HARNACK INEQUALITIES FOR SDES DRIVEN BY
TIME-CHANGED FRACTIONAL BROWNIAN MOTIONS
arXiv:1609.00885v1 [math.PR] 4 Sep 2016
CHANG-SONG DENG AND RENÉ L. SCHILLING
Abstract. We establish Harnack inequalities for stochastic differential equations
(SDEs) driven by a time-changed fractional Brownian motion with Hurst parameter H ∈ (0, 1/2). The Harnack inequality is dimension-free if the SDE has a drift
which satisfies a one-sided Lipschitz condition; otherwise we still get Harnacktype estimates, but the constants will, in general, depend on the space dimension.
Our proof is based on a coupling argument and a regularization argument for the
time-change.
1. Introduction
Throughout this paper, (Ω, A , P) is a probability space. Consider the following
d-dimensional SDE
Z t
(1)
Xt (x) = x +
bs Xs (x) ds + Ut , t ≥ 0, x ∈ Rd ,
0
d
d
where b : [0, ∞) × R → R is measurable, locally bounded in the time variable
t ≥ 0 and continuous in the space variable x ∈ Rd ; the driving noise U = (Ut )t≥0 is
a locally bounded measurable process on Rd starting at zero U0 = 0. Let us assume,
for the time being, that this SDE has a unique non-explosive solution.
In this paper, we want to establish for the solution to the SDE (1) a dimension-free
Harnack inequality with power, first introduced by Wang [17] for diffusions on Riemannian manifolds, and a log-Harnack inequality, considered in [16] for semi-linear
SDEs. Both, the power-Harnack and log-Harnack inequalities have been thoroughly
studied for various finite- and infinite-dimensional SDEs and SPDEs driven by Brownian noise; the main tool was a coupling method and the Girsanov transformation,
see [18] and the references mentioned there. If the noise is a jump process, it is
usually very difficult to construct a successful coupling, and the methods from diffusion processes cannot be directly applied. One exception are drivers which are
subordinate diffusion processes.
Let Σ : [0, ∞) → Rd ⊗ Rd be a measurable and locally bounded deterministic
function, and assume that U is of the following form:
Z t
Ut =
Σs− dWS(s) + Vt , t ≥ 0,
0
2010 Mathematics Subject Classification. 60H10, 60G15.
Key words and phrases. Harnack inequality, fractional Brownian motion, random time-change,
stochastic differential equation.
The first-named author gratefully acknowledges support through the Alexander-von-Humboldt
foundation, the National Natural Science Foundation of China (11401442) and the Specialized
Research Fund for the Doctoral Program of Higher Education of China (20130141120036). Part
of this work was completed during his stay at TU Dresden as Humboldt fellow. He is grateful for
the hospitality and the excellent working conditions.
1
2
C.-S. DENG AND R. L. SCHILLING
where W = (Wt )t≥0 is a standard d-dimensional Brownian motion, S = (S(t))t≥0
is a subordinator (i.e. a non-decreasing process on [0, ∞) with stationary and independent increments a.k.a. increasing Lévy process) and V = (Vt )t≥0 is a locally
bounded (B[0, ∞) ⊗ A /B(Rd )-)measurable process on Rd with V0 = 0; we will, in
addition, assume that the processes W, S and V are stochastically independent.
For such drivers, Wang & Wang [19] were able to obtain Harnack and log-Harnack
inequalities, using an approximation of the subordinator (as in [21]) and a coupling
argument. The following assumptions turned out to be crucial: The coefficient b
has to satisfy a so-called one-sided Lipschitz condition, i.e. there exists a locally
bounded measurable function k : [0, ∞) → R such that
(H)
hbt (x) − bt (y), x − yi ≤ k(t)|x − y|2 ,
x, y ∈ Rd , t ≥ 0;
moreover, Σ−1
exists for each t ≥ 0, and there exists a non-decreasing function
t
λ : [0, ∞) → [0, ∞) such that kΣ−1
t k ≤ λt for all t ≥ 0.
