and (6). By (7), (8) and (9), the firsttwo become

W−,+(u, v) = 1 − (b(u) + σ1

2B(u/σ))/2 + (c(v) − σ1

2C(v/σ))/2 − z−,+(u, v)(11)

W+,+(u, v) = 1 + (c(u) + σ1

2C(u/σ))/2 + (c(v) − σ1

2C(v/σ))/2 − z+,+(u, v)(12)

and the last two are

W+,−(u, v) = 1 + (c(u) + σ1

2C(u/σ))/2− (b(v) − σ1

2B(v/σ))/2 − z+,−(u, v)(13)

W−,−(u, v) = 1 − (b(u) + σ1

2B(u/σ))/2 − (b(v) − σ1

2B(v/σ))/2 − z−,−(u, v),(14)

12

where

W±,±(u, v) = W (√σ ± u,

√σ ± v), z±,±(u, v) = z(

√σ ± u,

√σ ± v).

For example, in case (11), we would apply Lemma 2.3 with h1 = h2 = h,

f(u) = −(b(u) + σ1

2B(u/σ))/2, g(u) = (c(v) − σ1

2C(v/σ))/2,

and γ(u, v) = z(√σ − u,

√σ + v). Condition (1) is satisfied since by (10), for small h,

the left hand-side of the inequality is < −4√h/2 + 2

√h/2 +

√h = 0. The other cases

are similar and are left to the reader. ♣

Remark 2.9 The result of Theorem 2.6 remains valid if one approaches the boundaryof C0 along smooth monotone curves, say. This can be checked using the same methodsas above.

13

3 Local geography of the level set

In order to understand the geography of L(1) in the neighborhood of (S, 1), where

S = infs > 0 : W (s, 1) = 1,

we consider W (S − s, 1 + t) for small s and t. It is clearly smaller than 1 if t = 0and 0 < s < S, and by continuity it must be smaller than 1 in some neighborhoodof the segment [0, S[×1. What does this neighborhood look like? It can be rathercomplicated—it will not be simply connected, for instance—but we can ask whether ornot there is a neighborhood of a simpler character. The class of convex neighborhoodsis not appropriate, for any convex neighborhood of the segment intersects W > 1(see Theorem 3.1 (b) below). The natural neighborhoods to consider are shaped likea thorn, with the tip at (S, 1) (see the first picture in Figure 3). The purpose ofthis section is to determine the exact shape of these particular neighborhoods and todescribe the size and position of the connected components of W > 1 just beyond thethorn. The shape of individual clusters and components is described in the companionpaper [5].

If Λ is an event in σb (resp. σB), then P c(Λ) denotes the conditional probabilityof Λ given that b(0) = c (resp. B(0) = c).

3.1 The thorn: horizontal lines

It is easiest to visualize neighborhoods of [0, S[×1 if we translate the origin to thepoint (S, 1) and reverse the s-axis: let W (s, t) = W (S − s, 1 + t/S) − 1. Clearly,W > 1 corresponds to W > 0. By (3),

W (s, t) = B(t) − b(s) − x(s, t/S), s > 0, t > 0 ,(15)

where B is a standard Brownian motion starting from 0, b is a Bessel(3) processstarting at 0 and independent of B and x is comparatively small by (2). By (6), asimilar decomposition holds for s > 0 and t < 0. We only study the first case, since itis then not difficult to check that the results are not affected by the additional errorterm in (6), and so it is sufficient to consider thorns which are symmetric with respectto the segment [0, S[×1.

Let τ(s) be a continuous non-decreasing function on RI + with τ(0) = 0 and τ(s) > 0if s > 0. Let Tτ be the thorn-shaped neighborhood Tτ = (s, t) : s ≥ 0, 0 ≤ t ≤ τ(s)and let γτ be the upper boundary curve of Tτ , i.e. the graph of t = τ(s). We saythat Tτ is initially in W < 0 if with probability one, there exists η > 0 such thatTτ ∩ (s, t) : 0 ≤ s ≤ η, t ≥ 0 ⊂ W < 0.

14

Figure 3: A thorn, and the sets Tψβiand curves γβi

= γψβi, β1 < β2 < β3.

Theorem 3.1 Suppose that s 7→ τ(s)/s is increasing for s > 0. Then(a) Tτ is initially in W < 0 if and only if

I(τ) =∫

0+

(

τ(s)

s

)1

2 ds

s<∞ .

(b) If this integral is infinite, then γτ is not initially in W < 0. Moreover, for all

κ > 0, there is a sequence (sn) ↓ 0 such that W (sn) > κ τ(sn)1

2 , for all n.

Remark 3.2 (a) For β ≥ 0, let

ψβ(s) = s(

log1

s

)−2 (

log log1

s

)−β

.

