Differential equations driven by Hölder
continuous functions of order greater than 1/2
Yaozhong Hu and David Nualart
Department of Mathematics, University of Kansas, 405 Snow Hall, Lawrence,
Kansas 66045-2142. Y. Hu is supported in part by the National Science
Foundation under Grant No. DMS0204613 and DMS0504783.
Summary. We derive estimates for the solutions to differential equations driven
by a Hölder continuous function of order β > 1/2. As an application we deduce the
existence of moments for the solutions to stochastic differential equations driven by
a fractional Brownian motion with Hurst parameter H > 12 .
1 Introduction
We are interested in the solutions of differential equations on Rd of the form
Z t
xt = x0 +
f (xr )dyr ,
(1)
0
where the driving force y : [0, ∞) → Rm is a Hölder continuous function of
order β > 1/2. If the function f : Rd → Rmd has bounded partial derivatives
which are Hölder continuous of order λ > β1 − 1, then there is a unique
solution x : Rd → R, which has bounded β1 -variation on any finite interval.
These results have been proved by Lyons in [2] using the p-variation norm
and the technique introduced by Young in [8]. The integral appearing in (1)
is a Riemann-Stieltjes integral.
In [9] Zähle has introduced a generalized Stieltjes integral using the techniques of fractional calculus. This integral is expressed in terms of fractional derivative operators and it coincides with the Riemann-Stieltjes integral
RT
f dg, when the functions f and g are Hölder continuous of orders λ and µ,
0
respectively and λ + µ > 1 (see Proposition 1 below). Using this formula for
the Riemann-Stieltjes integral, Nualart and Răşcanu have obtained in [3] the
existence of a unique solution for a class of general differential equations that
includes (1). Also they have proved that the solution of (1) is bounded on a
κ
finite interval [0, T ] by C1 exp(C2 kyk0,T,β ), where κ > β1 if f is bounded and
1
is f has linear growth. Here kyk0,T,β denotes the β-Hölder norm of y
κ > 1−2β
on the time interval [0, T ]. These estimates are based on a suitable application
2
Yaozhong Hu and David Nualart
of Gronwall’s lemma. It turns out that the estimate in the linear growth case
is unsatisfactory because κ tends to infinity as β tends to 1/2.
The main purpose of this paper is to obtain sharper estimates for the
solution xt in the case where f is bounded or has linear growth using a direct
approach based on formula (8). In the case where f is bounded we estimate
sup0≤t≤T |xt | by
1
β
C 1 + kyk0,T,β
and if f has linear growth we obtain the estimate
1
β
C1 exp C2 kyk0,T,β
.
In Theorem 2 we provide explicit dependence on f and T for the constants C,
C1 and C2 . We also establish estimates for the solution of a linear equation
with rough time dependent coefficient (Theorem 3.2).
Another novelty of this paper is that we establish stability type of results
for the solution xt to (1) on the initial condition x0 , the driving control y and
the coefficient f (Theorem 3.2).
As an application we deduce the existence of moments for the solutions to
stochastic differential equations driven by a fractional Brownian motion with
Hurst parameter H > 12 . We also discuss the regularity of the solution in the
sense of Malliavin Calculus, improving the results of Nualart and Saussereau
[4], and we apply the techniques of the Malliavin calculus to establish the
smoothness of the density of the solution under suitable non-degeneracy conditions. More precisely, Theorem 3.2 allows us to show that the solution of a
stochastic differential equation
2 Fractional integrals and derivatives
Let a, b ∈ R with a < b. Let f ∈ L1 (a, b) and α > 0. The left-sided and
right-sided fractional Riemann-Liouville integrals of f of order α are defined
for almost all x ∈ (a, b) by
Z t
1
α−1
α
Ia+ f (t) =
(t − s)
f (s) ds
Γ (α) a
and
−α
α
Ib−
f
(−1)
(t) =
Γ (α)
Z
b
α−1
(s − t)
f (s) ds,
t
R∞
−α
respectively, where (−1)
= e−iπα and Γ (α) = 0 rα−1 e−r dr is the Euler
α
α
gamma function. Let Ia+
(Lp ) (resp. Ib−
(Lp )) be the image of Lp (a, b) by the
α
α
α
p
α
operator Ia+ (resp. Ib− ). If f ∈ Ia+ (L ) (resp. f ∈ Ib−
(Lp )) and 0 < α < 1
then the Weyl derivatives are defined as
Differential equations driven by Hölder continuous functions
α
Da+
f
1
(t) =
Γ (1 − α)
and
α
α
Db−
f
(−1)
(t) =
Γ (1 − α)
f (t)
α +α
(t − a)
Z
f (t)
α +α
(b − t)
Z
t
f (t) − f (s)
α+1
(t − s)
a
b
f (t) − f (s)
α+1
(s − t)
t
3
!