The first-named author used in [6] the same approximation argument and a gradient estimate approach, in order to improve the Harnack inequalities derived in [19];
note that in [6] the function λ is only assumed to be measurable and not necessarily
non-decreasing. Recently, in [20] the approximation argument was also used to establish Harnack type inequalities for SDEs with non-Lipschitz drift and anisotropic
subordinate Brownian noise, i.e. with U having the form
(1)
(d)
Ut = WS (1) (t) , . . . , WS (d) (t) , t ≥ 0,
where W (1) , . . . , W (d) is a standard Brownian motion in Rd , and S (1) , . . . , S (d)
is an independent d-dimensional Lévy process such that each coordinate process
S (i) is a subordinator. Unfortunately, this gives only dimension-dependent Harnack
inequalities. Note that the techniques of [19, 6, 20] do not really need that the
time-change is a subordinator; we may, as we do here, assume that the time-change
is any non-decreasing process on [0, ∞) starting from zero and which is independent
of the original process.
It is a natural question to ask whether one can still get Harnack-type inequalities
if the driving noise U is a more general, maybe non-Markovian, process. As far
as we know, Harnack inequalities were established in [8, 10, 9] for SDEs driven
by fractional Brownian motions. Inspired by these papers as well as [19, 6], we will
combine general time-change and coupling arguments to obtain Harnack inequalities
for SDEs driven by time-changed fractional Brownian motions.
Recall that a fractional Brownian motion W H = (WtH )t≥0 on Rd with Hurst
parameter H ∈ (0, 1) is a self-similar, mean-zero Gaussian process with stationary
increments. The covariance function is given by
1 2H
(2)
E WtH,i WsH,j =
t + s2H − |t − s|2H δij , t, s ≥ 0, 1 ≤ i, j ≤ d,
2
(δij denotes Kronecker’s delta). If H = 1/2, then W H is the classical Brownian
motion which will be denoted as W ; if H 6= 1/2, then W H does not have independent
increments. One can deduce from (2) that W H is self-similar with index H, i.e. for
any constant c > 0, the processes (WctH )t≥0 and (cH WtH )t≥0 have the same finite
dimensional distributions. Let Z = (Z(t))t≥0 be a non-decreasing process on [0, ∞)
starting from 0, independent of W H , and introduce the (random) time-changed
H
process WZH = (WZ(t)
)t≥0 . Typically, Z can be a subordinator or the inverse of a
subordinator; since inverse subordinators are constant on some random intervals,
HARNACK INEQUALITIES FOR SDES
3
WZH is sometimes called a ‘delayed’ fractional Brownian motion. We refer to [12]
for small deviation probabilities of time-changed fractional Brownian motions, while
[13, 11] consider large deviations of fractional Brownian motions delayed by inverse
α-stable subordinators.
Assume that U = WZH + V where V is a locally bounded measurable process
on Rd starting from zero V0 = 0. In this paper, we restrict ourselves to the case
H ∈ (0, 1/2). In order to ensure the existence and uniqueness of the solution to
the SDE (1) and to construct a successful coupling, we assume that the coefficient
b satisfies the one-sided Lipschitz condition (H). As a direct consequence of the
log-Harnack inequality, we obtain a gradient estimate for the associated Markov
semigroup.
As in [20], we can also deal with the anisotropic case, i.e.
Hd ,d
1 ,1
+ Vt ,
,
.
.
.
,
W
Ut = WZH(1)
(d)
Z (t)
(t)
, t ≥ 0,
where, for each i = 1, . . . , d, W Hi ,i = (WtHi ,i )t≥0 is a real-valued fractional Brownian motion with Hurst index Hi ∈ (0, 1/2), Z (i) = (Z (i) (t))t≥0 is a one-dimensional
non-decreasing process such that Z (i) (0) = 0, and V = (Vt )t≥0 is a locally bounded
measurable process with values in Rd and V0 = 0; moreover, we assume that these
processes are independent. As in [20], we replace the Lipschitz condition for the
drift coefficient b by a Yamada–Watanabe type condition, which is, in general, incomparable with the one-sided Lipschitz condition.
The remaining part of this paper is organized as follows. We collect some basics on fractional Brownian motions in Section 2. In Section 3 we establish the
Harnack inequalities for SDEs driven by a time-changed fractional Brownian motion and with drift coefficient satisfying the one-sided Lipschitz condition (H). As
a direct consequence of the log-Harnack inequality we get a gradient estimate for
the corresponding transition semigroup. More explicit expressions in the Harnack
and log-Harnack inequalities are obtained if the time-change Z is (the inverse of) a
subordinator; this is a consequence of our moment estimates from [7]; if Z is the inverse of a subordinator, only the log-Harnack inequality holds, since the exponential
moment of Z(t)−θ is usually infinite for θ > 0. The last section is devoted to case of
an anisotropic driving noise; as one would expect from [20], the Harnack inequalities
turn out to be dimension-dependent.
