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Vol. 15 (2010), Paper no. 17, pages 484–525.
Journal URL
http://www.math.washington.edu/~ejpecp/
Support theorem for a stochastic Cahn-Hilliard equation ∗
Lijun Bo1 Kehua Shi2,† and Yongjin Wang3
1
Department of Mathematics, Xidian University, Xi’an 710071, China
bolijunnk@yahoo.com.cn
2
School of Mathematical Sciences, Xiamen University, Xiamen 361005, China
kehuashink@gmail.com
3
School of Mathematical Sciences, Nankai University, Tianjin 300071, China
yjwang@nankai.edu.cn
Abstract
In this paper, we establish a Stroock-Varadhan support theorem for the global mild solution to a
d (d ≤ 3)-dimensional stochastic Cahn-Hilliard partial differential equation driven by a spacetime white noise.
Key words: Stochastic Cahn-Hilliard equation, Space-time white noise, Stroock-Varadhan support theorem.
AMS 2000 Subject Classification: Primary 60H15, 60H05.
Submitted to EJP on November 7, 2009, final version accepted Aril 15, 2010.
∗
The research of K. Shi and Y. Wang was supported by the LPMC at Nankai University and the NSF of China (No.
10871103). The research of L. Bo was supported by the Fundamental Research Fund for the Central Universities (No.
JY10000970002).
†
Corresponding author. Email: kehuashink@gmail.com
484
1
Introduction and main result
In this paper, we consider the following stochastic Cahn-Hilliard equation:

 ∂ u/∂ t = −∆ ∆u + f (u) + σ(u)Ẇ , in [0, T ] × D,
u(0) = ψ,

∂ u/∂ n = ∂ [∆u] /∂ n = 0,
on [0, T ] × ∂ D,
(1.1)
where ∆ denotes the Laplace operator, the domain D = [0, π]d (d = 1, 2, 3), and f : R → R is a
polynomial of degree 3 with positive dominant coefficient (which is due to the background of the
equation from material science). Assume that σ : R → R is a bounded and Lipschitzian function and
Ẇ is a Gaussian space-time white noise on some complete probability space (Ω, F , P) satisfying
”
—
E Ẇ (x, t)Ẇ ( y, s) = δ(|t − s|)δ(|x − y|), (t, x), (s, y) ∈ [0, T ] × D.
Here δ(·) is the Dirac delta function concentrated at the point zero.
The (deterministic) Cahn-Hilliard equation (i.e., σ ≡ 0 in (1.1)) has been extensively studied (see,
e.g., [2; 3; 4; 5; 10; 15; 18]) as a well-known model of the macro-phase separation that occurs in an
isothermal binary fluid, when a spatially uniform mixture is quenched below a critical temperature
at which it becomes unstable. A stochastic version of the Cahn-Hilliard equation (when σ ≡ 1
in (1.1)) was first proposed by Da Prato and Debussche [8], and the existence, uniqueness and
regularity of the global mild solution were explored. In Cardon-Weber [6], the authors considered
this type of stochastic equation in a general case on σ, which is equivalent to the following form:
u(t, x) =
Z
G t (x, y)ψ( y)d y +
Z tZ
0
D
+
Z tZ
0
∆G t−s (x, y) f (u(s, y))d yds
D
G t−s (x, y)σ(u(s, y))W (d y, ds),
(1.2)
D
where G t (·, ∗) denotes the Green kernel corresponding to the operator ∂ /∂ t + ∆2 with the homogeneous Neumann’s boundary condition as in (1.1). Since Stroock and Varadhan [17] established
their famous support theorem for diffusion processes, there have been many research works on this
issue, for example, a variety of support theorems for 1-dimensional second-order parabolic and hyperbolic stochastic partial differential equations (abbr. SPDEs) have been discussed in the literature
(see, e.g., [1; 7; 13; 14]). Millet and Sanz-Solé [13] characterized the support of the law of the
solution to a class of hyperbolic SPDEs, which simplified the proof in [17]. In Bally et al. [1], the
authors proved a support theorem for a semi-linear parabolic SPDE. Moreover, a support result for
a generalized Burgers SPDE (containing a quadratic term) was established in Cardon-Weber and
Millet [7]. Herein, we are attempting to establish a support theorem of the law corresponding to
the solution to Equation (1.1) in C([0, T ], L p ([D])) for p ≥ 4. The main strategy used in this paper
is an approximation procedure by using a space-time polygonal interpolation for the white noise,
and we particularly adopt a localization argument, which was used in [7] for studying a support
theorem of a Burgers-type equation. However here we need more technical estimates concerning
the high-order Green kernel G t (·, ∗), which sharp the estimates in [6] (see Appendix).
485
In what follows, we introduce the main result of this paper. To do it, we define the following
Cameron-Martin space H by
¨
Z tZ
H
=
h(t, x) =
h(s, y)d yds; (t, x) = (t, (x 1 , . . . , x d )) ∈ [0, T ] × D,
Qd
i=1 [0,x i ]
0
«
h ∈ L 2 ([0, T ] × D) ,
and the corresponding norm by
s
khkH =
Z TZ
|h(s, y)|2 d yds,
0
for all h ∈ H .
D
Let Hb represent the subset of H , in which the first-order derivative h of h ∈ H is bounded. For
h ∈ H , consider the following skeleton equation:
S(h)(t, x) =
Z
G t (x, y)ψ( y)d y +
Z tZ
0
D
+
Z tZ
0
∆G t−s (x, y) f (S(h)(s, y))d yds
D
G t−s (x, y)σ(S(h)(s, y))h(s, y)d yds.
(1.3)
D
Recall Equations (1.1) and (1.2). We make the following assumptions throughout the paper:
(H1). Assume that σ : R → R is bounded and belongs to C 3 (R) with bounded first to third-order
partial derivatives, and
(H2).
The initial function ψ ∈ L p (D) for p ≥ 4, and ψ is % ∈]0, 1]–Hölder continuous.
Now we are at the position to state the main result of this paper.
Theorem 1.1. Under the assumptions (H1) and (H2), let u = (u(t, x))(t,x)∈[0,T ]×D be the unique
solution to Equation (1.2) in C([0, T ], L p (D)) with p ≥ 4 and P ◦ u−1 denote the law (a probability
measure) of the solution u. Recall the skeleton equation (1.3), and set SH = {S(h); h ∈ H }. Then we
have
%
(a) Let p > 6. Then for ᾱ ∈]0, min{ 21 (1 − d4 ), 4 }[ and β̄ ∈]0, min{2 − d2 , %}[, the topological support
supp(P ◦ u−1 ) in C ᾱ,β̄ ([0, T ] × D) of the law P ◦ u−1 is the closure of SH .
%
(b) Let p ≥ 4. Then for ᾱ ∈]0, min{ 12 (1 − d4 ), 4 }[, the topological support supp(P ◦ u−1 ) in
C ᾱ ([0, T ], L p (D)) of the law P ◦ u−1 is the closure of SH .
The rest of this paper is organized as follows: In the coming section, we give a difference approximation to the (d + 1)–dimensional space-time white noise Ẇ (x, t) and study some concrete properties
of the approximating noises. In Section 3, we introduce a localization framework as in [7], and
then switch to prove the support theorem by checking the conditions (C1) and (C2) below (see
Section 3). Sections 4 and 5 are devoted to checking the validity of the conditions (C1) and (C2),
respectively. In Section 6, we prove the continuity of the solution S(h) to the skeleton equation (1.3)
in Hb and finally we complete the proof of Theorem 1.1.
486
2
Difference approximation to white noise
In this section, we give a difference approximation to the (d +1)-dimensional space-time white noise
Ẇ , which is a space-time polygonal interpolation for Ẇ .
Let n ∈ N and t ∈ [0, T ], set
tn =
¦
max
j∈{0,1,...,2n }
©
j T 2−n ; j T 2−n ≤ t ,
”
—
and t n = t n − T 2−n ∨ 0.
Let k := (k1 , . . . , kd ) ∈ Idn := {0, 1, . . . , n − 1}d . Define a partition (4 j,k ) j=0,1,...,2n−1 ,k∈Id of O T :=
n
[0, T ] × D by
4 j,k = Dk ×] j T 2−n , ( j + 1)T 2−n ],
where Dk =
Qd
−1
j=1 ]k j πn , (k j
Qd
−1
j=1 ]k j πn , (k j
+ 1)πn−1 ]. For x j ∈]k j πn−1 , (k j + 1)πn−1 ] with j = 1, . . . , d, we
+ 1)πn−1 ]. Further, for each (t, x) ∈ O T , we define the following
set Dk (x) =
difference approximation to Ẇ by
( W (4 )
j−1,k
, (x, t) ∈ 4 j,k , j = 1, . . . , 2n − 1, k ∈ Idn ,
|4 j−1,k |
Ẇn (x, t) =
(2.1)
0,
(x, t) ∈ 40,k , k ∈ Idn ,
where 4 j,k = T πd (nd 2n )−1 is the volume of the partition 4 j,k for each j = 0, 1, . . . , 2n − 1 and
k ∈ Idn .
Next we suppose that
(H3). the mappings F, H, K : R → R are bounded, globally Lipschitzian and H ∈ C 3 (R) with
bounded first to third-order derivatives.
We now consider the following equations for h ∈ Hb ,
X n (t, x) = G t ∗ ψ(x)
Z tZ
+
G t−s (x, y) F (X n (s, y))W (d y, ds) + H(X n (s, y))Wn (d y, ds)
0
+
D
Z tZ
0
h
G t−s (x, y) K(X n (s, y))h(s, y)
D
i
−Ḣ(X n (s, y))[αn (s, y)F (X n (s, y)) + βn (s, y)H(X n (s, y))] d yds
Z tZ
+
∆ y G t−s (x, y) f (X n (s, y))d yds,
0
D
and
X (t, x) = G t ∗ ψ(x) +
Z tZ
0
+
Z tZ
0
G t−s (x, y)[F + H](X (s, y))W (d y, ds)
D
G t−s (x, y)K(X (s, y))h(s, y)d yds
D
487
(2.2)
+
Z tZ
0
∆ y G t−s (x, y) f (X (s, y))d yds,
(2.3)
D
where
G t ∗ ψ(x) :=
Z
G t (x, y)ψ( y)d y,
D
and for each n ∈ N,
d −1
d n
αn (t, x) := n 2 (T π )
βn (t, x) := nd 2n (T πd )−1
Z t nZ
G t−s (x, y)d yds,
t n Dk (x)
Z tZ
G t−s (x, y)d yds.
t n Dk (x)
For αn (t, x) and βn (t, x), by virtue of (A.4) in Lemma A.1, we claim that
sup |αn (t, x)| ≤ C nd , and
(2.4)
(t,x)∈O T
sup |βn (t, x)| ≤ C nd .
(2.5)
(t,x)∈O T
Indeed, using (A.4), we have for each t ∈ [0, T ],
(
sup |αn (t, x)| ≤ C nd 2n max 1 Dk (x)
k∈Idn
x∈D
)
Z t nZ
tn
|G t−s (x, y)|d yds
Dk
≤ C nd 2n |t n − t n |
≤ C nd ,
and
sup |βn (t, x)| ≤ C nd 2n |t − t n | ≤ C nd ,
x∈D
follows from the equality (A.19).
In the following, let F = (F t )0≤t≤T be the natural filtration generated by W , i.e.,
F t = σ{W (B × [0, s]); s ∈ [0, t], B ∈ B(D)}.
Then for every t ∈ [0, T ] and n ∈ N fixed, (Ẇn (x, t)) x∈D given by (2.1) is F t -adapted. More
precisely, it is F t n -adapted and which is independent of the information F t n .
Lemma 2.1. For each fixed n ∈ N and p ≥ 1, we have
”
—
dp
np
sup E |Ẇn (x, t)| p ≤ C p n 2 2 2 .
(t,x)∈O T
488
Proof. By virtue of the definition (2.1),
”
—
sup E |Ẇn (x, t)| p
(t,x)∈O T
=
n
2X
−1 X
sup E
(t,x)∈O T
W (4 j−1,k )
|4 j−1,k |
j=1 k∈Id
n

≤ C p max
 W (4 j−1,k )
E

|4 j−1,k |
p
14 j−1,k (x, t)

p
; j = 1, . . . , 2n − 1, k ∈ Idn

.