ds
(1)
!
ds
(2)
where a ≤ t ≤ b (the convergence of the integrals at the singularity s = t
holds point-wise for almost all t ∈ (a, b) if p = 1 and moreover in Lp -sense if
1 < p < ∞).
For any λ ∈ (0, 1), we denote by C λ (a, b) the space of λ-Hölder continuous
functions on the interval [a, b]. We will make use of the notation
kxka,b,β =
|xr − xθ |
,
β
a≤θ<r≤b |r − θ|
sup
and
kx||a,b,∞ = sup |xr |,
a≤r≤b
d
where x : R → R is a given continuous function.
Recall from [6] that we have:
•
If α <
1
p
and q =
p
1−αp
then
α
α
Ia+
(Lp ) = Ib−
(Lp ) ⊂ Lq (a, b) .
•
If α >
1
p
then
1
α
α
Ia+
(Lp ) ∪ Ib−
(Lp ) ⊂ C α− p (a, b) .
The following inversion formulas hold:
α
α
α
Ia+
Da+
f = f,
∀f ∈ Ia+
(Lp )
α
α
α
Ia−
Da−
f = f,
∀f ∈ Ia−
(Lp )
(3)
(4)
and
α
α
Da+
Ia+
f = f,
α
α
Da−
Ia−
f = f,
∀f ∈ L1 (a, b) .
(5)
1
On the other hand, for any f, g ∈ L (a, b) we have
b
Z
α
Ia+
f (t)g(t)dt = (−1)α
b
Z
a
α
f (t)Ib−
g(t)dt ,
(6)
a
α
α
and for f ∈ Ia+
(Lp ) and g ∈ Ia−
(Lp ) we have
Z
a
b
α
Da+
f (t)g(t)dt
−α
Z
= (−1)
a
b
α
f (t)Db−
g(t)dt.
(7)
4
Yaozhong Hu and David Nualart
Suppose that f ∈ C λ (a, b) and g ∈ C µ (a, b) with λ + µ > 1. Then, from
Rb
the classical paper by Young [8], the Riemann-Stieltjes integral a f dg exists.
The following proposition can be regarded as a fractional integration by parts
Rb
formula, and provides an explicit expression for the integral a f dg in terms
of fractional derivatives (see [9]).
Proposition 1. Suppose that f ∈ C λ (a, b) and g ∈ C µ (a, b) with λ + µ > 1.
Rb
Let λ > α and µ > 1 − α. Then the Riemann Stieltjes integral a f dg exists
and it can be expressed as
Z
b
α
Z
b
f dg = (−1)
a
a
1−α
α
Da+
f (t) Db−
gb− (t) dt,
(8)
where gb− (t) = g (t) − g (b).
3 Estimates for the solutions of differential equations
Suppose that y : [0, ∞) → Rm is a Hölder continuous function of order β >
1/2. Fix an initial condition x0 ∈ Rd and consider the following differential
equation
Z
t
xt = x0 +
f (xr )dyr ,
(1)
0
where f : Rd → Rmd is given function. Lyons has proved in [2] that Equation
(1) has a unique solution if f is continuously differentiable and it has a derivative f 0 which is bounded and locally Hölder continuous of order λ > β1 − 1.
Our aim is to obtain estimates on xt which are better than those given by
Nualart and Răşcanu in [3].
