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Vol. 8 (2003) Paper no. 12, pages 1–39.
Journal URL
http://www.math.washington.edu/˜ejpecp/
Large Deviation Principle for a Stochastic Heat Equation
With Spatially Correlated Noise 1
David M´
arquez-Carreras and M`
onica Sarr`
a
Facultat de Matem`atiques, Universitat de Barcelona
Gran Via 585, 08007-Barcelona, Spain
E-mails: marquez@mat.ub.es and sarra@mat.ub.es
Abstract. In this paper we prove a large deviation principle (ldp) for a
perturbed stochastic heat equation defined on [0, T ] × [0, 1]d . This equation
is driven by a Gaussian noise, white in time and correlated in space. Firstly,
we show the H¨older continuity for the solution of the stochastic heat equation. Secondly, we check that our Gaussian process satisfies a ldp and some
requirements on the skeleton of the solution. Finally, we prove the called

Freidlin-Wentzell inequality. In order to obtain all these results we need
precise estimates of the fundamental solution of this equation.
Keywords and phrases: Stochastic partial differential equation, stochastic heat equation, Gaussian noise, large deviation principle.
Mathematics Subject Classification (2000): 60H07, 60H15, 35R60.
Submitted to EJP on December 4, 2002. Final version accepted on June 20,
2003.

1

Partially supported by the grant BFM2000-0607

1

Introduction

Consider the following perturbed d-dimensional spatial stochastic heat equation on the compact set [0, 1]d


Luε (t, x) = εα(uε (t, x))F˙ (t, x) + β(uε (t, x)), t ≥ 0, x ∈ [0, 1]d ,




uε (t, x) = 0,
x ∈ ∂([0, 1]d ),



 uε (0, x) = 0,
x ∈ [0, 1]d ,

(1.1)

∂
− ∆ where ∆ is the Laplacian on IRd and ∂([0, 1]d ) is the
with ε > 0, L = ∂t
boundary of [0, 1]d . We consider null initial conditions and the compact set
[0, 1]d instead of [0, ζ]d , ζ > 0, for the sake of simplicity.
Assume that the coefficients satisfy the following assumptions:

(C) the functions α and β are Lipschitz.

The noise F = {F (ϕ), ϕ ∈ D(IRd+1 )} is an L2 (Ω, F, P )-valued Gaussian
process with mean zero and covariance functional given by
Z
Z
Z
dy ϕ(s, x)f (x − y)ψ(s, y),
(1.2)
dx
ds
J(ϕ, ψ) =
IR+

IRd

IRd

and f : IRd → IR+ is a continuous symmetric function on IRd − {0} such
that there exists a non-negative tempered measure λ on IRd whose Fourier
transform is f . The functional J in (1.2) is said to be a covariance functional
if all these asumptions are satisfied. Then, in addition,

Z
Z
ds
λ(dξ)Fϕ(s, ·)(ξ)Fψ(s, ·)(ξ),
J(ϕ, ψ) =
IR+

IRd

where F is the Fourier transform and z is the conjugate complex of z. In (1.2)
we could also work with a non-negative and non-negative definite tempered
measure, therefore symmetric, instead of the function f but, in this case, all
the notation over the sets appearing in the integrals is becoming tedious. In
this paper, moreover, we assume the following hypothesis on the measure λ:
Z
λ(dξ)
< ∞,
(Hη )
2 η
IRd (1 + kξk )

for some η ∈ (0, 1]. For instance, the function f (x) = kxk−κ , κ ∈ (0, d), satisfies (Hη ). As in Dalang [4] (see also Dalang and Frangos [5]) the Gaussian
2

process F can be extended to a worthy martingale measure, in the sense
given by Walsh [23],
M = {Mt (A), t ∈ IR+ , A ∈ Bb (IRd )}.
Then, following the approach of Walsh [23], one can give a rigorous meaning
to (1.1) by means of a weak formulation. Assumptions (C) and (H1 ) will
ensure the existence and uniqueness of a jointly measurable adapted process
{uε (t, x), (t, x) ∈ IR+ × [0, 1]d } such that
Z tZ
ε
G(t − s, x, y)α(uε (s, y))F (ds, dy)
u (t, x) = ε
0
[0,1]d
Z t Z
(1.3)
ε
dy G(t − s, x, y)β(u (s, y)),

ds
+
0

[0,1]d

where ε > 0 and G(t, x, y) denotes the fundamental solution of the heat
equation on [0, 1]d :

∂

t ≥ 0, x, y ∈ [0, 1]d ,

∂t G(t, x, y) = ∆x G(t, x, y),


G(t, x, y) = 0,
x ∈ ∂([0, 1]d ),




 G(0, x, y) = δ(x − y).

The stochastic integral in (1.3) is defined with respect to the Ft -martingale
measure M . Denote DTd = [0, T ] × [0, 1]d . If d > 1, the evolution equation
can not be driven by the Brownian sheet because G does not belong to
L2 (DTd ) and we need to work with a smoother noise. The study of existence
and uniqueness of solution to this sort of equation (1.3) on IRd has been
analyzed by Dalang in [4]. Many other authors have also studied existence
and uniqueness of solution to d-dimensional spatial stochastic equations, in
particular, wave and heat equations (see, for instance, Dalang and Frangos
[5], Karszeswka and Zabszyk [12], Millet and Morien [15], Millet and SanzSol´e [16], Peszat and Zabszyk [17], [18]).
Assuming the above-mentioned hypothesis, (C) and (Hη ) for some η ∈ [0, 1],
we will also check that the trajectories of the process are (γ1 , γ2 )-H¨older
continuous with respect to the parameters t and x, satisfaying γ1 ∈ (0, 1−η
2 )
and γ2 ∈ (0, 1 − η). The H¨older continuity for the stochastic heat equation
on IRd has been studied by Sanz-Sol´e and Sarr`a [20], [21]. On the other
hand, the wave case has been dealt in [15], [16] and [20].
Under (C) and (Hη ) for some η ∈ (0, 1), we will prove the most important

result of this paper: the existence of a large deviation principle (ldp) for
3

the law of the solution uε to (1.3) on C γ,γ (DTd ), with γ ∈ (0, 1−η
4 ). This
means that we check the existence of a lower semi-continuous function I :
C γ,γ (DTd ) → [0, ∞], called rate function, such that {I ≤ a} is compact for
any a ∈ [0, ∞), and
ε2 log P {uε ∈ O} ≥ −Λ(O),
2

ε

ε log P {u ∈ U } ≤ −Λ(U ),

for each open set O,
for each closed set U,

where, for a given subset A ∈ C γ,γ (DTd ),
Λ(A) = inf I(l).

l∈A

The proof of this goal is based on a classical result given by Azencott in [1]
(see also Priouret [19]), that allow us to go beyond a ldp from εF to u ε .
Azencott’s method is the following:
Theorem 1.1 Let (Ei , di ), i = 1, 2, be two Polish spaces and Xi : Ω → Ei ,
ε > 0, i = 1, 2, be two families of random variables. Suppose the following
requirements:
1. {X1ε , ε > 0} obeys a ldp with the rate function I1 : E1 → [0, ∞].
2. There exists a function K : {I1 < ∞} → E2 such that, for every
a < ∞, the function
K : {I1 ≤ a} → E2
is continuous.
3. For every R, ρ, a > 0, there exist θ > 0 and ε0 > 0 such that, for
h ∈ E1 satisfying I1 (h) ≤ a and ε ≤ ε0 , we have
µ
¶
n
o
R

ε
ε
P d2 (X2 , K(h)) ≥ ρ, d1 (X1 , h) < θ ≤ exp − 2 .
(1.4)
ε
Then, the family {X2ε , ε > 0} obeys a ldp with the rate function
I2 (φ) = inf{I1 (h) : K(h) = φ}.
Consequently, we will need to check that our initial Gaussian process satisfies
a ldp, the existence of a function K and, finally, to prove (1.4). We will
follow the approach of Freidlin and Wentzell [10] for diffusion process (see
also Dembo and Zeitouni [6]). Another remarkable article is Chenal and
Millet [3] where they prove the existence of a ldp for a one-dimensional
4

stochastic heat equation. Sowers has also studied this last equation but
using a different method. For more information about the study of onedimensional stochastic heat equations, we refer to Chenal and Millet [3].
Finally, Chenal [2] has checked a ldp for a stochastic wave equation on IR2 .
The paper is organized as follows. In Section 2 we establish a ldp for our
Gaussian process (first point of Theorem 1.1). In Section 3 we state some
properties on uε (t, x), mainly the H¨older continuity. Section 4 contains the
proofs of some requirements on the skeleton of uε (second point of Theorem 1.1). Section 5 is devoted to the proof of called Freidlin-Wentzell’s
inequality (third point of Theorem 1.1). All the arguments of this paper
need precise estimates of the fundamental solution G which will be enunciated and proved in an appendix. Moreover, this appendix also contains an
exponential inequality used in Section 5.
In this paper we fix T > 0 and all constants will be denoted independently
of its value. We finish this introduction by giving some basic notations. We
write, for functions φ, φ′ : DTd → IR,
kφk∞ = sup{|φ(t, x)|, (t, x) ∈ DTd },
kφkγ1 ,γ2 = sup

o
n |φ(t, x) − φ(s, y)|
d
,
(t,
x)
=
6
(s,
y)
∈
D
T ,
|t − s|γ1 + kx − ykγ2

dγ1 ,γ2 (φ, φ′ ) = kφ − φ′ k∞ + kφ − φ′ kγ1 ,γ2 .