2. Basics of fractional Brownian motion
In this section, we recall briefly some basic facts on fractional Brownian motion
(fBM) which will be used later on. For further details of fBM and proofs we refer
the readers, for instance, to [2, 5] or [14].
Denote by Γ(·), resp., B(·, ·), the Euler Gamma and Beta functions, and write
1
2 F1 for Gauss’ hypergeometric function. Let a, b ∈ R with a < b. For f ∈ L [a, b]
and α > 0, the left fractional Riemann-Liouville integral of f of order α on (a, b) is
given by the expression
Z x
1
α
Ia+ f (x) :=
(x − y)α−1 f (y) dy, x ∈ (a, b).
Γ(α) a
4
C.-S. DENG AND R. L. SCHILLING
Let W H = (WtH )t≥0 be a fractional Brownian motion on Rd with Hurst parameter
H ∈ (0, 1/2) ∪ (1/2, 1) and define for 0 < s < t the kernel
1
1
(t − s)H− 2 2 F1 H − 12 , 21 − H, H + 21 , 1 − st
KH (t, s) :=
1
Γ H+2
Rt
Fix T > 0. It is known that the integral operator KH f (t) := 0 KH (s, t)f (s) ds
is well-defined on L2 ([0, T ]; Rd ) and defines an isomorphism KH : L2 ([0, T ]; Rd ) →
H+1/2
I0+ (L2 ([0, T ]; Rd )). Moreover, fractional Brownian motion has the following integral representation with respect to the usual d-dimensional standard Brownian
motion W = (Wt )t≥0 :
Z t
H
Wt =
KH (t, s) dWs .
0
H+1/2
1
2
In particular, if 0 < H < and h ∈ I0+
ous, the inverse operator is given by
(L2 ([0, T ]; Rd )) is absolutely continu-
1/2−H 1/2−H ′
−1
(KH
h)(s) = sH−1/2 I0+
(3)
s
h (s),
cf. [15, Eq. (13), p. 108].
3. SDEs driven by delayed fractional Brownian motions
Consider the following SDE on Rd
Z t
H
(4)
Xt (x) = x +
bs Xs (x) ds + WZ(t)
+ Vt ,
t ≥ 0, x ∈ Rd ,
0
d
where b : [0, ∞) × R → Rd , (t, x) 7→ bt (x) is measurable, locally bounded as a
function of t ≥ 0 and continuous in x. The processes W H = (WtH )t≥0 , Z = (Zt )t≥0
and V = (Vt )t≥0 are stochastically independent and satisfy
 H
d

 W is a fBM on R with Hurst parameter H ∈ (0, 1/2);
Z is a time-change, i.e. a non-decreasing process on [0, ∞) with Z(0) = 0;
(5)


V is a locally bounded measurable process on Rd with V0 = 0.
Moreover, we assume that the coefficient b satisfies the one-sided Lipschitz condition
(H).
1/2
Throughout this section, we write |x| := |x(1) |2 +· · ·+|x(d) |2
for the Euclidean
(1)
(d)
d
norm of x = x , . . . , x
∈ R . It is easy to see that under (H), there exists a
unique non-explosive solution to the SDE (4). Set
(6)
t ≥ 0, f ∈ Bb (Rd ), x ∈ Rd .
Pt f (x) := Ef (Xt (x)),
3.1. Statement of the main result. In order to state our main result, we need
the following notation:
Z t
(7)
K(t) :=
k(s) ds, t ≥ 0,
0
where k(s) is the constant appearing in (H),
(8)
1
ΘH :=
4(1 − H)
B
3
2
− H, 12 − H
Γ 21 − H
!2
,
HARNACK INEQUALITIES FOR SDES
5
and we denote for any function f : Rd → R the local Lipschitz constant at the point
x by
|f (y) − f (x)|
∈ [0, ∞], x ∈ Rd .
|∇f |(x) := lim sup
|y
−
x|
y→x
Theorem 3.1. Denote by Xt (x) the unique solution of the SDE (4), and assume
that (5) and (H) hold.
i) For T > 0, x, y ∈ Rd and any bounded Borel function f : Rd → [1, ∞)
"
#
Z(T )2−2H
2
PT log f (y) ≤ log PT f (x) + ΘH E R T
2 |x − y| .