Then s 7→ ψβ(s)/s is increasing and I(ψβ) < +∞ if and only if β > 2.(b) If we had set W (s, t) = W (1 − s, 1 + t) − 1, Theorem 3.1 would remain valid.

Indeed, (15) would become W (s, t) = B(St) − b(s) − x(s, t). But the distribution of

B(S ·) is the same as that of S1

2B(·), and the factor S1

2 clearly does not affect thecalculation below.

(c) For later reference, we point out that the proof of Theorem 3.1 (b) shows in factthat for all κ > 0, the events

W (S − νn, τ(νn)) > κψβ(e−n)

1

2 , b(νn) < ψβ(e−n)

1

2occur infinitely often, where νn is the unique time in [e−n, e1−n] such that b(νn) =minu∈[e−n,e1−n] b(u).

15

The proof of Theorem 3.1 uses the following property of the Bessel(3) process b.

Lemma 3.3 There is a positive constant k such that Pmin1≤s≤e b(s) < a ≤ k a, forall a > 0, and a positive constant k′ such that Pmin1≤s≤e b(s) < a ≥ k′ a for allsufficiently small a.

Proof. The first statement follows immediately from [11, Chapter 6, Corollary (3.4)],since using the explicit density of b(1) (see (22)), the probability in question is

≤ Pmins≥1

b(s) < a = aE1/b(1) <∞.

Concerning the second statement, let α be a random variable with a uniform distri-bution on [0, c], c > 0, and set T (a) = inft ≥ 1 : B(t) = a, where B is a Brownianmotion independent of α. We apply a path decomposition theorem [11, Chapter 6,Theorem (3.11)] to see that

P c min0≤s≤e−1

b(s) ≥ a = P c(α ≥ a ∪ α ≤ a, T (a) > e− 1)= 1 − a/c+ (a/c)P cT (a) > e− 1≤ 1 − a/c+ (a/c)P cT (0) > e− 1.

Thus P cmin0≤s≤e−1 b(s) ≤ a ≥ (a/c)P cT (0) < e − 1. Integrating with respect tothe density of b(1) and using the Markov property of b gives the conclusion. ♣

Proof of Theorem 3.1. First suppose that the integral is finite. We claim that Tτis initially in W < 0. Set φ(s) = τ(s)/s, put εn = e(1−n)(1−δ), where 0 < δ < 1/4,and let

mn = minb(s) : e−n ≤ s ≤ e1−n, An = mn ≤ B∗(τ(e1−n)) + εn .

Now t < s in a neighborhood Tτ , so by (2), for all large enough n and s ∈ [e−n, e1−n],we have |x(s, t)| ≤ (s ∨ t)1−δ ≤ εn. If An does not occur, then

s ∈ [e−n, e1−n], t ≤ τ(e1−n) =⇒ B(t) − b(s) + εn < 0

since τ is non-decreasing. Thus W (s, t) ≤ B(t) − b(s) + εn < 0. It follows that

Tτ ∩ ([e−n, e1−n] × [0, τ(e1−n)]) ⊂ W < 0 on Ω \ An.

So it will be sufficient to prove that An occurs only finitely often. Since b and B areindependent by Lemma 2.4, Brownian scaling yields

P (An) = Pe−n/2m0 ≤ τ(e1−n)1

2 B∗(1) + εn= Pm0 ≤ Kφ(e1−n)

1

2 B∗(1) + en/2εn

16

where K denotes a constant whose value may change from line to line, and we haveused the fact that enτ(e1−n) = eφ(e1−n). Since δ < 1/4, en/2εn ≪ e−n/4 for large n, sothat by Lemma 3.3, the last expression is

≤ K∫ ∞

0φ(e1−n)

1

2 a PB∗(1) ∈ da +Ke−n/4

= K φ(e1−n)1

2 EB∗(1) +Ke−n/4 .

Now sum this over n. The sum of the exponentials is finite and φ is monotone, so wecan estimate the sum as an integral to see that

∞∑

n=1

P (An) <∞ ⇐⇒∫ ∞

1φ(e1−x)

1

2 dx <∞ ⇐⇒∫ 1

0φ(s)

1

2 s−1 ds <∞ ,

which, thanks to the Borel-Cantelli Lemma, proves our claim. This implies the “if ”part of (a).

Now consider the process along the top boundary γτ of Tτ . Suppose the integral in(a) is infinite, and fix κ > 0. We claim there exist sn → 0 such that W (sn, τ(sn)) >

κ τ(sn)1

2 . This will prove (b) and the “only if ” part of (a) at the same time. Define

b(u) = eu/2b(e−u). Then b(s) = s1

2 b(log(1/s)) and

W (s, τ(s)) > κ τ(sn)1

2 ⇐⇒ B(τ(s)) > κ τ(sn)1

2 + s1

2 b(log(1/s)) + x(s, τ(s)).