Theorem 2. Let f be a continuously differentiable function such that f 0 is
bounded and locally Hölder continuous of order λ > β1 − 1.
(i) Assume kf 0 k∞ > 0. There is a constant k depending only on β, such that
for all T ,
sup |xt | ≤ 21+kT [kf
0
1/β
k∞ ∨|f (0)|]
1/β
kyk0,T ,β
(|x0 | + 1) .
(2)
0≤t≤T
(ii) Assume that f is bounded. Then, there is a constant k, which depends
only on β, such that for all T ,
1−β
1
β
β
β
0
sup |xt | ≤ |x0 | + kkf k∞ T kyk0,T,β ∨ T kf k∞ kyk0,T,β .
(3)
0≤t≤T
Differential equations driven by Hölder continuous functions
5
Proof. Without loss of generality we assume that d = m = 1. Set kykβ =
kyk0,T,β . We can assume that kykβ > 0, otherwise the inequalities are obvious.
Let α < 1/2 such that α > 1 − β. Henceforth k will denote a generic constant
depending only on β.
Step 1. Assume first that f is bounded. It suffices to assume that kf 0 k∞ >
0. First we use the fractional integration by parts formula given in Proposition
1 to obtain for all s, t ∈ [0, T ],
Z t
Z t
1−α
α
|
f (xr )dyr | ≤
|Ds+
f (xr ) Dt−
yt− (r)|dr.
s
s
From (1) and (2) it is easy to see
1−α
|Dt−
yt− (r)| ≤ kkykr,t,β (t − r)α+β−1 ≤ kkykβ (t − r)α+β−1
(4)
α
|Ds+
f (xr )| ≤ k kf k∞ (r − s)−α + kf 0 k∞ kxks,t,β (r − s)β−α .
(5)
and
Therefore
Z t
Z t
|
f (xr )dyr | ≤ kkykβ
kf k∞ (r − s)−α (t − r)α+β−1
s
s
+ kf 0 k∞ kxks,t,β (r − s)β−α (t − r)α+β−1 dr
≤ kkykβ kf k∞ (t − s)β + kf 0 k∞ kxks,t,β (t − s)2β .
Consequently, we have
kxks,t,β ≤ kkykβ kf k∞ + kf 0 k∞ kxks,t,β (t − s)β .
Choose ∆ such that
∆=
1
0
2k kf k∞ kykβ
β1
.
Then, for all s and t such that t − s ≤ ∆ we have
kxks,t,β ≤ 2kkykβ kf k∞ .
(6)
kx||s,t,∞ ≤ |xs | + kxks,t,β (t − s)β ≤ |xs | + 2kkykβ kf k∞ ∆β .
(7)
Therefore,
If ∆ ≥ T we obtain the estimate
kx||0,T,∞ ≤ |x0 | + 2kkykβ kf k∞ T β .
(8)
Assume ∆ < T . Then, from (7) we get
−1
kx||s,t,∞ ≤ |xs | + kf k∞ kf 0 k∞ .
(9)
6
Yaozhong Hu and David Nualart
Divide the interval [0, T ] into n = [T /∆] + 1 subintervals (where [a] denotes
the largest integer bounded by a). Applying the inequality (9) for s = 0 and
t = ∆ we obtain
−1
sup |xt | ≤ |x0 | + kf k∞ kf 0 k∞ .
0≤t≤∆
Then, applying the inequality (9) on the intervals [∆, 2∆], . . . , [(n − 1)∆, n∆]
recursively, we obtain
−1
−1
sup |xt | ≤ |x0 | + n kf k∞ kf 0 k∞ ≤ |x0 | + ∆−1 (T + ∆) kf k∞ kf 0 k∞
0≤t≤T
1−β
1
≤ |x0 | + T k kf k∞ kf 0 k∞β kykββ .
(10)
The inequality (3) follows from (8) and (10).