Then, we define the topology of (γ1 , γ2 )-H¨older convergence on DTd by means
of dγ1 ,γ2 . Let C γ1 ,γ2 (DTd ) be the set of functions φ : DTd → IR such that
kφk∞ + kφkγ1 ,γ2 < ∞.
Finally, for θ > 0, a set A ⊂ IRn , n ≥ 1 and φ : A → IR, we introduce the
following notation,
kφkθ,A = sup |φ(w)| + sup
w∈A

2

n |φ(w) − φ(w ′ )|
|w − w′ |θ

o
; w 6= w′ , w, w′ ∈ A .

(1.5)

Large deviation principle for the Gaussian process

Let E be the space of measurable functions ϕ : IRd → IR such that
Z
Z
dz |ϕ(y)|f (y − z)|ϕ(z)| < ∞,
dy
IRd

IRd

5

endowed with the inner product
Z
Z
dy
hϕ, ψiE =
IRd

IRd

dz ϕ(y)f (y − z)ψ(z).

Let H be the completation of (E, h, iE ). For T > 0, let HT = L2 ([0, T ]; H).
This space is a real separable Hilbert space such that, if ϕ, ψ ∈ D([0, T ] ×
IRd ),
Z T
hϕ(s, ·), ψ(s, ·)iH ds = hϕ, ψiHT ,
(2.1)
E (F (ϕ)F (ψ)) =
0

where F is the noise introduced in Section 1.
For (t, x), (t′ , x′ ) ∈ DTd , set
˜
Γ((t,
x), (t′ , x′ )) =

Z

t∧t′

ds
0

Z

dy
R(x)

Z

R(x′ )

dz f (y − z),

where R(x) is the rectangle [0, x] (here [0, x] is a product of rectangles). We
˜
have Γ((t,
x), (t′ , x′ )) = hϕt,x , ϕt′ ,x′ iHT with
ϕt,x (s, y) =1l([0,t]×R(x)) (s, y).
One can easily check the following three conditions:
˜ is symmetric.
(i) Γ
(ii) For any c1 , . . . , cn ∈ IR, (t1 , x1 ), . . . , (tn , xn ) ∈ DTd ,
n
n X
X
i=1 j=1

˜ i , xi ), (tj , xj )) ≥ 0.
ci cj Γ((t

(iii) For any (t, x), (t′ , x′ ) ∈ DTd ,
˜
˜ ′ , x′ ), (t′ , x′ ))−2Γ((t,
˜
Γ((t,
x), (t, x))+Γ((t
x), (t′ , x′ )) ≤ CT (|t−t′ |+kx−x′ k).
Then, Example 1.2 in Watanabe [24] implies that, for any γ ′ ∈ [0, 21 ), there
′ ′
exists a Gaussian measure ν on C γ ,γ (DTd ; IR) such that
Z
˜
w(t, x)w(t′ , x′ )dν(w) = Γ((t,
x), (t′ , x′ )).
Cγ

′ ,γ ′

d ;IR)
(DT

6

³ ′ ′
´
Moreover, if H denotes the autoreproducing space of ν, C γ ,γ (DTd ; IR), H, ν

is an abstract Wiener space. For any (t, x) ∈ DTd , set
Z tZ
WF (t, x) =
F (ds, dy).
0

R(x)

˜ The trajectories
Then, WF is a Gaussian process with covariance function Γ.
′ ,γ ′
d
γ
(DT ; IR), this means that the law of the process WF
of WF belong to C
˜ such that, for h ∈ HT and for
is ν. The space H is the set of functions h
d
any (t, x) ∈ DT ,
˜ x) = h1l([0,t]×R(x)) , hiH ,
h(t,
T
with the following scalar product
˜ ki
˜ H = hh, kiH .
hh,
T

H is a Hilbert space isomorphic to HT .
Classical results on Gaussian processes (see, for instance, Theorem 3.4.12
in [7]) show that, for γ ′ ∈ [0, 12 ), the family {εWF , ε > 0} satisfies a large
′ ′
deviation principle on C γ ,γ (DTd ; IR) with rate function
( 1 ˜
˜
2 khkH , if h ∈ H,
˜ =
˜ h)
I(
+∞,
otherwise.
˜ = I(h) for h
˜ ∈ H.
˜ h)
Remark that, if I(h) = 12 khkHT for h ∈ HT , then I(

3

Properties on the solution

In this section we analyze the existence and uniqueness of solution (1.3) and
the H¨older continuity with respect to the parameters.
Proposition 3.1 Assume (C) and (H1 ). Then (1.3) has a unique solution.
Moreover, for any T > 0, p ∈ [1, ∞),
sup

sup

0 0, p ∈ [2, ∞), 0 < ε ≤ 1, t, t′ ∈ [0, T ], x, x′ ∈ [0, 1]d , γ1 ∈ (0, 1−η
2 ) and
γ2 ∈ (0, 1 − η),
¡
¢
E |uε (t, x) − uε (t′ , x)|p ≤ C|t − t′ |γ1 p ,
(3.2)
¡
¢
E |uε (t, x) − uε (t, x′ )|p ≤ Ckx − x′ kγ2 p .
(3.3)

Moreover, the trajectories of uε are a.s. (γ1 , γ2 )-H¨
older continuous in (t, x) ∈
1−η
d
DT , for γ1 ∈ (0, 2 ) and γ2 ∈ (0, 1 − η).
Proof: We at first prove (3.2). For 0 ≤ t′ ≤ t ≤ T , x ∈ [0, 1]d , we have
4
X
¡
¢
Ai ,
E |uε (t, x) − uε (t′ , x)|p ≤ C
i=1

with
A1
A2
A3
A4

¯p !
ïZ ′ Z
¯
¯ t
h
i
¯
¯
′
ε
=E ¯
ε G(t − s, x, y) − G(t − s, x, y) α(u (s, y))F (ds, dy)¯ ,
¯
¯ 0 [0,1]d
¯p !
ïZ Z
¯
¯ t
¯
¯
=E ¯
εG(t − s, x, y)α(uε (s, y))F (ds, dy)¯ ,
¯
¯ t′ [0,1]d
¯p !
ïZ ′ Z
¯
¯ t
h
i
¯
¯
ds
dy G(t − s, x, y) − G(t′ − s, x, y) β(uε (s, y))¯ ,
=E ¯
¯ 0
¯
d
[0,1]
¯
ïZ
!
¯ t Z
¯p
¯
¯
ε
= E ¯ ds
dy G(t − s, x, y)β(u (s, y))¯ .
¯
¯ t′
[0,1]d

Burkholder’s and H¨older’s inequalities, (3.1) and (6.1.7) imply
i
h
A1 ≤ C 1 + sup sup E(|uε (t, x)|p )
0 0 and
˜ → S h˜ is continuous from {I˜ ≤ a} to C γ1 ,γ2 (Dd ), (γ1 , γ2 ) ∈
a > 0. The map h
T
1−η
1−η
(0, 4 ) × (0, 2 ), where {I˜ ≤ a} is endowed with the topology of uniform
convergence.
˜

˜

Proof: In order to prove the continuity of S h we decompose dγ1 ,γ2 (S h , S g˜)
into two parts:
˜
sup kS h (·, x) − S g˜(·, x)kγ1 ,[0,T ] ,
(4.8)
x∈[0,1]d

and

˜

sup kS h (t, ·) − S g˜(t, ·)kγ2 ,[0,1]d .

