Thick Points for the Cauchy Process

Ann. I. H. Poincaré – PR 41 (2005) 953–970 www.elsevier.com/locate/anihpb

Thick points for the Cauchy process

Olivier Daviaud 1

Department of Mathematics, Stanford University, Stanford, CA 94305, USA Received 13 June 2003; received in revised form 11 August 2004; accepted 15 October 2004 Available online 23 May 2005

Abstract Let µ(x, ) denote the occupation measure of an interval of length 2 centered at x by the Cauchy process run until it hits −∞ − ]∪[ ∞ 2 → → ( , 1 1, ). We prove that sup|x|1 µ(x,)/((log ) ) 2/π a.s. as 0. We also obtain the multifractal spectrum 2 for thick points, i.e. the Hausdorff dimension of the set of α-thick points x for which lim→0 µ(x,)/((log ) ) = α>0.  2005 Elsevier SAS. All rights reserved. Résumé Soit µ(x, ) la mesure d’occupation de l’intervalle [x − ,x + ] parleprocessusdeCauchyarrêtéàsasortiede(−1, 1). 2 → → Nous prouvons que sup|x|1 µ(x, )/((log ) ) 2/π p.s. lorsque 0. Nous obtenons également un spectre multifractal 2 de points épais en montrant que la dimension de Hausdorff des points x pour lesquels lim→0 µ(x, )/((log ) ) = α>0est égale à 1 − απ/2.  2005 Elsevier SAS. All rights reserved.

MSC: 60J55

Keywords: Thick points; Multi-fractal analysis; Cauchy process

1. Introduction

Let X = (Xt ,t 0) be a Cauchy process on the real line R, that is a process starting at 0, with stationary independent increments with the Cauchy distribution: s dx P X + − X ∈ (x − dx,x + dx) = ,s,t>0,x∈ R. t s t π(s2 + x2)

E-mail address: [email protected] (O. Daviaud). 1 Research partially supported by NSF grant #DMS-0072331.

Next, let θ¯ X := µθ¯ (A) 1A(Xs) ds 0 ¯ be the occupation measure of a measurable subset A of R by the Cauchy process run until θ := inf{s: |Xs| 1}. Let I(x,)denote the interval of radius centered at x. Our ﬁrst theorem follows:

Theorem 1.1. X µ ¯ (I (x, )) lim sup θ = 2/π a.s. (1.1) → + 2 0 x∈R (log )

This should be compared to the following analog of Ray’s result [12]: for some constant 0

Theorem 1.2. For any a 2/π, X µ ¯ (I (x, )) dim x ∈ R: lim θ = a = 1 − aπ/2 a.s. (1.2) →0+ (log )2

The results obtained here are the analogues of those in [6], when replacing the planar Brownian motion (which is a stable process of index 2 in dimension 2) by the Cauchy process (which is a stable process of index 1 in dimension 1). Theorems 1.1 and 1.2 answer the first part of open problem (6) of that paper (also implicitly present in [11]), the second part being solved in [7]. Our work relies heavily on the techniques developed in [6] and [7], and therefore owes a substantial debt to these papers. The Cauchy process is a symmetric stable process of index α = 1. Thick points for one-dimensional stable processes have been studied in several papers. In [14], a multifractal spectrum of thick points is obtained for stable subordinators of index α<1. A one-sided version of this result also follows from [10, Theorem A], which deals with fast points of local times (recall that by [15] all stable subordinators of index α<1 appear as inverses of appropriate local times; in this analogy fast points of local times correspond to thick points of subordinators). The case of transient symmetric stable processes (i.e. symmetric stable processes with index α<1) is treated in [4] and can be seen as a generalization of the results on thick points for spatial Brownian motion obtained in [5]. Note that all these processes are transient (unlike the Cauchy process), and therefore the techniques used in these papers O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 955 differ significantly from the ones we shall use to study the case of the Cauchy process. When α>1 the process is recurrent, and thick points are easier to understand, since for such processes there exists a bi-continuous local time (e.g. see [3]). Thus in that case Theorem 1.1 would hold, but with a different scaling (simply ) and 2/π would be replaced by a random variable (more precisely: the supremum of the local time). In conclusion, our findings apply only at the border of transience and recurrence of stable processes. The main difficulty in obtaining results similar to those in [6] is that the Cauchy process is not continuous. Indeed, the proof of the lower bounds in [6] relies on the idea that unusually high occupation measures in the neighborhood of a point x are the result of an unusually high number of excursions of all scales around this point. But defining the notion of excursion is not clear when it comes to a non-continuous process. Our proof avoids this problem essentially by working with the Brownian representation of the Cauchy process: up to a time-change, the Cauchy process can be seen as the intersection of a two-dimensional Brownian motion and, say, the x-axis. Using this framework we obtain lower bounds by adapting the strategy in [7]. The same strategy could be used to derive upper bound results. However, because of its independent interest we use the following proposition as the key to our proof of the upper bounds: ¯ Proposition 1.3. Fix r>0, and let θ = inf{t: |Xt | r}. For any x0 ∈ (−r, r) and any bounded Borel measurable function f : [−r, r]→R, θ¯ r

x0 E f(Xs) ds = f(x)G(x0,x)dx (1.3) 0 −r where G is given by − 1 h(x/r) h(x0/r) G(x0,x)=− log , (1.4) 2π 1 − h(x/r)h(x0/r) and √ (1 + x)/(1 − x) − 1 h(x) = √ . (1.5) (1 + x)/(1 − x) + 1

Remarks.