Step 2. In the general case, assuming kf 0 k∞ > 0, instead of (5) we have
α
f (xr )| ≤ k (|f (0)| + kf 0 k∞ |xr |) (r − s)−α + kf 0 k∞ kxks,t,β (r − s)β−α .
|Ds+
As a consequence,
h
i
kxks,t,β ≤ kkykβ |f (0)| + kf 0 k∞ kxks,t,∞ + kf 0 k∞ kxks,t,β (t − s)β .
Suppose that ∆ satisfies
∆≤
1
3k kf 0 k∞ kykβ
β1
.
(11)
Then, for all s and t such that t − s ≤ ∆ we have
3
kxks,t,β ≤ kkykβ |f (0)| + kf 0 k∞ kxks,t,∞ .
2
Therefore,
3
|xt | ≤ |xs | + kkykβ |f (0)| + kf 0 k∞ kxks,t,∞ ∆β ,
2
and
3
kxks,t,∞ ≤ |xs | + kkykβ |f (0)| + kf 0 k∞ kxks,t,∞ ∆β .
2
Using again (11) we get
kxks,t,∞ ≤ 2|xs | + 2kkykβ |f (0)|∆β .
Assume also that
∆≤
Then
1
k |f (0)| kykβ
β1
.
(12)
Differential equations driven by Hölder continuous functions
7
kxks,t,∞ ≤ 2 (|xs | + 1) .
Hence,
sup |xr | ≤ 2
0≤r≤t
sup |xr | + 1 .
(13)
0≤r≤s
As before, divide the interval [0, T ] into n = [T /∆] + 1 subintervals, and use
the estimate (13) in every interval to obtain
sup |xt | ≤ 2n (|x0 | + 1) .
(14)
0≤t≤T
Choose
1
−β
∆ = (kkykβ (kf 0 k∞ ∨ |f (0)|))
,
in such a way that (11) and (12) hold. Then, (14) implies
sup |xt | ≤ 21+kT [kf
0
1/β
k∞ ∨|f (0)|]
1/β
kykβ
(|x0 | + 1) .
0≤t≤T
The proof of the theorem is now complete.
Consider now the following system of equations
Z t
xt = x0 +
f (xr )dyr ,
0
Z t
zt = z0 +
g(xr )zr dyr ,
0
where y : [0, ∞) → Rm is a Hölder continuous function of order β > 1/2.
2
f : Rd → Rmd and g : Rd → RM d are given functions and x0 ∈ Rm , z0 ∈ RM .
We make the following assumptions:
H1) f is bounded with a bounded derivative f 0 which is locally Hölder continuous of order λ > β1 − 1.
H2) g is bounded with bounded derivative.
Theorem 3. Assume conditions H1) and H2). Then, there is a constant k
depending only on β, such that for all T ,
sup |zt | ≤ 2
”i1/β
h
“
√
1/β
kyk0,T ,β
1+kT kf 0 k∞ ∨ kgk∞ + kg 0 k∞ kf k∞
|z0 | .
(15)
0≤t≤T
Proof. Without loss of generality we assume that d = m = M = 1. Set
kykβ = kyk0,T,β . We can assume that kykβ > 0, otherwise the inequality is
obvious. Let α < 1/2 such that α > 1 − β.
If we choose ∆ such that
8
Yaozhong Hu and David Nualart
∆β ≤
1
,
2k kf 0 k∞ kykβ
by (6) for all s and t such that t − s ≤ ∆ we have
kxks,t,β ≤ 2kkykβ kf k∞ .
(16)
On the other hand, using the fractional integration by parts formula we
obtain for all s, t ∈ [0, T ],
Z t
Z t
1−α
α
|
g(xr )zr dyr | ≤
|Ds+
(g(xr )zr ) Dt−
yt− (r)|dr.
(17)
s
s
From (1) we get
α
Ds+ (g(xr )zr ) ≤ k
kgk∞ kzks,r,∞ (r − s)−α +
Z
s
r
|g(xr )zr − g(xθ )zθ |
dθ
.
|r − θ|α+1
Now if 0 ≤ s ≤ r ≤ t ≤ T , then
Z r
Z r
|g(xr )zr − g(xθ )zθ |
dθ ≤ k kgk∞
kzks,r,β |r − θ|β−α−1 dθ
|r − θ|α+1
s
s
Z r
0
+ k kg k∞
kzks,r,∞ kxks,r,β |r − θ|β−α−1 dθ
s
≤ k (kgk∞ kzks,t,β + kg 0 k∞ kzks,r,∞ kxks,r,β ) |r − s|β−α .
