On the volume of intersection of three independent Wiener sausages

www.imstat.org/aihp 2010, Vol. 46, No. 2, 313–337

DOI:10.1214/09-AIHP316

© Association des Publications de l’Institut Henri Poincaré, 2010

On the volume of intersection of three independent Wiener sausages

M. van den Berg

Department of Mathematics, University of Bristol, University Walk, Bristol BS8 1TW, United Kingdom. E-mail:M.vandenBerg@bris.ac.uk Received 6 December 2008; revised 9 February 2009; accepted 10 February 2009

Abstract. LetKbe a compact, non-polar set inR^m, m≥3 and letS_Kⁱ (t)= {Bⁱ(s)+y: 0≤s≤t, y∈K}be Wiener sausages associated to independent Brownian motionsBⁱ, i=1,2,3 starting at 0. The expectation of volume of₃

i=1S_Kⁱ (t)with respect to product measure is obtained in terms of the equilibrium measure ofKin the limit of larget.

Résumé. SoitKun ensemble compact, non-polaire dansR^m(m≥3)et soitS_Kⁱ (t)= {Bⁱ(s)+y: 0≤s≤t, y∈K}des saucisses de Wiener associées à des processus Browniens indépendantsBⁱ, i=1,2,3 initalisés à 0. L’espérance des volumes de₃

i=1S_Kⁱ (t) par rapport à la mesure produit est obtenue en termes de la mesure d’équilibre deKlorsquettend vers l’infini.

MSC:35K20; 60J65; 60J45

Keywords:Wiener sausage; Equilibrium measure

1. Introduction

Let K be a compact, non-polar set in Euclidean space R^m (m=2,3, . . .), and let (B(s), s≥0;Px, x ∈R^m) be Brownian motion associated to the parabolic operator−+_∂t^∂. The Wiener sausage associated toK, and generated byBup to timetis the random setSK(t)defined by

S_K(t)=

B(s)+y: 0≤s≤t, y∈K .

Its volume, denoted by|S_K(t)|, is a simple example of a non-Markovian functional of Brownian motion. It plays a key role in the study of stochastic phenomena like trapping in random media, random Schrödinger operators, and diffusion of matter [15]. The expectation of|S_K(t)|has been the subject of extensive investigation. Spitzer [14], Le Gall [8–10] and Port [12] analyzed its asymptotic behaviour for larget, while van den Berg and Le Gall [17] initiated the study of the asymptotic behaviour for smallt. See in particular Chapter 2 of [3] for an up to date account of the smalltbehaviour in the more general setting of a Riemannian manifold.

LetS_Kⁱ

i(t), i=1, . . . , ndenote Wiener sausages associated to compact, non-polar setsK_i, i=1, . . . , n, and gen- erated by independent Brownian motions Bⁱ, i=1, . . . , n respectively. The random setn

i=1S_Kⁱ

i(t) shows up in numerous places in the physical sciences. In quantum field theory one is interested in estimates for the probability that this random set is empty, in particular in the casen=2, m=4, andK1= · · · =K_n=K[1,2]. The phenomenon of loop condensation [6] for the intersection ofnindependent random walks onZ^m(or Wiener sausages inR^m) with

1Supported by The Leverhulme Trust, Research Fellowship 2008/0368.

n > m/(m−2)initiated the study of the large deviations for the volume of intersection on the scale of their mean [18] in the case where theKi’s are balls with equal radius. In polymer physics one wishes to obtain properties of the volume of intersection in either the random walk approximation onZ³[7,11], or in the Brownian motion approxima- tion inR³. Below we calculate the precise expected volume of intersection in the physically relevant casesm=3, and n=2, orn=3. While we only consider identical polymersK₁= · · · =K_n=K, and of equal lengtht, extensions can easily be obtained to include expressions for the expected volume ofS_Kⁱ

i(ait ), i=1, . . . , n, where theai’s are strictly positive and not all of them infinite. In the proofs of Theorems1–3below we will see that the physically relevant case m=3 is mathematically the most challenging yielding a non-trivial leading term.

