Tài liệu Đề tài " On the volume of the intersection of two Wiener sausages " ppt

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744 M. VAN DEN BERG, E. BOLTHAUSEN, AND F. DEN HOLLANDER
Theorem 1. Let d ≥ 3 and a>0. Then, for every c>0,
lim
t→∞
1
t
(d−2)/d
log P

|V
a
(ct)|≥t

= −I
κ
a
d
(c),(1.8)
where
I
κ
a
d
(c)=c inf
φ∈Φ
κ
a
d
(c)


R
d
|∇φ|
2
(x)dx

(1.9)
with
Φ
κ
a
d
(c)=

φ ∈ H
1
(R
d
):

R
d
φ
2
(x)dx =1,

R
d

1 − e
−κ
a
cφ
2
(x)

2
dx ≥ 1

.
(1.10)
Theorem 2. Let d =2and a>0. Then, for every c>0,
lim
t→∞
1
log t
log P

|V
a
(ct)|≥t/ log t

= −I
2π
2
(c),(1.11)
where I
2π
2
(c) is given by (1.9) and (1.10) with (d, κ
a
) replaced by (2, 2π).
Note that we are picking a time horizon of length ct and are letting t →∞
for fixed c>0. The sizes of the large deviation, t respectively t/ log t, come
from the expected volume of a single Wiener sausage as t →∞, namely,
E|W
a
(t)|∼

κ
a
t if d ≥ 3,
2πt/ log t if d =2,
(1.12)
as shown in Spitzer [14]. So the two Wiener sausages in Theorems 1 and 2 are
doing a large deviation on the scale of their mean.
The idea behind Theorem 1 is that the optimal strategy for the two Brow-
nian motions to realise the large deviation event {|V
a
(ct)|≥t} is to behave
like Brownian motions in a drift field xt
1/d
→ (∇φ/φ)(x) for some smooth
φ: R
d
→ [0, ∞) during the given time window [0,ct]. Conditioned on adopting
this drift:
– Each Brownian motion spends time cφ
2
(x) per unit volume in the neigh-
bourhood of xt
1/d
, thus using up a total time t

R
d
cφ
2
(x)dx. This time
must equal ct, hence the first constraint in (1.10).
– Each corresponding Wiener sausage covers a fraction 1 − e
−κ
a
cφ
2
(x)
of
the space in the neighbourhood of xt
1/d
, thus making a total intersection
volume t

R
d
(1 − e
−κ
a
cφ
2
(x)
)
2
dx. This volume must exceed t, hence the
second constraint in (1.10).
The cost for adopting the drift during time ct is t
(d−2)/d

R
d
c|∇φ|
2
(x)dx. The
best choice of the drift field is therefore given by minimisers of the variational
problem in (1.9) and (1.10), or by minimising sequences.
ON THE VOLUME OF THE INTERSECTION OF TWO WIENER SAUSAGES
745
Note that the optimal strategy for the two Wiener sausages is to “form a
Swiss cheese”: they cover only part of the space, leaving random holes whose
sizes are of order 1 and whose density varies on space scale t
1/d
(see [3]). The
local structure of this Swiss cheese depends on a. Also note that the two
Wiener sausages follow the optimal strategy independently. Apparently, under
the joint optimal strategy the two Brownian motions are independent on space
scales smaller than t
1/d
.
1
A similar optimal strategy applies for Theorem 2, except that the space
scale is

t/ log t. This is only slightly below the diffusive scale, which explains
why the large deviation event has a polynomial rather than an exponential cost.
Clearly, the case d =2iscritical for a finite time horizon. Incidentally, note
that I
2π
2
(c) does not depend on a. This can be traced back to the recurrence
of Brownian motion in d = 2. Apparently, the Swiss cheese has random holes
that grow with time, washing out the dependence on a (see [3]).
There is no result analogous to Theorems 1 and 2 for d = 1: the variational
problem in (1.9) and (1.10) certainly continues to make sense for d = 1, but it
does not describe the Wiener sausages: holes are impossible in d =1.
1.4. Analysis of the variational problem. We proceed with a closer
analysis of (1.9) and (1.10). First we scale out the dependence on a and c.
Recall from Theorem 2 that κ
a
=2π for d =2.
Theorem 3. Let d ≥ 2 and a>0.
(i) For every c>0,
I
κ
a
d
(c)=
1
κ
a
Θ
d
(κ
a
c),(1.13)
where Θ
d
:(0, ∞) → [0, ∞] is given by
Θ
d
(u) = inf