So it will be sufficient to show that there is a sequence un ↑ ∞ such that

B(τ(e−un)) > κ τ(e−un)1

2 + e−un/2 b(un) + x(e−un , τ(e−un)).(16)

Fix 0 < C < D and let ρ0 = infu > 0 : b(u) = D and, having defined ρ0, σ1, . . ., σn,ρn, put

σn+1 = infu > ρn : b(u) = C, ρn+1 = infu > σn+1 : b(u) = D.

Since b is a stationary Markov process, (σn+1 − σn, n ∈ NI ) and (ρn+1 − ρn, n ∈ NI )are both i.i.d. sequences with finite mean [10, Sect. 5.4 p.145], and, in particular,σn < ρn < ∞, for all n ∈ NI . Let νn be the (unique) time at which b takes on itsminimum in [σn, ρn] and define

Fn = minσn≤u≤ρn

b(u) ≤ φ(e−νn)1

2 − e−νn( 1

2−δ),

Gn = B(τ(e−νn)) ≥ (κ+ 1)τ(e−νn)1

2.

On Fn ∩Gn, there is un ∈ [σn, ρn] such that (16) holds, namely the time un = νn. Sowe only need to show that Fn ∩ Gn occurs infinitely often. This is just a property of

17

the law of the independent processes b and B, which we now prove, without using thefact that these two processes come from a Brownian sheet.

We can assume that b and B are canonically defined on the product of two canonicalprobability spaces (Ω1,F1, P1) and (Ω2,F2, P2), i.e. b(ω1, ω2) (resp. B(ω1, ω2)) doesnot depend on ω2 (resp. ω1). We can also assume that Fn ∈ F1, and replace Fn ∩Gn

by (Fn × Ω2) ∩Gn.Observe that the events Fn depend only on b and are independent of each other,

and, by the strong law of large numbers, that limn→∞ ρn/n = α > 0. In addition,setting m0 = min0≤u≤ρ1 b(u), letting PC

1 denote conditional probability given b0 = Cand using Lemma 3.3, we see that

P1(Fn) = PC1 m0 ≤ φ(e−νn)

1

2 − e−νn( 1

2−δ)

≥ PC1 m0 ≤ φ(e−(α+1)n)

1

2 − eα( 1

2−δ)n

≥ K(φ(e−(α+1)n)1

2 − eα( 1

2−δ)n).

Since the sum of the exponentials is finite, it follows that

∑

n∈NI

P1(Fn) ≥∫ ∞

1Kφ(e−(α+1)u)

1

2 du+K ′ = K∫

0+φ(s)

1

2 s−1 ds+K ′ = +∞.

By the converse of the Borel-Cantelli Lemma, P1(lim supn∈NI Fn) = 1.Now fix ω1 ∈ lim supn∈NI Fn. There is a sequence (nk(ω1), k ∈ NI ) increasing to +∞

such that ω1 ∈ Fnk(ω1), for all k ∈ NI . The ω1-section of Gnk(ω1) is

Gnk(ω1) = ω2 ∈ Ω2 : B(τ(e−νnk

(ω1))) ≥ (κ + 1)τ(e−νnk(ω1))

1

2.

Now, P2(Gnk(ω1)) = P2B(1) ≥ κ + 1 > 0 and τ(e−νnk

(ω1)) ↓ 0. Since

P2(lim supk∈NI

Gnk(ω1)) ≥ lim sup

k∈NIP2(Gnk

(ω1)) > 0,

it follows by the 0-1 Law that P2(lim supk∈NI Gnk(ω1)) = 1. Thus P1-almost all sections

of lim supn∈NI (Fn × Ω2) ∩Gn have full P2-measure, hence the P1 × P2-measure of thisset is 1. The proof is complete. ♣

3.2 The thorn: other lines

We would like to know whether the results of Section 2.3 hold when we approach theset W > 1 along lines which are not horizontal or vertical. Specifically, we want toknow about thorns along lines with positive slope. We will use the setting and notationof Section 2.3. As mentioned there, we need only consider the main diagonal.

Let us rotate coordinates by setting ξ = 2−1

2 (t+ s) and η = 2−1

2 (t− s), so that theξ-axis lies along the main diagonal. Define W by W (ξ, η) = W (s, t), let S = infξ ≥0 : W (ξ, 0) = 1 and let W (ξ, η) = W (S − ξ, η) − 1.

18

Let τ be a continuous non-decreasing function on RI + with τ(0) = 0 and τ(ξ) > 0if ξ > 0. Let Tτ be the thorn-shaped neighborhood Tτ = (ξ, η) : 0 < ξ < S, 0 ≤η ≤ τ(ξ) and let γτ be the upper boundary curve of Tτ . Then we have the followinganalogue of Theorem 3.1.

Theorem 3.4 Suppose that ξ 7→ τ(ξ)/ξ is increasing for ξ > 0. Then(a) Tτ is initially in W < 0 if and only if

I(τ) ≡∫

0+

(

τ(u)

u

)1

2 du

u<∞ .