Define fort >0 the expectation with respect to the product law by N_n,m(t)=E¹₀⊗ · · · ⊗Eⁿ₀

n i=1

S_Kⁱ (t)

. Letu:(R^m−K)×(0,∞)→Rbe the solution of

u=∂u

∂t, x∈R^m−K, t >0, (1)

with initial condition

u(x;0)=0, x∈R^m−K, (2)

and boundary condition

u(x;t )=1, x∈∂K, t >0, (3)

where∂Kis the boundary ofK. Equations (1)–(3) are to be understood in the weak sense. (The pointwise limit in (3) holds only at the regular points of∂K.) It is well known that the solution of (1)–(3) is given by

u(x;t )=Px[TK< t], (4)

whereT_K is the first hitting time ofK, T_K=inf

s≥0: B(s)∈K

. (5)

We adopt the usual convention in (5) that the infimum over the empty set equals+∞. We extenduto all ofR^m× (0,∞)by puttingu(x;t )=1 onK×(0,∞). By Fubini’s theorem we have that

N_n,m(t)=

R^mdx

Px[T_K< t]n

. (6)

In the last paragraph of this section we will show that

N2,m(t)=2N1,m(t)−N1,m(2t ), t >0. (7)

The asymptotic behaviour ofN_2,m(t)ast→ ∞ort→0 can then be read-off from the results obtained in [3,8,9,12, 14,17]. No such recursive formulae are known forn=3,4, . . . .In this paper we obtain the asymptotic bahaviour in the case wheren=3, m=3,4, . . . ,andt→ ∞. The strong dimension dependence in our main results, Theorems1–3 below, is directly related to the integrability properties ofPx[T_K <∞]for large|x|. These integrability properties improve as the dimensionmincreases. Ifm=3,4, . . . ,then (Theorem 2 in [16])

tlim→∞Nn,m(t)=

R^mdx

cm μK(dy)|x−y|^2−m n

<∞, (8)

if and only ifn > m/(m−2), whereμ_K denotes the equilibrium measure supported onK, and cm=4⁻¹π^−m/2

(m−2)/2 .

Throughout the paper we denote the Newtonian capacity ofKby C(K)=μ_K(K).

Euler’s constant is denoted byγ, Catalan’s constant is denoted byG, and H= ^1/

√3

0

1+θ²₋1

logθdθ. (9)

Theorem 1. Letm=3.Then fort→ ∞ N_3,3(t)=2⁻¹(4π)⁻²C(K)³logt

−(4π)⁻²C(K)² μ_K(dy)log|y| +2(4π)⁻³(2π−πγ−12G−6H )C(K)³ +(4π)⁻³

R³dx

μK(dy1)μK(dy2)μK(dy3)

|x−y1|⁻¹|x−y2|⁻¹− |x|⁻²

|x−y3|⁻¹ +3(4π)^−7/2C(K)⁴t^−1/2logt+O

t^−1/2 . Theorem 2. Letm=4.Then fort→ ∞

N_3,4(t)=

R⁴dx

c₄ μ_K(dy)|x−y|⁻² 3

−3(4π)⁻⁴C(K)³t⁻¹logt +6(4π)⁻⁴C(K)t⁻¹

μ_K(dy1)μ_K(dy2)log|y₁−y₂| +3(4π)⁻⁴(γ−2−log 4+log 3)C(K)³t⁻¹

+3(4π)⁻²

R⁴dx

c4 μK(dy)|x−y|⁻² 3

C(K)t⁻¹+O

t⁻¹logt2 . Theorem 3. Letm=5,6,7, . . . .Then fort→ ∞

N_3,m(t)=

R^mdx

c_m μ_K(dy)|x−y|^2−m 3

−3(4π)^−m2^m−3(m−2)⁻¹

(m−4)/2 C(K)

μ_K(dy1)μ_K(dy2)|y₁−y₂|^4−mt^(2−m)/2 +6(m−2)⁻¹(4π)^−m/2C(K)

R^mdx

c_m μ_K(dy)|x−y|^2−m 3

t^(2−m)/2

+

⎧⎨

⎩

3⁻¹(4π)⁻⁵

12+π−2√ 3

C(K)³t⁻²+O t^−5/2

, m=5, 2⁻¹(4π)⁻⁶C(K)³t⁻³logt+O

t⁻³

, m=6,

O t^−m/2

, m≥7.