∇ψ
2
2
: ψ ∈ H
1
(R
d
), ψ
2
2
= u,

(1 − e
−ψ
2
)
2
≥ 1

.(1.14)
(ii) Define u

= min
ζ>0
ζ(1 − e
−ζ
)
−2
=2.45541 Then Θ
d
= ∞ on
(0,u

] and 0 < Θ
d
< ∞ on (u

, ∞).
(iii) Θ
d
is nonincreasing on (u

, ∞).
(iv) Θ
d
is continuous on (u

, ∞).
(v) Θ
d
(u)  (u −u

)
−1
as u ↓ u

.
Next we exhibit the main quantitative properties of Θ
d
.
1
To prove that the Brownian motions conditioned on the large deviation event {|V
a
(ct)|
≥ t} actually follow the “Swiss cheese strategy” requires substantial extra work. We will not
address this issue here.
746 M. VAN DEN BERG, E. BOLTHAUSEN, AND F. DEN HOLLANDER
Theorem 4. Let 2 ≤ d ≤ 4. Then u → u
(4−d)/d
Θ
d
(u) is strictly decreas-
ing on (u

, ∞) and
lim
u→∞
u
(4−d)/d
Θ
d
(u)=µ
d
,(1.15)
where
µ
d
=







inf

∇ψ
2
2
: ψ ∈ H
1
(R
d
), ψ
2
=1, ψ
4
=1

if d =2, 3,
inf

∇ψ
2
2
: ψ ∈ D
1
(R
4
), ψ
4
=1

if d =4,
(1.16)
satisfying 0 <µ
d
< ∞.
2
Theorem 5. Let d ≥ 5 and define
η
d
= inf{∇ψ
2
2
: ψ ∈ D
1
(R
d
),

(1 − e
−ψ
2
)
2
=1}.(1.17)
(i) There exists a radially symmetric, nonincreasing, strictly positive min-
imiser ψ
d
of the variational problem in (1.17), which is unique up to transla-
tions. Moreover, ψ
d

2
2
< ∞.
(ii) Define u
d
= ψ
d

2
2
. Then u → θ
d
(u) is strictly decreasing on (u

,u
d
)
and
Θ
d
(u)=η
d
on [u
d
, ∞).(1.18)
0
s
u

u
d
η
d
(iii)
0 u

µ
4
(ii)
0 u

(i)
Figure 1 Qualitative picture of Θ
d
for: (i) d =2, 3; (ii) d = 4; (iii) d ≥ 5.
Theorem 6. (i) Let 2 ≤ d ≤ 4 and u ∈ (u

, ∞) or d ≥ 5 and u ∈ (u

,u
d
].
Then the variational problem in (1.14) has a minimiser that is strictly positive,
radially symmetric (modulo translations) and strictly decreasing in the radial
component. Any other minimiser is of the same type.
(ii) Let d ≥ 5 and u ∈ (u
d
, ∞). Then the variational problem in (1.14)
does not have a minimiser.
2
We will see in Section 5 that µ
4
= S
4
, the Sobolev constant in (4.3) and (4.4).
ON THE VOLUME OF THE INTERSECTION OF TWO WIENER SAUSAGES
747
We expect that in case (i) the minimiser is unique (modulo translations).
In case (ii) the critical point u
d
is associated with “leakage” in (1.14); namely,
L
2
-mass u − u
d
leaks away to infinity.
1.5. Large deviations for infinite-time intersection volume. Intuitively,
by letting c →∞in (1.8) we might expect to be able to get the rate constant
for an infinite time horizon. However, it is not at all obvious that the limits
t →∞and c →∞can be interchanged. Indeed, the intersection volume might
prefer to exceed the value t on a time scale of order larger than t, which is not
seen by Theorems 1 and 2. The large deviations on this larger time scale are
a whole new issue, which we will not address in the present paper.
Nevertheless, Figure 1(iii) clearly suggests that for d ≥ 5 the limits can
be interchanged:
Conjecture. Let d ≥ 5 and a>0. Then
lim
t→∞
1
t
(d−2)/d
log P