(b) If I(τ) is infinite, then γτ is not initially in W < 0.

Remark 3.5 This states that the thorn has exactly the same shape as in the horizontaland vertical cases. This might seem surprising in light of the strong horizontal-verticalorientation of the set W > 0 which is so visible in Figure 1. However, Theorem 3.4is only an order-of-magnitude result: it tells us that the tip of the thorn has the sameorder of sharpness in the two cases, but it does not distinguish between one thornand another which is, for example, twice as sharp. Finer results might well show adifference.

Proof of Theorem 3.4. We can reduce the proof of this theorem to the calculationswe did in Theorem 3.1. Let B1, B2, X and Y be defined as in Section 2.3 and note thatalong the first diagonal W (ξ, 0) = X(ξ2/2) = B1(ξ2/2) +B2(ξ2/2), and that equation(7) becomes

W (ξ, η) = B1((ξ − η)2/2) +B2((ξ + η)2/2) − z(ξ, η)

=1

2

(

(X + Y )((ξ − η)2/2) + (X − Y )((ξ + η)2/2))

− z(ξ, η) ,

where, by (10), z(ξ, η) is small compared to X and Y if η is small. We will expressthis in terms of Bessel processes and Brownian motions. Let σ = S2/2 and recall that

b(u) = X(σ) −X(σ − u), B(u) = σ− 1

2 (Y (σ) − Y (σ − σu))

are, respectively, a Bessel and an independent Brownian motion (Lemma 2.8). ThusW (ξ, η) is equal to

1

2[σ

1

2

(

B(1 − (ξ + η)2/(2σ)) −B(1 − (ξ − η)2/(2σ)))

− b(σ − (ξ + η)2/2) − b(σ − (ξ − η)2/2)] − z(ξ, η).

19

Let u = S − ξ, and v = η. Then 1 − (ξ ± η)2/(2σ) = (2S (u∓ v) − (u∓ v)2)/(2σ) andσ − (ξ ± η)2/2 = S (u∓ v) − (u∓ v)2/2.

When u and v are small, we can neglect z(ξ, η) and the second order terms. UsingBrownian scaling, we see that

W (ξ, η) ≈ S1

2

2(B(u− v) −B(u+ v) − b(u− v) − b(u+ v)) .

But when u is small and 0 < v ≪ u, as we have near the tip of the thorn, thedistribution of (b(u + v) + b(u− v))/2 is essentially the same as that of b(u), whereasthe distribution of (B(u + v) − B(u − v))/2 is exactly that of B(v)/

√2. This means

that up to negligeable terms, W has essentially the same decomposition as W in (15).Therefore, the remainder of the proof of this theorem will be essentially identical tothe proof of Theorem 3.1 concerning the case of horizontal lines. The details are leftto the reader. ♣

3.3 Supergeometric spacing of bubbles near the thorn

As we move away from the thorn Tβ , we encounter components of W > 0. We aregoing to describe the position and shape of these components. Intuitively, the firstcomponents we encounter will be small, and then we will encounter larger and largercomponents. We are going to make these statements precise. Set

b+(s) = b(s) + s3/4, b−(s) = b(s) − s3/4.

Notice that the thorn Tβ is contained in (s, t) : s ≥ t. By (15) and (2), the followingimplications are true for all small s and t such that s ≥ t:

b+(s) < B(t) =⇒ W (s, t) > 0 =⇒ b−(s) < B(t)(17)

b−(s) > B(t) =⇒ W (s, t) < 0 =⇒ b+(s) > B(t).(18)

Observe that b+(s) > 0 for all s, and, by [10, Example 5.4.7], s 7→ s3/4 is a lowerescape function for b, so there is a random variable η− > 0 such that b−(s) > 0 when0 < s < η−. In particular, by (18), W (s, t) < 0 on the horizontal segments where s ≥ tand B(t) = 0. Set

Q = (s, t) : s > t ≥ 0, W (s, t) > 0.(19)

Remark 3.6 For small s and t, any point (s, t) ∈ Q is contained in a horizontalstrip of the form RI + × [t1, t2], where [t1, t2] is an interval where B is accomplishingan excursion above 0. Points in Q which are sufficiently near the origin and are indifferent horizontal strips are in distinct connected components of Q. Indeed, assumeA1 = (u1, v1) ∈ Q, A2 = (u2, v2) ∈ Q, with u1 < η−, u2 < η− and

t1 < v1 < t2 < t3 < v2 < t4 < τ(η−),

20

where τ = ψa (defined in Remark 3.2), a > 2, and both [t1, t2] and [t3, t4] are intervalswhere B is accomplishing an excursion above 0. Then by Theorem 2.1, any curve Γin Q with extremities A1 and A2 must be contained in Q ∩ (Tτ )

c. But if i = 3, 4and 0 < s < η−, then b−(s) > B(ti) = 0, so by (18), W (s, t) < 0 on the horizontalsegment ]0, η−[×ti, i = 3, 4. Now Γ must pass through one of these two segments, acontradiction.