Therefore
α
Ds+ (g(xr )zr ) ≤ k(kgk
∞
kzks,r,∞ (r − s)−α
+ (kgk∞ kzks,t,β + kg 0 k∞ kzks,r,∞ kxks,r,β ) |r − s|β−α ).
(18)
Substituting (18) and (4) into (17) yields
Z t
|
g(xr )zr dyr | ≤ kkykβ kgk∞ kzks,t,∞ (t − s)β
s
!
0
2β
+ (kgk∞ kzks,t,β + kg k∞ kzks,t,∞ kxks,t,β ) (t − s)
Consequently, is t − s ≤ ∆, applying (16) yields
kzks,t,β ≤ kkykβ kgk∞ kzks,t,∞
0
+ (kgk∞ kzks,t,β + kg k∞ kzks,t,∞ kxks,t,β ) ∆
≤ kkykβ kgk∞ kzks,t,∞
+ (kgk∞ kzks,t,β
β
+ kg k∞ kf k∞ kykβ kzks,t,∞ ) ∆ .
0
β
.
Differential equations driven by Hölder continuous functions
9
Supposse that ∆ is sufficiently small such that
∆≤
1
2kkgk∞ kykβ
β1
.
(19)
Then we have
kzks,t,β ≤ 2kkykβ kzks,t,∞ kgk∞ + kg 0 k∞ kf k∞ kykβ ∆β .
This implies that
kzks,t,∞ ≤ |zs | + kkykβ ∆β kzks,t,∞ kgk∞ + kg 0 k∞ kf k∞ kykβ ∆β .
If ∆ satisfies
kgk∞ ∆β + kg 0 k∞ kf k∞ kykβ ∆2β ≤
1
2kkykβ
(20)
then we have
kzks,t,∞ ≤ 2|zs |
Hence,
sup |zr | ≤ 2 sup |zr |.
0≤r≤t
(21)
0≤r≤s
As before, divide the interval [0, T ] into n = [T /∆] + 1 subintervals, and use
the estimate (21) in every interval to obtain
kzk0,T,∞ ≤ 2n |z0 | .
(22)
Notice that for (20) to hold it suffices that
q
kgk2∞ + k2 kg 0 k∞ kf k∞ − kgk∞
∆β kykβ ≤
2kg 0 k∞ kf k∞
1
= q
.
2
k
kgk2∞ + k kg 0 k∞ kf k∞ + kgk∞
If we choose
− β1
q
0
0
,
∆ = kkykβ max kf k∞ , kgk∞ + kg k∞ kf k∞
then (22) yields
h
“
”i1/β
√
1/β
1+kT kf 0 k∞ ∨ kgk∞ + kg 0 k∞ kf k∞
kyk0,T ,β
kzk0,T,∞ ≤ 2
The proof is now complete.
|z0 | .
10
Yaozhong Hu and David Nualart
Suppose now that we have two differential equations of the form
Z t
xt = x0 +
f (xs )dys ,
0
and
Z
x̃t = x̃0 +
t
f˜(x̃s )ỹs ,
0
where y and ye are Hölder continuous functions of order β > 1/2, and f
and fe are two functions which are continuously differentiable with locally
Hölder continuous derivatives of order λ > β1 − 1. Then, we have the following
estimate.
Theorem 4. Suppose in addition that f is twice continuously differentiable
and f 00 is bounded. Then there is a constant k such that
sup |xr − x̃r | ≤ k2kD
1/β
1/β
kyk0,T ,β T
0≤r≤T
(
h
i
× |x0 − x̃0 | + kyk0,T,β kf − f˜k∞ + kxk0,T,β kf 0 − f˜0 k∞
h
i
)
+ kf˜k∞ + kf˜0 k∞ kxk0,T,∞ ky − ỹk0,T,β ,
where
D = kf 0 k∞ ∨ kf 0 k∞ + kf 00 k∞ (kxk0,T,β + kx̃k0,T,β )T β .