|V
a
|≥t

= −I
κ
a
d
,(1.19)
where
I
κ
a
d
= inf
c>0
I
κ
a
d
(c)=I
κ
a
d
(c
∗
)=
η
d
κ
a
(1.20)
with c
∗
= u
d
/κ
a
.
The idea behind this conjecture is that the optimal strategy for the two
Wiener sausages is time-inhomogeneous: they follow the Swiss cheese strategy
until time c
∗
t and then wander off to infinity in different directions. The
critical time horizon c
∗
comes from (1.13) and (1.18) as the value above which
c → I
κ
a
d
(c) is constant (see Fig. 1(iii)). During the time window [0,c
∗
t] the
Wiener sausages make a Swiss cheese parametrised by the ψ
d
in Theorem
5; namely, (1.9) and (1.10) have a minimising sequence (φ
j
) converging to
φ =(c
∗
κ
a
)
−1/2
ψ
d
in L
2
(R
d
).
We see from Figure 1(ii) that d =4iscritical for an infinite time horizon.
In this case the limits t →∞and c →∞apparently cannot be interchanged.
Theorem 4 shows that for 2 ≤ d ≤ 4 the time horizon in the optimal
strategy is c = ∞, because c → I
κ
a
d
(c) is strictly decreasing as soon as it
is finite (see Fig. 1(i–ii)). Apparently, even though lim
t→∞
|V
a
(t)| = ∞ P -
almost surely (recall (1.7)), the rate of divergence is so small that a time of
order larger than t is needed for the intersection volume to exceed the value
t with a probability exp[−o(t
(d−2)/d
)] respectively exp[−o(log t)]. So an even
larger time is needed to exceed the value t with a probability of order 1.
1.6. Three or more Wiener sausages. Consider k ≥ 3 independent
Wiener sausages, let V
a
k
(t) denote their intersection up to time t, and let
748 M. VAN DEN BERG, E. BOLTHAUSEN, AND F. DEN HOLLANDER
V
a
k
= lim
t→∞
V
a
k
(t). Then the analogue of (1.7) reads (see e.g. Le Gall [11])
P (|V
a
k
| < ∞)=

0if1≤ d ≤
2k
k−1
,
1ifd>
2k
k−1
.
(1.21)
The critical dimension 2k/(k −1) comes from the following calculation:
E|V
a
k
| =

R
d
P

σ
B
a
(x)
< ∞

k
dx =

R
d

1 ∧

a
|x|

d−2

k
dx,(1.22)
where σ
B
a
(x)
= inf{t ≥ 0: β(t) ∈ B
a
(x)}. The integral converges if and only
if (d − 2)k>d.
It is possible to extend the analysis in Sections 1.3 and 1.4 in a straight-
forward manner, leading to the following modifications (not proved in this
paper):
(1) Theorems 1 and 2 carry over with:
– V
a
replaced by V
a
k
;
– c replaced by kc/2 in (1.9);
–

R
d
(1 − e
−κ
a
cφ
2
(x)
)
2
dx replaced by

R
d
(1 − e
−κ
a
cφ
2
(x)
)
k
dx in (1.10).
(2) Theorems 3, 4 and 5 carry over with:
–

(1 − e
−ψ
2
)
2
replaced by

(1 − e
−ψ
2
)
k
in (1.14) and (1.17);
– u

= min
ζ>0
ζ(1 −e
−ζ
)
−k
;
– ψ
4
replaced by ψ
2k
in (1.16).
For k = 3, the critical dimension is d = 3, and a behaviour similar to that
in Figure 1 shows up for: (i) d = 2; (ii) d = 3; (iii) d ≥ 4, respectively. For
k ≥ 4 the critical dimension lies strictly between 2 and 3, so that Figure 1(ii)
drops out.
1.7. Back to simple random walks. We expect the results in Theorems 1
and 2 to carry over to the discrete space-time setting as introduced in Section
1.1. (A similar relation is proved in Donsker and Varadhan [6] for a single
random walk, respectively, Brownian motion.) The only change should be
that for d ≥ 3 the constant κ
a
needs to be replaced by its analogue in discrete
space and time:
κ = P(S(n) =0∀n ∈ N ),(1.23)
the escape probability of the simple random walk. The global structure of the
Swiss cheese should be the same as before; the local structure should depend
on the underlying lattice via the number κ.
ON THE VOLUME OF THE INTERSECTION OF TWO WIENER SAUSAGES
749
1.8. Outline. Theorem 1 is proved in Section 2. The idea is to wrap
the Wiener sausages around a torus of size Nt
1/d
, to show that the error com-
mitted by doing so is negligible in the limit as t →∞followed by N →∞,
and to use the results in [3] to compute the large deviations of the intersection
volume on the torus as t →∞for fixed N. The wrapping is rather delicate
because typically the intersection volume neither increases nor decreases under
the wrapping. Therefore we have to go through an elaborate clumping and re-
flection argument. In contrast, the volume of a single Wiener sausage decreases
under the wrapping, a fact that is very important to the analysis in [3].
Theorem 2 is proved in Section 3. The necessary modifications of the
argument in Section 2 are minor and involve a change in scaling only.
Theorems 3–6 are proved in Sections 4–7. The tools used here are scaling
and Sobolev inequalities. Here we also analyse the minimers of the variational
problems in (1.14) and (1.17).
2. Proof of Theorem 1
By Brownian scaling, V
a
(ct) has the same distribution as tV
at
−1/d
(ct
(d−2)/d
).
Hence, putting
τ = t
(d−2)/d
,(2.1)
we have
P