(4.9)

t∈[0,T ]

We first examine (4.8). According to (1.5), if we want to study (4.8), we
need to deal with the following two functions, for t ∈ [0, T ],
ψ(t) =

˜

sup
(s,x)∈[0,t]×[0,1]d

|S h (s, x) − S g˜(s, x)|,
˜

ψγ1 (t) = sup

x∈[0,1]d

˜

|S h (s, x) − S g˜(s, x) − S h (r, x) + S g˜(r, x)|
.
|s − r|γ1

sup
s, r ∈ [0, t]
s 6= r

We start the proof studying ψ(T ). For (s, x) ∈ DTd , we have
Z s Z
h
i
˜
˜
h
g
˜
dy G(s − r, x, y) β(S h (r, y)) − β(S g˜(r, y))
dr
S (s, x) − S (s, x) =
[0,1]d

0

D
h
i E
˜
+ G(s − ·, x, ∗) α(S h (·, ∗)) − α(S g˜(·, ∗)) , g

+

3
X

Ai (s, x),

i=1

11

HT

with
D
E
˜
A1 (s, x) = (G − Gn )(s − ·, x, ∗)α(Snh (·, ∗)), h − g
,
iHT
E
D
h
˜
˜
,
A2 (s, x) = G(s − ·, x, ∗) α(S h (·, ∗)) − α(Snh (·, ∗)) , h − g
HT
D
E
˜
A3 (s, x) = Gn (s − ·, x, ∗)α(Snh (·, ∗)), h − g
.
HT

The usual argument based on H¨older’s inequality together with (C) and the
˜ g˜ ∈ {I˜ ≤ a},
estimates (6.1.5) and (6.1.6) imply that, for t ∈ [0, T ] and h,
2

ψ(t) ≤ C(1 +

kgk2HT )

Z

t

3
X

2

ψ(s) ds + C

sup sup A2i (s, x).

d
i=1 s≤t x∈[0,1]

0

Then, Gronwall’s lemma yields
ψ(T ) =

sup
d
(t,x)∈DT

˜
h

g˜

|S (t, x) − S (t, x)| ≤ C

3
X

sup

Ai (t, x).

(4.10)

d
i=1 (t,x)∈DT

The rest of the study of ψ(T ) consists in dealing with Ai , for i = 1, 2, 3.
For ξ1 ∈ (0, 1−η
2 ), from hypothesis (C) and the estimates (4.2), (6.1.28) and
(6.1.29) it follows that
sup
d
(t,x)∈DT

|A1 (t, x)| ≤ Ckh − gkHT

sup
d
(t,x)∈DT

k(G − Gn )(t − ·, x, ∗)kHT

≤ C2−ξ1 (n−1) .

For ξ2 ∈ (0,
we have
sup
d
(t,x)∈DT

1−η
2 )

(4.11)
and ξ3 ∈ (0, 1−η), using (C), (4.3), (4.4), (4.7) and (6.1.5),

|A2 (t, x)| ≤ Ckh − gkHT
≤

C(2−ξ2 (n−1)

sup
d
(t,x)∈DT

+

˜

˜

|S h (t, x) − Snh (t, x)|

(4.12)

2−ξ3 n ).

For i1 , . . . , id ∈ {0, . . . , 2n − 1}, denote
½
¾
ij + 1
d ij
Ri1 ,...,id = y = (y1 , . . . , yd ) ∈ [0, 1] , k ≤ yj ≤
, j = 1, . . . , d .
2n
2

12

Then,
n −1
2X

A3 (t, x) =

˜

G(2−n T, x, (i1 , . . . , id ))α(S h (0, (i1 , . . . , id )))

i1 ,...,id =0

D
E
× 1l[0,t]×Ri1 ,...,i , h − g
d

n −1
2X

"

+

HT

1l{t 0.
We now analyze ψγ1 (T ). Fix x ∈ [0, 1]d , assume t, t′ ∈ [0, T ], t′ < t. Then,
from the Lipschitz property of β, it follows
7
¯
¯
X
˜
¯
¯ h˜
g˜
h
g˜
′
Bi ,
¯(S − S )(t, x) − (S − S )(t , x)¯ ≤ C

(4.16)

i=1

with
B1 =

Z

t

ds

B2

˜

[0,1]d

t′

Z

Z

t′

dy G(t − s, x, y)|S h (s, y) − S g˜(s, y)|

Z

˜

dy |G(t − s, x, y) − G(t′ − s, x, y)||S h (s, y) − S g˜(s, y)|,
ds
+
[0,1]d
0
¯Dh
i E ¯
ih
˜
˜
¯
¯
= ¯ G(t − ·, x, ∗) − G(t′ − ·, x, ∗) α(S h (·, ∗)) − α(Snh (·, ∗)) , h
¯,
HT

13

¯Dh
i
¯
B3 = ¯ G(t − ·, x, ∗) − Gn (t − ·, x, ∗) − G(t′ − ·, x, ∗) + Gn (t′ − ·, x, ∗)
E ¯
˜
¯
×α(S h (·, ∗)), h
¯,
HT
¯Dh
E ¯
i
˜
¯
¯
B4 = ¯ Gn (t − ·, x, ∗) − Gn (t′ − ·, x, ∗) α(Snh (·, ∗)), h − g
¯,
HT
¯Dh
ih
i E ¯
˜
¯
¯
B5 = ¯ Gn (t − ·, x, ∗) − Gn (t′ − ·, x, ∗) α(Snh (·, ∗)) − α(Sng˜ (·, ∗)) , g
¯,
HT

˜ for g and g˜,
B6 and B7 are the same terms as B2 and B3 , changing h and h
respectively.
Applying (4.15), (6.1.9) and (6.1.10), we have, for all γ˜1 ∈ (0, 12 ),
i
h
3d
˜ − g˜k∞ |t − t′ |2˜γ1 ,
(4.17)
B1 ≤ C 2−nξ + 2n( 2 +1) kh
for some ξ > 0.
In the sequel, ξ, γ1′ ∈ (0, 1−η
2 ). Schwarz’s inequality, the Lipschitz property
of α, and the estimates (6.1.7), (6.1.8), (4.3), (4.4) and (4.7), imply
′

B2 + B6 ≤ C2−nξ |t − t′ |γ1 ,

ξ > 0.

(4.18)

From (4.2), (6.1.37) and (6.1.39),
B3 + B7 ≤ C2−ξ(n−1) |t − t′ |

′
γ1
2

,

Using (4.15), (6.1.38) and (6.1.40), we obtain
h
i
3d
˜ − g˜k∞ |t − t′ |γ1′ ,
B5 ≤ C 2−nξ + 2n( 2 +1) kh

ξ > 0.

ξ > 0.

(4.19)

(4.20)

Finally, we bound B4 as follows

B4 ≤ C (B4,1 + B4,2 ) ,
with
B4,1

¯Z t Z
¯
ds
= ¯¯
′
t

B4,2

¯Z
¯
= ¯¯

0

dy

[0,1]d

t′

ds

Z

Z

[0,1]d

˜
dz G(t\
− sn , x, yn )α(Snh (s, y))f (y − z)

¯
¯
×[h(s, z) − g(s, z)]¯¯,
dy

[0,1]d

Z

[0,1]d

h
i
′ − s , x, y )
dz G(t\
− sn , x, yn ) − G(t\
n
n

˜
×α(Snh (s, y))f (y

¯
¯
− z)[h(s, z) − g(s, z)]¯¯.

14

Computing as in (4.14) we can get
3d

˜ − g˜k∞ .
B4,1 ≤ C2n( 2 +1) kh
Then, Schwarz’s inequality and (6.1.27) imply
1

1

2
2
B4,1 = B4,1
B4,1
1

1

n 3d
2
2
˜ − g˜k∞
B4,1
≤ C2 2 ( 2 +1) kh

(4.21)

1

n
1
2
˜ − g˜k∞
|t − t′ | 4 .
≤ C3 2 (2d+1) kh

By (6.1.40), B4,2 can be easily bounded as follows
′

B4,2 ≤ 2−nξ |t − t′ |γ1 ,

ξ > 0.