• By the scaling property of the Cauchy process, for any deterministic 0

It is quite natural to consider also the discrete analogues of the results presented here. For example, let (Xi) be a sequence of i.i.d. variables with distribution: C P(Xi = n) = ,n∈ Z 1 + n2 956 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 = n where C is a normalizing constant. Let Sn i=1 Xi , S := { = } Ln(x) # i: Si x, 0 i n , be the number of visits to x ∈ Z during the first n steps of the walk and X := X Tn max Ln (x), x∈Z its maximal value. Then we conjecture that there exists a constant α such that T X lim n = α, a.s. n→∞ (log n)2 The source of the difficulty here is the absence of strong approximation theorems (such results were used to prove the Erdos–Taylor˝ conjecture in [6]). Another integer-valued random variable for which we expect similar = 1 2 Z2 := 1 asymptotic results is the following one: if Xn (Xn,Xn) is a simple random walk in , then we define Yn Xtn 2 where tn is the time at which 0 is visited for the n-th time by X . This is the discrete time analog of the Brownian representation of the Cauchy process, so our techniques should apply here. More generally, we suspect the existence of similar results for random variables in the domain of attraction of the Cauchy distribution. In the next section, we prove Proposition 1.3 using the Brownian representation of the Cauchy process and the solution to some Dirichlet problem. In Section 3, we use this result to prove upper bounds for both theorems. In Section 4 we prove the lower bounds, using a well defined system of excursions analogous to the one which appears in [7]. Finally Section 5 establishes the connection between occupation measure and excursions.

2. Green function for the Cauchy process

This section will be devoted to proving Proposition 1.3. The proof is based on the Brownian representation 1 2 of the Cauchy process: if (B ,B ) is a planar Brownian motion, and τt is the (right-continuous) inverse of the 2 1 ∈ R c := local time of B at 0, then Bτt is a Cauchy process. In what follows whenever x0 we let x0 (x0, 0).Let ¯ := { 2 = | 1| } 2 a σ inf t: Bt 0, Bt r and L be the local time of B at 0 (more generally we will use the notation L for the local time of B2 at a ∈ R, and will typically omit the superscript when denoting the local time at 0). We start with the following lemma:

¯B := { | 1 | } [− ]→R Lemma 2.1. Let θ inf s: Bτs r . A.s. for every bounded measurable function f : r, r , θ¯B σ¯ 1 = 1 f(Bτs ) ds f(Bu) dLu. (2.1) 0 0

¯B Proof. We ﬁrst show that θ = Lσ¯ almost surely. Since neither the Cauchy process nor the planar Brownian motion hit points, we have ¯B := | 1 | ¯ := 2 = | 1| θ inf s: Bτs >r and σ inf t: Bt 0, Bt >r a.s. We need the following result, taken from [13, p. 242]:

{ 2 = } Lemma 2.2 [13]. Let Z be the random set t 0: Bt 0 . Then P(∀s 0,B2 = B2 = 0) = 1. τs τs−

Conversely, for any u ∈ Z, either u = τs or u = τs− . O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 957

Therefore with probability one 2 1 1 1 σ¯ = inf t: B = 0, |B | >r = inf τu: |B | >r ∧ inf τ − : |B | >r . (2.2) t t τu u τu− 1 Now (by definition) τ − = lim → − τs . So by continuity of the Brownian motion, for all u such that |B | >r u s u τu− | 1 | there exists sr. Thus, the first infimum in (2.2) is less than or equal to the second one, and 1 ¯ = { | | } ¯ = ¯B therefore σ inf τu: Bτu >r . Since τ is non-decreasing and right-continuous, this implies that σ τθ , which = ¯B = in turn gives Lσ¯ θ (since it follows from the definition of τ and the continuity of L that for any x 0, Lτx x). In particular this yields

¯B θ Lσ¯ 1 = 1 f(Bτs ) ds f(Bτs ) ds a.s. (2.3) 0 0 Next, note that since τ has at most a countable number of discontinuities (τ is monotonous and ﬁnite),

Lσ¯ Lσ¯ 1 1 f(B ) ds = f(B )1{ = } ds a.s. (2.4) τs τs τs− τs 0 0

Now recall that by Lemma 2.2, any u ∈ Z is of the form τs or τs− for some s. Since τs− = inf{t: Lt s},wehave = = = − = that in both cases Lu s. Hence τLu u provided that τs τs . Therefore, the change of variable s Lu gives

Lσ¯ σ¯ 1 1 f(B )1{τ − =τ } ds = f(B )1{τ − =τ } dLu a.s. (2.5) τs s s u Lu Lu 0 0 The same change of variable and countability argument gives that

Lσ¯ σ¯

0 = 1{τ − =τ } ds = 1{τ − =τ } dLu a.s. s s Lu Lu 0 0 and therefore (2.5) becomes

Lσ¯ σ¯ 1 1 f(B )1{ = } ds = f(B ) dL a.s. (2.6) τs τs− τs u u 0 0 Combining (2.3), (2.4) and (2.6) ﬁnishes the proof of Lemma 2.1. 2