−K(t)
e
dZ(t)
0
ii) For T > 0, x ∈ Rd and any bounded Borel function f : Rd → R
"
#
o
n
2−2H
Z(T
)
2
2ΘH E R T
|∇PT f |2 (x) ≤ PT f 2 (x) − PT f (x)
.
−K(t) dZ(t) 2
e
0
iii) For T > 0, x, y ∈ Rd , p > 1 and any bounded Borel function f : Rd → [0, ∞)
" pΘH
#!p−1
Z(T )2−2H |x − y|2
p
(p−1)2
p
PT f (y) ≤ PT f (x) · E exp
.
RT
2
e−K(t) dZ(t)
0
3.2. Proof of Theorem 3.1. For the proof of Theorem 3.1, we need a few preparations. Let ℓ : [0, ∞) → [0, ∞) be a non-decreasing and càdlàg function with ℓ(0) = 0,
and v : [0, ∞) → R a locally bounded measurable function with v(0) = 0. Because
of (H), the following SDE has a unique non-explosive solution
Z t
ℓ,v
H
(9)
Xt (x) = x +
bs Xsℓ,v (x) ds + Wℓ(t)
+ vt , t ≥ 0, x ∈ Rd .
0
Set for any bounded Borel function f : Rd → R
(10)
Ptℓ,v f (x) := Ef Xtℓ,v (x) , t ≥ 0, x ∈ Rd .
We want to transform the equation (9) into an SDE driven by a fractional Brownian motion which will allow us to establish Harnack inequalities using a combination
of coupling and the Girsanov transformation, cf. [8, 10, 9]. First, however, we have to
approximate the (deterministic) time-change ℓ by an absolutely continuous function.
Consider the following regularization of ℓ:
Z
Z 1
1 t+ǫ
ǫ
ℓ (t) :=
ℓ(s) ds + ǫt =
ℓ(ǫs + t) ds + ǫt, t ≥ 0, ǫ ∈ (0, 1).
ǫ t
0
By construction, for each ǫ ∈ (0, 1) the function ℓǫ is absolutely continuous, strictly
increasing and satisfies for any t ≥ 0
ℓǫ (t) ↓ ℓ(t) as ǫ ↓ 0.
(11)
ǫ
Let Xtℓ ,v (x) be the unique non-explosive solution to the SDE
Z t
ǫ
ℓǫ ,v
(12)
Xt (x) = x +
bs Xsℓ ,v (x) ds + WℓHǫ (t)−ℓǫ (0) + vt , t ≥ 0, x ∈ Rd ,
0
and define P
ℓǫ ,v
by (10) with ℓǫ instead of ℓ.
6
C.-S. DENG AND R. L. SCHILLING
ǫ
Lemma 3.2. Fix ǫ ∈ (0, 1) and let ℓǫ and Xtℓ ,v (x) be as above; in particular we
assume (H).
i) For T > 0, x, y ∈ Rd and any bounded Borel function f : Rd → [1, ∞)
ǫ
Ptℓ ,v
log f (y) ≤
ǫ
log Ptℓ ,v f (x)
ΘH [ℓǫ (T ) − ℓǫ (0)]2−2H
2
+
RT
2 |x − y| .
−K(t)
ǫ
e
dℓ (t)
0
ii) For T > 0, x, y ∈ Rd , p > 1 and any bounded Borel function f : Rd → [0, ∞)
#
" pΘ ǫ
ǫ
2−2H
2
H
[ℓ
(T
)
−
ℓ
(0)]
|x
−
y|
ǫ
ǫ
p
p−1
.