When β ≤ 2 and we follow the curve γψβtowards the origin, we will encounter

components of W > 0 infinitely many times. Due to the irregularity of the samplepaths of W , these components will never be isolated. But they will tend to occurin clusters, and the space between clusters will get larger and larger relative to theremaining distance to (S, 1). The following theorem makes this statement precise.

Let T be the class of continuous increasing functions τ defined in some interval[0, η[ —where η may depend on τ—such that τ(0) = 0, τ(s) > 0 if s > 0, and s 7→τ(s)/s is increasing on ]0, η[ . Let Q be defined as in (19). If a given point (s, τ(s)) isin Q, let Qs denote the connected component of Q which contains (s, τ(s)).

Theorem 3.7 Consider τ1 ∈ T such that I(τ1) = +∞. Fix 0 < c < 1 and letτ2, τ3 ∈ T be such that s3/2 ≤ τ1(s) ≤ τ2(s) ≤ τ3(s) ≤ c s and s 7→ τ2(s)/τ3(s) isincreasing and τ3(s)/τ3(c s) is bounded in a neighborhood of the origin. Assume

I(τ1, τ2, τ3) =∫

0+

(

τ1(s)

s

τ2(s)

τ3(s)

)1

2 ds

s< +∞.(20)

Then with probability 1, there exists η > 0 such that for 0 < s < η,

Qs 6= φ =⇒ ([τ3(s), c s] × [0, τ2(s)]) ⊂ W < 0.

Remark 3.8 (a) Consider the function ψβ defined in Remark 3.2. If τi(s) = ψβi(s),

i = 1, 2, 3, with β3 < β2 < β1 ≤ 2, then the conditions of Theorem 3.7 are satisfied andI(τ1, τ2, τ3) < +∞ if and only if β1 + β2 − β3 > 2.

(b) If s1 > s2 > · · · denote the s-coordinates of one point from each cluster encoun-tered along γψβ

, β ≤ 2, as we approach the origin, then sn/sn+1 ≥ sn/ψβ(sn) ↑ ∞ asn→ ∞. For a geometric sequence, this ratio is constant: this justifies the title of thissubsection.

(c) Theorem 3.7 shows in particular that there are no long bubbles parallel to γψ2.

Rather, the shape of the thorn is due to isolated short bubbles.(d) In view of Theorem 3.1, the assumption τ1(s) > s3/2 is not a serious restriction.

21

Proof of Theorem 3.7. We shall show that there are only finitely many s for which

b−(s) < B∗(τ1(s)) and minτ3(s)<u<c s

b−(u) < B∗(τ2(s)) .

By (18), this will complete the proof. Set δ = 1/4, c =√c, ε′n = c(n−2)(1−δ) and

An = A′n ∩ A′′

n, where

A′n = min

cn−1<u<cn−2

b(u) ≤ B∗(τ1(cn−2)) + ε′n,

A′′n = min

τ3(cn)<u<cnb(u) ≤ B∗(τ2(c

n−2)) + ε′n.

Since c s < cn for s ∈ ]cn−1, cn−2[ , it will be sufficient to show that P (lim supAn) = 0,and, by the Borel-Cantelli Lemma, we only need to show that

∑

P (An) < ∞. ByBrownian scaling, P (An) = P (C ′

n ∩ C ′′n), where

C ′n = min

c−1<u<c−2

b(u) ≤ KB∗(1)φ1(cn−2)

1

2 + ε′nc−n/2,

C ′′n = min

φ3(cn)<u<1b(u) < KB∗(1)φ2(c

n−2)1

2 + ε′nc−n/2,

and φi(s) = τi(s)/s, i = 1, 2, 3, and K denotes a constant. Thus

P (C ′n ∩ C ′′

n) =∫ ∞

0P (D′

n(a) ∩D′′n(a)) ν(da),

where ν(da) is the density of B∗(1) and D′n(a) and D′′

n(a) are defined by replacingB∗(1) by a in the definition of C ′

n and C ′′n, respectively. Let

αi(n, a) = Kaφi(cn−2)

1

2 + ε′nc−n/2, i = 1, 2.