|V
a
(ct)|≥t

= P

|V
aτ
−1/(d−2)
(cτ)|≥1

.(2.2)
The right-hand side of (2.2) involves the Wiener sausages with a radius that
shrinks with τ. The claim in Theorem 1 is therefore equivalent to
lim
τ→∞
1
τ
log P

|V
aτ
−1/(d−2)
(cτ)|≥1

= −I
κ
a
d
(c).(2.3)
We will prove (2.3) by deriving a lower bound (§2.2) and an upper bound
(§2.3). To do so, we first deal with the problem on a finite torus (§2.1) and
afterwards let the torus size tend to infinity. This is the standard compactifi-
cation approach. On the torus we can use some results obtained in [3].
2.1. Brownian motion wrapped around a torus. Let Λ
N
be the torus
of size N>0, i.e., [−
N
2
,
N
2
)
d
with periodic boundary conditions. Let β
N
(s),
s ≥ 0, be the Brownian motion wrapped around Λ
N
, and let W
aτ
−1/(d−2)
N
(s),
s ≥ 0, denote its Wiener sausage with radius aτ
−1/(d−2)
.
Proposition 1. (|W
aτ
−1/(d−2)
N
(cτ)|)
τ>0
satisfies the large deviation prin-
ciple on R
+
with rate τ and with rate function
J
κ
a
d,N
(b, c)=
1
2
c inf
ψ∈Ψ
κ
a
d,N
(b,c)


Λ
N
|∇ψ|
2
(x)dx

,(2.4)
750 M. VAN DEN BERG, E. BOLTHAUSEN, AND F. DEN HOLLANDER
where
Ψ
κ
a
d,N
(b, c)=

ψ ∈ H
1
(Λ
N
):

Λ
N
ψ
2
(x)dx =1,

Λ
N

1 − e
−κ
a
cψ
2
(x)

dx ≥ b

.
(2.5)
Proof. See Proposition 3 in [3]. The function ψ parametrises the optimal
strategy behind the large deviation: (∇ψ/ψ)(x) is the drift of the Brownian
motion at site x, cψ
2
(x) is the density for the time the Brownian motion spends
at site x, while 1 − e
−κ
a
cψ
2
(x)
is the density of the Wiener sausage at site x.
The factor c enters (2.4) and (2.5) because the Wiener sausage is observed over
a time cτ.
Proposition 1 gives us good control over the volume |W
aτ
−1/(d−2)
N
(τ)|.In
order to get good control over the intersection volume



V
aτ
−1/(d−2)
N
(cτ)



=



W
aτ
−1/(d−2)
1,N
(cτ) ∩ W
aτ
−1/(d−2)
2,N
(cτ)



(2.6)
of two independent shrinking Wiener sausages, observed until time cτ, we need
the analogue of Proposition 1 for this quantity, which reads as follows.
Proposition 2. (|V
aτ
−1/(d−2)
N
(cτ)|)
τ>0
satisfies the large deviation prin-
ciple on R
+
with rate τ and with rate function

J
κ
a
d,N
(b, c)=c inf
φ∈

Φ
κ
a
d,N
(b,c)


Λ
N
|∇φ|
2
(x)dx

,(2.7)
where

Φ
κ
a
d,N
(b, c)=

φ ∈ H
1
(Λ
N
):

Λ
N
φ
2
(x)dx =1,

Λ
N

1 − e
−κ
a
cφ
2
(x)