Define the last exit time ofKby LK=sup

s≥0:B(s)∈K

, (10)

andL_K= +∞if the supremum is over the empty set. The law ofL_Kis given by (see [13,15]) Px[L_K< t] = ^t

0

ds μ_K(dy)p(x, y;s), (11)

wherep(x, y;s), x∈R^m, y∈R^m, s >0 is the transition density for Brownian motion (associated to−+_∂s^∂ ) given by

p(x, y;s)=(4πs)^−m/2e^{−|x−y|2/(4s)}. (12)

In particular

Px[L_K<∞] =Px[T_K<∞] =c_m μ_K(dy)|x−y|^2−m. (13) This reproves (8) by the monotone convergence theorem, (6) and (13).

The first step in the proofs of Theorems1–3is to substitute

Px[T_K< t] =Px[L_K< t] +Px[T_K< t < L_K] (14)

in (6), and to bound higher order contributions from terms involving e.g. (Px[TK < t < LK])². These bounds are given in Proposition4below. The proof is deferred to Section2. The asymptotic behaviour of

R^mdx (Px[LK< t])³ ast→ ∞is given in Propositions6–8below for dimensionsm=3, m=4, andm≥5 respectively. The correspond- ing calculations are deferred to Sections 4–6 respectively. Finally, the asymptotic behaviour of

R^mdx (Px[L_K <

∞])²Px[T_K< t < L_K]ast→ ∞is given in Proposition5. The proof relies on an application of the strong Markov property atT_Kand is deferred to Section3.

Proposition 4. Letm=3,4, . . . .DefineR_m(t)by N3,m(t)=

R^mdx

Px[LK< t]3

+3

R^mdx

Px[LK<∞]2

Px[TK< t < LK] +Rm(t).

Then fort→ ∞ R₃(t)=O

t^−1/2 , R4(t)=O

t⁻²logt and form=5,6, . . .

R_m(t)=O t^2−m

.

Proposition 5. Letm=3,4, . . . .Then fort→ ∞ (i) m=3

R³dx

Px[LK<∞]2

Px[TK< t < LK] =(4π)^−7/2C(K)⁴t^−1/2logt+O t^−1/2

, (ii) m=4

R⁴dx

Px[LK<∞]2

Px[TK< t < LK] =(4π)⁻²

R⁴dx

Px[LK<∞]3

C(K)t⁻¹+O

t⁻¹logt2 , (iii) m=5,6, . . .

R^mdx

Px[L_K<∞]2

Px[T_K< t < L_K]

=(4π)^−m/2 m

2 −1 ₋1

R⁴dx

Px[L_K<∞]3

C(K)t^(2−m)/2+O t^−m/2

.

Proposition 6. Letm=3.Then fort→ ∞

R³dx

Px[L_K< t]3

=2⁻¹(4π)⁻²C(K)³logt−(4π)⁻²C(K)² μK(dy)log|y| +2(4π)⁻³(2π−πγ−12G−6H )C(K)³

+(4π)⁻³

R³dx

μ_K(dy1)μ_K(dy2)μ_K(dy3)

|x−y1|⁻¹|x−y2|⁻¹− |x|⁻²

|x−y3|⁻¹ +O

t^−1/2 .

Proposition 7. Letm=4.Then fort→ ∞

R⁴dx

Px[LK< t]3

= R⁴dx

Px[LK<∞]3

−3(4π)⁻⁴C(K)³t⁻¹logt +6(4π)⁻⁴C(K)t⁻¹

μ_K(dy1)μ_K(dy2)log|y₁−y₂| +3(4π)⁻⁴(γ−2−log 4+log 3)C(K)³t⁻¹+O

t⁻²logt . Proposition 8. Letm=5,6, . . . .Then fort→ ∞

R^mdx

Px[L_K< t]3

= R^mdx

Px[L_K<∞]3

−3(4π)^−m2^m−3(m−2)⁻¹

(m−4)/2 C(K)

μK(dy1)μK(dy2)|y1−y2|^4−mt^(2−m)/2

+

⎧⎨

⎩

3⁻¹(4π)⁻⁵

12+π−2√ 3

C(K)³t⁻²+O t^−5/2

, m=5, 2⁻¹(4π)⁻⁶C(K)³t⁻³logt+O

t⁻³

, m=6,

O t^−m/2

, m≥7.