Ptℓ ,v f (y) ≤ Ptℓ ,v f p (x) · exp
RT
−K(t) dℓǫ (t) 2
e
0
Proof. Fix T > 0, x, y ∈ Rd and denote by (Yt )t≥0 a solution of the equation
Z t
Z t
ǫ
Xsℓ ,v (x) − Ys
H
(13) Yt = y +
bs (Ys ) ds + Wℓǫ (t)−ℓǫ (0) + vt + ξ
1[0,τ ) (s) ℓǫ,v
dℓǫ (s),
|Xs (x) − Ys |
0
0
where
ξ := R T
0
and
|x − y|
e−K(r) dℓǫ (r)
ǫ
τ := inf t ≥ 0 : Xtℓ ,v (x) = Yt
is the coupling time. Since
Rd × Rd ∋ (z, z ′ ) 7→ 1{z6=z ′ }
z − z′
∈ Rd
′
|z − z |
ǫ
is locally Lipschitz continuous off the diagonal, the coupling (Xtℓ ,v (x), Yt ) is wellǫ
defined and unique for t ∈ [0, τ ). If τ < ∞, we set Yt = Xtℓ ,v (x) for all t ≥ τ . In
this way, we can construct a unique solution (Yt )t≥0 to (13).
Let us show that the coupling time satisfies τ ≤ T . Let t < τ and observe
that d|η| = 1{η6=0} |η|−1 hη, dηi; therefore, the differential versions of (12) and (13)
together with (H) yield
ǫ
|Xtℓ ,v (x) − Yt |e−K(t)
Z t
ℓǫ ,v
1
ǫ
= |x − y| +
Xs (x) − Ys , bs (Xsℓ ,v (x)) − bs (Ys ) e−K(s) ds
ℓǫ ,v
0 |Xs (x) − Ys |
Z t
Z t
ǫ
−K(s)
ǫ
−ξ
e
dℓ (s) −
|Xsℓ ,v (x) − Ys |k(s)e−K(s) ds
0
0
Z t
≤ |x − y| − ξ
e−K(s) dℓǫ (s).
0
Now assume that τ > T . Taking t = T in the above inequality, we get
ǫ
|XTℓ ,v (x) − YT |e−K(T ) ≤ |x − y| − |x − y| = 0,
ǫ
and so XTℓ ,v (x) = YT . But this is a contradiction to the assumption τ > T . Thereǫ
fore, we have τ ≤ T and XTℓ ,v (x) = YT .
Denote by γ ǫ : [ℓǫ (0), ∞) → [0, ∞) the inverse function of ℓǫ . By definition,
ℓǫ (γ ǫ (t)) = t for t ≥ ℓǫ (0), γ ǫ (ℓǫ (t)) = t for t ≥ 0, and t 7→ γ ǫ (t) is absolutely
HARNACK INEQUALITIES FOR SDES
7
continuous and strictly increasing. Let W = (Wt )t≥0 be a d-dimensional standard
Brownian motion and define
Z t
f
Wt := Wt +
ηs ds, t ≥ 0,
0
R•
−1
g
dr
(s), s ≥ 0, and
where ηs := KH
r
0
ǫ
gr := ξ1[0,ℓǫ(τ )−ℓǫ (0)) (r)
ǫ
ǫ
Xγℓ ǫ,v
(r+ℓǫ (0)) (x) − Yγ (r+ℓ (0))
ǫ
ǫ
ǫ
|Xγℓ ǫ,v
(r+ℓǫ (0)) (x) − Yγ (r+ℓ (0)) |
,
r ≥ 0.
The following stochastic integral defines a martingale
Z t
Mt := − hηs , dWs i, t ≥ 0.
0
Because of (3) we see
−1
KH
Z
•
0
1
1
−H
1
2
gr dr (s) = sH− 2 I0+
s 2 −H gs ,
and this yields for any s ∈ [0, ℓǫ (T ) − ℓǫ (0)]
Z s
1
1
1
1
sH− 2
|ηs | = 1
r 2 −H (s − r)−H− 2 gr dr Γ 2 − H
0
Z s
1
1
1
ξ
sH− 2
≤
r 2 −H (s − r)−H− 2 dr
1
Γ 2 −H
0
3
1
B 2 − H, 2 − H
1
−H
2
|x
−
y|
s
=
R
T
1
−K(t)
ǫ
dℓ (t)
Γ 2 −H 0 e
1
=: CT,H |x − y| s 2 −H .
Thus, the compensator of the martingale M satisfies
Z ℓǫ (T )−ℓǫ (0)
hMiℓǫ (T )−ℓǫ (0) =
|ηs |2 ds
0
(14)
≤
=
2
CT,H
|x
− y|
2
Z
ℓǫ (T )−ℓǫ (0)
s1−2H ds
0
2
CT,H
|x
2
− y| ǫ
[ℓ (T ) − ℓǫ (0)]2−2H .