Note that for each j = 1, . . . , 2n − 1 and k ∈ Idn ,
W (4 j−1,k )
|4 j−1,k |
∼ N (0, |4 j−1,k |−1 ).
For any random variable Z ∼ N (0, σ2 ), it holds that
1 p p 2p
E|Z| = p
2 σ Γ
π
p
p
2
+
1
2
,
where Γ denotes the Gamma function. This yields that
o
np
”
—
sup E |Ẇn (x, t)| p ≤ C p max
|4 j−1,k |−p ; j = 1, . . . , 2n − 1, k ∈ Idn ,
(t,x)∈O T
and which proves the lemma.
ƒ
Let n ∈ N be fixed. For α > 0 and t ∈]0, T ], we now define an event Ω̄αn,t by
Ω̄αn,t
¨
= ω ∈ Ω;
d
sup
(s, y)∈[0,t]×D
Ẇ ( y, s; ω) ≤ αn 2
n
2
«
.
(2.6)
For this event, we have
Lemma 2.2. If choose α > 2
q
log 2
,
T πd
then
lim P
n→∞
h
ic ‹
Ω̄αn,T
= 0.
Proof. Let Z ∼ N (0, 1) be a standard normal random variable. Then according to the definition
(2.1) for Ẇn (x, t),
‚
¨
«
Œ
h
ic ‹
W
(4
)
n
j−1,k
P
Ω̄αn,T
= P
max
≥ αnd 2 2
|4 j−1,k |
( j,k)∈{1,...,2n −1}×Idn

‹
p
≤ nd 2n P |Z| ≥ α T πd nd/2
489
‚
= nd 2n P
|Z|2
4
–
d n
≤ n 2 exp −
≥
α2 T πd
4
α2 T πd
4
Œ
nd
™ –
n
d
E exp
‚
|Z|2
4
Ϊ
.
(2.7)
h
2 i p
|Z|
Note that E exp 4
= 2. Then (2.7) further yields that
h
ic ‹
0≤P
Ω̄αn,T
≤
≤
–
™
p d
α2 T πd d
2n exp n log 2 −
n
4
Ϊ
– ‚
p d
α2 T πd
d
2n exp n log 2 −
4
→ 0, as n → ∞,
if α > 2
3
q
log 2
.
T πd
Thus the proof of the lemma is complete.
(2.8)
ƒ
Localization framework
In this section, we adopt a localization method used in [7] to deal with Equation (1.1). In addition,
we will prove a key proposition, which is useful in the proof of Theorem 1.1.
Proposition 3.1. Under the assumptions (H1) and (H2), let X = (X (t, x))(t,x)∈[0,T ]×D (resp. X n ) be
the unique solution to Equation (2.3) (resp. (2.2)) in C([0, T ], L p (D)) with p ≥ 4. Recall the skeleton
equation (1.3), and set SH = {S(h); h ∈ H }. Then we have
%
(i) Let p > 6. Then for ᾱ ∈]0, min{ 12 (1 − d4 ), 4 }[ and β̄ ∈]0, min{2 − d2 , %}[, the sequence X n
converges in probability to X in C ᾱ,β̄ ([0, T ] × D).
%
(ii) Let p ≥ 4. Then for ᾱ ∈]0, min{ 21 (1 − d4 ), 4 }[, the sequence X n converges in probability to X in
C ᾱ ([0, T ], L p (D)).
Next we give a sketch for the proof of the conclusion (ii) in Proposition 3.1. The similar argument
can also be used to prove the part (i). For (t, x) ∈ O T , set
Yn (t, x) := X n (t, x) − X (t, x).
From (2.2) and (2.3), it follows that
Yn (t, x) =
3
X
Γin (t, x) + Λn (t, x),
i=1
where
Γ1n (t, x)
Z tZ
G t−s (x, y) (F + H)(X n (s, y)) − (F + H)(X (s, y)) W (d y, ds),
:=
0
Γ2n (t, x) :=
D
Z tZ
0
G t−s (x, y) K(X n (s, y)) − K(X (s, y)) h(s, y)d yds,
D
490
(3.1)
Γ3n (t, x)
Z tZ
∆ y G t−s (x, y) f (X n (s, y)) − f (X (s, y)) d yds,
:=
0
D
and
Z tZ
Λn (t, x) :=
0
G t−s (x, y)H(X n (s, y)) Wn (d y, ds) − W (d y, ds)
D
Z tZ
G t−s (x, y)Ḣ(X n (s, y))
−
0
D
× αn (s, y)F (X n (s, y)) + βn (s, y)H(X n (s, y)) d yds.
(3.2)
Introduce an auxiliary F t n -adapted process
X n− (t, x) := G t−t n x, X n (t n , ·) ,
for (t, x) ∈ O T .
(3.3)
Recall the localization argument adopted in [7]. For γ ∈ (0, 1) and p ≥ 4, define
Φnp,γ (t) := sup kYn (t, ·)k p +
s∈[0,t]
kYn (s, ·) − Yn (s0 , ·)k p
sup
|s − s0 |γ
s6=s0 ∈[0,t]
,
where k · k p corresponds to the norm of L p (D) and for δ > 0,
¦
©
τδn := inf t > 0; Φnp,γ (t) ≥ δ ∧ T.
For M > δ, define the following events
¨
A t (M − δ) :=
«
ω ∈ Ω; sup kX (s, ·)k p ≤ M − δ ,
(3.4)
s∈[0,t]
¨
AM
n (t)
«
ω ∈ Ω; sup kX n (s, ·)k p ∨ sup kX (s, ·)k p ≤ M .
:=
s∈[0,t]
(3.5)
s∈[0,t]
Then for t ∈]0, T ],
A t (M − δ) ∩ {t ≤ τδn } ⊆ AM
n (t).
(3.6)
In fact, from the inequality | y| ≤ |x − y| + |x|, it follows that
A t (M − δ) ∩ {t ≤ τδn }
¨
«
¨
sup kX (s, ·)k p ≤ M − δ ∩
⊆
s∈[0,t]
sup kX n (s, ·) − X (s, ·)k p ≤ δ
s∈[0,t]
¨
«
sup kX (s, ·)k p ≤ M
⊆
«
¨
«
sup kX n (s, ·)k p ≤ M
∩
s∈[0,t]
s∈[0,t]
= AM
n (t).
Recall the
event Ω̄αn,t
p
q
log 2
defined by (2.6) in Section 2 and that α > 2 T πd . For each fixed δ > 0 and
V ∈ C γ ([0, T ]; L (D)), set
kV kγ,p,τδ :=
n
sup
s∈[0,T ∧τδn ]
kV (s, ·)k p +
sup
s6=s0 ∈[0,T ∧τδn ]
491
kV (s, ·) − V (s0 , ·)k p
|s − s0 |γ
.
Then for q ≥ 1,
€
Š
P Φnp,γ (T ) ≥ δ
h
ic ‹
€
Š
≤ P Φnp,γ (T ) ≥ δ, A T (M − δ) ∩ Ω̄αn,T + P AcT (M − δ) + P
Ω̄αn,T
h
ic ‹
€
Š
α
c
α
Ω̄n,T
= P kYn kγ,p,τδ ≥ δ, A T (M − δ) ∩ Ω̄n,T + P A T (M − δ) + P
n
•
˜

‹
h
i
€
Š
c
q
c
α
≤ δ−q E 1AT (M −δ)∩Ω̄α kYn k
+
P
A
(M
−
δ)
+
P
Ω̄
,
δ
T
n,T
γ,p,τn
n,T
However, Lemma 2.2 and Lemma 3.1 (the latter will be proved below) yield that
h
ic ‹
Š
€
Ω̄αn,T
→ 0, as M → ∞, n → ∞.
P AcT (M − δ) + P
Therefore, by Lemma A.1 in [7] and (3.6), in order to prove
Φnp,γ (T ) → 0, in probability, as n → ∞,
(3.7)
it suffices to check that there exist q ≥ p and θ > qᾱ (we have set γ = ᾱ, where ᾱ is the exponent
presented in Proposition 3.1) such that
h
i
(C1) ∀ t ∈ [0, T ], lim E 1ĀMn (t) kYn (t, ·)kqp = 0;
n→∞
h
i
(C2) ∀ s < t ∈ [0, T ], E 1ĀMn (t) kYn (t, ·) − Yn (s, ·)kqp ≤ C|t − s|1+θ .
Here the event ĀM
n (t) is defined by
M
α
ĀM
n (t) = An (t) ∩ Ω̄n,t ,
t ∈ [0, T ],
(3.8)
and which satisfies the order relation:
M
ĀM
n (t) ⊂ Ān (r), if r ≤ t.
Lemma 3.1. Let the event A T (M ) be defined by (3.4). Then
€
c Š
lim P A T (M )
= 0.
M →∞
Proof.
Note that for p ≥ 4, β ∈ [p, ∞[ and β ∈]p, 6p/(6 − p)+ [ (if d = 3),
‚
Œ
–
™
€
c Š
P A T (M )
=P
sup kX (t, ·)k p ≥ M ≤ M −β E sup kX (t, ·)kβp .
t∈[0,T ]
t∈[0,T ]
Therefore, it remains to prove
‚
E
sup
t∈[0,T ]
kX (t, ·)kβp
Œ
< ∞.
Define for (t, x) ∈ O T ,
L1 (u)(t, x) :=
Z tZ
0
G t−s (x, y)[F + H](u(s, y))W (d y, ds);
D
492
(3.9)
L2 (u)(t, x) :=
Z tZ
0
G t−s (x, y)K(u(s, y))h(s, y)d yds,
D
and set Z = X − L(X ) with L = L1 + L2 . Then

∂Z
2
 ∂ t + ∆ Z − ∆ f ([Z + L(X )]) = 0,
Z(0) = ψ,

∂ Z/∂ n = ∂ [∆Z]/∂ n = 0, on ∂ D.
(3.10)
Further, the Garsia-Rodemich-Rumsey lemma (see, e.g., Theorem B.1.1 and Theorem B.1.5 in [9])
yields that, if for any q, δ ∈]1, ∞[ and some γ0 , γ00 ∈]0, 1],
”
—
(a)
sup E |L(u)(t, x)|2qδ < ∞,
(t,x)∈O T
”
—
”
00
0 —q
(b) E |L(u)(t, x) − L(u)(t 0 , x 0 )|2q ≤ C |t − t 0 |γ + |x − x 0 |γ , q > 1,
then (3.9) holds. So we only need to prove (a) and (b). Note that K is bounded and d ≤ 3. Then in
light of (A.4),
sup E
(t,x)∈O T
h
L2 (u)(t, x)
2qδ
i
≤ kKk2qδ
sup
∞
(t,x)∈O T
2qδ
Z tZ
0
|G t−s (x, y)h(s, y)|d yds
D
d
(1− 4 )qδ
≤ kKk2qδ
< ∞.
∞ khkH T
qδ
If set t > t 0 , then by (A.7)–(A.9) in Lemma A.2, we have for γ0 ∈ [0, 1−d/4[ and γ00 ∈ [0, 2∧(4−d)[,
”
—
”
00
0 —q
E |L2 (u)(t, x) − L2 (u)(t 0 , x 0 )|2q
≤ C |t − t 0 |γ + |x − x 0 |γ .
The estimate of L1 is similar to that of L2 (or see [6]). Thus the proof the lemma is complete.
4
ƒ
Auxiliary lemmas
In this section, we present a sequence of auxiliary lemmas for checking the conditions (C1) and
(C2) under (H1)–(H3) given in Section 1 and 2. Throughout Sections 4–6, (H1)–(H3) are assumed
to be satisfied.
The following lemma tells us that, to check (C1), it suffices to show (C1) holds with Λn (t, x) instead
of Yn (t, x).
Lemma 4.1. Assume p ≥ 4, and q ≥ p if d = 1, 2, and q ∈]p, 6p/(6 − p)+ [ if d = 3. Then for each
n ∈ N,
h
h
i
i
sup E 1ĀMn (t) kYn (t, ·)kqp ≤ C sup E 1ĀMn (t) kΛn (t, ·)kqp ,
(4.1)
t∈[0,T ]
t∈[0,T ]
where Λn (t, x) is defined by (3.2).
493
Proof. Note that for each t ∈ [0, T ],


3
h
i
X
E kYn (t, ·)kq ≤ CE 
1ĀM (t) kΓi (t, ·)kq + 1ĀM (t) kΛn (t, ·)kq  ,
p
n
n
p
p
n
i=1
where Γi (t, x) (i = 1, 2, 3) are defined in (3.1). Therefore by (A.16) in Lemma A.3, for
1 ∈ [0, 1],
i
h
E 1ĀMn (t) kΓ1n (t, ·)kqp
Z h
i
1
r
=
2
q
−
2
p
+
E 1ĀMn (t) |Γ1n (t, x)|q dx
≤ C
D

q
2
Z tZ

≤ CE 
0
D
2
G t−s
(·, y)1ĀMn (s) |Yn (s, y)|2 d yds
t
–Z
(t − s)
≤ CE
d
− d2
4r
0
t
–Z
(t − s)
≤ C
d
− d2
4r
™
q
2



q
2
1ĀMn (s) kYn (s, ·)k2p ds
™ q−2 –Z
t
2
(t − s)
ds
d
− d2
4r
r∈[0,s]
0
0
™
i
h
q
sup E 1ĀMn (r) kYn (r, ·)k p ds .
Similarly, we have
h
i
E 1ĀMn (t) kΓ2n (t, ·)kqp

Z tZ

q
2
≤ CkhkH E 
G t−s
(·, y)1ĀMn (s) |Yn (s, y)|2 d yds
0
t
–Z
(t − s)
≤ C
D
d
− d2
4r
™ q−2 –Z
t
2
(t − s)
ds
d
− d2
4r
As for Γ3n (t, x), using (A.13) with
1
r2
=
1
q
−
1
ρ
q
2



™
h
i
q
sup E 1ĀMn (r) kYn (r, ·)k p ds .
r∈[0,s]
0
0
q
2
+ 1 ∈ [0, 1],
h
i
E 1ĀMn (t) kΓ3n (t, ·)kqp