We conclude thisIntroductionwith a short proof of (7). Lett >0, and letB(s)=B(t+s)−B(t ), andB (s)= B(t−s)−B(t )for everys∈ [0, t]. ThenB andB are two independent Brownian motions on the time interval[0, t]. LetS_K(t)andS_K (t)respectively denote the corresponding Wiener sausages associated toK over the time interval [0, t]. Let S_K(t,2t )= {B(s)+y: t ≤s≤2t, y∈K}be the Wiener sausage generated by B over the time interval [t,2t]. By translation invariance of Lebesgue measureE[|S_K(t)∩S_K(t,2t )|] =E[|S_K(t)∩S_K (t)|] =N_2,m(t).On the other hand,N1,m(2t )=E[|S_K(2t )|] =E[|S_K(t)|] +E[|S_K(t,2t )|] −E[|S_K(t)∩S_K(t,2t )|] =2N1,m(t)−N2,m(t), which is (7).

2. Proof of Proposition4

To prove Proposition4we defineL_KandT_Kas in (10) and (5) respectively. Then Px[TK< t]3

=

Px[LK< t] +Px[TK< t < LK]3

≥

Px[LK< t]3

+3

Px[LK< t]2

Px[TK< t < LK]. On the other hand, sincePx[T_K< t < L_K] =Px[T_K< t < L_K<∞] ≤Px[t < L_K<∞], we have that

Px[LK< t]

Px[TK< t < LK]2

≤Px[LK<∞]Px[t < LK<∞]Px[TK< t < LK]

and

Px[T_K< t < L_K]3

≤Px[L_K<∞]Px[t < L_K<∞]Px[T_K< t < L_K]. Hence

Px[T_K< t]3

≤

Px[L_K< t]3

+3

Px[L_K< t]2

Px[T_K< t < L_K] +4Px[L_K<∞]Px[t < L_K<∞]Px[T_K< t < L_K] and

0≤Rm(t)≤4

R^mdxPx[LK<∞]Px[t < LK<∞]Px[TK< t < LK]. (15) Lemma 9. Letm=3,4, . . . .Then forx∈R^m, t >0

Px[t < L_K<∞] ≤ m

2 −1 ₋1

(4π)^−m/2C(K)t^(2−m)/2≤C(K)t^(2−m)/2. (16) Proof. By (11) and (12)

Px[t < L_K<∞] = ^∞

t

ds μ_K(dy)p(x, y;s)

≤ ^∞

t

ds μ_K(dy)(4πs)^−m/2

= m

2 −1 ₋1

(4π)^−m/2C(K)t^(2−m)/2. (17)

By (15) and Lemma9 0≤R_m(t)≤C(K)t^(2−m)/2

R^mdxPx[L_K<∞]Px[T_K< t < L_K]. (18) Le Gall showed that form=3 andt→ ∞(Lemma 2 in [8])

R³dxPx[LK<∞]Px[TK< t < LK] =(4π)⁻²C(K)³+o(1), (19) and that form=4 andt→ ∞((9) in [8])

R⁴dxPx[LK<∞]Px[TK< t < LK] =2(4π)⁻⁴C(K)³t⁻¹(logt )

1+o(1)

. (20)

Proposition4follows for m=3 and m=4 from (18)–(20). The proof of Proposition4 for m≥5 relies on some independent estimates ((32), Lemmas10and14) which will be proved in Sections3and4below. By (15) and (32) below we have that fort≥2T

R_m(t)≤4C

R^mdxPx[L_K<∞]Px[t < L_K<∞]Px[T_K< t]t^(2−m)/2 +4C

m 2 −1

R^mdxPx[L_K<∞]Px[t < L_K<∞] ^t−T

0

dsPx[s < T_K< t](t−s)^−m/2, (21)

where C=

m 2 −1

₋1

(4π)^−m/2C(K) (22)

and

T =C^2/(m−2). (23)

By Lemma9we have that the first term in the right-hand side of (21) is bounded by 4CC(K)

R^mdx

Px[L_K<∞]2

t^2−m.