(4.22)

Then, for γ1 ∈ (0, 1−η
4 ), (4.15)-(4.22) yield
h
1 i
˜
2
˜ − g˜k∞
,
sup kS h (·, x) − S g˜(·, x)kγ1 ×[0,T ] ≤ C 2−nξ + 2k1 (d)n kh

(4.23)

x∈[0,1]d

for some ξ > 0 and a positive constant k1 (d) depending on the spatial
dimension.
We have just studied (4.8) and we now have to examine (4.9). From (6.1.19),
(6.1.20), (6.1.53) and (6.1.54), we can study (4.9) by means of similar arguments as before and to prove that, for γ2 ∈ (0, 1−η
2 ),
i
h
′
˜
sup kS h (t, ·) − S g˜(t, ·)kγ2 ,[0,1]d ≤ C 2−nξ + 2k2 (d)n kh − gk∞ , (4.24)
t∈[0,T ]

for some ξ ′ ∈ (0, 1−η
2 ) and k2 (d) > 0.
Finally, fixing ρ > 0 and choosing n0 such that, for any n ≥ n0 , C(2−nξ +
′
2−nξ ) < ρ2 , and δ ≤ ρ4 (2−nk1 (d) ∧ 2−nk2 (d) ), inequalities (4.23) and (4.24)
imply
˜
sup dγ1 ,γ2 (S h , S g˜) ≤ ρ.
˜ g k∞ ≤δ
kh−˜

This conclude the proof of this theorem.
¤

15

5

The Freidlin-Wentzell inequality

In this section we will prove the inequality (1.4) of Theorem 1.1.
Proposition 5.1 Assume (C) and (Hη ) for some η ∈ (0, 1). For all R, ρ, a >
′
0, γ ′ ∈ (0, 12 ) and γ ∈ (0, 1−η
4 ∧ γ ), there exists δ > 0 and ε0 > 0 such that,
˜ ∈ H satisfying I(
˜ ≤ a and ε < ε0 , we have
˜ h)
for any h
µ
¶
n
o
R
˜
ε
h
˜
P dγ,γ (u , S ) ≥ ρ, dγ ′ ,γ ′ (εWF , h) < δ ≤ exp − 2 .
ε
Section 3, Theorem 4.2 and Proposition 5.1, by means of Theorem 1.1, allow
us to deduce the next theorem.
Theorem 5.2 Assume (C) and (Hη ) for some η ∈ (0, 1). Then, the law of
the solution uε satisfies on C γ,γ (DTd ), γ ∈ (0, 1−η
4 ), a large deviation principle
with the rate function

˜ S h˜ = g}, if g ∈ Im(S · ),
˜ h);
 inf{I(
S(g) =
 +∞,
otherwise.

Proof of Proposition 5.1: We need to prove that,
′
∀R, ρ, a > 0, γ ′ ∈ (0, 12 ) and γ ∈ (0, 1−η
4 ∧ γ ),

˜ ∈ H satisfying I(
˜ ≤ a and ∀ε < ε0 ,
˜ h)
∃δ, ε0 > 0, such that ∀h

(5.1)

we have
¶
µ
o
n
R
˜
ε
h
˜
P ku − S kγ,γ ≥ ρ, kεWF − hkγ ′ ,γ ′ < δ ≤ exp − 2 .
ε

(5.2)

In all the following steps of this proof we will assume the above-mentioned
conditions (5.1). We follow the classical proof of this kind of inequalities,
which follows several different steps.
Step 1. Using the stopping time
(
)
¯
¯
˜
¯
¯
τ ε = inf t, sup sup ¯uε (s, x) − S h (s, x)¯ ≥ ρ ∧ T,
s≤t x∈[0,1]d

and by applying a usual localization procedure (see, for instance, Proposition
3.9 in [14]) we can assume that the coefficients α and β are bounded.
16

˜ ∈ H such that I(
˜ ≤ a and ε ≤ ε0 ,
˜ h)
Step 2. Given h
Z tZ
˜
˜
G(t − s, x, y)α(v ε,h (s, y))F (ds, dy)
v ε,h (t, x) = ε
[0,1]d

0

D
E
˜
+ G(t − ·, x, ∗)α(v ε,h (·, ∗)), h
+

Z

t

ds

Z

[0,1]d

0

HT
˜

dy G(t − s, x, y)β(v ε,h (s, y)).

An extension of Girsanov’s Theorem (see, for instance, Section IV.5 in [2])
allows us to reduce the proof of (5.2) to establishing the following
¶
µ
n
o
R
˜
˜
(5.3)
P kv ε,h − S h kγ,γ ≥ ρ, kεWF kγ ′ ,γ ′ < δ ≤ exp − 2 .
ε
Step 3. We will now observe that (5.3) is equivalent to
¶
µ
n
o
R
˜
P kεKε,h kγ,γ ≥ ρ, kεWF kγ ′ ,γ ′ < δ ≤ exp − 2 ,
ε

(5.4)

where
K

˜
ε,h

(t, x) =

Z tZ
0

˜

[0,1]d

G(t − s, x, y)α(v ε,h (s, y))F (ds, dy),

given (t, x) ∈ DTd . In order to check this equivalence we proceed as in
Theorem 4.2. We will give the basic ideas of the proof. Schwarz’s inequality,
the Lipschitz property on α and β, (6.1.5), (6.1.6) and Gronwall’s Lemma
yield
˜
˜
˜
kv ε,h − S h k∞ ≤ CkεKε,h k∞ .
(5.5)
It only remains to study the H¨older property, that means to deal with
¯
¯
˜
˜
˜
¯
¯ ε,h˜
¯v (t, x) − S h (t, x) − v ε,h (t′ , x′ ) + S h (t′ , x′ )¯ ,

for (t, x) 6= (t′ , x′ ) ∈ DTd . Then, Schwarz’s inequality, the Lipschitz property,
(5.5), (6.1.7)-(6.1.10), (6.1.19) and (6.1.20) imply the equivalence between
(5.3) and (5.4).
˜
Step 4. Here the stochastic integral K ε,h will be discretized as follows
Z tZ
˜
˜
ε,h
G(t\
− sn , x, yn )α(v ε,h (sn , yn ))F (ds, dy).
Kn (t, x) =
0

[0,1]d

17

We will also consider the sets
o
n
˜
˜
Aεn = kε(Kε,h − Knε,h )kγ,γ > ρ2 ,

n
o
˜
Bnε = kεKnε,h kγ,γ > ρ2 , kεWF kγ ′ ,γ ′ < δ .

It is immediate to check that
n
o
˜
kεKε,h kγ,γ ≥ ρ, kεWF kγ ′ ,γ ′ < δ ⊂ Aεn ∪ Bnε .

In order to conclude the proof of this proposition we need to prove the next
two facts:
1) There exists n0 such that, for n ≥ n0 ,
¶
µ
R
ε
P (An ) ≤ exp − 2 .
ε
2) For any n > 0, we can choose δ > 0 such that, considering
n
o
ρ
˜
Bnε (δ) = kεKnε,h kγ,γ > , kεWF kγ ′ ,γ ′ < δ ,
2

(5.6)

(5.7)

then Bnε (δ) = ∅.

We first show (5.6). Before we need to see that, for any µ > 0, there exists
n
˜ 0 such that, for n ≥ n
˜0,
)
(
µ
¶
¯
¯
R
˜
¯
¯ ε,h˜
ε,h
(5.8)
P
sup ¯v (t, x) − v (tn , xn )¯ ≥ µ ≤ exp − 2 .
ε
(t,x)∈D d
T

Set

3
¯ X
¯
˜
˜
¯
¯ ε,h˜
ε,h
Li,n
(t, x),
¯v (t, x) − v ε,h (tn , xn )¯ ≤
i=1

with
˜
ε,h
L1,n
(t, x)

=

Z

t

ds
0

Z

[0,1]d

¯
¯
˜
¯
¯
dy ¯G(t − s, x, y) − G(tn − s, xn , y)¯β(v ε,h (s, y)),

¯Dh
i
E ¯¯
¯
˜
˜
ε,h
ε,
h
¯
¯,
L2,n (t, x) = ¯ G(t − ·, x, ∗) − G(tn − ·, xn , ∗) α(v (·, ∗)), h
HT ¯
¯
¯
˜
˜
˜
¯
¯
h
ε,h
ε,h
ε(K
(t,
x)
−
K
(t
,
x
))
(t,
x)
=
Lε,
¯
n n ¯.
3,n
18

Since β is bounded, from (6.1.9), (6.1.10), (6.1.20), (4.4) and (4.7), we obtain
sup
d
(t,x)∈DT

˜

ε,h
(t, x) ≤ C2−θ1 n kβk∞ ,
L1,n

for θ1 ∈ (0, 1).
On the other hand, Schwarz’s inequality together with (6.1.7), (6.1.8), (6.1.19),
(4.4) and (4.7) imply
°
°
˜
°
°
ε,h
(t, x) ≤ Ckαk∞ °G(t − ·, x, ∗) − G(tn − ·, xn , ∗)°
L2,n
HT