Z tZ

≤ CE 1ĀMn (t)
∆ y G(t − s, ·, y)[ f (X n (s, ·)) − f (X (s, ·))]d yds
0

–Z
≤ CE 1ĀMn (t)
D
t
0
p
ku2 (s, ·) − v 2 (s, ·)kρ



q
™q 
d
d+2
−
(t − s) 4r2 4 k f (X n (s, ·)) − f (X (s, ·))kρ ds  .
Note that u(s, ·), v(s, ·) ∈ L p (D), for each s ∈ [0, T ], we have for ρ = 3 ,
ku(s, ·) − v(s, ·)kρ
q
≤ Cku(s, ·) − v(s, ·)k p ,
”
—
≤ Cku(s, ·) − v(s, ·)k p ku(s, ·)k p + kv(s, ·)k p
494
(4.2)
ku3 (s, ·) − v 3 (s, ·)kρ
≤ Cku(s, ·) − v(s, ·)k p
i
h
× ku(s, ·)k2p + kv(s, ·)k2p + ku(s, ·)k p kv(s, ·)k p .
Hence from (4.2), it follows that
i
h
E 1ĀMn (t) kΓ3n (t, ·)kqp
–Z t
(t − s)
≤ CE
–Z
0
t
≤ C
(t − s)
d
− d+2
4r2
4
d
− d+2
4r2
4
™q
1ĀMn (s) kYn (s, ·)k p ds
™q−1 –Z
ds
t
(t − s)
d
− d+2
4r2
4
r∈[0,s]
0
0
™
i
h
sup E 1ĀMn (r) kYn (r, ·)kqp ds .
Note that the following equivalent relations holds:
d
−
d
+1>0 ⇔
4r 2
d
d +2
−
+1>0 ⇔
4r2
4
1
1
− ,
q
p 6
1 1 1 2
> + − .
q
p 2 d
>
1
Then the desired result follows from the Gronwall’s lemma.
ƒ
Recall the F t n -adapted process X n− (t, x) defined by (3.3). Then we have
Lemma 4.2. Let q ≥ p ≥ 6. Then there exists a constant C := C M > 0 such that
h
i
sup E 1ĀMn (t) |X n (t, x) − X n− (t, x)|q ≤ C2−nqι ,
(t,x)∈O T
where ι := 12 (1 − d4 ).
Proof. Recall (2.2) and (3.3), and we get
X n (t, x) − X n− (t, x) =
4
X
Tnk (t, x),
k=1
where
Tn1 (t, x)
Z tZ
:=
Tn2 (t, x) :=
Tn3 (t, x) :=
Tn4 (t, x) :=
tn D
Z tZ
tn D
Z tZ
tn D
Z tZ
G t−s (x, y)F (X n (s, y))W (d y, ds),
G t−s (x, y)H(X n (s, y))Wn (d y, ds),
∆ y G t−s (x, y) f (X n (s, y))d yds,
G t−s (x, y)Kn (s, y)d yds,
tn D
495
(4.3)
and
Kn (s, y) = K(X n (s, y))h(s, y)
−(F Ḣ)(X n (s, y))αn (s, y) − (H Ḣ)(X n (s, y))βn (s, y).
From (2.4) and (2.5), it follows that for h ∈ Hb ,
sup |Kn (s, y)| ≤ C nd .
(4.4)
(s, y)∈O T
On the other hand, using (A.4) and the boundedness of F ,
Z Z
q
t
—
”
1
q
= E
E |Tn (t, x)|
G t−s (x, y)F (X n (s, y))W (d y, ds) 
tn D

q/2
Z tZ

≤ CE 
2
G t−s
(x, y)F 2 (X n (s, y))d yds
tn D



Z Z
q/2
t
2
≤ C
G t−s
(x, y)d yds
tn D
q/2
Z
t
d
−
≤ C  (t − s) 4 ds
tn
≤ C2
− 12 nq(1− d4 )
.
(4.5)
Further, the Hölder inequality, Lemma 2.1 and the boundedness of H jointly imply that
”
—
E |Tn2 (t, x)|q
Z Z
q
t
= E
G t−s (x, y)H(X n (s, y))Wn (d y, ds) 
tn D
Z Z
q
t
|G t−s (x, y)||Ẇn ( y, s)|d yds 
≤ CE 
tn D

Z tZ

≤ CE 
tn D
 Z Z
q−1 
t

|G t−s (x, y)||Ẇn ( y, s)|q d yds · 
|G t−s (x, y)|d yds 
tn D
Z Z
q
t
”
—
≤ C
|G t−s (x, y)|d yds sup E |Ẇn (x, t)|q
(t,x)∈O T
tn D
dq
2
nq
2
dq
−nq
2
≤ C n 2 |t − t n |q
≤ Cn 2 2
.
(4.6)
Note that f is a polynomial of degree 3. Then by virtue of (A.13) with κ =
h
i
E 1ĀMn (t) |Tn3 (t, x)|q
496
1
− 3p +1
∞
= 1− 3p ∈ [0, 1],

= E 1ĀMn (t)
Z
∆ y G t−s (x, y) f (X n (s, y))d yds 
tn D
q 
t
(t − s)
≤ E 
q
Z tZ
d
κ− d+2
4
4
Z
t
(t − s)
≤ E 
1ĀMn (s) k f (X n (s, ·))k p ds 
3
tn
d
(1− 3p )− d+2
4
4
tn
q 
i
h
1ĀMn (s) kX n (s, ·)k p + kX n (s, ·)k2p + kX n (s, ·)k3p ds 
Z
q
t
1 3d
−(
+
)
≤ C  (t − s) 2 4p ds
tn
≤ C2
)
−nq( 21 − 3d
4p
.
(4.7)
Thanks to (4.4), we conclude that
Z Z
q
t
”
—
E |Tn4 (t, x)|q
= E
G t−s (x, y)Kn (s, y)d yds 
tn D
Z Z
q
t
≤ C ndq 
|G t−s (x, y)|d yds
tn D
dq
≤ C n |t − t n |q
≤ C ndq 2−nq .
Thus the estimate (4.3) follows from (4.5)–(4.8).
Remark 4.1. We can easily check that there exists a constant C M > 0 such that for q ≥ p ≥ 4,
h
i
sup E 1ĀMn (t) kX n (t, ·) − X n− (t, ·)kqp ≤ C M 2−nqι ,
(4.8)
ƒ
(4.9)
t∈[0,T ]
where ι = 21 (1 − d4 ). In fact, for the proof of (4.9), the only estimate that needs to be checked is that of
Tn4 . Apply (A.13) with κ =
1
p
−
3
p
+1=1−
2
p
∈ [0, 1] to Tn4 and we can get (4.9).
Next we prove a useful lemma, which will be used frequently later. For (t, x), (s, y) ∈ O T , set
η(t, s, x, y) (·, ∗) := 1[0,t] (·)G t−· (x, ∗) − 1[0,s] (·)Gs−· ( y, ∗),
(4.10)
e s, x, y) (·, ∗) := 1[0,t] (·)∆G t−· (x, ∗) − 1[0,s] (·)∆Gs−· ( y, ∗).
η(t,
(4.11)
Let V be an F = (F t )0≤t≤T -predictable process. For 0 ≤ s ≤ t ≤ T and x, y ∈ D, define
λkn (V )(t, s, x,
Z
(k+1)T
2n
∧tZ
η(t, s, x, y) (r, z)V (r, z)Wn (dz, dr),
y) :=
kT
2n
∧t
D
with k = 0, 1, . . . , 2n − 1. Then we have
497
(4.12)
Lemma 4.3. Let V : Ω × O T → R be an F-predictable process. If for each p ∈ [1, ∞[, there exist some
constants α̂ > 0, a > 0 and C > 0 such that
sup
t∈[0,T ]
kV (t, ·)k L r1 p (D)
L r1 p (Ω)
≤ C2−nα̂ na , ∀ r1 ≥ 2.
(4.13)
Then for θ < 1− d4 , θ 0 < 4− d and θ 0 ≤ 2, there exists C > 0 such that for 0 ≤ s ≤ t ≤ T and x, y ∈ D,
[2n X
t/T ]−1
λkn (V )(t, s, x, y)
2
k=0
0
≤ C n(d+2a) 2−nα
”
0—
|t − s|θ + |x − y|θ ,
(4.14)
L p (Ω)
1
p
where α0 = 2α̂ − 2 +
2
γβ
+
> 0 with γ, β ≥ 1 such that 1 < γβ < 2.
Proof. As in (4.10), define
(k+1)T
2n
Z
λkn (V )(t, x,
Z
y) :=
kT
2n
G t−r (x, z) − G t−r ( y, z) V (r, z)Wn (dz, dr).
(4.15)
D
Applying Hölder inequality thrice with the respective exponents (2pδ, γ), (δ, δ0 ) and (β,
by (4.13) and Lemma 2.1,
i
h
2p
E λkn (V )(t, x, y)


2p
Z Tn (k+1)Z
 2


= E
[G
(x,
z)
−
G
(
y,
z)]V
(r,
z)
Ẇ
(z,
r)dzdr
t−r
t−r
n
 T

2n

– Z

≤ E 

Z

×

D
T
2n
k
2pδ
|V (r, z)|
D
γ
Z
k
γ
D
γ
δ
D
×|G t−r (x, z) − G t−r ( y, z)| dzdr

≤ E


×
Z
k
T
(k+1)
2n
T
2n
k
D
T
(k+1)
2n
hZ
Z
T
2n
i 2pδ0
γ
δ
Z
T
2n
E
1
T
(k+1)
2n
Z
–

|V (r, z)|2pδ dzdr 
γ

™

|Ẇn (z, r)| |G t−r (x, z) − G t−r ( y, z)| dzdr 
1
Z
k

dzdr 
 2p
T
(k+1)
2n
T
2n
δ
Z
T
(k+1)
2n
Z
1
T
(k+1)
2n
T
2n

≤ E
k
2pδ0
).
γ
–

|V (r, z)|2pδ dzdr 
™
k
|Ẇn (z, r)|γ
D
1
δ0


E
Z

T
(k+1)
2n
Z
T
2n
k
D
0

|Ẇn (z, r)|2pδ dzdr 
 2pδ0
γβ
Z
D

|G t−r (x, z) − G t−r ( y, z)|γβ dzdr 
498
™ 10
δ
Then

≤ Cn
d p+2ap np−2npα̂−n 
2

 2p
γβ
T
(k+1)
2n
Z
Z
T
2n
k
γβ
D

|G t−r (x, z) − G t−r ( y, z)| dzdr 
.
(4.16)
In light of (A.24) in Lemma A.4, there exists a θ 0 ∈]0, 4 − d[ and θ 0 < 2 such that
[2n X
t/T ]−1
λkn (V )(t, x, y)
2
≤ C nd+2a 2
k=0
2
+ 1p )
−n(2α̂−2+ γβ
0
|x − y|θ .
(4.17)
L p (Ω)
Next we turn to the time increment. Set
Z (k+1)T
Z
n ∧t
2
λkn (V )(t, s, x) :=
kT
2n
∧s
[G t−r (x, z) − Gs−r (x, z)]V (r, z)Wn (dz, dr).
(4.18)
D
Then for 0 ≤ s ≤ t ≤ T and x ∈ D,
λkn (V )(t, s, x)
Z Tn (k+1)∧sZ
2
[G t−r (x, z) − Gs−r (x, z)]V (r, z)Ẇn (z, r)dzdr
≤
Tk
∧s
2n
+
Z
D
T
(k+1)∧t
2n
Z
Tk
∨s
2n
G t−r (x, z)V (r, z)Ẇn (z, r)dzdr
D
=: Ak1,n (V )(t, s, x) + Ak2,n (V )(t, s, x),
(4.19)
with the definition
Ak2,n (V )(t, s, x) = 0, whenever [(k + 1)T 2−n ∧ t] < [kT 2−n ∨ s].
By the similar proof to that of the space increment, there exists a θ ∈]0, 1 − d4 ] such that
[2n X
t/T ]−1
Ak1,n (V )(t, s, x)
k=0
2
≤ C nd+2a 2
2
−n(2α̂−2+ 1p + γβ
)
|t − s|θ .
(4.20)
L p (Ω)
Using the proof of (4.15), we have for each k = 0, 1, . . . , [2n t/T ],
h
i
E |Ak2,n (V )(t, s, x)|2p

1
(d+2a)p −np(2α̂−1+ p ) 
≤ Cn
2

Z
γβ
Z
Tk
∨s
2n
Ak2,n (V )(t, s, x)
k=0
2
L p (Ω)
499
γβ
D
Then from (A.4) and the Hölder inequality, it follows that
[2n X
t/T ]−1
 2p
T
(k+1)∧t
2n

|G t−r (x, z)| dzdr 
.
(4.21)

≤ C nd+2a 2
≤ Cn
−n(2α̂−1+ 1p ) 
Z

T
(k+1)∧t
2n
Tk
∨s
2n
2
1
d+2a −n(2α̂−1+ p ) −n( γβ −1)

d

(t − r)− 4 (γβ−1) dr 
–Z
2
2
2
γβ
t
− d4 (γβ−1)
(t − r)
™
dr
s
≤ C nd+2a 2
2
)
−n(2α̂−2+ 1p + γβ
d
|t − s|− 4 (γβ−1)+1 .
(4.22)
Thus from (4.17), (4.20) and (4.22), we get (4.14). Hence the proof of the lemma is complete. ƒ
Remark 4.2. From the
• proof of Lemma 4.3, ˜the conclusion of the lemma still holds if we replace
λkn (V )(t, s, x, y) by E λkn (V )(t, s, x, y)|F (k−1)T .
2n
Lemma 4.4. For (s, y) ∈ O T , it holds that

Z Z

s
E Ẇn ( y, s)
= (T πd )−1 nd 2n
Gs−r ( y, z)F (X n− (r, z))W (dz, dr)|Fsn 
sn D
Z s nZ
sn
Dk ( y)
Gs−r ( y, z)F (X n− (r, z))dzdr.
Proof. For u ≥ sn , set
Nu (s, y) := (T πd )−1 nd 2n W (Dk ( y)×]sn , u]),
Z uZ
Gs−r ( y, z)F (X n− (r, z))W (dz, dr).
Mu (s, y) :=
sn
D
Then the martingale property of {Mu (s, y); u ≥ sn } and Itô formula jointly imply that