Remark 5. The above inequality is valid only when each term appeared on the
right hand side is finite.
Proof. Fix s, t ∈ [0, T ]. Set
xt − x̃t − (xs − x̃s ) = I1 + I2 + I3 ,
where
Z
t
[f (xr ) − f (x̃r )] dyr
I1 =
s
I2 =
Z th
i
f (x̃r ) − f˜(x̃r ) dyr
s
Z
I3 =
t
f˜(e
xr )d [yr − ỹr ] .
s
The terms I2 and I3 can be estimated easily. In fact, we have
h
i
|I2 | ≤ kkykβ kf − f˜k∞ (t − s)β + kf 0 − f˜0 k∞ kx̃ks,t,β (t − s)2β
Differential equations driven by Hölder continuous functions
and
11
i
h
|I3 | ≤ kky − ỹkβ kf˜k∞ (t − s)β + kf˜0 k∞ kx̃ks,t,β (t − s)2β ,
where kykβ = kyk0,T,β and ky − ỹkβ = ky − ỹk0,T,β . The term I1 is a little
more complicated.
Z t
1−α
α
|I1 | ≤
|Ds+
[f (xr ) − f (x̃r )] ||Dt−
yt− (r)|dr
s
Z t
≤k
kyks,t,β (t − r)α+β−1 |f (xr ) − f (x̃r )|(r − s)−α
s
+kf 0 k∞ kx − x̃ks,r,β (r − s)β−α
+kf 00 k∞ kx − x̃ks,r,∞ [kxks,r,β + kx̃ks,r,β ] (r − s)β−α dr
≤ kkykβ kf 0 k∞ kx − x̃ks,t,∞ (t − s)β + kf 0 k∞ kx − x̃ks,t,β (t − s)2β
+kf 00 k∞ kx − x̃ks,t,∞ [kxks,t,β + kx̃ks,t,β ] (t − s)2β .
Therefore
kx − x̃ks,t,β ≤ kkykβ kf 0 k∞ kx − x̃ks,t,∞ + kf 0 k∞ kx − x̃ks,t,β (t − s)β
+kf 00 k∞ kx − x̃ks,t,∞ [kxks,t,β + kx̃ks,t,β ] (t − s)β
o
+kf − f˜k∞ + kf 0 − f˜0 k∞ kx̃ks,t,β (t − s)β
h
i
+kky − ỹkβ kf˜k∞ + kf˜0 k∞ kx̃ks,t,β (t − s)β .
Rearrange it to obtain
(
"
kx − x̃ks,t,β ≤ k(1 − kkf 0 k∞ kykβ (t − s)β )−1 kykβ kf 0 k∞ kx − x̃ks,t,∞
+kf 00 k∞ kx − x̃ks,t,∞ [kxks,t,β + kx̃ks,t,β ] (t − s)β
#
0
0
β
˜
˜
+kf − f k∞ + kf − f k∞ kx̃ks,t,β (t − s)
h
˜0
+kky − ỹkβ kf˜k∞ + kf k∞ kx̃ks,t,β (t − s)
Set ∆ = t − s, and A = kkf 0 k∞ kykβ . Then
β
i
)
.
12
Yaozhong Hu and David Nualart
kx − x̃ks,t,∞ ≤ |xs − x̃s | + kx − x̃ks,t,β (t − s)β
(
β −1
≤ |xs − x̃s | + k(1 − A∆ )
∆
β
"
kykβ kf 0 k∞ kx − x̃ks,t,∞
+kf 00 k∞ kx − x̃ks,t,∞ [kxks,t,β + kx̃ks,t,β ] ∆β
#
0
0
β
+kf − f˜k∞ + kf − f˜ k∞ kx̃ks,t,β ∆
+kky − ỹkβ
h
kf˜k∞ + kf˜0 k∞ kx̃ks,t,β ∆β
i
)
.