≤ C −θ2 n kαk∞ ,

for θ2 ∈ (0, 1−η
2 ). ¡
¢
Then, assuming C 2−θ1 n kβk∞ + 2−θ2 n kαk∞ ≤ µ2 and using these two last
bounds, we have
)
(
(
)
¯
¯
˜
µ
˜
¯
¯ ε,h˜
ε,h
(t, x) ≥
⊂
sup L3,n
sup ¯v (t, x) − v ε,h (tn , xn )¯ ≥ µ
2
d
d
(t,x)∈DT
(t,x)∈DT
n
o
µ
˜
,
⊂ kεKε,h kθ3 ,θ3 ≥
3 · 21−nθ3
with θ3 ∈ (0, 1−η
2 ).

n
˜
So, (5.8) will follows from the bound of this last set kεKε,h kθ3 ,θ3 ≥

For any (t, x), (t′ , x′ ) ∈ DTd , (6.1.7), (6.1.8) and (6.1.19) yield

°2
°h
i
°
°
° G(t − ·, x, ∗) − G(t′ − ·, x′ , ∗) α(v ε,h˜ (·, ∗))°
°
°
HT
h
i
≤ Ckαk2∞ |t − t′ |2θ3 + kx − x′ k2θ3 ,

o
.
3·21−nθ3
µ

1
µ
˜ 3 ) and 0 < ε < 1, Lemma 6.2.1 for
≥ 2C 2 kαk∞ C(θ3 )C(θ
then, if 3·21−nθ
3
τ = T ensures
(
)
¯
¯
˜
˜
¯
¯
P
sup ¯v ε,h (t, x) − v ε,h (tn , xn )¯ ≥ µ
d
(t,x)∈DT
µ
¶
µ2
.
≤ K exp − 2 4
3 · 2 · Ckαk2∞ C 2 (θ3 )2−2nθ3 ε2

One can choose n
˜ 0 such that for any n ≥ n
˜ 0 , (5.8) is satisfied.
19

Now, we will use (5.8) to prove (5.6). Let
˜

˜

˜

˜

ε,h
ε,h
(t, x),
(t, x) + K2,n
Kε,h (t, x) − Knε,h (t, x) = K1,n

with
˜
ε,h
(t, x)
K1,n

=

Z

t

ds

0

Z

−
˜

ε,h
(t, x) =
K2,n

Z

0

t

ds

[0,1]d

h
˜
G(t − s, x, y) α(v ε,h (s, y))

˜
α(v ε,h (s

Z

[0,1]d

h
˜

n , yn ))

i

F (ds, dy),

i
G(t − s, x, y) − G(t\
− sn , x, yn )

× α(v ε,h (sn , yn ))F (ds, dy).
Define

n
o
˜
ε,h
kγ,γ ≥ ρ4 ,
Aεn,1 = kεK1,n
o
n
˜
ε,h
Aεn,2 = kεK2,n
kγ,γ ≥ ρ4 ,
)
(
¯
¯
˜
¯
¯ ε,h˜
ε,h
ε
Dn =
sup ¯v (t, x) − v (tn , xn )¯ ≤ µ .
d
(t,x)∈DT

It is obvious that

¡
¢
Aεn ⊂ Aεn,1 ∪ Aεn,2 ⊂ Aεn,1 ∩ Dnε ∪ (Dnε )c ∪ Aεn,2 .

ˆ0
The set (Dnε )c has alredy bounded in (5.8). Proving that there exists n
such that, for any n ≥ n
ˆ 0,
µ
¶
¢
¡ ε
R
ε
(5.9)
P An,1 ∩ Dn ≤ exp − 2 ,
ε
¶
µ
¡
¢
R
P Aεn,2 ≤ exp − 2 ,
(5.10)
ε
we can conclude (5.6) for any n ≥ n0 being n0 = n
˜0 ∨ n
ˆ0.
Set
n
o
˜
˜
τnε = inf t ≥ 0, ∃(s, x) ∈ DTd , s ≤ t, |v ε,h (s, x) − v ε,h (sn , xn )| ≥ µ ∧ T.

20

For any (t, x), (t′ , x′ ) ∈ Dτdnε , by (6.1.7), (6.1.8) and (6.1.19) we have
°h
i°
ih
°2
°
° G(t − ·, x, ∗) − G(t′ − ·, x′ , ∗) α(v ε,h˜ (·, ∗)) − α(vnε,h˜ (·, ∗)) °
°
°
HT
h
i
2
′
2γ
′
2γ
≤ Cµ |t − t | + kx − x k ,

1
˜
˜
ε,h
ε,h
˜
(sn , yn ). Then, if ρ4 ≥ 2C 2 µC(γ)C(γ)
for γ ∈ (0, 1−η
2 ) and with vn (s, y) = v
and 0 < ε < 1, Lemma 6.2.1 gives
¶
µ
¡ ε
¢
ρ2
ε
.
P An,1 ∩ Dn ≤ K exp − 8 2 2
2 Cµ C (γ)ε2

For any R > 0 we can choose µ > 0 such that (5.9) is proved.
For any (t, x), (t′ , x′ ) ∈ DTd , (6.1.39) and (6.1.53) yield
°2
°h
i
°
°
° G(t − ·, x, ∗) − Gn (t − ·, x, ∗) − G(t′ − ·, x′ , ∗) + Gn (t′ − ·, x′ , ∗) α(vnε,h˜ (·, ∗))°
°
°
HT
h
i
≤ Ckαk2∞ |t − t′ |γ0 + kx − x′ kγ0 2−γ0 (n−1) ,
γ

1
ρ
− 20 (n−1)
˜
2
and
for γ0 ∈ (0, 1−η
2 ). Then, if 4 ≥ 2C kαk∞ C(γ, γ0 )C(γ, γ0 )2
0 < ε < 1, Lemma 6.2.1 implies
¶
µ
¡ ε ¢
ρ2
,
P An,2 ≤ K exp − 8
2 Ckαk2∞ C(γ, γ0 )2−γ0 (n−1) ε2

′
for γ ∈ (0, γ20 ) what means γ ∈ (0, 1−η
4 ). Choosing n0 large enough, for any
′
n ≥ n0 , we have (5.10).
Consequently, (5.6) is completely proved. In order to conclude the proof of
Proposition 5.1 we have to check (5.7), that means that, for any n > 0, we
can find δ > 0 such that
½
¾
δ
˜
Bnε (δ) = kεKnε,h kγ,γ > , kεWF kγ ′ ,γ ′ < δ = ∅.
(5.11)
2

The proof of (5.11) is similar in some aspects to the argument used in Theorem 4.2. We give a brief sketch of this proof. We have to work with the
difference
³
´
˜
˜
ε Knε,h (t, x) − Knε,h (t′ , x′ ) ,

for x, x′ ∈ [0, 1]d , t, t′ ∈ [0, T ]. We assume x = x′ and t′ < t for the sake of
simplicity. Then,
³
´
˜
˜
˜
˜
ε,h
ε,h
,
+ M2,n
ε Knε,h (t, x) − Knε,h (t′ , x) = M1,n
21

with
˜
ε,h
M1,n
˜

= ε

ε,h
= ε
M2,n

Z tZ
t′

Z

t′

0

˜

[0,1]d

Z

[0,1]d
˜
ε,h

×α(v

G(t\
− sn , x, yn )α(v ε,h (sn , yn ))F (ds, dy),
h
i
′ − s , x, y )
G(t\
− sn , x, yn ) − G(t\
n
n

(sn , yn ))F (ds, dy).

Decompositions similar to (4.13) for stochastic integrals allow us to study
these two terms and to obtain
¯ ³
h
´¯
i
′
′
˜
˜
¯
¯
¯ε Knε,h (t, x) − Knε,h (t′ , x′ ) ¯ ≤ C2k(d)n |t − t′ |γ + kx − x′ kγ kεWF kγ ′ ,γ ′ ,

for a positive constant k(d) depending on the spatial dimension. In this last
inequality we can not get an estimation in terms of kεWF k∞ , that is the
unique difference between the proof of (4.13) and the study of (5.11).
This argument finishes the proof of Proposition 5.1.
¤

6
6.1

Appendix
Study of the kernel

Here we will enunciate and prove all the results on the kernel G used in this
paper. Recall that G denotes the fundamental solution of the heat equation

∂

G(t, x, y) = ∆x G(t, x, y),
t ≥ 0, x, y ∈ [0, 1]d ,

 ∂t

(6.1.1)
G(t, x, y) = 0,
x ∈ ∂([0, 1]d ),



 G(0, x, y) = δ(x − y).