To estimate the contribution from the second term in the right-hand side of (21) we consider the contributions from [0, T],[T , t/2]and[t /2, t−T]to the integral with respect tos. Since

T 0

dsPx[s < T_K< t](t−s)^−m/2≤T (t−T )^−m/2Px[L_K<∞]

we have by (17) that[0, T]contributes at most C(K)²

R^mdx

Px[L_K<∞]2

T (t−T )^−m/2t^(2−m)/2=O t^1−m

. To estimate the contribution from[T , t /2]we note that

Px[s < T_K< t](t−s)^−m/2≤(t/2)^−m/2Px[s < L_K<∞]. (24) Hence[T , t/2]contributes, by (24), the first equality in (16) and Lemma14below, at most

4C m

2 −1

2^m/2t^−m/2

t /2 T

ds R^mdxPx[LK<∞]Px[t < LK<∞]Px[s < LK<∞]

≤C(K)⁴t^−m/2

t /2 T

ds

∞ 0

ds1

∞ t

ds2

∞ s

ds3(s₁s₂+s₂s₃+s₃s₁)^−m/2

≤C(K)⁴t^−m/2

t /2 T

ds

∞ t

ds2

∞ s

ds3(s2+s3)⁻¹(s2s3)^(2−m)/2

≤C(K)⁴t^−m/2

t /2 T

ds

∞ t

ds2s₂^−m/2

∞ T

ds3s₃^(2−m)/2

≤C(K)⁴T^(4−m)/2t^2−m.

To estimate the contribution from the interval[t /2, t−T]we have, by Lemma10below, that

R^mdxPx[L_K<∞]Px[t < L_K<∞] ^t−T

t /2

dsPx[s < T_K< t](t−s)^−m/2

≤C1

R^mdxPx[LK<∞]Px[t < LK<∞] ^t−T

t /2

ds s^−m/2(t−s)^(2−m)/2

≤C₁(t/2)^−m/2

R^mdxPx[L_K<∞]Px[t < L_K<∞] ^t−T

−∞ ds (t−s)^(2−m)/2

≤C12^(2+m)/2T^(4−m)/2

R^mdxPx[LK<∞]Px[t < LK<∞]t^−m/2. (25)

By the first equality in (16)

R^mdxPx[L_K<∞]Px[t < L_K<∞]

= R^mdx μ_K(dy1)

∞ 0

ds1p(x, y₁;s₁) μ_K(dy2)

∞ t

ds2p(x, y₂;s₂)

=

μ_K(dy1)μ_K(dy2)

∞ 0

ds1

∞ t

ds2p(y1, y2;s1+s2)≤C(K)²t^(4−m)/2. Hence (25) is O(t^2−m).

3. Proof of Proposition5

Letc∈R^m, r >0, and letB(c;r)= {x: |x−c| ≤r}, Rc=inf{r >0: K⊂B(c;r)}, and letR=inf{Rc: c∈R^m}. Without loss of generality we may assume that the infimum in the latter is attained at the origin.

The main ingredient in the proof of Proposition5is to use the strong Markov property forPx[T_K< t < L_K]at the stopping timeT_K. So

Px[T_K< t < L_K] =Ex

1_{TK≤t}PB(TK)[t−T_K< L_K<∞]

. (26)

For anyz∈Kandy∈K, we have that|y−z| ≤2R. Hence Pz[t−s < LK<∞] = ^∞

t−s

dτ (4πτ )^−m/2 μK(dy)e^{−|y−z|2/(4τ )}

≥ ^∞

t

dτ (4πτ )^−m/2 μ_K(dy)e^−R2/t

= m

2 −1 ₋1

(4π)^−m/2C(K)e^−R2/tt^(2−m)/2. (27)

It follows that

R^mdx

Px[L_K<∞]2

Px[T_K< t < L_K]

≥ m

2 −1 ₋1

(4π)^−m/2C(K)e^−R2/tt^(2−m)/2

R^mdx

Px[L_K<∞]2

Px[T_K< t]. (28)

To prove the lower bound in Proposition5form≥5 we note that

Px[TK< t] ≥Px[LK<∞] −Px[t < LK<∞]. (29)

Hence form≥5 we have, by Lemma9, that

R^mdx

Px[LK<∞]2

Px[TK< t] ≥

R^mdx

Px[LK<∞]3

−C(K)t^(2−m)/2

R^mdx

Px[LK<∞]2

. We conclude that form≥5 the left-hand side of (28) is bounded from below by

m 2 −1

₋1

(4π)^−m/2C(K)

R^mdx

Px[L_K<∞]3

t^(2−m)/2+O t^−m/2

.