Using the Markov property of b, we obtain

P (D′n(a) ∩D′′

n(a)) = EP (D′′n(a)|b(1))P b(1)(D′

n(a))≤ P (D′′

n(a))P0(D′

n(a)),

where D′n(a) = mind1<u<d2 b(u) ≤ α1(n, a) and di = c−i−1 > 0, i = 1, 2. By Lemma

3.3, P 0(D′n(a)) ≤ Kα1(n, a). In order to estimate P (D′′

n(a)), we again use the scalingproperty of b and Lemma 3.3 to observe that

P (D′′n(a)) = P min

1<u<1/φ3(cn)b(u) < α2(n, a)φ3(c

n)−1

2

≤ K(a + 1)(φ2

φ3

(cn−2))1

2

22

(the last inequality is valid for large n and uses the fact that ε′nc−n/2 < φ2(c

n−2)1

2 sinces3/2 < τ2(s) by assumption, and that Kφ3(c

n) ≥ φ3(cn−2), also by assumption). Thus,

for large n,

P (An) ≤∫ ∞

0[K(a+ 1)ε′nc

−n/2(φ2

φ3(cn−2))

1

2 +Ka(a+ 1)(φ1φ2

φ3(cn−2))

1

2 ] ν(da)

≤ KEB∗(1) + 1ε′nc−n/2 +KEB∗(1)2(φ1φ2

φ3

(cn−2))1

2 ,

since φ2/φ3 is bounded for large n by assumption. Summing over n, using the mono-tonicity of φ1φ2/φ3 to replace the sum by an integral and applying (20), we concludethat

∑

P (An) <∞. ♣

3.4 Height of bubbles near the thorn

The following theorem shows that bubbles that meet a particular curve near the thornwill not be very tall. Let Q be defined as in (19). As before, K denotes a constantwhose value may change from line to line.

Theorem 3.9 Let τ1, τ2 ∈ T be such that s5/4 ≤ τ1(s) ≤ τ2(s), s 7→ τ1(s)/τ2(s)is increasing with limit 0 as s ↓ 0 and, for each c > 1, τ2(cs)/τ2(s) is bounded in aneighborhood of the origin. Suppose

J(τ1, τ2) =∫

0+

τ1(s)

(s τ2(s))1

2

ds

s<∞ .

Then with probability one, there is η > 0 such that if 0 < s < η and (s, τ1(s)) ∈ Q,then Qs is contained in RI + × [0, τ2(s)] .

Remark 3.10 (a) If τ1 = ψβ, β ≤ 2, as defined in Remark 3.2, and β ′ ≤ β, thenJ(ψβ , ψβ′) < +∞ if and only if β ′ < 2β − 2.

(b) If I1 ≡ ∫

0+(τ1(s)/s)1

2 s−1 ds < ∞, the integral in (i) is always finite, but thetheorem is empty since Q does not intersect γ1. Thus the theorem is only of interestwhen I1 diverges.

Proof of Theorem 3.9. Suppose the integral is finite. Let γi be the graph of τi.Both τ1 and τ2 are defined near 0, so for all large n we can define

An = ∃ s ∈ [e−n, e1−n] : (s, τ1(s)) ∈ Q, Qs ∩ (RI + × [τ2(s),∞[ ) 6= φ.

We must show An occurs only finitely often. Let δ = 1/4, εn = e(1−n)(1−δ) and

mn = infe−n≤s≤e1−n

b(s), Mn = supτ1(e−n)≤t≤τ1(e1−n)

B(t) .

23

By (18), if mn ≥Mn + εn, then Q does not intersect γ1 between s = e−n and s = e1−n.On the other hand, since Qs ⊂ (Tψ3

)c by Theorem 3.1 and Remark 3.2 (a), if there issome t ∈ [τ1(e

1−n), τ2(e−n)] such that B(t) ≤ −εn, then W (s, t) < 0 for all small s, so

Qs ⊂ RI + × [0, t]. Since Mn ≤ B∗(τ1(e1−n)), it follows that

P (An) ≤ P

mn < B∗(τ1(e1−n)) + εn , inf

τ1(e1−n)≤t≤τ2(e−n)B(t) > −εn

.

Let λn = τ2(e−n)/τ1(e

1−n) and observe that en/2εn < e−n/4. Set φi(s) = τi(s), i = 1, 2.Using Brownian scaling and the independence of b and B, the last probability above is

= P

m0 ≤ Kφ1(e1−n)

1

2 B∗(1) + en/2εn , τ1(e1−n)

1

2 inf1≤t≤λn

B(t) > −εn

≤∫ ∞

0P

m0 ≤ Kφ1(e1−n)

1

2a+ e−n/4

P

B∗(1) ∈ da, inf1≤t≤λn

B(t) > −e−n/8

,

since εnτ1(e1−n)−

1

2 < e−n/8 because s5/4 < τ1(s) by assumption. By Lemma 3.3, the

first probability is dominated by K(φ1(e1−n)

1

2 a+ e−n/4) . To bound the second factor,apply the reflection principle at T (a), the first hit of a by B:

P inf1≤u≤λn

B(u) > −e−n/8 | T (a) ≤ 1 = P sup1≤u≤λn

B(u) < 2a+ e−n/8 | T (a) ≤ 1

≤ P 0(λn − 1)1

2 B∗(1) < a+ e−n/8by the strong Markov property at T (a) and scaling. Applying the reflection principleagain and since B(1) is standard Normal, this is

= P 0

|B(1)| < (a+ e−n/8)(λn − 1)−1

2

≤ K(a + e−n/8)(λn − 1)−1

2 .