The details about the construction of the fundamental solution G can be
found in Chapter 1 of [11], more specifically in Section 4. This fundamental
solution G is non-negative and can be decomposed into different terms as
follows:
X
G(t − s, x, y) = H(t − s, x, y) + R(t − s, x, y) +
Hi (t − s, x − y), (6.1.2)
i∈Id

where H is the heat kernel on IRd
¶
¶d
µ
µ
kxk2
1
√
,
exp −
H(t, x) =
4t
4πt
22

©
ª
R is a Lipschitz function, Id = i = (i1 , . . . , id ) ∈ {−1, 0, 1}d − {(0, . . . , 0)}
and
Hi (t − s, x − y) = (−1)k H(t − s, x − y i ),
with k =

d
X
j=1

|ij | and


if ij = 0,
yji = yj ,



yji = −yj ,
if ij = 1,



 y i = 2 − y , if i = −1.
j
j
j

This decomposition is given in [9].
Finally, it is well-known that
Z
¡
¢
FH(t, ·)(ξ) =
e−2πix·ξ H(t, x)dx = exp −4π 2 tkξk2 ,

(6.1.3)

Rd

where F is the Fourier transform.
The proof of following bounds of G, Dt G and Dx G can be found in Theorem
1.1 of [8].
Lemma 6.1.1 There exists a positive constant C such that
G(t − s, x, y) ≤ CH(t − s, x − y),
|Dt G(t − s, x, y)| ≤ C(t − s)

− d+2
2

kx − yk2
exp −
4(t − s)
µ

(6.1.4)
¶

,

¶
µ
°
°
kx − yk2
°
°
− d+1
.
°Dx G(t − s, x, y)° ≤ C(t − s) 2 exp −
4(t − s)

In the sequel we give new results on G.

Lemma 6.1.2 Assume (H1 ). There exist positive constants C1 and C2 such
that, for any 0 ≤ s ≤ t ≤ T , x ∈ [0, 1]d ,
°2
°
°
°
≤ C1 ,
(6.1.5)
°G(t − ·, x, ∗)°
HT

Z

[0,1]d

dy G(t − s, x, y) ≤ C2 .
23

(6.1.6)

Proof: The first bound (6.1.5) is a consequence of (6.1.4) and Example 8
in [4]. The estimate (6.1.6) becomes from (6.1.4).
¤
Lemma 6.1.3 Assume (Hη ) for some η ∈ (0, 1). There exists a positive
constant C such that, for any 0 ≤ t′ ≤ t ≤ T , x ∈ [0, 1]d , γ1 ∈ (0, 1−η
2 ) and
γ2 ∈ (0, 12 ),
Z

t′

t

t′
Z t′

°2
°
°
°
ds °G(t − s, x, ∗)° ≤ C|t − t′ |2γ1 ,
ds

t

t′

(6.1.8)

H

0

Z

(6.1.7)

H

0

Z

°
°2
°
°
ds °G(t − s, x, ∗) − G(t′ − s, x, ∗)° ≤ C|t − t′ |2γ1 ,

ds

Z

Z

[0,1]d

[0,1]d

¯
¯
¯
¯
dy ¯G(t − s, x, y) − G(t′ − s, x, y)¯ ≤ C|t − t′ |2γ2 , (6.1.9)

dy G(t − s, x, y) ≤ C|t − t′ |.

(6.1.10)

Remark. Recall that the fundamental solution G of (6.1.1) satisfies
°2
°
°
°
°G(t − ·, x, ∗) − G(t′ − ·, x, ∗)°

HT

+

Z

0

t′

=

Z

t
t′

ds kG(t − s, x, ∗)k2H

°2
°
°
°
ds °G(t − s, x, ∗) − G(t′ − s, x, ∗)° .
H

Proof of Lemma 6.1.3: Observe that (6.1.2) implies
°
°2
°
°
°G(t − ·, x, ∗) − G(t′ − s, x, ∗)°

H

·°
°2
°
°
d
′
≤ (3 + 1) °H(t − s, x − ∗) − H(t − s, x − ∗)°

H

°
°2
°
°
+°R(t − s, x − ∗) − R(t′ − s, x − ∗)°

(6.1.11)

HT

°2 ¸
X°
°
°
′
+
°Hi (t − s, x − ∗) − Hi (t − s, x − ∗)° .
H

i∈Id

In order to check (6.1.7) we only need to bound the right-hand side of
(6.1.11). The terms Hi can be studied using the same arguments as H.
24

Let L be the Lipschitz constant of the function R. Clearly
Z

t′

°2
°
°
°
ds°R(t − s, x − ∗) − R(t′ − s, x − ∗)° ≤ L2 Θ T |t − t′ |2 ,
H

0

where

Θ=

Z

dy
[0,1]d

Z

[0,1]d

(6.1.12)

dzf (y − z),

and f is defined in Section 1.
Now we analyze the term with H. First of all, from h·, ·iH is a scalar product,
we have
Z t′ °
°2
°
°
ds°H(t − s, x − ∗) − H(t′ − s, x − ∗)° ≤ 2(A1 + A2 ),
(6.1.13)
H

0

with
A1

A2

¶d
1
dzf (y − z)
dy
ds
=
4π(t − s)
[0,1]d
[0,1]d
0
·
µ
¶
µ
¶¸
kx − yk2
kx − yk2
× exp −
− exp − ′
4(t − s)
4(t − s)
¶
µ
¶¸
·
µ
kx − zk2
kx − zk2
− exp − ′
,
× exp −
4(t − s)
4(t − s)
"µ
¶d µ
¶ d #2
Z
Z t′ Z
2
2
1
1
dzf (y − z)
dy
ds
=
−
′
4π(t − s)
4π(t − s)
[0,1]d
[0,1]d
0
¶
µ
¶
µ
kx − zk2
kx − yk2
exp − ′
.
× exp − ′
4(t − s)
4(t − s)
Z

t′

µ

Z

Z

Since t′ < t, using basic tools of mathematical calculus and (6.1.3), we obtain
A1 ≤ 2

Z

t′

ds
0

Z

IRd

dy

Z

IRd

dzH(t − s, x − y)f (y − z)H(t − s, x − z)

¶
t′ − s d
dz
dy
ds
−2
H(t′ − s, x − y)
t−s
IRd
IRd
0
×f (y − z)H(t′ − s, x − z)
Ã
¶d !
µ ′
Z t′ Z
¡
¢
t
−
s
λ(dξ) exp −4π 2 (t′ − s)kξk2 1 −
ds
≤ 2
d
t−s
IR
0
Z

t′

Z

Z

µ

25

= 2

Z

t′
0

µ
¶
Z
¡
¢
t′ − s
λ(dξ) exp −4π 2 (t′ − s)kξk2 1 −
ds
d
t−s
!
à IR
¶
¶
µ
µ
t′ − s d−1
t′ − s 2
t′ − s
.
+ ... +
+
× 1+
t−s
t−s
t−s

′
For γ1 ∈ (0, 1−η
2 ) and s < t , we have

µ

t − t′
t−s

¶2γ1

<

t − t′
< 1,
t−s

and then,
A1 ≤ 2 d |t − t′ |2γ1
≤ C|t −

Z

t′ |2γ1 Γ(1

t′

ds
0

Z

IRd

− 2γ1 )

Z

¡
¢
λ(dξ) exp −4π 2 (t′ − s)kξk2 (t′ − s)−2γ1

IRd

λ(dξ)

≤ C|t − t′ |2γ1 ,

1

kξk2(1−2γ1 )
(6.1.14)

where Γ is the Gamma function.
On the other hand, (6.1.3) again yields
A2 ≤

Z

t′

ds

0

Z

′

IRd

λ(dξ)(t − s)

d

õ

¡
¢
× exp −4π 2 (t′ − s)kξk2

1
t−s

¶d
2

−

µ

1
t′ − s

¶ d !2
2

µ√ ′
¶2
t −s
√
λ(dξ) exp −4π (t − s)kξk
ds
=
−1
t−s
IRd
0
Ã
√
¶2
¶d−1 !2
µ√ ′
µ√ ′
t′ − s
t −s
t −s
× 1+ √
+ ··· + √
.
+ √
t−s
t−s
t−s
Z

t′

Z

¡

2

′

2

¢

The next inequality
√
√
√
t − s − t′ − s ≤ t − t ′ ,
and similar computations to the study of A1 show that
A2 ≤ C|t − t′ |2γ1 ,
for all γ1 ∈ (0, 1−η
2 ).
26

(6.1.15)

Then, (6.1.7) follows from (6.1.12)-(6.1.15).
Now we prove (6.1.8). Set
Z
with

t
t′

°2
°
°
°
ds°G(t − s, x, ∗)° ≤ (3d + 1)[B1 + B2 + B3 ],

(6.1.16)

H

t

B1 =

Z

t

B2 =

Z

t′

t′

B3 =

°2
°
°
°
ds°R(t − s, x − ∗)° ,
H

°2
°
°
°
ds°H(t − s, x − ∗)° ,
H

XZ

i∈Id

t

t′

°
°2
°
°
ds°Hi (t − s, x − ∗)° .
H

As before the study of B1 is obvious and we can deal with B3 by means of
B2 . Then, we will only need to work with B2 .
From (6.1.3), by integrating we obtain

Since

B2 ≤

Z

=

Z

t

ds

t′

IRd

Z

IRd

¡
¢
λ(dξ) exp −4π 2 (t − s)kξk2

¡
¡
¢¢
λ(dξ) 1 − exp −4π 2 (t − t′ )kξk2
1 − e−x ≤ x,

∀x > 0,

1
4π 2 kξk2

the hypothesis (Hη ), implies, for γ1 ∈ (0, 1−η
2 ),
Z
¡
¡
¢¢2γ1 1
1
λ(dξ) 1 − exp −4π 2 (t − t′ )kξk2
B2 ≤ 2
4π IRd
kξk2
Z
1
|t − t′ |2γ1 kξk4γ1 (4π 2 )2γ1
≤ 2
λ(dξ)
4π IRd
kξk2

.