In particular, since Bτt is a Cauchy process, the above result implies that

θ¯ σ¯ c Ex0 = Ex0 1 f(Xs) ds f(Bu) dLu. (2.7) 0 0 From here on, without loss of generality we assume f to be continuous with compact support in (−r, r) (by the monotone class theorem, Proposition 1.3 will then follow for any bounded measurable function). We now need the following 958 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970

+ Lemma 2.3. Let gδ : R → R be a family of continuous functions with support in (−δ,δ) and such that gδ = 1. Then σ¯ σ¯ c c Ex0 1 0 = Ex0 1 2 f(Bu) dLu lim f(Bu)gδ(Bu) du. (2.8) δ→0 0 0

Let us postpone the proof of this lemma, and continue with the proof of Proposition 1.3. We ﬁrst rewrite the quantity in the right-hand side of (2.8). Using for example [9, Exercise 2.25, p. 253], we know it is equal to uδ(x0, 0) where uδ is deﬁned as follows:

Lemma 2.4. The unique solution to the partial differential equation: −(1/2) uδ(x1,x2) = gδ(x2)f (x1) on Dr , uδ(x1,x2) = 0 on ∂Dr 2 where Dr := R \ ((−∞,r]∪[r, ∞)), is given by

uδ(x) = 2gδ(z2)f (z1)G(x, z) dz (2.9)

Dr where G is the Green function of Dr . G is given by a complex analog of (1.4),i.e. − 1 h(z/r) h(z0/r) G(z0,z)=− log , (2.10) 2π 1 − h(z/r)h(z0/r) where, as in (1.5), √ (1 + z)/(1 − z) − 1 h(z) = √ . (2.11) (1 + z)/(1 − z) + 1

Proof of Lemma 2.4. h(z) can be written u(v(z)) where

1 + z z − 1 v(z) := , and u(z) := . 1 − z z + 1 v is a conformal mapping of D1 to the upper half-plane, and u is a conformal mapping of the upper-half plane to the unit disk. Thus h maps D1 conformally to the unit disk, and sends 0 to 0. Hence for any z0 ∈ Dr ,

h(z/r) − h(z0/r)

1 − h(z/r)h(z0/r) is a conformal mapping of Dr to the unit disk, which sends z0 to 0. Now the Green function of the unit disk with pole at 0 is −(1/2π)log z. Since Green functions are conformally invariant (e.g. see [1, p. 257]), G(z0,z) as deﬁned in (2.10) is indeed the Green function of Dr with pole at z0. Then (2.9) is simply the Green’s representation formula (e.g. see [8, Chapter 2]). 2

So the expectation in the right-hand side of (2.8) becomes ∞ r c = c uδ(x0) 2 gδ(z2) f(z1)G x0,(z1,z2) dz1 dz2. −∞ −r O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 959

Now by dominated convergence, the second integral is a continuous function of z2. Indeed, since h is one-to-one and analytic, for z in a compact subset K of Dr there exists M such that − c h(z/r) h(x0/r) c M|z − x |. − c 0 1 h(z/r)h(x0/r) Thus, 1 1 G(xc,z) − log(M) − log |x − z | (2.12) 0 2π 2π 0 1 which does not depend on z2 and is integrable as a function of z1.Wehaveprovedthat σ¯ r c Ex0 1 2 = c lim f(Bu)gδ(Bu) du 2 f(z1)G x0,(z1, 0) dz1. δ→0 0 −r This, together with (2.7) and Lemma 2.3 yields

θ¯ r x0 E f(Xs) ds = 2 f(z1)G (x0, 0), (z1, 0) dz1 0 −r which completes the proof of Proposition 1.3. 2

Proof of Lemma 2.3. We will prove the following stronger version of (2.8):

σ¯ σ¯ 1 0 = 1 2 f(Bu) dLu lim f(Bu)gδ(Bu) du (2.13) δ→0 0 0 where the limit is taken with respect to the norm · 2. To this aim, we ﬁrst notice that

σ¯ σ¯ u σ¯ ∞ 1 2 = 1 2 = 1 a f(Bu)gδ(Bu) du f(Bu) d gδ(Bt ) dt f(Bu) d gδ(a)Lu da 0 0 0 0 −∞ ∞ σ¯ = 1 a gδ(a) f(Bu) dLu da −∞ 0 where the second step follows from the occupation time formula [13, Corollary (1.6) p. 224], and the last one by Fubini’s theorem. Therefore (2.13) will follow once we prove that

σ¯ σ¯ 1 0 = 1 a f(Bu) dLu lim f(Bu) dLu (2.14) a→0 0 0 in · 2.Nowleta>0. By Tanaka’s formula (see e.g. [13], p. 222), for any t 0 t t 1 a 2 + 2 + 2 2 L = (B − a) − (B − a) − 1 2 dB =− 1 2 dB . 2 t t 0 (Bs >a) s (Bs >a) s 0 0 960 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970

Therefore for any t 0 t a 0 2 L − L = 2 1 2 dB , t t (00 and deﬁne

∞ −r− −r+ r− G z, (x, y) dx = G z, (x, y) dx + G z, (x, y) dx + G z, (x, y) dx −∞ −∞ −r− −r+ r+ ∞ + G z, (x, y) dx + G z, (x, y) dx r− r+

= I1 + I2 + I3 + I4 + I5.