Thus, the above integral is

≤ K∫ ∞

0

(

φ1(e1−n)

1

2 a+ e−n/4)

(a+ e−n/8)(λn − 1)−1

2 PB∗(1) ∈ da .

Since φ1 is increasing, this is

≤ K(

e−3n/8 + e−n/8EB∗(1) + φ1(e1−n)

1

2EB∗(1)2)

(λn − 1)−1

2 .

It follows from our assumptions that (λn − 1)−1

2 is ≤ K(φ1/φ2(e1−n))

1

2 for large n,which is decreasing in n by hypothesis, hence the last expression is bounded by

K(

e−3n/8 + e−n/8 + φ1(e1−n)φ2(e

1−n)−1

2

)

.

Summing over n and dominating the sum by an integral, we conclude that∑

n P (An) <∞ since

∫ ∞

0φ1(e

−x)φ2(e−x)−

1

2 dx =∫ 1

0φ1(y)φ2(y)

− 1

2 y−1dy <∞by hypothesis. Thus An only happens finitely often by the Borel-Cantelli Lemma,proving the theorem. ♣

24

3.5 Width of bubbles

Recall that Q = (s, t) : s > t ≥ 0, W (s, t) > 0. K and c are constants whose valuesmay change from line to line. We work with the functions ψβ defined in Remark 3.2.

Theorem 3.11 Let β ≤ 2 and fix β ′ < 2β − 2 and ε > 0. For s > 0, let V εs =

]s−εψβ′(s), s+εψβ′(s)[×[0,∞[ . Then with probability one, there exists an η > 0 suchthat if 0 < s < η and if Qs is the connected component of Q which contains the point(s, ψβ(s)), then Qs is contained in the vertical strip V ε

s .

There is one calculation to be made before we prove this. We separate it out as alemma.

Lemma 3.12 Let a > 0, λ > 0, and put L = aλ−1

2 . Set q(a, λ) = P ab ever has anexcursion below a of duration ≥ 2λ. Then there exists a constant C, independent ofa and λ, such that

q(a, λ) ≤ C(L+ 1)4e−1/(CL2).(21)

Proof. By Brownian scaling, q(a, λ) = q(L), where q(L) = PLb has an excursionbelow L of duration ≥ 2. Let T (L) be the first time b hits L and let S(L) be the firsttime after t = 1 that b hits L. Consider time 1. If b(1) < L, then b must stay below Lat least one unit longer to complete its excursion, and if b(1) > L, b must first return toL, then make its excursion. Thus q(L) ≤ p1 + p2, where p1 = PLb(1) < L, S(L) ≥ 2and

p2 = PLS(L) <∞, there is an excursion below L of duration ≥ 2 after S(L).

Applying the Markov property of b at time 1, we have

p1 ≤ PLb(1) < LP 0T (L) ≥ 1.

Clearly

PLb(1) ≤ L ≤ P 0b(1) ≤ L = K∫ L

0x2e−x

2/2 dx,(22)

where we have used the density of b(1) (see e.g. [11, Chapter 11 §1]). By [4, Theorem2], the second probability on the right hand side is bounded by c exp(−1/(cL2)). Withthis, we see that p1 ≤ cL3 exp(−1/(cL2)). Turning to p2, we note that if S(L) < ∞,then b(S(L)) = L, and the probability of having the excursion after S(L) is just q(L).Thus p2 = PLS(L) <∞q(L), hence

q(L) ≤ cL3e−1/(cL2) + PLS(L) <∞q(L),

or equivalently,q(L) ≤ cL3e−1/(cL2)(1 − PLS(L) <∞)−1.(23)

25

Now by [11, Corollary (3,4)]

P xT (L) <∞ = P xmins≥0

b(s) ≤ L = L/x .(24)

It follows thatPLS(L) = ∞ =

∫ ∞

L(1 − L/x) PLb(1) ∈ dx.

Notice that for x > L, PLb(1) ≤ x ≤ PLB(1) ≤ x, where B is a standard Brow-nian motion (since b has the same distribution as the modulus of a three-dimensionalBrownian motion) so that this is

≥∫ ∞

L(1 − L/x) ce−(x−L)2/2 dx = c

∫ ∞

0(√

2u+ L)−1 e−u du

after the change of variable u = (x− L)2/2. It is easy to see that this is a continuousdecreasing function of L which is asymptotic to 1/L for large L. Thus there is someconstant K such that PLS(L) = ∞ ≥ K/(L + 1) for all L, which, together with(23), proves the lemma. ♣

Proof of Theorem 3.11. Fix β ′′ such that β ′ < β ′′ < 2β − 2 (which is ≤ β sinceβ ≤ 2). Let In = [e−n, e1−n] and let An be the event on which there exists a componentQs of Q which contains a point (s, ψβ(s)) for some s ∈ In and which is not containedin the corresponding strip V ε

s . Then there exist u and v such that the points (s, ψβ(s))and (u, v) are in Qs while |u− s| > εψβ′(s) ≥ εψβ′(e−n).