Z Z
s
Gs−r ( y, z)F (X n− (r, z))W (dz, dr)|Fsn 
E Ẇn ( y, s)
sn D
i
h
= (T πd )−1 nd 2n E Ns n (s, y)Ms (s, y)|Fsn
h h
i
i
= (T πd )−1 nd 2n E E Ns n (s, y)Ms (s, y)|Fs n Fsn
i
h
= (T πd )−1 nd 2n E Ns n (s, y)E[Ms (s, y)|Fs n ]Fsn
h
i
= (T πd )−1 nd 2n E Ns n (s, y)Ms n (s, y)|Fsn
Z s nZ
= (T πd )−1 nd 2n
sn
Dk ( y)
Gs−r ( y, z)F (X n− (r, z))dzdr,
follows from the fact that F (X n− (r, z)) is Fsn -measurable when r ≤ s. This proves the lemma.
ƒ
The following lemma shows the local Hölder continuity of F-adapted process X n (t, x) defined by
(2.2). Recall the assumption (H2), in which the initial function ψ is %-Hölder continuous (% ∈
]0, 1]).
500
%
Lemma 4.5. Let q ≥ p ≥ 6. Then for ᾱ ∈]0, min{ 12 (1 − d4 ), 4 }[ and β̄ ∈]0, min{2 − d2 , %}[, there
exists a C > 0 such that
h
i
h
iq
q
E 1ĀMn (t)∩ĀMn (s) X n (t, x) − X n (s, y)
≤ C ndq |t − s|ᾱ + |x − y|β̄ ,
%
with (t, x), (s, y) ∈ O T . In addition, if q ≥ p ≥ 4, then for ᾱ ∈]0, min{ 12 (1 − d4 ), 4 }[, there exists a
C > 0 such that
h
i
q
E 1ĀMn (t)∩ĀMn (s) X n (t, ·) − X n (s, ·) p ≤ C ndq |t − s|ᾱq .
Proof. Recall (2.2), (4.11) and (4.12). We have for (t, x), (s, y) ∈ O T ,
X n (t, x) − X n (s, y) =
5
X
Ji (s, t, x, y),
i=1
where
J1 (t, s, x, y) := G t ∗ ψ(x) − Gs ∗ ψ( y),
Z tZ
h
J2 (t, s, x, y) :=
[η(t, s, x, y)](r, z) F (X n (r, z))W (dz, dr)
0
J3 (t, s, x, y) :=
i
+H(X n− (r, z))Wn (dz, dr) ,
Z tZ
h
[η(t, s, x, y)](r, z) (H(X n (r, z)) − H(X n− (r, z)))Wn (dz, dr)
0
J4 (t, s, x, y) :=
D
i
−Ḣ(X n (r, z)) αn (r, z)F (X n (r, z)) + βn (r, z)H(X n (r, z)) dzdr ,
Z tZ
[η(t, s, x, y)](r, z)K(X n (r, z))ḣ(r, z)dzdr,
0
J5 (t, s, x, y) :=
D
D
Z tZ
0
η̃(t, s, x, y) (r, z) f (X n (r, z))dzdr.
D
Note that the initial function ψ(x) is %-Hölder continuous in x ∈ Rd . Then by Lemma 2.2 in [6],
”
”
—q
q—
E J1 (s, t, x, y)
≤ C |t − s|%/4 + |x − y|% .
(4.23)
Since F, H are bounded, by Burkhölder’s inequality and Lemma A.2, there exist γ0 ∈]0, 4 − d[, γ0 ≤ 2
and γ00 ∈]0, 1 − d/4[ such that
q
”
0
00 —
E |J2 (t, s, x, y)|q ≤ C |x − y|γ + |t − s|γ 2 .
(4.24)
On the other hand, since K is bounded, the Schwarz’s inequality yields that for h ∈ Hb ,
q
”
0
00 —
E |J4 (t, s, x, y)|q ≤ C |x − y|γ + |t − s|γ 2 ,
(4.25)
where γ0 , γ00 are presented in (4.24). Similar as in the proof of (4.7), by (A.14) and (A.15) with
1
1
=∞
− 3p + 1 ∈ [0, 1], we have for α ∈]0, 1/2 − 3d
[ and β ∈]0, min{2 − 3d
, 2 , 1}],
r
4p
p d
h
i
”
—q
E 1ĀMn (t)∩ĀMn (s) |J5 (t, s, x, y)|q ≤ C |x − y|β + |t − s|α .
501
(4.26)
and using (A.14) with κ =
1
p
−
3
p
+1=1−
2
p
∈ [0, 1], for α ∈]0, 1/2 −
d
[,
2p
h
i
E 1ĀMn (t)∩ĀMn (s) kJ5 (t, s, ·, ·)kqp ≤ C [|t − s|α ]q .
(4.27)
In the following, we are going to estimate J3 (t, s, x, y). Using the Taylor expansion of H at point
X n− (t, x),
”
—
H(X n (t, x)) − H(X n− (t, x)) = Ḣ(X n− (t, x)) X n (t, x) − X n− (t, x) + ρn (t, x),
(4.28)
and
ρn (t, x) ≤ C|X n (t, x) − X n− (t, x)|2 .
Recall the above J3 , and we have
J3 (t, s, x, y) =
5
X
J3i (t, s, x, y),
i=1
where
J31 (t, s, x,
Z tZ
h
[η(t, s, x, y)](r, z) Tn1 (r, z)Ḣ(X n− (r, z))Ẇn (z, r)
y) :=
0
J32 (t, s, x, y) :=
i
−Ḣ(X n (r, z))αn (r, z)F (X n (r, z)) dzdr,
Z tZ
h
[η(t, s, x, y)](r, z) Tn2 (r, z)Ḣ(X n− (r, z))Ẇn (z, r)
0
J33 (t, s, x, y) :=
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))Tn3 (r, z)Ẇn (z, r)dzdr,
D
Z tZ
0
J35 (t, s, x, y) :=
D
i
−Ḣ(X n (r, z))βn (r, z)F (X n (r, z)) dzdr,
Z tZ
0
J34 (t, s, x, y) :=
D
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))Tn4 (r, z)Ẇn (z, r)dzdr,
D
Z tZ
0
[η(t, s, x, y)](r, z)ρn (r, z)Ẇn (z, r)dzdr.
D
Next we decompose J31 (t, s, x, y) as follows:
J31 (t, s, x, y) =
5
X
1, j
J3 (t, s, x, y),
j=1
with
1,1
J3 (t, s, x,
y) =
Z tZ
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))Ẇn (z, r)
0
ZD Z

r
”
—
×
G r−u (z, v) F (X n (u, v)) − F (X n− (u, v)) W (dv, du) dzdr,
rn D
502
1,2
J3 (t, s, x,
y) =
Z tZ
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))
0
– ZD rZ
G r−u (z, v)F (X n− (r, z))W (dv, du)Ẇn (z, r)
×
r
D
Zn Z
™
r
−E 
G r−u (z, v)F (X n− (u, v))W (dv, du)Ẇn (z, r)|F rn  dzdr
rn D
1,3
J3 (t, s, x, y) =
Z tZ
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))
0
D

– Z r Z
× E
G r−u (z, v)F (X n− (u, v))W (dv, du)Ẇn (z, r)|F rn 
rn
n −d
−2 n
T
D
−1
−d
π
r nZ
Z
Dk (z)
rn
1,4
J3 (t, s, x,
n
y) = 2 T
−1 −d
n
−d
π
Z tZ
0
Z
rn
1,5
J3 (t, s, x, y) =
Z tZ
0
dzdr,
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))
D

r nZ
×
™
G r−u (z, v)F (X n− (u, v))dvdu
Dk (z)
G r−u (z, v)F (X n− (u, v))dvdu − αn (r, z)F (X n (r, z)) dzdr,
”
—
[η(t, s, x, y)](r, z) Ḣ(X n− (r, z)) − Ḣ(X n (r, x))
D
×αn (r, z)F (X n (r, z))dzdr.
1,1
We first estimate the term J3 . Define
V (r, z) = 1ĀMn (r) Ḣ(X n− (r, z))
Z rZ
”
—
G r−u (z, v) F (X n (u, v)) − F (X n− (u, v)) W (dv, du).
×
(4.29)
rn D
Then
1,1
J3 (t, s, x,
y) =
=
Z tZ
[η(t, s, x, y)](r, z)V (r, z)Ẇn (dz, dr)
0 D
[2n X
t/T ]−1
λkn (V )(t, s, x, y),
(4.30)
k=0
where λkn (V )(t, s, x, y) is defined by (4.12). By virtue of the Burkhölder inequality, (A.16) with
κ = 2q − 2q + 1 = 1 and Lemma 4.2,
h
i
h
i
d
E kV (r, ·)kqq
≤ C2−nq(1− 4 ) sup E 1ĀMn (r) kX n (r, ·) − X n− (r, ·)kqq
r∈[0,T ]
≤ C2
−nq(1− d4 +σ)
.
503
(4.31)
Then using Lemma 4.3, there exist θ 0 < 4 − d and θ 0 ≤ 2, θ < 1 −
E
•
1,1
1Ān (t)∩Ān (s) J3 (t, s, x,
M
M


nq

≤ 2 2 E 1Ān
n
M (t)∩Ā M (s)
dq

k=0
nq
2
− 2 (1+2σ− d2 + 1q + γβ
≤ Cn 2 2
)
such that
q˜
y)
[2n X
t/T ]−1
d
4
q 
2
k
2 
|λn (t, s, x, y)|

”
—q
0
|x − y|θ + |t − s|θ 2 .
(4.32)
1,2
Next we turn to estimate the term J3 (t, s, x, y). If we set
V (r, z) =
1ĀMn (r) Ḣ(X n− (r, z))
Z rZ
G r−u (z, v)F (X n− (u, v))W (dv, du),
(4.33)
rn D
then
1,2
J3 (t, s, x,
y) =
[2n X
t/T ]−1 •
λkn (V )(t, s, x,
y) − E
•
λkn (V )(t, s, x,
y)|F (k−1)T
˜˜
2n
k=0
.
(4.34)
Applying the discrete Burkhölder inequality and Jensen’s inequality to conclude that
” 1,2
—
E |J3 (t, s, x, y)|q
q
 n
[2 X
t/T ]−1
•
˜2 2

≤ CE 
λk (V )(t, s, x, y) − E λk (V )(t, s, x, y)|F (k−1)T
n
n
2n
k=0

≤ CE 
[2n X
t/T ]−1

q
2
2
λkn (V )(t, s, x, y) 
k=0

+CE 
[2n X
t/T ]−1
E
•
λkn (V )(t, s, x,
y)|F (k−1)T
2n
k=0
˜
2
q
2
 .
(4.35)
According to a similar proof to that of (4.31), for V (r, z) defined by (4.33), one gets
h
i
sup E kV (r, ·)kqq ≤ C2−nq(1−d/4) .
r∈[0,T ]
Also using Lemma 4.3,
•
E 1Ān (t)∩Ān
M (s)
M
where δ0 = − d2 +
1
q
+
2
γβ
1,2
J3 (t, s, x, y)
q˜
dq
≤ C n 2 2−
nqδ0
2
”
—q
0
|x − y|θ + |t − s|θ 2 ,
> 0. On the other hand, Lemma 4.4 yields that

E Ẇn (z, r)
Z rZ

G r−u (z, v)F (X n− (u, v))W (dv, du)|F rn 
rn D
504
(4.36)
n
= 2 T
−1 d
−d
n π
r nZ
Z
Dk (z)
rn
G r−u (z, v)F (X n− (u, v))dvdu.
This implies that
1,3
J3 (t, s, x, y) ≡ 0.
Since Ḣ and F are bounded, from (2.4), (A.4) and (A.7)–(A.9), it follows that
–Z tZ
™q
1,4
J3 (t, s, x, y)
q
≤ C ndq
[η(t, s, x, y)](r, z) dzdr
0
D

× 2n sup
Z
(r,z)∈O T
≤ Cn
d
2
where 0 < γ0 < 2 −
κ=
1
∞
−
1
p
dq
q
r nZ
rn
Dk (z)
|G r−u (z, v)|dvdu
”
00
0 —q
|t − s|γ + |x − y|γ ,
and 0 < γ00 < 21 (1 − d4 ). From (2.4), Lemma 4.2 and (A.11)–(A.12) with
+ 1 = 1 ∈ [0, 1], we conclude that there exist α ∈]0, 1 −
E
•
(4.37)
d
[
4p
and β ∈]0, 1[ such that
q˜
1,5
1ĀMn (t)∩ĀMn (s) J3 (t, s, x,
y)
h
i
”
—q
≤ C ndq |t − s|α + |x − y|β
sup E 1ĀMn (r) kX n (r, ·) − X n− (r, ·)kqq
r∈[0,T ]
”
—q
≤ C ndq 2−nqσ |t − s|α + |x − y|β .
(4.38)
From the above estimations, it follows that there exist ᾱ ∈]0, min{ 12 −
]0, min{2 −
3d
, 2 − d2 , d2 , ρ}[
p
3d 1
, (1
4p 2
ρ
− d4 ), 4 }[ and β̄ ∈
such that for (t, x), (s, y) ∈ O T ,
h
i
h
iq
q
E 1ĀMn (t)∩ĀMn (s) J31 (t, s, x, y)
≤ C ndq |t − s|ᾱ + |x − y|β̄ .
Now we turn to estimate the term J32 (t, s, x, y).
Replace F (X n (r, z)) by H(X n (r, z)), F (X n− (r, z))
αn (r, z, X n (r, z)) by βn (r, z, X n (r, z)), respectively.
that
h
i
q
E 1ĀMn (t)∩ĀMn (s) J32 (t, s, x, y)
(4.39)
The procedure is similar to that of J31 (t, s, x, y).
by H(X n− (r, z)), W (dz, dr) by Wn (dz, dr), and
Then there exist ᾱ, β̄ presented in (4.39) such
h
iq
≤ C nq |t − s|ᾱ + |x − y|β̄ .
(4.40)
As for the term J33 (t, s, x, y), we have
h
i
q
E 1ĀMn (t)∩ĀMn (s) J33 (t, s, x, y)

q
Z tZ
= E 1ĀM (t)∩ĀM (s)
[η(t, s, x, y)](r, z)Ḣ(X − (r, z))T 3 (r, z)Ẇn (z, r)dzdr 
n
n
n
0
n
D
= Bn1 (t, s, x, y) + Bn2 (t, s, x, y),
(4.41)
where
–
Bn1 (t, s, x,
y) = E 1ĀMn (t)∩ĀMn (s)
Z tZ
0
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))Ẇn (z, r)
D
505