2
dx ≥ b

.
(2.8)
Proof. The extra power 2 in the second constraint (compare (2.5) with
(2.8)) enters because (1−e
−κ
a
cφ
2
(x)
)
2
is the density of the intersection of the two
Wiener sausages at site x. The extra factor 2 in the rate function (compare
(2.4) with (2.7)) comes from the fact that both Brownian motions have to
follow the drift field ∇φ/φ. The proof is a straightforward adaptation and
generalization of the proof of Proposition 3 in [3]. We outline the main steps,
while skipping the details.
Step 1. One of the basic ingredients in the proof in [3] is to approximate the
volume of the Wiener sausage by its conditional expectation given a discrete
skeleton. We do the same here. Abbreviate
W
i
(cτ)=W
aτ
−1/(d−2)
i,N
(cτ) ,i=1, 2,(2.9)
V (cτ)=W
1
(cτ) ∩ W
2
(cτ) .
ON THE VOLUME OF THE INTERSECTION OF TWO WIENER SAUSAGES
751
Set
X
i,cτ,ε
= {β
i
(jε)}
1≤j≤cτ /ε
,i=1, 2,(2.10)
where β
i
(s), s ≥ 0, is the Brownian motion on the torus Λ
N
that generates
the Wiener sausage W
i
(cτ). Write E
cτ,ε
for the conditional expectation given
X
i,cτ,ε
, i =1, 2. Then, analogously to Proposition 4 in [3], we have:
Lemma 1. For al l δ>0,
lim
ε↓0
lim sup
τ→∞
1
τ
log P




|V (cτ)|−E
cτ,ε
(|V (cτ)|)



≥ δ

= −∞.(2.11)
Proof. The crucial step is to apply a concentration inequality of Talagrand
twice, as follows. First note that, conditioned on X
i,cτ,ε
, W
i
(cτ) is a union of
L = cτ/ε independent random sets. Call these sets C
i,k
,1≤ k ≤ L, and write
V (cτ)=

L

k=1
C
1,k

∩

L

k=1
C
2,k

.(2.12)
Next note that, for any measurable set D ⊂ Λ
N
, the function
{C
k
}
1≤k≤L
→






L

k=1
C
k

∩ D





(2.13)
is Lipschitz-continuous in the sense of equation (2.26) in [3], uniformly in D.
From the proof of Proposition 4 in [3], we therefore get
lim
ε↓0
lim sup
τ→∞
1
τ
log P




|V (cτ)|−E

|V (cτ)||X
1,cτ,ε
,β
2




≥ δ | β
2

= −∞,
(2.14)
uniformly in the realisation of β
2
. On the other hand, the above holds true
with β
1
and β
2
interchanged, and so we easily get
lim
ε↓0
lim sup
τ→∞
1
τ
log P




E

|V (cτ)||X
1,cτ,ε
,β
2

− E
cτ,ε
(|V (cτ)|)



≥ δ

= −∞,
(2.15)
uniformly in the realisation of β
2
. Clearly, (2.14) and (2.15) imply (2.11).
Step 2. We fix ε>0 and prove an LDP for E
cτ,ε
(|V (cτ)|), as follows. As
in equation (2.43) in [3], define I
(2)
ε
: M
+
1
(Λ
N
× Λ
N
) → [0, ∞]by
I
(2)
ε
(µ)=

h (µ | µ
1
⊗ π
ε
)ifµ
1
= µ
2
,
∞ otherwise,
(2.16)
where h(·|·) denotes relative entropy between measures, µ
1
,µ
2
are the two
marginals of µ on Λ
N
, and π
ε
(x, dy)=p
ε
(y−x)dy with p
ε
the Brownian transi-
752 M. VAN DEN BERG, E. BOLTHAUSEN, AND F. DEN HOLLANDER
tion kernel on Λ
N
.Forη>0, define Φ
η
: M
+
1
(Λ
N
× Λ
N
)×M
+
1
(Λ
N
× Λ
N
) →
[0, ∞)by
Φ
η
(µ
1
,µ
2
)=

Λ
N
dx

1 − exp

−ηκ
a

Λ
N
×Λ
N
ϕ
ε
(y −x, z −x) µ
1
(dy, dz)

×

1 − exp

−ηκ
a

Λ
N
×Λ
N
ϕ
ε
(y −x, z −x) µ
2
(dy, dz)

,
(2.17)
where ϕ
ε
is defined by
ϕ
ε
(y, z)=

ε
0
ds p
s
(−y) p
ε−s
(z)
p
ε
(z −y)
.(2.18)
Lemma 2. (E
cτ,ε
(|V (cτ)|))
τ>0
satisfies the LDP on R
+
with rate τ and
with rate function
(2.19)
J
ε
(b)
= inf

c
ε

I
(2)
ε
(µ
1
)+I
(2)
ε
(µ
2
)