By choosing appropriately, I2 and I4 can be made arbitrarily small. And by (2.12), I3 is a continuous function of y. Finally, on |x| r + , G(z, (x, y)) can be seen to be dominated by C/x2 for some constant C>0 uniformly on y (y small enough). Thus by dominated convergence, I1 +I5 tends to 0 as y → 0. These facts put together prove the continuity and the ﬁniteness of the inner integral in (2.15). Thus the right-hand side of (2.15) tends to 0 when a → 0+. The case a → 0− being similar, we have proved (2.14), and hence (2.13), which completes the proof of Lemma 2.3. 2

3. Upper bounds

Throughout this section, ﬁx 0 0: |Xt | r3} and deﬁne

θ¯ X := µθ¯ (r1) 1I(0,r1)(Xs) ds. 0 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 961

Lemma 3.1. There exists c>0 such that for all r1 r3/2 and |x0| r3 we have 2 r3 Ex0 X + µθ¯ (r1) r1 c log (3.1) π r1 and for all k 0 k k 2 r3 Ex0 X ! k + µθ¯ (r1) k r1 c log . (3.2) π r1

Proof of Lemma 3.1. Recall that by Proposition 1.3,

θ¯ r 1 Ex0 = c 1|Xs |r1 ds 2 G x0,(z1, 0) dz1.

0 −r1 To prove Lemma 3.1 we therefore need to ﬁnd an upper bound for the right-hand side of the above equation. We have r r 1 1 − c =− 1 h(z1/r3) h(x0/r3) 2 G x0,(z1, 0) dz1 log dz1 π 1 − h(z1/r3)h(x0/r3) −r1 −r1 r 1 1 z1 x0 2 − logh − h dz1 + r1 log 2. π r3 r3 π −r1 because h takes values in the unit disk. We will treat the cases |x0| (3/4)r3 and |x0| (3/4)r3 independently, and start with the former. The function h, when restricted to the compact set [−3/4, 3/4] is smooth and its derivative does not vanish. Since |z1|/r3 1/2 when z1 ∈ (−r1,r1), this yields − z1 x0 z1 x0 h − h M r3 r3 r3 for some constant M>0. Hence r r r 1 1 − 1 1 z1 x0 1 z1 x0 1 2 − logh − h dz1 − log dz1 − log M dz1 r1 log r3 + A π r3 r3 π r3 π π −r1 −r1 −r1 for some A uniformly in |x0| (3/4)r3. It remains to treat the case |x0| (3/4)r3. But then |x0/r3| 3/4 while again |z1/r3| 1/2. Therefore, h being continuous and one-to-one, there exists a constant B>0 uniform in r3 such that when x0 (3/4)r3, |h(z1/r3) − h(x0/r3)| B. So that in that case r 1 1 z1 x0 − logh − h dz1 Cr1 π r3 r3 −r1 for some constant C uniformly in r3. This ﬁnishes the proof of equation (3.1). (3.2) will then follow from the strong Markov property for the Cauchy process. Indeed, k k Ex0 X = !Ex0 ··· µθ¯ (r1) k 1I(0,r1)(Xsi ) ds1 dsk ¯ i=1 0s1···skθ 962 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 k−1 2 r3 !Ex0 + ··· k 1I(0,r1)(Xsi )r1 c log ds1 dsk−1 = π r1 ··· ¯ i 1 0 s1 sk−1 θ r3 k−1 = + Ex0 X r1 c log k µθ¯ (r1) , r1 proving (3.2) by induction on k. 2

This leads us to

−1 Lemma 3.2. We use the same notation as in Lemma 3.1.For0 λ<[(2/π)r1 log(r3/r1) + cr1] , −1 X 2 r x0 λµ ¯ (r1) 3 E e θ 1 − λr1 log + c , (3.3) π r1 which implies that for t>0 −1 −1 X 2 r3 + − 2 r3 + P µθ¯ (r1) t t r1 log c exp 1 t r1 log c . (3.4) π r1 π r1

Proof of Lemma 3.2. (3.4) follows from (3.3) by Chebyshev’s inequality. (3.3) is a straightforward consequence of (3.2). 2

In the remainder of this section, we use Lemma 3.1 to prove the upper bounds in Theorems 1.1 and 1.2. Namely if we deﬁne µX(I (x, )) θ¯ Thick := x ∈ I( , ) a , a 0 1 : lim sup 2 →0 (log ) ¯ (where θ = inf{t: |Xt | 1}), then we will show that for any a ∈ (0, 2/π],

dim(Thicka) 1 − aπ/2, a.s., (3.5) and X µ ¯ (I (x, )) 2 θ , lim sup sup 2 a.s. (3.6) →0 |x|<1 (log ) π Note that (3.5) will give the correct upper-bound for Theorem 1.2, since µX(I (x, )) θ¯ Thicka ⊃ Thicka := x ∈ I(0, 1): lim = a . →0 (log )2 Our proof follows [6]. Set h() = | log |2 and µ ¯(I (x, )) z(x, ) := θ . h() −2 Fix 0 <δ<1 and choose a sequence ˜n ↓ 0asn →∞in such a way that ˜n

h(ñ+1) = (1 − δ)h(ñ), (3.7) implying that n is monotone decreasing in n. Since, for ñ+1 ñ we have ˜ ˜ h(n+1) µθ¯(I (x, n)) z(x, ñ) = (1 − δ)z(x,), (3.8) h(ñ) h(ñ+1) O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 963 it is easy to see that for any a>0, Thicka ⊆ Da := x ∈ I(0, 1) lim sup z(x, ñ) (1 − δ)a . n→∞