Let J be the interval with endpoints u and s. Since Qs is connected, there is acontinuous curve (g1(x), g2(x)), 0 ≤ x ≤ 1 in Qs with (g1(0), g2(0)) = (s, ψβ(s)) and(g1(1), g2(1)) = (u, v). By (17), it follows that on the set An, we have B(g2(x)) >b−(g1(x)) for all x ∈ [0, 1]. Now ψβ′′(s) ≤ ψβ′′(e1−n) so by Theorem 3.9 and Remark3.10 (a), if n is large enough, then Qs ⊂ RI + × [0, ψβ′′(e1−n)]. This implies thatB∗(ψβ′′(e1−n)) ≥ b−(g1(x)) for each x. Since g1 runs through the interval J ,

B∗(ψβ′′(e1−n)) ≥ b−(u), ∀u ∈ J.

Thus, J is contained in an excursion interval of b− below the level B∗(ψβ′′(e1−n)). Letus simplify notation by setting δ = 1/4, ε′n = e(2−n)(1−δ) and

τ ′ = ψβ′(e−n), τ ′′ = ψβ′′(e1−n) .

For large n, if there is s ∈ In and a component Qs of Q which contains (s, ψβ(s)) andintersects (V ε

s )c, then

there exists an interval J of length at least ετ ′ which intersects In(25)

and is an excursion interval of b below the level B∗(τ ′′) + ε′n .

26

Let Dn be the event on which (25) happens and let us estimate P (Dn). Now P (Dn) ≤P (Dn,1) + P (Dn,2), where

Dn,1 = Dn ∩ b(e−n) ≤ (log n)β′′/2B∗(τ ′′),

Dn,2 = Dn ∩ b(e−n) > (logn)β′′/2B∗(τ ′′).

Using scaling, the independence of b and B, the density of b(1) (see (22)) and thedefinition of ψβ′′ in Remark 3.2 (a), we obtain

P (Dn,1) = Pb(1) ≤ KB∗(1)/n ≤ KEB∗(1)3/n3.

Thus∑

n P (Dn,1) < ∞. Note that (logn)β′′/2 ≥ 1 for n ≥ 3, so e−n 6∈ J on Dn,2. It

follows that the excursion interval J starts after time s = e−n. Use again independenceand scaling to see that P (Dn,2) is the probability that b(1) > B∗(1)/n and b has anexcursion below en/2

√τ ′′B∗(1) + ε′ne

n/2 after time s = 1 of duration ≥ εenτ ′. Settingα(a, n) = aen/2

√τ ′′ + ε′ne

n/2, conditioning on B∗(1) and using the strong Markovproperty of b at time S(α(a, n)), this probability is

≤∫ ∞

0P a/nmin

u≥1b(u) ≤ α(a, n) q(α(a, n), εenτ ′) ν(da),

where ν(da) is the density of B∗(1) and q(·, ·) is defined in Lemma 3.12. Let β ′′ − β ′ =

2d > 0 and L = α(a, n)(εenτ ′)−1

2 = ε−1

2 (a(log n)−d + ε′n(τ′)−

1

2 ). For large n, since

q(L) ≤ K(a + 2)4 exp(−(a+ 1)−2(log n)2d/C)

by Lemma 3.12, Lemma 3.3 implies that

P (Dn,2) ≤ K∫ ∞

0α(a, n)(a+ 2)4 exp(−(a + 1)−2(logn)2d/C) ν(da).

Sum this over n ≥ 2, estimate the sum by an integral and interchange order to see that∑∞n=2 P (Dn,2) is

≤ c+∫ ∞

0ν(da)(a+ 2)4a

∫ ∞

22x−1(log x)−β

′′/2 exp(−(a + 1)−2(log x)2d/C) dx.

Let y = (log x)/(a+ 1)1/d to see that the inner integral is

≤ (a+ 1)(2−β′′)/(2d)∫ ∞

0y−β

′′/2 exp(−y2d/C) dy .

This integral converges at 0 since β ′′ < 2 and clearly converges at ∞ since d > 0.Consequently the double integral is finite. Thus

∑

n P (Dn) < ∞, and the conclusionfollows from the Borel-Cantelli Lemma. ♣

Acknowledgement. The first author thanks Jim Pitman for several stimulatingdiscussions.

27

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