Denote
B = kkykβ kf 0 k∞ + kf 00 k∞ (kxk0,T,β + kx̃k0,T,β )T β .
Then
kx − x̃ks,t,∞ ≤ 1 − (1 − A∆β )−1 ∆β B
(
−1
× |xs − x̃s | + k(1 − A∆β )−1 ∆β
"
h
i
× kykβ kf − f˜k∞ + kf 0 − f˜0 k∞ kx̃ks,t,β ∆β
+ky − ỹkβ
h
kf˜k∞ + kf˜0 k∞ kx̃ks,t,β ∆β
i
#)
Let ∆ satisfy
A∆β ≤ 1/3 ,
B∆β ≤ 1/3
Namely, we take
∆=
1
3 (A ∨ B)
1/β
.
Then
kx − x̃ks,t,∞ ≤ 2 |xs − x̃s | + C∆β ,
where
"
h
i
3
C = k kykβ kf − f˜k∞ + kf 0 − f˜0 k∞ kx̃ks,t,β ∆β
2
#
h
i
0
β
˜
˜
+ky − ỹkβ kf k∞ + kf k∞ kx̃ks,t,β ∆
.
Applying the above estimate recursively we obtain
.
Differential equations driven by Hölder continuous functions
13
sup |xr − x̃r | ≤ 2n |x0 − x̃0 | + C∆β ,
0≤r≤T
where n = [T /∆] + 1. Or we have
sup |xr − x̃r | ≤ k2k(kf
0
1/β
k∞ ∨(kf 0 k∞ +kf 00 k∞ (kxk0,T ,β +kx̃k0,T ,β )T β ))1/β kyk0,T ,β T
0≤r≤T
(
h
i
× |x0 − x̃0 | + kyk0,T,β kf − f˜k∞ + kx̃k0,T,β kf 0 − f˜0 k∞
h
i
˜0
)
+ kf˜k∞ + kf k∞ kx̃k0,T,∞ ky − ỹk0,T,β .
4 Stochastic differential equations driven by a fBm
Let B = {Bt , t ≥ 0} be an m-dimensional fractional Brownian motion (fBm)
with Hurst parameter H > 1/2. That is, B is a Gaussian centered process
with the covariance function E(Bti Bsj ) = RH (t, s)δij , where
RH (t, s) =
1 2H
t + s2H − |t − s|2H .
2
Consider the stochastic differential equation on Rd
Z t
Xt = X0 +
σ(Xs )dBs ,
(1)
0
where X0 is a fixed d-dimensional random variable and the stochastic integral
is is a path-wise Riemann-Stieltjes integral. ([1]). This equation has a unique
solution (see [2] and [3]) provided σ is continuously differentiable, and σ 0 is
1
bounded and Hölder continuous of order λ > H
− 1.
Then, using the estimate (2) in Theorem 2 we obtain the
following estimate
for the solution of Equation (1), if we choose β ∈ 12 , H . Notice that β1 < 2.
0
1/β
sup |Xt | ≤ 21+kT (kσ k∞ ∨|σ(0)|)kBk0,T ,β (|X0 | + 1) .
(2)
0≤t≤T
If σ is bounded and kσ 0 k =
6 0 we can make use of the estimate (3) and we
obtain
1−β
1
1
β
β
sup |Xt | ≤ |X0 | + kkσk∞ T β kBk0,T,β
∨ T kσ 0 k∞β kBk0,T,β
.
(3)
0≤t≤T
These estimates improve those obtained by Nualart and Răşcanu in [3]
based on a suitable version of Gronwall’s lemma. The estimates (2) and (3)
allow us to establish the following integrability properties for the solution of
Equation (1).
14
Yaozhong Hu and David Nualart
Theorem 6. Consider the stochastic differential equation (1), and assume
that E(|X0 |p ) < ∞ for all p ≥ 2. If σ 0 is bounded and Hölder continuous of
1
− 1, then
order λ > H
p
E sup |Xt | < ∞
(4)
0≤t≤T
for all p ≥ 2. If furthermore σ is bounded and E (exp(λ|X0 |γ )) < ∞ for any
λ > 0 and γ < 2H, then
γ
E exp λ sup |Xt |
<∞
(5)
0≤t≤T
for any λ > 0 and γ < 2H.