Let {xj : j = 1,...,Kn} denote a maximal collection of points in I(0, 1) such that inf =j |x − xj | δ˜n.Let ¯ θ2 := inf{t: |Xt | 2} and An be the set of 1 j Kn, such that X + ˜ − ˜ µθ¯ I xj ,(1 δ)n (1 2δ)ah(n). (3.9)

Applying (3.4) with r1 = (1 + δ)ñ and r3 = 2gives x a(1−5δ)π/2 P µ ¯ I 0,(1 + δ)˜ (1 − 2δ)ah(˜ ) c˜ , θ2 n n n for some c = c(δ) < ∞, and any x ∈ I(0, 1). Note that for all x ∈ I(0, 1) and ,b 0 −x P µ ¯ I(x,) b P µ ¯ I(0,) b . θ θ2 Thus for any j and a>0, a(1−5δ)π/2 P(j ∈ An) cñ , implying that a(1−5δ)π/2−1 E|An| c ñ (3.10) V = ˜ ∈ ∈{ } ∈ V (by definition of Kn). Let n,j I(xj ,δn). For any x I(0, 1) there exists j 1,...,Kn such that x n,j , hence I(x,˜ ) ⊆ I(x ,(1 + δ)˜ ). Consequently, V forms a cover of D by sets of maximal n j n nm j∈An n,j a diameter 2δ˜m.Fixa ∈ (0, 2/π]. Since Vn,j have diameter 2δñ, it follows from (3.10) that for γ = 1−πa(1−6δ)/ 2 > 0, ∞ ∞ γ γ δaπ/2 E |Vn,j | c (2δ) ñ < ∞. = = n m j∈An n m ∞ Thus, |V |γ is finite a.s. implying that dim(D ) γ a.s. Taking δ ↓ 0 completes the proof of the n=m j∈An n,j a upper bound (3.5). Turning to prove (3.6), set a = 2(1 + δ)/(π(1 − 5δ)) noting that by (3.10) ∞ ∞ ∞ |A | E|A | ˜δ ∞ P n 1 n c n < . n=1 n=1 n=1

By Borel–Cantelli, it follows that a.s. An is empty for all n>n0(ω) and some n0(ω) < ∞. By (3.8) we then have µ ¯(I (x, )) 1 − 2δ θ a a, sup sup 2 ˜ |x|<1 (log ) 1 − δ n0(ω) and (3.6) follows by taking δ ↓ 0. 2

4. Lower bounds

In this section we adapt the proof of [7, Section 3]. While its authors studied the intersection local time for two independent Brownian motions, we are interested in the same quantity but for the intersection of a Brownian motion with a line. Throughout what follows we use notation similar to that in [7, Section 3]. Fixing a<2/π, c>0 and δ>0, let 964 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 µ ¯ (I (x, )) ¯ θc θc := inf t>0: |Xt | c ,Γc = Γc(ω) := x ∈ I(0, 1): lim = a , →0 (log )2 and Ec := {ω:dim(Γc(ω)) 1 − aπ/2 − δ}. In view of the results of Section 3, we will obtain Theorem 1.2 once we show that P(E1) = 1 for any a<2/π and δ>0. Indeed, since Thicka ⊂ Thicka and since we have seen that aπ dim(Thick ) 1 − , a.s., a 2 proving P(E1) = 1 will imply aπ dim Thick = 1 − , a.s. a 2 Moreover, the inequality µ ¯(I (x, )) µ ¯(I (x, )) lim inf sup θ sup lim inf θ → 2 → 2 0 |x|<1 (log ) |x|<1 0 (log ) then implies that for any η>0, µ ¯(I (x, )) lim inf sup θ 2/π − η, a.s. → 2 0 |x|<1 (log ) In view of (3.6), these lower bounds establish Theorem 1.1. The bulk of this section and the next will be dedicated to showing that P(E1)>0. Assuming this for the moment, let us show that this implies P(E1) = 1. Let us momentarily assume that X is the canonical version of the process. := c := −1 ¯ c = { | −1 | }= ¯ If ω X denotes the sample path, and Xt c Xct we have that cθ(ω ) inf ct: c Xct 1 θc(ω), and hence θ(ω¯ c) θ(ω¯ c) cθ(ω¯ c) Xc 1 1 X µ ¯ I(x,) = 1{|Xc−x|} ds = 1{|X −cx|c} ds = 1{|X −cx|c} ds = µ ¯ I(cx,c) . θ s cs c s c θc 0 0 0 c Consequently, Γc(ω) = cΓ1(ω ), so the Cauchy process’ scaling property implies that p = P(Ec) is independent of c>0. Let E := E lim sup n−1 , n→∞ so that P(E) p. The Cauchy process is a Feller process; hence if we let (Ft ) be the usual augmentation of the natural ﬁltration, it can be shown that (F ) is right-continuous (e.g. see [13, III-2]). Therefore, since E ∈ F ¯ , t c θc E ∈ F0 which implies P(E) ∈{0, 1}. Thus, p>0 yields P(E) = 1. We will see momentarily that the events Ec are essentially increasing in c,i.e.