(6.1.17)

(6.1.18)

≤ C|t − t′ |2γ1 .

Then, (6.1.8) follows from (6.1.16) and (6.1.18).
Finally, the proof of (6.1.9) is similar to (6.1.7) and the estimate (6.1.10) is
an immediate consequence of (6.1.4).
¤

27

Lemma 6.1.4 Assume (Hη ) for some η ∈ (0, 1). There exists a positive
constant C such that, for any 0 ≤ t ≤ T , x, x′ ∈ [0, 1]d , γ1 ∈ (0, 1 − η) and
γ2 ∈ (0, 21 ),
t

°2
°
°
°
′
(6.1.19)
ds°G(t − s, x, ∗) − G(t − s, x , ∗)° ≤ Ckx − x′ k2γ1 ,
H
0
Z t Z
¯
¯
¯
¯
dy ¯G(t − s, x, ∗) − G(t − s, x′ , ∗)¯ ≤ Ckx − x′ k2γ2 . (6.1.20)
ds
Z

H

[0,1]d

0

Proof: As in (6.1.11) we only need to check (6.1.19) replacing G by H. Let
D=

Z

0

t

°2
°
°
°
ds°H(t − s, x − ∗) − H(t − s, x′ − ∗)° .
H

If γ1 ∈ (0, 1 − η), then

D ≤ C(D1 + D2 ),

(6.1.21)

with
D1 =

Z

D2 =

Z

0

0

t

°¯
¯1−γ1 °
¯ γ1 ¯
°2
°¯
¯
¯ ¯
′
° ,
°
ds°¯H(t − s, x − ∗) − H(t − s, x − ∗)¯ ¯H(t − s, x − ∗)¯
°
H

°¯
¯1−γ1 °
¯ γ1 ¯
t
°2
°¯
¯
¯ ¯
′
′
° .
°
ds°¯H(t − s, x − ∗) − H(t − s, x − ∗)¯ ¯H(t − s, x − ∗)¯
°
H

(6.1.22)

We first analyze D1 . We have
D1

¶d
1
dzf (y − z)
dy
ds
=
4π(t − s)
[0,1]d
[0,1]d
0
¯
µ
µ
¶
µ
¶ ¯γ 1
¶
¯
kx − yk2
(1 − γ1 )kx − yk2
kx′ − yk2 ¯¯
¯
ׯ exp −
exp
−
− exp −
¯
¯
4(t − s)
4(t − s) ¯
4(t − s)
¯
¯
µ
¶
µ
¶ γ1
¶
µ
¯
(1 − γ1 )kx − zk2
kx′ − zk2 ¯¯
kx − zk2
¯
− exp −
.
ׯ exp −
¯ exp −
¯
4(t − s)
4(t − s) ¯
4(t − s)
Z

t

Z

µ

Z

It is well-known that, for any x > 0,

x ≤ ex ,

−x

e

≤ 1.

28

(6.1.23)
(6.1.24)

By the mean-value theorem, (6.1.23), (6.1.24) and (Hη ), we get
′ 2γ1

D1 ≤ Ckx − x k

Z

t

0

¶d

ds

Z

dy

[0,1]d

Z

[0,1]d

dz(t − s)−γ1 f (y − z)

¶
µ
¶
(1 − γ1 )kx − yk2
(1 − γ1 )kx − zk2
exp −
exp −
×
4(t − s)
4(t − s)
Z t Z
¡
¢
λ(dξ)(t − s)−γ1 exp −4π 2 (t − s)kξk2
ds
≤ Ckx − x′ k2γ1
µ

1
4π(t − s)

µ

0

≤ Ckx −

x′ k2γ1 Γ(1

≤ Ckx − x′ k2γ1 .

IRd

− γ1 )

Z

IRd

λ(dξ)
kξk2(1−γ1 )

(6.1.25)

Analogously,
D2 ≤ Ckx − x′ k2γ1 .

(6.1.26)

Then, (6.1.19) follows from (6.1.21)-(6.1.26).
The proof of (6.1.20) is very similar to (6.1.19).
¤
Consider the same discretization in time and space as in the Section 4.
By (6.1.4) and (4.5) we have
d

d

|Gn (t − s, x, y)| = |G(t\
− sn , x, yn )| ≤ C|t\
− sn |− 2 ≤ C2n 2 ,

(6.1.27)

for any 0 ≤ s < t ≤ T , x, y ∈ [0, 1]d .
Lemma 6.1.5 Assume (Hη ) for some η ∈ (0, 1). There exists a positive
constant C such that, for any (t, x) ∈ DTd and γ ∈ (0, 1−η
2 ),
Z
Z

t

0

H

0

t

°2
°
°
° \
ds°G(t − sn , x, ∗) − G(t − s, x, ∗)° ≤ C2−2γ(n−1) ,

°2
°
°
°
− sn , x, ∗)° ≤ C2−2γn .
ds°Gn (t − s, x, ∗) − G(t\
H

(6.1.28)

(6.1.29)

Proof: As in (6.1.7) we can prove (6.1.28). In order to check (6.1.29) we
will only need the next inequality, for all γ ∈ (0, 1−η
2 ),
E=

Z

t
0

°
°2
°
°
ds°Hn (t − s, x − ∗) − H(t\
− sn , x − ∗)° ≤ C2−2γn ,
H

29

(6.1.30)

where Hn (t − s, x − y) = H(t\
− sn , x − yn ).
If γ ∈ (0, 1), then
Z t °¯
¯γ
°¯
¯
\
ds°
E ≤
H
(t
−
s,
x
−
∗)
−
H(
t
−
s
,
x
−
∗)
¯
¯
n
° n
0

2
¯1−γ °
¯
¯ °
¯
− sn , x − ∗)¯ °
ׯHn (t − s, x − ∗) − H(t\
°

(6.1.31)

H

≤ C(E1 + E2 ),
where
t

E1 =

Z

t

E2 =

Z

0

0

°¯
¯1−γ °2
¯γ ¯
°¯
¯ °
¯ ¯ \
\
H(
t
−
s
,
x
−
∗)
H
(t
−
s,
x
−
∗)
−
H(
t
−
s
,
x
−
∗)
ds°
¯ ° ,
¯
¯
¯
n
n
° n
H

°¯
¯1−γ °2
¯γ ¯
°¯
¯ °
¯ ¯
ds°¯Hn (t − s, x − ∗) − H(t\
− sn , x − ∗)¯ ¯Hn (t − s, x − ∗)¯ ° .
H

Similar arguments to the computation of the estimate (6.1.25) together with
(4.7) imply
E1 ≤ C2−2γn ,
(6.1.32)
for all γ ∈ (0, 1−η
2 ).
Now we examine E2 . First of all, we can ensure the existence of a positive
constant C such that
kyn − xk2 ≥ Cky − xk2 ,

∀y ∈ B(x, 3 · 2−n )c .

(6.1.33)

Then, the mean-value theorem and (6.1.23) ensure
E2 ≤ C(E21 + E22 ),

(6.1.34)

where
E21 ≤

C2−2γn

Z

t

ds
0

Z

dy
B(x,3·2−n )c

Z

B(x,3·2−n )c

dz (t\
− sn )−γ f (y − z)

!d
!
Ã
!
Ã
(1 − γ)kx − yn k2
(1 − γ)kx − zn k2
1
exp −
,
×
exp −
4π(t\
− sn )
4(t\
− sn )
4(t\
− sn )
Z
Z t Z
−2γn
dz (t\
− sn )−γ f (y − z)
dy
≤ C2
ds
Ã

E22

B(x,3·2−n )

0

×

Ã

1
4π(t\
− sn )

!d

B(x,3·2−n )

Ã

(1 − γ)kx − yn k2
exp −
4(t\
− sn )
30

!