: µ
1
,µ
2
∈M
+
1
(Λ
N
× Λ
N
), Φ
c/ε
(µ
1
,µ
2
)=b

.
Proof. The proof is a straightforward extension of the proof of Proposi-
tion 5 in [3]. The basis is the observation that
(2.20)
E
cτ,ε
(|V (cτ)|)
=

Λ
N
dx P
cτ,ε
(x ∈ W
1
(cτ)) P
cτ,ε
(x ∈ W
2
(cτ))
=

Λ
N
dx

1 − exp

cτ
ε

Λ
N
×Λ
N
log

1 − q
τ,ε
(y −x, z −x)

L
1,cτ,ε
(dy, dz)

×

1 − exp

cτ
ε

Λ
N
×Λ
N
log

1 − q
τ,ε
(y −x, z −x)

L
2,cτ,ε
(dy, dz)

,
where
q
τ,ε
(y, z)=P
y

∃0 ≤ s ≤ ε with β
s
∈ B
aτ
−1/(d−2)
(0) | β
ε
= z

,(2.21)
and L
i,cτ,ε
is the bivariate empirical measure
L
i,cτ,ε
=
ε
cτ
cτ/ε

k=1
δ
(β
i
((k−1)ε),β
i
(kε))
,i=1, 2.(2.22)
Through a number of approximation steps we prove that
lim
τ→∞


E
cτ,ε
(|V (cτ)|) − Φ
c/ε
(L
1,cτ,ε
,L
2,cτ,ε
)


∞
=0 ∀ε>0.(2.23)
ON THE VOLUME OF THE INTERSECTION OF TWO WIENER SAUSAGES
753
This then proves our claim, since we can apply a standard LDP for Φ
c/ε
(L
1,cτ,ε
,L
2,cτ,ε
).
The proof of (2.23) runs as in the proof of Proposition 5 in [3] via the following
telescoping. Set
f
i
(x) = exp

cτ
ε

Λ
N
×Λ
N
log

1 − q
τ,ε
(y −x, z −x)

L
i,cτ,ε
(dy, dz)

,(2.24)
g
i
(x) = exp

−
cκ
a
ε

Λ
N
×Λ
N
ϕ
ε
(y −x, z −x) L
i,cτ,ε
(dy, dz)

.
Then
(2.25)
E
cτ,ε
(|V (cτ)|) − Φ
c/ε
(L
1,cτ,ε
,L
2,cτ,ε
)
=

Λ
N
dx [1 − f
1
(x)] [1 −f
2
(x)] −

Λ
N
dx [1 − g
1
(x)] [1 −g
2
(x)]
=

Λ
N
dx [g
1
(x) − f
1
(x)] [1 −f
2
(x)] +

Λ
N
dx [1 − g
1
(x)] [g
2
(x) − f
2
(x)] ,
and hence


E
cτ,ε
(|V (cτ)|) − Φ
c/ε
(L
1,cτ,ε
,L
2,cτ,ε
)


(2.26)
≤

Λ
N
dx |g
1
(x) − f
1
(x)| +

Λ
N
dx |g
2
(x) − f
2
(x)|.
We can therefore do the approximations on L
1,cτ,ε
and L
2,cτ,ε
separately, which
is exactly what is done in [3]. In fact, the various approximations on pp. 371–
377 in [3] have all been done by taking absolute values under the integral sign,
and so the argument carries over.
Step 3. The last step is a combination of the two previous steps to obtain
the limit ε ↓ 0 in the LDP. If f : R
+
→ R is bounded and continuous, then
from the two previous steps we get
lim
τ→∞
1
τ
log E (exp [τ|V (cτ) |])(2.27)
= lim
ε↓0
sup
µ
1
,µ
2

f

Φ
c/ε
(µ
1
,µ
2
)

−
c
ε

I
(2)
ε
(µ
1
)+I
(2)
ε
(µ
2
)

.
Now set, for ν
1
,ν
2
∈M
1
(Λ
N
),
(2.28)
Ψ
c/ε
(ν
1
,ν
2
)=

Λ
N
dx

1 − exp

−
cκ
a
ε

ε
0
ds

Λ
N
p
s
(x − y) ν
1
(dy)

×

1 − exp

−
cκ
a
ε

ε
0
ds

Λ
N
p
s
(x − y) ν
2
(dy)

,