Z Z
™
r
q
−
,
×
∆ v G r−u (z, v)[ f (X n (u, v)) − f (X n (u, v))]dvdu dzdr
rn D
–
Bn2 (t, s, x,
y) = E 1ĀMn (t)∩ĀMn (s)
Z tZ
[η(t, s, x, y)](r, z)Ḣ(X n− (r, z))Ẇn (z, r)
0
D
Z Z

™
r
q
−
×
∆ v G r−u (z, v) f (X n (u, v))dvdu dzdr
.
rn D
By virtue of (A.11) and (A.12) with κ =
and (4.9), we have for α ∈]0, 1 −
d
[
4p
1
∞
− 1p + 1 = 1 − 1p ∈ [0, 1], (A.13) with κ1 =
rn D
≤ Cn 2
1
nq
2
− 1p + 1 = 1
and β ∈]0, 1[,
Bn1 (t, s, x, y)
”
—q
1
≤ C ndq 2 2 nq |t − s|α + |x − y|β

Z rZ

×E 
|∆ v G r−u (·, v)|1ĀMn (u) |X n (u, v) − X n− (u, v)|dvdu
dq
1
p
p



q
h
i
”
—q 1
|t − s|α + |x − y|β 2− 2 nq sup E kX n (r, ·) − X n− (r, ·)kqp
r∈[0,T ]
dq −nqι
≤ Cn 2
”
—q
|t − s|α + |x − y|β .
(4.42)
For r ≤ t, set
V (r, z) :=
Z rZ
rn D
∆ v G r−u (z, v)1ĀMn (u) f (X n− (u, v))dvdu.
(4.43)
Using the B-D-G inequality

Bn2 (t, s, x, y) ≤ E 
[2n X
t/T ]−1
q
2
2
λkn (V )(t, s, x, y)  .
(4.44)
k=0
Since X n− (u, v) = Gu−un (v, X n (un , ·)), it is obvious that
kX n− (u, ·)kq ≤ CkX n (u, ·)kq .
By virtue of (4.7),
h
i
)
−nq( 21 − 3d
4q .
sup E kV (r, ·)kqq ≤ C2
(4.45)
r∈[0,T ]
Again by Lemma 4.3, there exist θ 0 < 4 − d and θ 0 ≤ 2, θ < 1 −
1
0
Bn2 (t, s, x, y) ≤ C n 2 dq 2−nqα
where α0 = −1 −
3d
2q
+
1
q
+
2
.
γβ
d
4
such that
”
0 —q
|t − s|θ + |x − y|θ
,
(4.46)
So that for α ∈]0, 12 (1 − d4 )[ and β ∈]0, (2 − d/2) ∧ 1[,
h
i
—q
0 ”
q
E 1ĀMn (t)∩ĀMn (s) J33 (t, s, x, y)
≤ C ndq 2−nqδ |t − s|α + |x − y|β .
506
(4.47)
with δ0 = ι ∧
κ :=
1
∞
−
1
p
α0
.
2
Finally we consider J34 (t, s, x, y) and J35 (t, s, x, y). From (A.11) and (A.12) with
+ 1 ∈ [0, 1] and (4.8), it follows that for α ∈]0, 1 −
d
[
4p
and β ∈]0, 1[,
h
i
q
E 1ĀMn (t)∩ĀMn (s) J34 (t, s, x, y)
h
i
”
—q
1
sup E 1ĀMn (r) kTn4 (r, ·)kqp
≤ C ndq 2 2 nq |t − s|α + |x − y|β
r∈[0,T ]
≤ C n2dq 2
− 12 nq
”
—q
|t − s|α + |x − y|β .
As for J35 (t, s, x, y), using (A.11) and (A.12) with κ :=
we get for α ∈]0, 1 −
d
[
2p
and β ∈]0, min{4 −
2d
, 1}[,
p
1
∞
−
(4.48)
2
p
+ 1 := 1 −
2
p
∈ [0, 1] and Lemma 4.2,
h
i
q
E 1ĀMn (t)∩ĀMn (s) J35 (t, s, x, y)

q
Z tZ
= E 1ĀMn (t)∩ĀMn (s)
[η(t, s, x, y)](r, z)ρn (r, z)Ẇn (z, r)dzdr 
0

≤ E 1ĀMn (t)∩ĀMn (s)
dq
2
≤ Cn 2
D
q
Z tZ
( 12 − d4 )nq
0
[η(t, s, x, y)](r, z)|X n (r, z) − X n− (r, z)|2 Ẇn (z, r)dzdr 
D
”
—q
|t − s|α + |x − y|β .
When d = 3, we can obtain a more precise estimate than the cases of d = 1, 2, which will be
concluded in Lemma 6.1 of Section 6. Thus we complete the proof of the lemma.
ƒ
5
The proof of (C1)
The condition (C1) presented by Section 3 shall be verified in this section. By Lemma 4.1, to check
(C1), we only need to prove the right hand side of (4.1) approaches 0, when n → ∞. Note that for
each fixed n ∈ N,



 14 j,k (x, t)
; j = 0, 1, . . . , 2n − 1, k ∈ Idn
Æ


4 j,k
forms a CONS of L 2 ([0, T ] × D). Let πn be the orthogonal projection of above basis and for any
mapping g : R → R, define
€
Š
τn g(s) = g (s + T 2−n ) ∧ T , s ∈ [0, T ].
Then for each t ∈ [0, T ] and F-predictable process (ψ(t, x); x ∈ D)0≤t≤T ,
Z tZ
0
ψ(s, y)Wn (d y, ds) =
D
Z TZ
0
” €
Š—
πn τn 1[0,t] (·)ψ(·, ∗) (s, y)W (d y, ds).
D
507
Recall Λn (t, x) defined by (3.2) and we have
Λn (t, x) =
3
X
Λ̃in (t, x),
i=1
where
Λ̃1n (t, x)
Z tZ
G t−s (x, y)H(X n− (s, y))[Wn (d y, ds) − W (d y, ds)],
:=
0
Λ̃2n (t, x) :=
D
Z tZ
0
Λ̃3n (t, x) :=
G t−s (x, y)[H(X n− (s, y)) − H(X n (s, y))]W (d y, ds),
D
Z tZ
0
G t−s (x, y)[H(X n (s, y)) − H(X n− (s, y))]Wn (d y, ds)
D
Z tZ
G t−s (x, y)Ḣ(X n (s, y))
−
0
D
× αn (s, y)F (X n (s, y)) + βn (s, y)H(X n (s, y)) d yds.
We begin with the estimation of the term Λ̃1n (t, x). Note that
Z tZ h
” €
Š—
1
Λ̃n (t, x) =
πn τn 1[0,t] (·)G t−· (x, ∗)H(X n− (·, ∗)) (s, y)
0
D
i
”
—
−πn 1[0,t] (·)G t−· (x, ∗)H(X n− (·, ∗)) (s, y) W (d y, ds),
Z tZ h
”
—
πn 1[0,t] (·)G t−· (x, ∗)H(X n− (·, ∗)) (s, y)
+
0
D
i
−1[0,t] (s)G t−s (x, y)H(X n− (s, y)) W (d y, ds)
1,2
=: Λ̃1,1
n (t, x) + Λ̃n (t, x).
Then by the Hölder inequality and Burkhölder’s inequality, and note that πn is an orthogonal projection of L 2 ([0, t] × D), we have for q ≥ p,
h
i
q
E 1ĀMn (t) Λ̃1,1
(t,
·)
n
p
Z TZ Z –
”
—
≤ C
E 1ĀMn (t)
τn 1[0,t] (s)G t−s (x, y)H(X n− (s, y))
0
D
−1[0,t] (s)G t−s (x,
–
≤
CE 1ĀMn (t)
D
y)H(X n− (s,
™
q
y)) W (d y, ds)
dx
Z TZ h
”
—
τn 1[0,t] (s)G t−s (•, y)H(X n− (s, y))
0
D
i2
−1[0,t] (s)G t−s (•, y)H(X n− (s, y)) d yds
–
=
CE 1ĀMn (t)
Z TZ 0
q
2
™
q
2
€
Š
€
Š
1[0,t] s + T 2−n G t−(s+T 2−n ) (•, y)H X n− (s + T 2−n , y)
D
508
y)H(X n− (s,
−1[0,t] (s)G t−s (•,
–
≤
CE 1ĀMn (t)
×H
€
2
0
X n− (s +
–
×
+CE 1ĀMn (t)

≤
Z

CE 
+CE
T2
−n
(t−T 2−n )+Z
T2
Z
−n
=: C
G t−s (•, y)
2
,
Z
(t−T 2−n )+ D
”
2
G t−s
(•,
—2
y)) d yds
q
2
™
q
2
y)H
2
€
Š
X n− (s, y) d yds
™
q
2
q
2
q
2
(t−T 2−n )+Z



2
G t−s (•, y)
D
X n−
– Z
q
2
—2
G t−(s+T 2−n ) (•, y) − G t−s (•, y) d yds
D
– Z
3
X
™
y)) − H(X n− (s,
t
0
+CE
q
2
D
0
×1ĀMn (s)
—2
G t−(s+T 2−n ) (•, y) − G t−s (•, y)
q
2
(t−T 2−n )+Z
”
q
2
Š
, y) d yds
0
H(X n− (s +
–
”
™
D
Z
+CE 1ĀMn (t)
”
(t−T 2−n )+Z
Z
q
2
2
y)) d yds
€
Š
—2
s + T 2−n , y − X n− (s, y) d yds
t
Z
(t−T 2−n )+ D
2
G t−s
(•,
y)H
2
€
X n− (s,
q
2
™
q
2
Š
y) d yds
q
2
™
q
2
Λ̃1,1,i
(t).
n
i=1
Take the boundedness of the mapping H into account, from (A.8) and (A.9) in Lemma A.2, it follows
that for γ00 < 1 − d4 ,
sup
t∈[0,T ]
”
—
1 00
|Λ̃1,1,1
(t)| + |Λ̃1,1,3
(t)| ≤ C2− 2 γ nq .
n
n
(5.1)
On the other hand, applying (A.16) with κ = 2q − 2q + 1 = 1, the Hölder inequality, Lemma 4.2, and
Lemma 6.1 (in Section 6) to conclude that
q
€
Š
1,1,2
−
−n
−
Λ̃n (t) ≤ C sup E 1ĀMn (s)∩ĀMn (s+T /2n ) X n s + T 2 , · − X n (s, ·)
q
s∈[0,T ]
•
≤ C sup E 1ĀMn (s+T /2n )
(s, y)∈O T
q˜
€
Š
X n− s + T 2−n , y − X n (s + T 2−n , y)
•
q˜
€
Š
+C sup E 1ĀMn (s)∩ĀMn (s+T /2n ) X n s + T 2−n , y − X n (s, y)
(s, y)∈O T
509
i
h
q
+C sup E 1ĀMn (s) X n s, y − X n− (s, y)
(s, y)∈O T
—
”
≤ C 2−ιnq + 2−ξnq + 2−ιnq .
Therefore for q ≥ p >
(5.2)
3d
,
2
h
sup E 1ĀMn (t) Λ̃1,1
n (t, ·)
q
p
t∈[0,T ]
i
→ 0,
n → ∞.
(5.3)
As for the term Λ̃1,2
n (t, x), note that
Z tZ
”
—
πn 1[0,t] (·)G t−· (x, ∗)H(X n− (·, ∗)) (s, y)W (d y, ds)
=
0 D
n
Z Z
[2
Xt] X t
0
j=0 k∈Id
n
14 j,k ( y, s)
4 j,k
D
Z