To prove the lower bound in Proposition5form=4 we note that by (28), (29)

R⁴dx

Px[L_K<∞]2

Px[T_K< t < L_K]

≥(4π)⁻²C(K)e^−R2/tt⁻¹

R⁴dx

Px[L_K<∞]3

−C(K)t⁻¹

R⁴dx

Px[L_K<∞]2

Px[t < L_K<∞]. Lemma15in Section5implies that

R⁴dx

Px[LK<∞]2

Px[t < LK<∞] =O

t⁻¹logt .

We conclude that form=4 the left-hand side of (28) is bounded from below by (4π)⁻²C(K)

R⁴dx

Px[LK<∞]3

t⁻¹+O

t⁻²logt .

Finally form=3 we have that the left-hand side of (28) is bounded from below by 2(4π)^−3/2C(K)e^−R2/tt^−1/2

R³dx

Px[L_K<∞]2

Px[L_K< t]. Lemma12in Section4implies that

R³dx

Px[L_K<∞]2

Px[L_K< t] =2⁻¹(4π)⁻²C(K)³logt+O(1). (30) We conclude that form=3 the left-hand side of (28) is bounded from below by

(4π)^−7/2C(K)⁴t^−1/2logt+O t^−1/2

.

This completes the proof of the lower bound in Proposition5.

To prove the upper bound in Proposition5we note that by the first equality in (27) Pz[t−s < L_K<∞] ≤

m 2 −1

₋1

(4π)^−m/2

C(K)(t−s)^(2−m)/2∧1

. (31)

By (26), (31) and the identity(t−T_K)^(2−m)/2=t^(2−m)/2+2⁻¹(m−2)_TK

0 ds (t−s)^−m/2it follows that Px[TK< t < LK] ≤CPx[TK< t]t^(2−m)/2+C

m 2 −1

t−T 0

dsPx[s < TK< t](t−s)^−m/2, (32) whereCandT are given by (22) and (23) respectively. Form≥4 we have, by (32), that

R^mdx

Px[LK<∞]2

Px[TK< t < LK]

≤C

R^mdx

Px[LK<∞]3

t^(2−m)/2 (33)

+C(K)

R^mdx

Px[L_K<∞]2 t−T 0

dsPx[s < T_K< t](t−s)^−m/2.

The first term in the right-hand side of (33) jibes with the leading term in Proposition5(form≥4). To estimate the second term in the right-hand side of (33) we need the following lemma.

Lemma 10. Letm≥3,and letT be given by(23).There exists a constantC₁depending onKonly such that for all s≤t−T andx∈R^m

Px[s < T_K< t] ≤C₁s^−m/2(t−s).

Proof. By the Markov property atswe have that Px[s < T_K< t] =

R^mdy p_R^m₋K(x, y;s)Py[T_K< t−s]

≤ R^mdy p(x, y;s)Py[TK< t−s] ≤(4πs)^−m/2N1,m(t−s), (34) wherep_R^m₋K(x, y;s), x∈R^m, y∈R^m, s >0 is the transition density with killing onK. SinceN1,m(t)=C(K)t+ o(t)ast→ ∞ we have that there existsT₁such thatt≥T₁impliesN_1,m(t)≤2C(K)t. LetT >0 be arbitrary. If T ≥T₁thenN_1,m(t) <2C(K)tfor allt≥T. IfT < T₁andT ≤t≤T₁then we have by monotonicity oft→N_1,m(t) thatN1,m(t)≤N1,m(T1)≤2C(K)T1≤(2C(K)T1/T )t. Hence for allt≥T

N_1,m(t)≤2C(K) T1

T ∨1

t, (35)

and the lemma holds with C₁=2(4π)^−m/2C(K)

T₁ T ∨1

. (36)

To complete the proof of Proposition5form≥5 we estimate the second term in the right-hand side of (33) as follows. The contribution froms∈ [0, T]to the integral is bounded byT (t−T )^−m/2C(K)

R^mdx (Px[L_K<∞])²= O(t^−m/2). The contribution froms∈ [T , t /2]is bounded, using (16), by

C(K)

R^mdx

Px[L_K<∞]2 t /2 T

dsPx[s < L_K<∞](t−s)^−m/2

≤C(K)²

R^mdx

Px[L_K<∞]2 t /2 T

ds s^(2−m)/2(t−s)^−m/2=O t^−m/2

, while the contribution froms∈ [t /2, t−T]is bounded, using Lemma10, by

C(K)C₁

R^mdx

Px[L_K<∞]2 t−T t /2

ds s^−m/2(t−s)^(2−m)/2=O t^−m/2

. This completes the proof of Proposition5form≥5.