∀0

So we just have to show that P(E1)>0. To achieve this goal we will use the Brownian representation of the Cauchy process, and follow the strategy of [7]. More precisely, moving to a Brownian setting, we will now focus our attention on the “projected intersection local time measures”: t I := 0 t (A) 1{Bu∈A} dLu, 0 = 1 2 0 2 where B (B ,B ) is a planar Brownian motion, L. is the local time of B at 0 and A is any measurable subset of R2. I is simply the amount of local time spent in A before t. To see how this relates to the Cauchy process, 2 2 1 note that, for example, for any set A ⊂ R , Iσ¯ (A) = µ ¯(A ∩ x−axis) where σ¯ := inf{t: B = 0 and |B | 1}, ¯ θ t t θ := inf{t: |Xt | 1}, X is the Cauchy process associated to the planar Brownian motion B and µ is the occupation measure for X. Indeed this follows from Lemma 2.1 2 We then reproduce the setting of [7, p. 248]: ﬁx a<2, 1 = 1/8 and the square S = S1 =[1, 21] ⊂ D(0, 1), 2 where for x ∈ R , ρ>0, D(x,ρ) denotes the closed disk of center x and radius ρ. Note that for all x ∈ S and ∈ ∪{ } ∈ ∈ ⊂ ⊂ = ! −3 = k −3 y S 0 both 0 / D(x,1) and 0 D(x,1/2) D(y,1) D(x,2).Letk 1(k ) 1 l=2 l .For ∈ x x S, k 2 and ρ>1,letNk (ρ) denote the number of excursions of B· from ∂D(x,k) to ∂D(x,k−1) prior to 2 hitting ∂D(x,ρ). Set nk = 3ak log k. We will say that a point x ∈ S is n-perfect if − x x + ∀ = nk k Nk (1/2) Nk (2) nk k, k 2,...,n. = 2 2 = n 6 = For n 2 we partition S into Mn 1 /(2n) (1/4) l=1 l non-overlapping squares of edge length 2n 3 21/(n!) , which we denote by S(n,i); i = 1,...,Mn with xn,i denoting the center of each S(n,i).LetY(n,i); i = 1,...,Mn be the sequence of random variables deﬁned by

Y(n,i)= 1ifxn,i is n-perfect and Y(n,i)= 0 otherwise. Deﬁne An = S(n,i), i: Y (n,i)=1 and F = F(ω)= An := Fm. (4.2) m nm m

Note that each x ∈ F is the limit of a sequence {xn} such that xn is n-perfect. We ﬁnally rotate this picture by 45 degrees clockwise. S now intersects the x-axis; let D be this intersection. The next lemma will be proved in the next section.

¯ Lemma 4.1. Let θ := inf{t: |Bt | 1}. A.s. for all x ∈ F ∩ D I ¯(D(x, )) 2 lim θ = a. →0 (log )2 π

Now Lemma 3.2 in [7] shows that for every a<1, and every δ>0 such that 1 − a − δ>0, P dim(F ∩ D) 1 − a − δ > 0. (4.3) This, together with Lemma 4.1 implies I ¯(D(x, )) 2 P dim D ∩ lim θ = a 1 − a − δ > 0. →0 (log )2 π 966 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970

Now if σ¯ denotes the ﬁrst time that the planar Brownian motion hits the complement of (−1, 1) (on the real axis), we have by the strong Markov property I ¯ (D(x, )) 2 P dim D ∩ lim σ = a 1 − a − δ → 2 0 (log ) π I ¯(D(x, )) 2 KP dim D ∩ lim θ = a 1 − a − δ →0 (log )2 π where K = inf Pa Brownian motion hits (−∞, −1]∪[1, ∞) before D . |a|=1

Since K>0, in view of Lemma 2.1 we have proved that P(E1)>0, which concludes the section.

5. From excursions to intersection local time

This section follows (very) closely the argument developed in [7, Section 4]. The sets F and D are the same as in the previous section, and h() := (log )2. Lemma 4.1 will follow from the next two lemmas.

Lemma 5.1. For every δ>0,ifx ∈ F ∩ D then 2 I ¯(D(x, )) a(1 − δ)6 lim inf θ . (5.1) π →0 h()

Lemma 5.2. For every δ>0,ifx ∈ F ∩ D then I ¯(D(x, )) 2 lim sup θ a(1 + δ)5. (5.2) →0 h() π

6 Proof of Lemma 5.1. We use the same notation as in [7]. Let k be as in the previous section, δk := k/k and let Dk be a δk-net of points in S.Let = 1/k6 = −2/k6 k ke ,k−1 k−1e , so that for k large enough, + − k k δk,k−1 k−1 δk. (5.3) ∈ D − We will say that a point x k is lower k-successful if there are at least nk k excursions of B from ∂D(x ,k) ¯ to ∂D(x ,k−1) prior to θ.Let −j/k k,j = k e ,j= 0, 1,...,3k log(k + 1), = −4/k3 = −j/k −4/k3−1/k6 ∈ D and let k,j k,j e k e e . By analogy with [7], we say that x k is lower k, δ-successful if it is lower k-successful and in addition, 2 (1 − δ) ρ D(x , ) , ∀j = 0,...,3k log(k + 1) (5.4) π k,j k,j where ρ denotes the measure supported on the real axis, and whose restriction to the real axis is 1/π times the Lebesgue measure. We recall Lemma 2.3 of [7], adapted to our situation. In what follows, ¯ θx,r := inf t: |Bt − x|=r , and we write a = b ± c as a shorthand for |a − b| c. O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 967