In [4] Nualart and Saussereau have proved that the random variable Xt
belongs locally to the space D∞ if the function σ is infinitely differentiable
and bounded together with all its partial derivatives. As a consequence, they
have derived the absolute continuity of the law of Xt for any t > 0 assuming that the initial condition is constant and the vector space spanned by
{(σ ij (x0 ))1≤i≤d , 1 ≤ j ≤ m} is Rd .
Applying Theorem 3.2 we can show that the derivatives of Xt possess
moments of all orders, and we can then derive the C ∞ property of the density.
Define the matrix
!
m
X
α(x) =
σ il (x)σ jl (x)
.
l=1
1≤i,j≤d
Theorem 7. Consider the stochastic differential equation (1), with constant
initial condition x0 . Suppose that σ(x) is bounded infinitely differentiable with
bounded derivatives of all orders, and α(x) is uniformly elliptic. Then, for any
t > 0 the probability law of Xt has an C ∞ density.
Proof. Let us first show that Xt belongs to the space D∞ . From Equation
(34) of [4] we have
Drj Xti = σ ij (Xr ) +
d Z tX
m
X
k=1
∂k σ il (Xu )Drj Xuk dBul .
(6)
0 l=1
As a consequence, (15) applied to the system formed by the equations (1)
and (6) yields
”i 1
h
“
√ 00
1/β
0
β
j i
kBk0,T ,β
Dr Xt ≤ 21+kT kσk∞ ∨ kσ k∞ + kσ k∞ kσk∞
kσk∞ .
This implies that for all p ≥ 2
p 

m Z tZ t
X
Dsj Xti Drj Xti |r − s|2H−2 dsdr  < ∞,
E 
j=1 0 0
Differential equations driven by Hölder continuous functions
15
and the random variable Xti belongs to the Sobolev space D1,p for all p ≥ 2.
In a similar way, writing down the linear equations satisfied by the iterated
derivatives, one can show that Xti belongs to the Sobolev space Dk,p for all
p ≥ 2 and k ≥ 2.
In order to show the nongeneracy of the density we use the notation of [4]
and follow the idea of [7]. By Itô’s formula we have
0
Drj Xti Drj 0 Xti
ij
= σ (Xr )σ
i0 j
(Xr0 ) +
d X
m Z
X
k=1 l=1
+
d X
m Z
X
t
0
k=1 l=1
0
t
0
∂k σ i l (Xu )Drj Xui Drj 0 Xuk dBul
0
∂k σ il (Xu )Drj 0 Xui Drj Xuk dBul .
Denote
0
βl (Xu ) = ∂k σ i l (Xu )
1≤i0 ,k≤d


m Z tZ t
X
0
Γt = 
|r − r02H−2 Drj Xti Drj 0 Xti drdr0 
j=1
0
0
.
1≤i,i0 ≤d
Then H(2H − 1)Γt is the Malliavin covariance matrix of the random vector
Xt , and we need to show that Γt−1 is in Lp for any p ≥ 1 and for all t > 0.
We have
m Z t
X
Γt = α0 +
βl (Xu )Γu + Γu βlT (Xu ) dBul ,
l=0
where
α0 =
m Z tZ
X
j=1
0
0
t
|r − r0 |2H−2 σij (Xr )σi0 j (Xr0 )drdr0 .
0
By using Itô formula again we have
Γt−1 = α0−1 −
m Z
X
l=0
t
Γu−1 βl (Xu ) + βlT (Xu )Γu−1 dBul .
(7)
0
By the estimate (15) applied to the equations (1) and (7) , we see that Γt−1
is in Lp for any p ≥ 1. This proves the theorem (see [5]).
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motions. Mem. Amer. Math. Soc. 175 (2005), no. 825.
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inequality of L. C. Young. Mathematical Research Letters 1 (1994) 451-464.
16
Yaozhong Hu and David Nualart
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