G t−r (x, z)H(X n− (r, z))dzdr  W (d y, ds).
4 j,k
Then the B-D-G inequality yields that for q ≥ p,
h
i
q
E 1ĀMn (t) Λ̃1,2
(t,
·)
n
p
Z –
Z TZ [2n t]
X X 14 j,k ( y, s)
≤ C
E 1ĀMn (t)
4 j,k
D
0 D j=0 k∈Id
n
Z
−
−
×
G t−r (x, z)H(X n (r, z)) − G t−s (x, y)H(X n (s, y)) dzdr
4 j,k
q
™
×W (d y, ds) dx
Z TZ
–
≤
CE 1ĀMn (t)
14 j,k ( y, s)
4 j,k
D j=0 k∈Id
n
2
2
Z
×
0
n
[2
Xt] X
q
2
–
+CE 1ĀMn (t)
×
™
G t−r (•, z) − G t−s (•, y) H(X n− (r, z))dzdr  d yds
4 j,k
Z
q
2
n
[2
Xt] X
Z TZ
0
D j=0 k∈Id
n
14 j,k ( y, s)
4 j,k
2
2
—
G t−s (•, y) H(X n− (r, z)) − H(X n− (s, y)) dzdr  d yds
q
2
”
4 j,k
q
2
”
—
=: C H n1 (t) + H n2 (t) .
By the boundedness of H and Dini’s theorem, we have for t ∈ [0, T ],
Z TZ [2n t]
X X 14 j,k ( y, s)
1
0 ≤ H n (t) ≤ C
2
4 j,k
0 D j=0 k∈Id
n
510
™
(5.4)
Z
×
2
G t−r (•, z) − G t−s (•, y) dzdr  d yds
4 j,k
Z tZ
=
q
2
2
πn G t−· (•, ∗) (s, y) − G t−s (•, y) d yds
C
0
D
2
→ 0,
0
q
2
πn [G t−· (x, ∗)](s, y) − G t−s (x, y) d yds
C sup
x∈D
q
2
q
2
Z tZ
≤
q
2
D
n → ∞,
follows from the fact that when n → ∞, it holds that
πn G t−· (x, ∗) − G t−· (x, ∗)
L 2 ([0,T ]×D)
→ 0,
for all (t, x) ∈ O T (a compact set in Rd+1 ). Because H is Lipschitzian continuous, using the Hölder
inequality

Z
–
Z TZ [2n t]
X X 14 j,k ( y, s)
2

G t−s
(•, y)dzdr 
H n2 (t) ≤ CE 1ĀMn (t)
2
4 j,k
4 j,k
0 D j=0 k∈Id
n
q
Z

™
2
|H(X n− (r, z)) − H(X n− (s, y))|2 dzdr  d yds
×
4 j,k
q
–
≤ CE 1ĀMn (t)
Z
×
2

Z
Z TZ [2n t]
X X 14 j,k ( y, s)
2

G t−s
(•, y)dzdr 
2
4 j,k
4 j,k
0 D j=0 k∈Id
n
q

™
2
|H(X n− (r, z)) − H(X n− (s, y))|2 dzdr  d yds
4 j,k
≤ C sup
q
2
n
[2
Xt] X
sup
(s, y)∈O T j=0
(z,r)∈4 j,k
k∈Idn
h
i
q
E 1ĀMn (r)∩ĀMn (s) X n− (r, z)) − X n− (s, y)
×14 j,k ( y, s).
(5.5)
Note that
|X n− (r, z) − X n− (s, y)| ≤ |X n− (r, z) − X n (r, z)| + |X n (r, z) − X n (s, y)|
+|X n (s, y) − X n− (s, y)|.
Then from Lemma 4.2, and Lemma 6.1 (in Section 6), it follows that
”
—
H n2 (t) ≤ C 2−ιnq + 2−ξnq + 2−ιnq ≤ C2−λnq ,
,
where λ = min{ι, ξ}. This further implies that for q ≥ p > 3d
2
h
i
q
sup E 1ĀMn (t) Λ̃1,2
n (t, ·) p → 0, n → ∞,
t∈[0,T ]
511
(5.6)
3d
,
2
and hence for q ≥ p >
h
sup E 1ĀMn (t) Λ̃1n (t, ·)
t∈[0,T ]
q
p
i
→ 0,
n → ∞.
(5.7)
Finally, we turn to the estimation of the term Λ̃2n (t, x). In light of the B-D-G inequality and (A.16)
with κ = 2p − 2p + 1 = 1,
h
E 1ĀMn (t) Λ̃2n (t, ·)
–Z t Z
≤ CE 1ĀMn (t)
t
–Z
i
™
q
2
2
(·, y)|H(X n (s, y)) − H(X n− (s, y))|2 d yds
G t−s
D
− d4
(t − s)
≤ CE
≤ C2
0
q
p
p
2
™
kX n (s, ·) − X n− (s, ·)kqp ds
0
−ιnq
.
(5.8)
Observe that
Λ̃3n (t, x) = J3 (t, 0, x, 0),
(5.9)
for the term J3 (t, s, x, y) defined in Lemma 4.5. Then there exists a λ > 0 such that for q ≥ p >
h
sup E 1ĀMn (t) Λ̃3n (t, ·)
t∈[0,T ]
q
p
i
≤ C2−λnq .
Thus we prove that the condition (C1) holds.
6
3d
,
2
(5.10)
ƒ
The proof of (C2)
The aim of this section is to check the validity of the condition (C2) presented in Section 3. Note
,
that for all s < t ∈ [0, T ], we have for q ≥ p > 3d
2
h
i
q
E 1ĀMn (t)∩ĀMn (s) Yn (t, ·) − Yn (s, ·) p
h
i
h
i
q
≤ CE 1ĀMn (t)∩ĀMn (s) X n (t, ·) − X n (s, ·) p + CE 1ĀMn (t)∩ĀMn (s) kX (t, ·) − X (s, ·)kqp .
Observe the forms of Equations (2.2) and (2.3), X is a particular case of X n . Hence in order to prove
that (C2) holds, it suffices to check that for all 0 ≤ s < t ≤ T , there exist q ≥ p and θ > qᾱ (ᾱ is
presented in Theorem 1.1) such that for each n ∈ N,
i
h
q
(C2)0 E 1ĀMn (t)∩ĀMn (s) X n (t, ·) − X n (s, ·) p ≤ C|t − s|1+θ .
From (2.2), it follows that for (s, y), (t, x) ∈ O T ,
X n (t, x) − X n (s, y)
512
= G t ∗ ψ(x) − Gs ∗ ψ( y)
Z TZ
+
[η(s, t, x, y)](r, z) F (X n (r, z))W (dz, dr) + H(X n (r, z)Wn (dz, dr))
0
+
D
Z TZ
0
[η(s, t, x, y)](r, z) K(X n (r, z))ḣ(r, z)
D
”
—
− αn (r, z)(F Ḣ)(X n (r, z)) + βn (r, z)(H Ḣ)(X n (r, z)) dzdr
+
Z TZ
0
D
e t, x, y)](r, z) f (X n (r, z))dzdr.
[η(s,
(6.1)
To prove (C2)0 , we will sharp the estimations in Lemma 4.5 by the following lemma.
Lemma 6.1. It holds that
%
(i) Let q ≥ p > 6. Then for ᾱ ∈]0, min{ 21 (1 − d4 ), 4 }[ and β̄ ∈]0, min{2 − d2 , %}[, there exists a
constant C > 0 such that
h
iq
h
iq
E 1ĀMn (s)∩ĀMn (t) X n (t, x) − X n (s, y)
≤ C ndq 2−nqξ |t − s|ᾱ + |x − y|β̄ ,
(6.2)
for some ξ ∈]0, min{ 21 , σ, 21 (−1 −
respectively.
3d
2q
+
1
q
+
2
)[
γβ
with σ, γ, β defined in Lemma 4.2 and Lemma 4.3,
%
(ii) Let q ≥ p ≥ 4. Then for ᾱ ∈]0, min{ 21 (1 − d4 ), 4 }[ there exists a C > 0 such that
h
E 1ĀMn (s)∩ĀMn (t) X n (t, ·) − X n (s, ·)
q
p
i
≤ C ndq 2−nqξ |t − s|ᾱq .
(6.3)
where ξ is the same as in (i).
Proof. Recall the entire proof of Lemma 4.5. We only need to re-estimate J31 (t, s, x, y), J32 (t, s, x, y)
and J35 (t, s, x, y) given in Lemma 4.5. To improve the estimation of J31 (t, s, x, y), we only need to
1,4
estimate the term J3 (t, s, x, y). Note that, for (s, y), (t, x) ∈ O T ,
1,4
J3 (t, s, x, y)
Z tZ
≤
n
[η(s, t, x, y)](r, z)Ḣ(X n− (r, z))
C2 2 nd
0
r nZ
Z
×
n
Z tZ
0
Z
[η(s, t, x, y)](r, z)Ḣ(X n− (r, z))
D
r nZ

G r−u (z, v) F (X n (u, v)) − F (X n (r, z)) dvdu dzdr
×
rn
=:

—
G r−u (z, v) F (X n− (u, v)) − F (X n (u, v)) dvdu dzdr
”
Dk (z)
rn
+C2 2 nd
D
Dk (z)
Kn1 (t, s, x,
y) + Kn2 (t, s, x, y).
513
(6.4)
Using a similar argument to that of (4.42), we get for α, β ∈ (0, 1),
h
i
”
—q
1
q
E 1ĀMn (t)∩ĀMn (s) Kn1 (t, s, x, y)
≤ C ndq 2−nq( 2 +ι) |t − s|α + |x − y|β .
(6.5)
In order to estimate Kn2 (t, s, x, y), we apply the Hölder inequality w.r.t the measures d y and
G r−u (z, v)dvdu, respectively. Then (A.11), (A.12) and Lemma 4.5 jointly imply that, for α, β ∈]0, 1[,
h
i
q
E 1ĀMn (t)∩ĀMn (s) Kn2 (t, s, x, y)
–
Z t Z r nZ
—q
”
nq
α
β
dq
E 1ĀMn (t)∩ĀMn (s)
≤ C2 2 n
|t − s| + |x − y|
G r−u (·, v)
0
q
× F (X n (u, v)) − F (X n (r, ·)) dvdu
rn
Dk (·)
™
dr
q
”
—q
nq
≤ C2 2 n2dq 2−(q−1)n |t − s|α + |x − y|β
–Z t Z r Z
™
h
iq
n
ᾱ
β̄
G r−u (z, v) |r − u| + |z − v|
dvdu dr
×
0
rn
D (z)
”k
—q
≤ C2
n
|t − s|α + |x − y|β
‚
–Z t Z r Z n
− d4
β̄q
|r − u| |v| exp −C
×
n−nq/2 2dq
0
+2
rn
−nᾱq
Rd
G r−u (z, v) dvdu dr
Œ
|r − u|1/3
™
”
—q
≤ C2n−nq/2 n2dq |t − s|α + |x − y|β
Z t Z

0
rn
β̄q
4

|r − u| du + 2−n−ᾱqn  dr
rn
”
—q
|t − s|α + |x − y|β ,
−nq( 12 +α0 ) 2dq
≤ C2
|v|4/3
n
(6.6)
β̄
where α0 = min{ᾱ, 4 }. The estimation of J32 (t, s, x, y) is similar to that of J31 (t, s, x, y).
Finally, we improve the estimation of the term J35 (t, s, x, y). Using the decomposition of X n (r, z) −
X n− (r, z),
h
i
q
E 1ĀMn (t)∩ĀMn (s) J35 (t, s, x, y)

q
Z Z
t
≤
:=
[η(t, s, x, y)](r, z)|X n (r, z) − X n− (r, z)|2 Ẇn (z, r)dzdr 
E 1ĀM (t)∩ĀM (s)
n
4 X
4
X
n
0
D
Ri,n j (t, s, x, y),
(6.7)
i=1 j=1
where
Ri,n j (t, s, x, y)

= E 1ĀMn (t)∩ĀMn (s)
q
Z tZ
0
[η(t, s, x, y)](r, z)Tni (r, z)Tnj (r, z)Ẇn (z, r)dzdr  .
D
514
Using (A.11) and (A.12) with κ :=
2d
]0, min{4 − p , 1}[, and γ > max
1
∞
−
2
p
+1 = 1−
2
p
∈ [0, 1], we get for α ∈]0, 1 −
1
,
d
β ,
1
1
,
1+ d4 (κ−1)−α 1+ d (κ−1)−
4
4
dq
nq
Ri,n j (t, s, x, y) ≤ C n 2 2 2
d
[,
2p
β ∈
1− 4 β
i1
h
”
—q
γ
|t − s|α + |x − y|β
sup E 1ĀMn (t)∩ĀMn (s) kTni (r, ·)kqγ
p
r∈[0,t]
h
i1
γ
× sup E 1ĀMn (t)∩ĀMn (s) kTnj (r, ·)kqγ
.
p
r∈[0,t]
Thanks to (4.5)–(4.8), we deduce that there exists δ > 0 such that for all (i, j) with either i or j not
belonging to {1, 4},
”
—q
(6.8)
Ri,n j (t, s, x, y) ≤ C2−nqδ |t − s|α + |x − y|β .
i, j
Next we improve the estimate of R n when i, j ∈ {1, 4}. We first deal with R1,4
n . By the Hölder
1
1
inequality with 2 + 2 = 1, for p > 6,
nq
dq
Ri,n j (t, s, x, y) ≤ C n 2 2 2
h
i1
”
—q
qγ γ
|t − s|α + |x − y|β
sup E 1ĀMn (t)∩ĀMn (s) kTni (r, ·)k2p
r∈[0,t]
h
i1
qγ γ
j
× sup E 1ĀMn (t)∩ĀMn (s) kTn (r, ·)k2p
r∈[0,t]
≤ C2
)
−nq( 12 − 3d
4p
”
—q
|t − s|α + |x − y|β .
(6.9)
1,4
1,1
The estimation of the term R4,4
n is similar to that of R n . To improve the estimation of R n (t, s, x, y),
we introduce the process
Z rZ
G r−u (z, v)F (X n− (u, v))W (dv, du).
Tn1− (r, z) =
(6.10)
rn D
Then
R1,1
n (t, s, x, y)

≤
E 1ĀMn (t)∩ĀMn (s)