Lemma 5.3. We can ﬁnd c<∞ such that for all k 1, r1 r2 r/2 1/2, x and x0 with |x0 − x|=r2, k x k r E 0 I ¯ D(x,r ) k! ρ D(x,r ) log + cr , (5.5) θx,r 1 1 1 r1 and x r E 0 I ¯ D(x,r ) = ρ D(x,r ) log ± cr . (5.6) θx,r 1 1 1 r2

The above lemma can be seen as an analog of Lemma 3.1; the main difference lies in the fact that the Brownian motion is now stopped when it leaves a disk of radius r. We are now in a position to prove Lemmas 5.1 and 5.2. We will derive Lemma 5.1 from the following lemma, which is an analog of Lemma 4.3 in [7].

Lemma 5.4. There exists a k0 = k0(δ, ω) such that for all k k0 and x ∈ Dk,ifx is lower k,δ-successful then 2 4 (1 − δ) h( ) I ¯ D(x , ) , ∀j = 0, 1,...,3k log(k + 1). (5.7) π k,j θ k,j

The derivation of Lemma 5.1 from Lemma 5.4 is exactly the same as the derivation of Lemma 4.1 from Lemma 4.3 in [7]. For the reader’s convenience, we repeat this argument here. First, for x ∈ F , there exists a sequence {yn} of n-perfect points which tends to x. Since n-perfect points are also k-perfect for k n, we can find a sequence {˜xk} of k-perfect points such that |x −˜xk| δk. Finally, by definition of Dk for any such point x˜k there exits xk ∈ Dk such that |˜xk − xk| δk. Since x˜k is k-perfect, (5.3) and the fact that δk is decreasing guarantee that −4/k3 xk is lower k-successful. Note that |x − xk| 2δk. Thus, since δk/(k+1 e ) → 0ask →∞, 2 (1 − δ) ρ D(x , ) , ∀j = 0,...,3k log(k + 1) π k,j k k,j as soon as k is greater than some k1(δ). Therefore, xk is lower k,δ-successful as soon as k k1. Consequently, by Lemma 5.4, 2 4 (1 − δ) h( ) I ¯ D(x , ) , ∀j = 0, 1,...,3k log(k + 1) (5.8) π k,j θ k k,j = ∨ + as soon as k is greater than k2(δ, ω) k1 k0. Now notice that for k large enough (say greater than k3), k,j 2δk k,j for all j in the range of interest. This implies 2k I ¯ D(x, ) I ¯ D x, + I ¯ D(x , ) . θ k,j θ k,j k6 θ k k,j Combined with (5.8) this gives 2 4 2 5 I ¯ D(x, ) (1 − δ) h( ) (1 − δ) h( ) θ k,j π k,j π k,j for all j = 0, 1,...,3k log(k +1) and k sufficiently large. Finally, for any k+1 k,letj be such that k,j+1 k,j . Then, using the monotonicity of h, we obtain that I ¯(D(x, )) I ¯(D(x, k,j+ )) I ¯(D(x, k,j+ )) 2 θ θ 1 θ 1 1 − , h() h(k,j ) h(k,j+1) k where the last step holds for k large enough. This completes the proof of Lemma 5.1. 2

Proof of Lemma 5.4. Here again most of the proof is identical to the one of [7, Lemma 4.3]. Let x ∈ Dk be lower = − k,δ-successful. Then B makes at least nk nk k excursions between ∂D(x ,k) and ∂D(x ,k−1). For such a 968 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970

⊂ point we let τl,k,j denote the projected intersection local time measure of D(x ,k,j ) D(x ,k) accumulated by B during its l-th excursion from ∂D(x ,k) to ∂D(x ,k−1).Let 2 4 A(x ,k,j):= I ¯ D(x , ) (1 − δ) h( ) . θ k,j π k,j We deﬁne n k 2 P := P A(x ,k,j),x is lower k,δ-successful P τ a(1 − δ)4h( ) x ,k,j x ,s l,k,j π k,j l=1 · := ·| El where Px ,s( ) P( x is lower k,δ-successful).Ifwelet x,s be the conditional expectation with respect to the measure Px,s and given the trajectory of B up to the starting time of the l-th excursion, then 2 r El − 2 x,s(τl,k,j) (1 δ) log r1 Px ,s a.s. (5.9) π r2 := := := − where r1 k,j , r2 k, r k−1, and where we have used (5.6) in conjunction with (5.4). The extra (1 δ) factor comes from the fact that log(r/r2) tends to inﬁnity as k becomes large. Note that the left-hand side of (5.9) is random, while the right-hand side is not. Now 4 4 log( ) log(k+1) + = log 1 − 3log(k + 1)!+ = 1 + o(1) 3k log k, (5.10) k,j k3 k3 = 2 − where the last step follows from Sterling’s formula. On the other hand, since nk 3ak log k k, r = − −6 = + 2 nk log nk(3logk 3k ) 1 o(1) a(3k log k) . (5.11) r2 Therefore, (5.9) (multiplied by nk) together with (5.10) and (5.11) gives