2(1 − H)
Set
1
R := exp Mℓǫ (T )−ℓǫ (0) − hMiℓǫ (T )−ℓǫ (0) .
2
Using Novikov’s criterion we get ER = 1, and by the Girsanov theorem, the process
ft )0≤t≤ℓǫ (T )−ℓǫ (0) is a d-dimensional Brownian motion under the new probability
(W
measure RP. This allows us to rewrite (12) and (13) as
Z t
Z ℓǫ (t)−ℓǫ (0)
ℓǫ ,v
ℓǫ ,v
Xt (x) = x +
bs Xs (x) ds +
KH (ℓǫ (t) − ℓǫ (0), s) dWs + vt
0
and
Yt = y +
Z
0
t
bs (Ys ) ds +
0
Z
0
ℓǫ (t)−ℓǫ (0)
fs + vt ,
KH (ℓǫ (t) − ℓǫ (0), s) dW
8
C.-S. DENG AND R. L. SCHILLING
ǫ
respectively. Thus, the distribution of (XTℓ ,v (y))0≤t≤T under P coincides with the
law of (Yt )0≤t≤T under RP; in particular, we get for all bounded Borel functions
f : Rd → R
(15)
ǫ
ǫ
Ef XTℓ ,v (y) = ERP f (YT ) = E [Rf (YT )] = E Rf XTℓ ,v (x) .
By the Jensen inequality, we get for any random variable F ≥ 1,
F
F
F
= ERP log
≤ log ERP
= log EF,
E R log
R
R
R
hence
E [R log F ] ≤ log EF + E [R log R] .
Combining this with (15) and the observation that we have
log R = −
Z
ℓǫ (T )−ℓǫ (0)
0
Z
1
hηs , dWs i −
2
ℓǫ (T )−ℓǫ (0)
Z
ℓǫ (T )−ℓǫ (0)
|ηs |2 ds
0
fs i + 1 hMiℓǫ (T )−ℓǫ (0)
hηs , dW
2
0
Z ℓǫ (T )−ℓǫ (0)
2
CT,H
|x − y|2 ǫ
f
[ℓ (T ) − ℓǫ (0)]2−2H ,
≤−
hηs , dWs i +
4(1 − H)
0
=−
we get for all bounded Borel functions f : Rd → [1, ∞) that
PTℓ
ǫ ,v
ǫ
log f (y) = E log f XTℓ ,v (y)
ǫ
= E R log f XTℓ ,v (x)
ǫ
≤ log Ef XTℓ ,v (x) + E[R log R]
ǫ
= log PTℓ ,v f (x) + ERP log R
ǫ
≤ log PTℓ ,v f (x) +
2
CT,H
|x − y|2 ǫ
[ℓ (T ) − ℓǫ (0)]2−2H .
4(1 − H)
This completes the proof of the log-Harnack inequality.
Let us now prove part ii) of the Lemma. For any bounded Borel function f :
R → [0, ∞) we find with (15) and the Hölder inequality
d
ǫ
PTℓ ,v f (y)
(16)
p
p
ǫ
= Ef XTℓ ,v (y)
p
ǫ
= E Rf XTℓ ,v (x)
p−1
ǫ
≤ PTℓ ,v f p (x) · E[Rp/(p−1) ]
.
HARNACK INEQUALITIES FOR SDES
Using (14) we get
9
p
p
Mℓǫ (T )−ℓǫ (0) −
hMiℓǫ (T )−ℓǫ (0)
R
= exp
p−1
2(p − 1)
p
= exp
hMiℓǫ (T )−ℓǫ (0)
2(p − 1)2
p2
p
Mℓǫ (T )−ℓǫ (0) −
hMiℓǫ (T )−ℓǫ (0)
× exp
p−1
2(p − 1)2
2
pCT,H
|x − y|2
2−2H
ǫ
ǫ
≤ exp
[ℓ (T ) − ℓ (0)]
4(p − 1)2 (1 − H)
p2
p
Mℓǫ (T )−ℓǫ (0) −
hMiℓǫ (T )−ℓǫ (0) .