+E 1ĀMn (t)∩ĀMn (s)
:=
R1,1,1
(t, s, x,
n
q
”
—
1
2
1−
2
[η(t, s, x, y)](r, z) (Tn (r, z)) − (Tn (r, z)) Ẇn (z, r)dzdr 
Z tZ
0
D
q
Z tZ
0
[η(t, s, x, y)](r, z)(Tn1− (r, z))2 Ẇn (z, r)dzdr 
D
y) + R1,1,2
(t, s, x,
n
(6.11)
y).
Similarly to the proof of the term J 1,1 (t, s, x, y) in (4.32), we conclude that there exists a δ0 :=
2
2 − d + 1q + γβ
> 0 with γβ ∈]1, 2[ such that
1
R1,1,1
(t, s, x, y) ≤ C2− 2 nqδ
n
0
”
—q
|t − s|α + |x − y|β .
515
(6.12)
As for the term R1,1,2
(t, s, x, y), we have
n
R1,1,2
(t, s, x, y)
n
–
Z tZ
≤
E 1ĀMn (t)∩ĀMn (s)
0
[η(t, s, x, y)](r, z)
D
q
™
”
—
× (Tn1− (r, z))2 Ẇn (z, r) − E[(Tn1− (r, z))2 Ẇn (z, r)|F rn ] dzdr

+E 1ĀMn (t)∩ĀMn (s)
:=
Zn1 (t, s, x,
y) +
q
Z tZ
0
[η(t, s, x, y)](r, z)E[(Tn1− (r, z))2 Ẇn (z, r)|F rn ]dzdr 
D
Zn2 (t, s, x,
(6.13)
y).
1,2
1
> 0 with αβ ∈]1, 2[
Similarly to the proof of J3 (t, s, x, y) in (4.36), there exists a δ := − d2 + 1q + αβ
such that
”
—q
1
(6.14)
Zn1 (t, s, x, y) ≤ C2− 2 nqδ |t − s|α + |x − y|β .
Since Ẇn (z, r) is F r n -measurable and we have

Z rZ

G r−u (z, v)F (X n− (u, v))W (dv, du)
E Ẇn (z, r)
D
rn

2

F rn 


Z Z
r
2
G r−u
(z, v)F 2 (X n− (u, v))dvdu F rn  |F rn  = 0,
= E Ẇn (z, r)E 

rn
D
and
Z Z

r
E Ẇn (z, r) 
G r−u (z, v)F (X n− (u, v))W (dv, du)
–
rn D
Z

r nZ
G r−u (z, v)F (X n− (u, v))W (dv, du) |F rn
×
rn
™
= 0.
D
Applying Itô rule to conclude that
”
—
E (Tn1− (r, z))2 Ẇn (z, r)|F rn

– Z r Z w Z
n
1
1

≤ C n 2 d 2 2 nE
G r−u (z, v)F (X − (u, v))W (dv, du)
n
rn
Z
×
+C n
rn
D

Dk (z)
1
d
2
2
G r−u (z, v)F (X n− (u, v))dv  dw
1
n
2
–Z
™
F rn
rn
W (Dk (z) × (rn , w])
E
rn
™
–Z
2
G r−u
(z, v)F 2 (X n− (u, v))dv
×
D
516
™
dw F rn
Zn2,1 (t, s, x, y) + Zn2,2 (t, s, x, y).
:=
(6.15)
However the Fubini’s theorem yields that
”
q—
E Zn2,1 (t, s, x, y)
 Z Z Z

q
w
rn
−

≤ E
G r−u (z, v)F (X (u, v))W (dv, du) dw 
n
rn
3
rn
D
d
≤ C2−nq( 2 − 8 ) .
R
2
Note that the integral D G r−u
(z, v)F 2 (X n− (u, v))dv is Fsn -measurable. Hence
(6.16)
Zn2,2 (t, s, x, y) = 0.
From (A.11) and (A.12) with κ :=
2d
, 1}[,
p
1
∞
−
1
p
+ 1, it follows that for α ∈]0, 1 −
d
[
2p
and β ∈]0, min{4 −
—q
d ”
Zn2 (t, s, x, y) ≤ C2−nq(1− 8 ) |t − s|α + |x − y|β .
(6.17)
Using the estimations (6.11)–(6.14) and (6.17), we conclude that there exists a δ > 0 such that
”
—
−nqδ
α
β q
R1,1
(t,
s,
x,
y)
≤
C2
|t
−
s|
+
|x
−
y|
.
(6.18)
n
Thus we complete the proof of the lemma.
ƒ
Now we prove the skeleton process S(h) to Equation (1.3) is continuous for all h ∈ H .
Lemma 6.2. Let α ∈]0, 1[ and q ≥ 4. Then for each a ≥ 0, there exists a C > 0 such that when
kh1 kH ∨ kh2 kH ≤ a,
S(h1 ) − S(h2 )
α,q
≤ C h1 − h2
H
.
Proof. A similar proof to that of Theorem 1.4 in [6] yields that, for q ≥ 4,
sup
sup kS(h)(t, ·)kq = Ca < ∞.
h∈{h∈H ;khkH ≤a} t∈[0,T ]
Recall the skeleton equation (1.3) and we have
=
S(h2 )(t, x) − S(h1 )(t, x)
Z tZ
∆G t−s (x, y) f (S(h2 )(s, y)) − f (S(h1 )(s, y)) d yds
0
+
D
Z tZ
0
+
G t−s (x, y)σ(S(h2 )(s, y)) h2 (s, y) − h1 (s, y) d yds
D
Z tZ
0
G t−s (x, y) σ(S(h2 )(s, y)) − σ(S(h1 )(s, y)) h1 (s, y)d yds
D
517
3
X
=:
R i (t, x)
i=1
Note that u(s, ·), v(s, ·) ∈ L q (D) for each s ∈ [0, T ], we have for ρ = q/3,
ku(s, ·) − v(s, ·)kρ
2
≤ Cku(s, ·) − v(s, ·)kq ,
—
”
≤ Cku(s, ·) − v(s, ·)kq ku(s, ·)kq + kv(s, ·)kq
2
ku (s, ·) − v (s, ·)kρ
ku3 (s, ·) − v 3 (s, ·)kρ
≤ Cku(s, ·) − v(s, ·)kq
i
h
× ku(s, ·)k2q + kv(s, ·)k2q + ku(s, ·)kq kv(s, ·)kq .
Then by virtue of (A.13) with κ =
R1 (t, ·)
Z
≤
q
1
q
−
3
q
t
+1=1−
d
− 12 − 2q
(t − s)
2
q
∈ [0, 1],
kS(h2 )(s, y) − S(h1 )(s, y)kq ds.
0
On the other hand, under (H1), the Schwarz’s inequality implies that
R2 (t, ·)
Z tZ
q
≤ C h2 − h1
H
0
[G t−s (·, y)]2 d yds
D
Again by the Schwarz’s inequality and (A.10) with κ =
kR3 (t, ·)kq ≤ C
Z
t
≤ Ckh2 − h1 kH .
q
1
q
−
1
q
+ 1 = 1 ∈ [0, 1],
d
(t − s)− 4 kS(h2 )(s, y) − S(h1 )(s, y)kq ds.
0
Hence the Gronwall’s lemma yields that
kS(h2 ) − S(h1 )kq ≤ Ckh2 − h1 kH .
While for (t, x) ∈ O T and t 0 ∈ [0, T ],
S(h2 )(t, x) − S(h1 )(t, x) − S(h2 )(t 0 , x) + S(h1 )(t 0 , x)
Z tZ
=
[η̃(t, t 0 , x, x)](s, y) f (S(h2 )(s, y)) − f (S(h1 )(s, y)) d yds
0
+
D
Z tZ
0
+
[η(t, t 0 , x, x)(s, y)σ(S(h2 )(s, y)) h2 (s, y) − h1 (s, y) d yds
D
Z tZ
0
[η(t, t 0 , x, x)](s, y) σ(S(h2 )(s, y)) − σ(S(h1 )(s, y)) h1 (s, y)d yds,
D
q
and then by (A.11) and (A.14) with ρ = q and ρ = 2 , we get for α ∈]0, 12 [,
S(h2 )(t, ·) − S(h1 )(t, ·) − S(h2 )(t 0 , ·) + S(h1 )(t 0 , ·)
α,q
≤ Ckh2 − h1 kH |t − t 0 |α ,
where the Hölder norm k · kα,q := k · kα,q,∞ , which is defined in Section 3. Thus we complete the
proof of the lemma.
ƒ
518
Now we are at the position to prove Theorem 1.1.
Proof of Theorem 1.1. We adopt the method used in Theorem 2.1 of [1]. We only provide a
sketch for this proof. For parsimony, we only prove the part (b) of Theorem 1.1, since the proof of
the part (a) is similar. Given h ∈ H b , let X nh be the solution to Equation (2.2) with h = F = K = 0
and H = σ. Define H n : Ω → H b by
Ḣ n (s, y) = Ẇn ( y, s) − σ̇(X n (s, y))βn (s, y).
By virtue of the uniqueness of the solution to (2.2), we have X n = S(H n ). Moreover, Proposition 3.1
yields that X n → u in probability in C ᾱ,0 ([0, T ], L p (D)). Using Proposition 2.1-(i) in [13], it holds
that
supp (P ◦ u−1 ) ⊂ S(H b ).
On the other hand, we fix h ∈ H b and let X n be the solution to (2.2) with F = σ, H = −σ and
K = σ. Set
K̇n (s, y) := ḣ(s, y) − σ̇(X n (s, y))(αn (s, y) − βn (s, y)),
where αn , βn are defined in (2.3). Let Γhn : Ω → Ω and
Γhn (ω) = ω − ωn + Kn (ω).
Then the Girsanov’s theorem implies that P ◦ (Γhn )−1 P. Note that X n = u ◦ Γhn and X = S(h) and
then Proposition 3.1 concludes that u ◦ Γhn converges in probability to S(h) in C ᾱ,0 ([0, T ]; L p (D)).
By Proposition 2.2-(ii) in [13],
S(H b ) ⊂ supp (P ◦ u−1 ).
Next we have to check that S(H b ) = S(H ). Since H b is dense in H , it suffices to check that for any
%
0 < M < ∞ and ᾱ ∈]0, min{ 21 (1 − d4 ), 4 }[, there exists a C > 0 such that
kS(h2 ) − S(h1 )kᾱ,p ≤ Ckh1 − h2 kH ,
(6.19)
when kh1 kH ∨ kh2 kH < M . Hence the desired result follows from Lemma 6.2. Thus we finish the
entire proof of Theorem 1.1.
ƒ
Acknowledgments. We thank the anonymous referee and the Associate Editor for the valuable
comments and suggestions on the earlier version of this paper. In addition, K. Shi would like to
thank Prof. Zenghu Li for his stimulating discussions.
Appendix
In this section, we present some new estimates on the Green kernel G t (·, ∗) corresponding to the operator ∂ /∂ t + ∆2 with
the homogeneous Neumann’s boundary condition in (1.1). As in Da Prato and Debussche [8], the Green kernel admits
the following expansion.
519
Let A = −∆ be defined on D(A) = {u ∈ H 2 (D); ∂ u/∂ n = 0, on ∂ D} and {Θk }k∈Nd be the basis of eigenfunctions of A in
L 2 (D), which can be written as
Θk (x) =
d
Y
θki (x i ),
(A.1)
i=1
where k = (k1 , . . . , kd ) ∈ Nd , x = (x 1 , . . . , x d ) ∈ D and
(
Æ
θki (x i ) = π2 cos(ki x i ), ki 6= 0,
θ0 (x i ) = p1π ,
ki = 0,
(A.2)
Pd
for i = 1, . . . , d. Here {λk = i=1 ki2 }k∈Nd are the eigenvalues corresponding to the eigenfunctions. Hence the Green
kernel G on [0, ∞) × D can be expressed as
X
2
G t (x, y) =
e−λk t Θk (x)Θk ( y),
(A.3)
k∈Nd
with (t, x, y) ∈ [0, ∞) × D2 .
The following estimates concerning the Green kernel G are quoted from [6].
Lemma A.1 There exist C, K > 0 such that for t ∈]0, T ] and x, y ∈ D,
G t (x, y)
∆ y G t (x, y)
∂ G t (x, y)
∂t
K
≤
d
4
exp
−C
|x − y| 3
≤
t
d+2
4
K
≤
t
d+4
4
,
1
4
exp
−C
|x − y| 3
−C
!
1
,
(A.5)
.
(A.6)
|t| 3
4
exp
(A.4)
|t| 3
t4
K
!
|x − y| 3
!
1
|t| 3
Lemma A.2 For γ0 < 4 − d and γ0 ≤ 2, γ00 < 1 − d4 , there exists C > 0 such that for 0 ≤ s < t ≤ T and x, z ∈ D,
Z tZ
0
≤
C|x − z|γ ,
|G t−u (x, y) − Gs−u (x, y)|2 d ydu
≤
C|t − s|γ ,
(A.7)
D
Z sZ
0
0
|G t−u (x, y) − G t−u (z, y)|2 d ydu
00
(A.8)
D
Z tZ
s
|G t−u (x, y)|2 d ydu
≤
00
C|t − s|γ .
(A.9)
D
The proofs of the following lemma are similar to that of Lemma 3.1 in Gyöngy [11], which further improve the estimates
in [6].
Lemma A.3 Assume that ρ ∈ [1, ∞[, p ∈ [ρ, ∞[, γ > 1 and κ :=
0 ≤ t 0 ≤ t ≤ T and x ∈ D, define
Z tZ
1
p
−
1
ρ
+ 1 ∈ [0, 1]. For v ∈ L γ ([0, T ], L ρ (D)),
H(t − r, x, y)v(r, y)d ydr.
J(v)(t 0 , t, x) :=
t0
D
Then J is a bounded operator from L γ ([0, T ], L ρ (D)) into L ∞ ([0, T ], L p (D)) such that the following conclusions hold.
(a) Let H(t, x, y) = G t (x, y). Then there exists C > 0 such that
520
(1) For all γ >
1
,
1+ d4 (κ−1)
t
Z
J(v)(t 0 , t, ∗)
p
≤
d
(t − r) 4 (κ−1) kV (r, ∗)kρ dr
C
t0
≤
(2) Let α ∈]0, 1 + d4 (κ − 1)[. For all γ >
1+ d4 (κ−1)− γ1
Ct
kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) .
(A.10)
1
,
1+ d4 (κ−1)−α
kJ(v)(0, t, ∗) − J(v)(0, s, ∗)k p ≤ C|t − s|α kv(·, ∗)k L γ ([0,T ],L ρ (D)) ,
(A.11)
with 0 ≤ s ≤ t ≤ T .
(3) Let β ∈]0, min{4 − (1 − κ)d, 1}[. For all γ > max
1
β
1+ d4 (κ−1)− 4
1
1− d4 β
,
,
kJ(v)(0, t, ∗) − J(v)(0, t, ∗ + z)k p ≤ C|z|β kv(·, ∗)k L γ ([0,T ],L ρ (D)) ,
(A.12)
with t ∈ [0, T ], where we set J(v)(t, y) := 0 when y ∈ D c .
(b) If H(t, x, y) = ∆ y G t (x, y) (if d = 3, we also need κ−1 < 3; if d = 2, κ−1 6= ∞), there exists C > 0 such that
(1) For all γ >
1
1 d
+ (κ−1)
2 4
,
t
Z
kJ(v)(t 0 , t, ∗)k p
≤
1
d
(t − r)− 2 + 4 (κ−1) kv(r, ∗)kρ dr
C
t0
1 d
+ (κ−1)− γ1
2 4
kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) .
(A.13)
kJ(v)(0, t, ∗) − J(v)(0, s, ∗)k p ≤ C|t − s|α kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) ,
(A.14)
≤
(2) Let α ∈]0, 12 + d4 (κ − 1)[. For all γ >
1
1 d
+ (κ−1)−α
2 4
Ct
,
with 0 ≤ s ≤ t ≤ T .
(3) Let β ∈]0, min{2 − (1 − κ)d, d2 , 1}[. For all γ > max
1
,
β
1 d
+ (κ−1)− 4
2 4
1
1 d
− β
2 4
,
kJ(v)(0, t, ∗) − J(v)(0, t, ∗ + z)k p ≤ C|z|β kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) ,
(A.15)
with t ∈ [0, T ], where we set J(v)(t, y) := 0 when y ∈ D .
c
(c). If H(t, x, y) = G 2 (t, x, y) (if d = 3, we also need κ−1 < 32 ; if d = 2, κ−1 6= ∞), there is a constant C > 0 such that
(1) For all γ >
1
,
1+ d4 (κ−2)
t
Z
kJ(v)(t 0 , t, ·)k p
≤
d
(t − r) 4 (κ−2) kV (r, ∗)kρ dr
C
t0
≤
1+ d4 (κ−2)− γ1
kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) .
(A.16)
kJ(v)(0, t, ·) − J(v)(0, s, ·)k p ≤ C|t − s|α kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) ,
(A.17)
(2) Let α ∈]0, 1 + d4 (κ − 2)[, for all γ >
Ct
1
,
1+ d4 (κ−2)−α
for all 0 ≤ s ≤ t ≤ T .
(3) Let β ∈]0, min{4 − (2 − κ)d, 4−d
, 1}[, for all γ > max
d
1
β
1+ d4 (κ−2)− 4
,
1
1− d4 (β+1)
,
kJ(v)(0, t, ·) − J(v)(0, t, · + z)k p ≤ C|z|β kv(·, ∗)k L γ ([t 0 ,T ],L ρ (D)) ,
for all t ∈ [0, T ], where we set J(v)(t, y) := 0 when y ∈ D .
c
521
(A.18)
Proof. We only prove the case (a), since the proofs of the cases (b) and (c) are similar. We first treat (A.10). By (A.4)
and the Minkowski’s inequality
Z t
kJ(v)(t 0 , t, ∗)k p
≤
G(t − r, ∗, y)v(r, y)d y
p
dr
t0
t
Z
≤
–
|t − r|−d/4
C
‚
exp −C
t0
Let κ =
1
p
−
1
ρ
Œ
| · |4/3
™
∗ |v(r, ·)| (·)
|t − r|1/3
dr.
p
+ 1 ∈ [0, 1]. From the Young’s inequality and Hölder’s inequality, it follows that, for γ >
t
Z
kJ(v)(t 0 , t, ∗)k p ≤ C
d
(κ−1)
4
|t − r|
kv(r, ∗)kρ dr ≤ C t
Z
1+ d4 (κ−1)− γ1
1/γ
t
kv(r, ∗)kγρ dr 