l 2 3 n E (τ ) a(1 − δ) h( ) P a.s. k x ,s l,k,j π k,j x ,s provided k is large enough. This results with

n 1 k P P τˆ 1 − δ , x ,k,j x ,s n l,k,j k l=1 ˆ := El El ˆ = ˜ := ˆ − El ˆ where τl,k,j τl,k,j/ x,s(τl,k,j). By deﬁnition, x,s(τl,k,j) 1, so that, with τl,k,j τl,k,j x,s(τl,k,j) we have n 1 k P P τ˜ −δ . x ,k,j x ,s n l,k,j k l=1 Since r − log = 3logk − 3k 6 r2 and r − − log 3logk(k + 1) − 2k 6 + 4k 3, r1 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970 969

El ˜2 Lemma 5.3 implies that x,s(τl,k,j) C for some C>0. Now remark that there exists D>0 such that for x 1, x 2 e 1 + x + Dx . Since τ˜l,k,j −1, it follows that for all 0 <θ<1, − ˜ 2 El θτl,k,j + 2El ˜2 + 2 CDθ x,s(e ) 1 Dθ x,s(τl,k,j) 1 CDθ e . Taking θ = δ/(2CD) (which is smaller than 1 provided that δ is small enough) and applying Chebyshev’s inequality (together with the Markov property) shows that for some λ = λ(a, δ) > 0, C1 < ∞ and all k,x ∈ Dk,j, −λk2 log k Px,k,j C1 e . C k log k Now since |Dk| e 2 for some C2 < ∞ and all k, it follows that ∞ 3k log(k+1) ∞ 2 C2k log k −λk log k Px,k,j 3C1k log(k + 1) e e < ∞. (5.12) k=1 j=0 x ∈Dk k=1 The Borel–Cantelli lemma completes the proof. 2

Proof of Lemma 5.2. Again, we use the same notation as in [7]: we let ¯ = −2/k6 ¯ = 1/k6 k k e , k−1 k−1 e , so that for k large enough ¯ − ¯ + k k δk, k−1 k−1 δk. ¯ := 4/k3 ∈ D := + Let k,j k,j e . We now say that x k is upper k-successful if there are at most nk nk k excursions ¯ ¯ ¯ of B from ∂D(x , k) to ∂D(x , k−1) prior to θ, and if in addition 1 ¯ ρ D(x , ¯ ) , ∀j = 0,...,3k log(k + 1). (5.13) π k,j k,j The constant 1/π on the left-hand side could be replaced by any positive constant smaller than 2/π.Usingthe same argument as in the previous case, Lemma 5.2 can be derived from

Lemma 5.5. There exists a k0 = k0(δ, ω) such that for all k k0 and x ∈ Dk,ifx is upper k-successful then 2 3 a(1 + δ) h(¯ ) I ¯ D(x , ¯ ) , ∀j = 1,...,3k log(k + 1). π k,j θ k,j

Proof of Lemma 5.5. Again, the proof is very similar to the one of [7, Lemma 4.4]. In a similar manner we let τl,k,j ¯ denote the projected intersection local time measure of D(x , k,j ) accumulated by B during its l-th excursion from ¯ ¯ ¯ ⊂ ¯ ∂D(x , k) to ∂D(x , k−1) (note that D(x , k,j ) D(x , k) for k large enough since we only consider j 1). Let El denote the conditional expectation given the trajectory of B up to the starting time of the l-th excursion. ¯ ¯ ¯ Using the same line of arguments as in the proof of Lemma 5.4 (where r, r1 and r2 are set to k−1, k,j and k respectively, and where (5.9) is replaced by 2 r El + x,s(τl,k,j) (1 δ)log r1 Px ,s a.s. (5.14) π r2 ¯ ¯ since ρ(D(x , k,j )) is now simply bounded by (2/π)k,j ) we obtain that if 2 3 B(x ,k,j):= I ¯ D(x , ¯ ) (1 + δ) h(¯ ) θ k,j π k,j then 970 O. Daviaud / Ann. I. H. Poincaré – PR 41 (2005) 953–970

n k 2 Q := P B(x ,k,j),x is upper k-successful = P τ a(1 + δ)3h(¯ ) x ,k,j l,k,j π k,j l=1 n 1 k P τ˜ δ . n l,k,j k l=1 l Here τ˜l,k,j := (τl,k,j/E τl,k,j) − 1 whenever there are at least l excursions and τ˜l,k,j := 0 otherwise. For any a,b 0, (a + b)n 2n(an + bn), and therefore by (5.5), (5.6) and (5.13) El ñ ! n (τl,k,j) n AB (5.15) for some A,B > 0. Thus, ∞ n ∞ ˜ λ 2 El(eλτl,k,j ) = 1 + El(τñ ) = 1 + A(λB)n 1 + Cλ2 eCλ n! l,k,j n=2 n=2 for some C>0, where the first inequality holds as soon as λ is small enough. As in the proof of Lemma 5.4, −aλk2 log k Chebyshev’s inequality then implies that Qx,k,j b e for some constants a,b. A last argument of the form (5.12) finishes the proof. 2

Acknowledgements

I am very grateful to Amir Dembo and Yuval Peres for suggesting this problem to me, and for many helpful discussions.

References

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