Ã

(1 − γ)kx − zn k2
exp −
4(t\
− sn )

!

.

By (6.1.33) we can easily study E21 . Indeed, (6.1.33) yields
Z
Z t Z
−2γn
dz (t\
− sn )−γ f (y − z)
dy
E21 ≤ C2
ds
×

B(x,3·2−n )c

B(x,3·2−n )c

0

Ã

1
4π(t\
− sn )

!d

Ã

(1 − γ)kx − yk2
exp −
4(t\
− sn )

!

Ã

(1 − γ)kx − zk2
exp −
4(t\
− sn )

Now, same computations as in (6.1.25) give, for all γ ∈ (0,

!

1−η
2 ),

E21 ≤ C2−2γn .

(6.1.35)

Let γ ∈ (0, 1−η
2 ). From (6.1.24) and (4.5), we obtain by integrating with
respect to s and changing to polar coordinates,
Z
Z t Z
−2γn
dzf (y − z)(t\
− sn )−d−γ
dy
E22 ≤ C2
ds
≤ C2−2γn+dn

Z

≤ C2−2γn+dn

Z

t
0

ds(t − s)−γ
dy

B(x,3·2−n )

= C2−2γn+dn−dn
≤

B(x,3·2−n )

B(x,3·2−n )

0

Z

1

Z

Z

dy

B(x,3·2−n )

B(x,3·2−n )

Z

B(x,3·2−n )

dzf (y − z)

dzf (y − z)

r−(d−1) f (r)dr

0

C2−2γn .
(6.1.36)

Hence, the estimates (6.1.31)-(6.1.36) prove (6.1.30).
¤
Lemma 6.1.6 Assume (Hη ) for some η ∈ (0, 1). There exists a positive
constant C such that, for any 0 ≤ t′ ≤ t ≤ T , x ∈ [0, 1]d and γ ∈ (0, 1−η
2 ),
Z t °
°2
°
°
ds°G(t − s, x, ∗) − Gn (t − s, x, ∗)° ≤ C2−2γ(n−1) |t − t′ |γ , (6.1.37)
H
t′
Z t °
°2
°
°
(6.1.38)
ds°Gn (t − s, x, ∗)° ≤ C|t − t′ |2γ ,
t′
t′

Z

0

H

°
°
ds°G(t − s, x, ∗) − Gn (t − s, x, ∗) − G(t′ − s, x, ∗)
°2
°
+Gn (t′ − s, x, ∗)° ≤ C2−γ(n−1) |t − t′ |γ ,
H

31

(6.1.39)

,

Z

t′

0

°2
°
°
°
ds°Gn (t − s, x, ∗) − Gn (t′ − s, x, ∗)° ≤ C|t − t′ |2γ .
H

(6.1.40)

Proof: As usual, we will only check the inequalities (6.1.37)-(6.1.40) changing G by H. We start proving (6.1.37). Clearly
Z t °
°2
°
°
(6.1.41)
ds°H(t − s, x − ∗) − Hn (t − s, x − ∗)° ≤ (F1 + F2 ),
H

t′

where

Z

t

F1 =

t

F2 =

Z

H

t′

t′

°2
°
°
°
− sn , x − ∗)° ,
ds°H(t − s, x − ∗) − H(t\

°2
°
°
°
− sn , x − ∗) − Hn (t − s, x − ∗)° .
ds°H(t\

(6.1.42)

H

Basic calculus as in Lemma 6.1.3 give

F1 ≤ C(F11 + F12 ),

(6.1.43)

where
F11

F12

!d
1
dz
dy
ds
=
f (y − z)
4π(t\
− sn )
[0,1]d
[0,1]d
t′
!
"
Ã
µ
¶#
kx − yk2
kx − yk2
− exp −
× exp −
4(t − s)
4(t\
− sn )
!
"
Ã
µ
¶#
kx − zk2
kx − zk2
− exp −
,
× exp −
4(t − s)
4(t\
− sn )
Ã
!d Ã
! d 2
Z
Z t Z
2
2
1
1

−
dz 
dy
ds
=
d
d
′
\
\
4π(
t
−
s
)
4π(
t
−
s
)
[0,1]
[0,1]
t
n
n
µ
¶
µ
¶
2
2
kx − yk
kx − zk
×f (y − z) exp −
exp −
.
4(t − s)
4(t − s)
Z

t

Z

Z

Ã

As in (6.1.7) we can obtain
Z t Z
¡
¢ |(t\
− sn ) − (t − s)|
.
λ(dξ) exp −4π 2 (t − s)kξk2
ds
F11 ≤ 2
t\
− sn
IRd
t′
Notice that, for all x > 0

e−x ≤ x−p ,
32

p > 0.

(6.1.44)

Since γ ∈ (0, 1−η
2 ), (6.1.44), by integrating with respect to the time and the
assumption (Hη ) ensure
C2d−γ(n−1)

Z

≤ C2d−γ(n−1)

Z

F11 ≤

t

ds

Z

ds

Z

t′
t
t′

IRd

IRd

¡
¢
λ(dξ) exp −4π 2 (t − s)kξk2 (t − s)−γ
(t − s)−γ

λ(dξ)

(4π 2 (t − s)kξk2 )1−2γ
Z t
ds(t − s)−1+γ

λ(dξ)
kξk2(1−2γ) t′
Z
λ(dξ)
d−γ(n−1)
′
γ
≤ C2
|t − t |
2(1−2γ)
IRd kξk

≤ C2d−γ(n−1)

Z

IRd

≤ C2−γ(n−1) |t − t′ |γ .

(6.1.45)

Analogously, we can prove
F12 ≤ C2−γ(n−1) |t − t′ |γ

(6.1.46)

F2 ≤ C2−γ(n−1) |t − t′ |γ .

(6.1.47)

and, moreover,
Then, the estimate (6.1.37) follows from (6.1.41)-(6.1.47).
We now prove (6.1.38). Using the set B(x, 3 · 2−n ), we have
Z

t

t′

where
J1 =

Z

t

ds

t′

Z

°
°2
°
°
ds°Gn (t − s, x, ∗)° ≤ J1 + J2 ,
H

dy
B(x,3·2−n )

Z

B(x,3·2−n )

(6.1.48)

dzf (y − z)G(t\
− sn , x, yn )

×G(t\
− sn , x, zn ),
J2 =

Z

t

ds

t′

Z

dy
B(x,3·2−n )c

Z

B(x,3·2−n )c

×G(t\
− sn , x, zn ).

33

dzf (y − z)G(t\
− sn , x, yn )

Now, (6.1.4), (6.1.24) and (4.5) yield
Z
Z t Z
dy
ds
J1 ≤ C
C2dn |t

≤

B(x,3·2−n )

B(x,3·2−n )

t′

−

t′ |

Z

dy

B(x,3·2−n )

dz(t\
− sn )−d f (y − z)

Z

B(x,3·2−n )

(6.1.49)

dzf (y − z)

≤ C|t − t′ |.
On the other hand, since sn < s, (6.1.4) again, (6.1.3), (6.1.44) and (Hη )
ensure, for all γ ∈ (0, 1−η
2 ),
J2 ≤

Z

t

ds

Z

dy

B(x,3·2−n )c

t′

Z

B(x,3·2−n )c

dzf (y − z)H(t\
− sn , x − y)f (y − z)

×H(t\
− sn , x − z)
≤C

Z

≤C

Z

≤C

Z

t

ds

Z

ds

Z

ds

Z

t′
t

t′
t
t′

IRd

IRd

IRd

≤ C|t − t′ |2γ .

¡
¢
λ(dξ) exp −4πkξk2 |t − sn |
¡
¢
λ(dξ) exp −4πkξk2 |t − s|
λ(dξ)
− s)1−2γ

kξk2(1−2γ) (t

(6.1.50)

Then, the estimates (6.1.48)-(6.1.50) imply (6.1.38).
Finally, we check (6.1.39). Let
K=

Z

0

t′

°
°2
°
°
ds°G(t−s, x, ∗)−Gn (t−s, x, ∗)−G(t′ −s, x, ∗)+Gn (t′ −s, x, ∗)° .
H

On the one hand, we have

˜1
K = K1 + K
where
K1 =

Z

˜1 =
K

Z

t′

H

0
t′
0