To complete the proof of Proposition5form=4 we note that Lemma10cannot be used to estimate the integral in (33) since

R⁴dx (Px[LK<∞])²is divergent. Instead we will use the first inequality in (34). First we note that by (16),Px[s < T_K< t] ≤Px[s < L_K<∞] ≤C(K)s⁻¹. Hence

B(0;2R)

dx

Px[L_K<∞]2

Px[s < T_K< t] ≤B(0;2R)C(K)s⁻¹. (37)

Furthermore for|x| ≥2Randy∈Kwe have that|x−y|²≥(|x| −R)²≥ |x|²/4. SinceK⊂B(0;R)we have that

{|x|>2R}dx

Px[L_K<∞]2

Px[s < L_K<∞]

≤ {|x|>2R}dx

Px[L_B(0;R)<∞]2 ∞ s

dτ μ_K(dy)p(x, y;τ ) (38)

≤C(K)

{|x|>2R}dx

1∧ R

|x| 4 ∞

s

dτ (4πτ )⁻²e^{−|x|2/(16τ )}

≤C(K)R⁴(8s)⁻¹

∞ R/(2√

s)

dρ ρ⁻³

1−e^−ρ2

≤C(K)R⁴s⁻¹ log

R⁻²s

∨1 . By (37) and (38)

R⁴dx

Px[LK<∞]2

Px[s < TK< t] ≤C2s⁻¹

log s

R²

∨1

(39) for someC₂depending onKonly. First of all

T 0

ds R⁴dx

Px[L_K<∞]2

Px[s < T_K< t](t−s)⁻²≤T (t−T )⁻²

R⁴dx

Px[L_K<∞]3

=O t⁻²

. By (39)

t /2 T

ds R⁴dx

Px[L_K<∞]2

Px[s < T_K< t](t−s)⁻²

≤4t⁻²C2 t /2 T

ds s⁻¹

log s

R²

∨1

=O

t⁻¹logt2 .

To estimate the contribution froms∈ [t /2, t−T]we have by the first inequality in (34) and Fubini’s theorem that

R⁴dx

Px[L_K<∞]2 t−T t /2

dsPx[s < T_K< t](t−s)⁻²

≤ ^t−T

t /2

ds R⁴dyPy[T_K< t−s](t−s)⁻²

R⁴dx

Px[L_K<∞]2

(4πs)⁻²e^{−|x−y|2/(4s)}

≤(2πt )⁻²

t−T t /2

ds R⁴dyPy[TK< t−s](t−s)⁻²

R⁴dx

Px[LB(0;R)<∞]2

e^{−|x−y|2/(4t )}. (40) By the Hardy–Littlewood rearrangement inequality [5] and (35), (36) we have that the right-hand side of (40) is bounded by

(2πt )⁻²

t−T t /2

ds R⁴dyPy[T_K< t−s](t−s)⁻²

R⁴dx

Px[L_B(0;R)<∞]2

e^{−|x|2/(4t )}

≤ C1

2t²

t−T t /2

ds (t−s)⁻¹

∞ 0

dρ ρ³

1∧R ρ

4

e^{−ρ2/(4t )}=O

t⁻¹logt2 . This completes the proof of Proposition5form=4.

Finally we prove the upper bound in Proposition5form=3. By (22) Px[T_K< t < L_K] ≤C

Px[L_K< t] +Px[T_K< t < L_K]

t^−1/2+C 2

t−T 0

dsPx[s < T_K< t](t−s)^−3/2.

It follows that fort≥(2T )∨(4C²) Px[T_K< t < L_K] ≤Ct^−1/2

1+2Ct^−1/2

Px[L_K< t] +C

t−T 0

dsPx[s < T_K< t](t−s)^−3/2.