t0
1
,
1+ d4 (κ−1)
,
t0
where we have used the equality
4
Z
−C
exp
|x| 3
Rd
t
!
d
dx = K t 4 ,
1
3
(A.19)
for some constant K > 0. This proves (A.10). Next we turn to the proof of (A.11). For any 0 ≤ s ≤ t ≤ T ,
kJ(v)(0, t, ∗) − J(v)(0, s, ∗)k p ≤ A1 + A2 ,
where
Z sZ
=
A1
0
,
D
Z tZ
=
A2
G t−r (∗, y) − Gs−r (∗, y) v(r, y)d ydr
s
p
G t−r (∗, y)v(r, y)d ydr
D
.
p
Note that for α ∈]0, 1[ and h1 , h2 ∈ R, it holds that
—
”
|h1 − h2 | ≤ |h1 − h2 |α |h1 |1−α + |h2 |1−α .
Then by the Mean-Value theorem, for θ ∈ [s, t],
s
Z
A1
Z
|G t−r (∗, y) − Gs−r (∗, y)|v(r, y)d y
≤
0
≤
D
C|t − s|
α
s
Z
Z
0
≤
( d4 +1)α
(θ − r)
D
|t − s|α
Z
s
€ ε Šε
`
|t − s|α
0
Apply the Young’s inequality for κ =
1
p
−
A1 ≤ |t − s|α
1
ρ
Z
s
D
s
Z
1
3
v(r, y)
d
(s − r) 4 (1−α)
dy
1
|θ − r| 3 ε
4
d
1
D
!
(s − r)α(1+ 4 )+ 4 (1−α) | ∗ − y| 3 ε
1
α+ d4 − 13 ε
(s − r)
4
| ∗ − y| 3 ε
–Z
1
4
| y|− 3κ ε d y
1
(s − r)α+ 4 − 3 ε
D
522
dr
v(r, y)d y
p
dr.
v(r, y)d y
p
+ 1 ∈ [0, 1] to conclude that
d
0
|θ − r|
1
d
Z
−C
exp
| ∗ − y| 3
. Then
Z
0
≤
4
1
For any ` ∈ R+ and ε > 0, we have exp{−`} ≤
A1
dr
p
™κ
kv(r, ∗)kρ dr.
dr.
p
3κd
.
4
[ be sufficiently close to
Let ε ∈]0, 3κd
4
A1 ≤ |t − s|
α
s
Z
1
α+ d4 − 13 ε
(s − r)
0
As for A2 , if γ >
1
d
(κ−1)+1
4
1
,
1+ d4 (κ−1)−α
For α ∈]0, 1 + d4 (κ − 1)[ and γ >
kv(r, ∗)kρ dr ≤ C|t − s|
α
we have
s
–Z
kv(r, ∗)kγρ dr
™1
γ
.
(A.20)
0
, then
Z
t
Z
G t−r (∗, y)v(r, y)d y
≤
A2
s
≤
C|t − s|
s
p
1+ d4 (κ−1)− γ1
d
|t − r| 4 (κ−1) kv(r, ∗)kρ dr
dr ≤ C
D
–Z
t
Z
t
kv(r, ∗)kγρ dr
™ γ1
.
(A.21)
0
The estimates (A.20) and (A.21) imply (A.11). Finally we prove (A.12). Note that
kJ(v)(t, ∗) − J(v)(t, ∗ + z)k p ≤ B1 + B2 + R,
where R admits the similar expression as that on the right hand side of (A.10) and
Z t –Z
=
B1
0
D
1{|x− y|≤|z|} |G t−r (x + z, y) − G t−r (z, y)|v(r, y)d y
Z
1{|x− y|>|z|} |G t−r (x + z, y) − G t−r (z, y)|v(r, y)d y
1{x+z∈D}
0
™1
p
p
dx
dr,
D
Z t –Z
=
B2
Z
1{x+z∈D}
D
™1/p
p
dx
d r.
D
By (A.4),
Z t –Z Z
≤
B1
D
+ exp
t
Z
≤
B1
C
1
r1
Z
B1
≤
1
r2
+
C
C|z|β
(t − r)
Z t
C|z|β
d
4
4
−C
0
|z|
β
–Z
dκ
.
r1
|x − y| 3
1
+ exp
1
4
−C
|x − y| 3
kv(r, ∗)kγρ dr
1
1− d4 β
B2
=
|t − r|
1
3
+ exp
−C
t
Z
|x + z − y| 3
|t − r|
1
d
0
(t − r) 4 β
™κ
! r2
κ
1
3
r2
dy
kv(r, ∗)kρ dr
kv(r, ∗)kρ dr
,
Z
D
kv(r, ·)kρ dr.
™ γ1
(A.22)
1{|x− y|>|z|} |G t−r (x + z, y) − G t−r (z, y)|β
1{x+z∈D}
0
dy
1
. Using the Mean-Value theorem, for β < 1 and φ ∈ [x, x + z],
Z t –Z
™κ
κ−1
|t − r| 3
0
for γ >
!
4
|t − r| 4 (1−β) kv(r, ∗)kρ dr ≤ C|z|β
t
|x + z − y| 3
!
d
d
4
−C
Then
|z| exp
4
!
|t − r| 3
0
(t − r)
–Z
D
|z| = 1 and set β =
1
!
+ 1 ∈ [0, 1] further yields that
exp
(t − r) 4
0
≤
1
ρ
–Z
1
t
0
≤
−
d
0
Let r1 , r2 ≥ 1 satisfy
1
p
4
|x − y| 3
1{|x− y|≤|z|} exp −C
1
(t − r)d/4
|t − r| 3
!
™
4
p
1/p
|x + z − y| 3
−C
|v(r,
y)|d
y
dx
dr.
1
|t − r| 3
C
0
The Young’s inequality for κ =
1
D
×|G t−r (x + z, y) − G t−r (z, y)|
1−β
™1
p
|v(r, y)|d y
523
p
dx
dr
Z t –Z Z
≤
1{|x− y|>|z|}
C
‚
1
β(d+1)
4
exp −C
|φ − y|4/3
|t − r|1/3
|t − r|
p ™ 1p
d
×|t − r| 4 (1−β) |v(r, y)|d y dx dr
0
≤
C|z|
β
D
D
t
Z
|t − r|
0
4
–Z Z
1
1{|x− y|>|z|} exp
β
d
+4
4
D
Œ
−C
D
|φ − y| 3
|t − r|
|z|β
!
|v(r, y)|d y
1
3
™1
p
p
dx
dr.
Note that if |x − y| > |z| and |x| ≤ |φ| ≤ |x + z|, then
1
1
|φ − y| ≤ p |x − z + y|, or |φ − y| ≤ p |x + z + y|.
2
2
1
1
,
Hence for β ∈]0, min{4 − (1 − κ)d, 1}[ and γ > max
d
β ,
d
1+ 4 (κ−1)− 4
β
Z
t
B2 ≤ C|z|
1
β
–Z
kv(r, ∗)kρ dr ≤ C|z|
β
d
|t − r| 4 (1−κ)+ 4
0
1− 4 β
t
kv(r, ∗)kγρ dr
™ γ1
.
(A.23)
0
This further yields (A.12).
ƒ
Lemma A.4 Let 1 < γ < 2 and ` denote the positive integer such that `2−n ≤ t ≤ (` + 1)2−n . Then for θ 0 < 4 − d and
θ 0 ≤ 2, θ < 1 − d4 , there exists C > 0 such that for 0 ≤ s < t ≤ T and x, z ∈ D,

Z
X̀ 

k=0
(k+1)T
2n
2
γ
Z
kT
2n
D
0

−n( 2 −1)
|G t−r (x, z) − G t−r ( y, z)|γ dzdr  ≤ C2 γ |x − y|θ
(A.24)
and

Z
X̀ 

k=0
(k+1)T
2n
kT
2n
2
γ
Z
D

−n( 2 −1)
|G t−r (x, z) − Gs−r (x, z)|γ dzdr  ≤ C2 γ |t − s|θ .
(A.25)
Proof. We only prove (A.24), since the proof of (A.25) is very similar. By (A.8), (A.9) and the Hölder inequality, we
have for θ 0 < 4 − d and θ 0 ≤ 2,

Z
X̀ 

k=0
(k+1)T
2n
2
γ
Z
kT
2n
D

≤
−n( γ2 −1)
C2
X̀


|G t−r (x, z) − G t−r ( y, z)|γ dzdr 
kT
2n
k=0
−n( γ2 −1)
≤
C2
≤
C2
‚Z tZ
0
−n( γ2 −1)
(k+1)T
2n
Z

Z
D

|G t−r (x, z) − G t−r ( y, z)|2 dzdr 
Œ
|G t−r (x, z) − G t−r ( y, z)|2 dzdr
D
0
|x − y|θ .
(A.26)
Thus we finish the proof of the lemma.
ƒ
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