× exp
p−1
2(p − 1)2
i
h
p2
p
Mℓǫ (t)−ℓǫ (0) − 2(p−1)
This, and the fact that exp p−1
2 hMiℓǫ (t)−ℓǫ (0) , 0 ≤ t ≤ T , is a
martingale with mean 1 – this is due to Novikov’s criterion – we get
2
p/(p−1) pCT,H
|x − y|2
2−2H
ǫ
ǫ
E R
≤ exp
[ℓ (T ) − ℓ (0)]
.
4(p − 1)2 (1 − H)
p/(p−1)
Inserting this expression into (16), completes the proof of the power-Harnack inequality.
Proof of Theorem 3.1. By [1, Proposition 2.3], ii) is a direct consequence of i).
Fix T > 0. By a standard approximation argument, it is enough to prove Theorem
the formulae in i) and iii) for f ∈ Cb (Rd ).
Step 1: Assume that bt : Rd → Rd is, uniformly for t in compact intervals, a
global Lipschitz function, i.e. for any t > 0 there is some Ct > 0 such that
(17)
|bs (x) − bs (y)| ≤ Ct |x − y|,
0 ≤ s ≤ t, x, y ∈ Rd .
This implies that for all x ∈ Rd and ǫ ∈ (0, 1)
Z T
ℓǫ ,v
H X (x) − X ℓ,v (x) ≤
bs X ℓǫ ,v (x) − bs X ℓ,v (x) ds + W Hǫ
s
s
ℓ (T )−ℓǫ (0) − Wℓ(T )
T
T
0
Z T
ℓǫ ,v
H X (x) − X ℓ,v (x) ds + W Hǫ
≤ CT
s
s
ℓ (T )−ℓǫ (0) − Wℓ(T ) .
0
ℓǫ ,v
Since Xt (x) and Xtℓ,v (x) are non-explosive, the integral in the above expression is
finite. Therefore, we can apply Gronwall’s inequality with g(ǫ, t) := WℓHǫ (t)−ℓǫ (0) −
H and find
Wℓ(t)
Z T
ℓǫ ,v
ℓ,v
X (x) − X (x) ≤ g(ǫ, T ) + CT
g(ǫ, s)e(T −s)CT ds.
T
T
0
From (11), we conclude that limǫ↓0 g(ǫ, s) = 0 for all s ≥ 0. Then we get from the
dominated convergence theorem that
ǫ
lim XTℓ ,v (x) = XTℓ,v (x),
ǫ↓0
hence
ǫ
lim PTℓ ,v f = PTℓ,v f,
ǫ↓0
x ∈ Rd ;
f ∈ Cb (Rd ).
10
C.-S. DENG AND R. L. SCHILLING
Since ℓ is of bounded variation, the limit ℓǫ ↓ ℓ also holds for the integrals
Z T
Z T
−K(t)
ǫ
lim
e
dℓ (t) =
e−K(t) dℓ(t).
ǫ↓0
0
0
We can now use Lemma 3.2 i) and let ǫ ↓ 0 to get
ΘH ℓ(T )2−2H
2
PTℓ,v log f (y) ≤ log PTℓ,v f (x) + R T
2 |x − y|
−K(t)
e
dℓ(t)
0
for x, y ∈ Rd and all f ∈ Cb (Rd ) with f ≥ 1. Similarly, Lemma 3.2 ii) yields
" p
#
Θ ℓ(T )2−2H
p
p−1 H
ℓ,v
ℓ,v p
2
PT f (y) ≤ PT f (x) · exp R T
2 |x − y|
e−K(t) dℓ(t)
0
for x, y ∈ Rd and all non-negative f ∈ Cb (Rd ).
i
h
ℓ,v
Since the processes W , V and Z are independent, PT f = E PT f (·) ℓ=Z holds
v=V
for all bounded Borel functions f : Rd → R. Thus, the Jensen inequality yields for
all x, y ∈ Rd and f ∈ Cb (Rd ) with f ≥ 1
i
h
ℓ,v
PT log f (y) = E PT log f (y) ℓ=Z
v=V
"
#
i
h
2−2H
Z(T
)
2
≤ E log PTℓ,v f (x) ℓ=Z + ΘH E R T
2 |x − y|
−K(t)
v=V
e
dZ(t)
0
"
#
Z(T )2−2H
2
≤ log PT f (x) + ΘH E R T
2 |x − y| .
−K(t)
e
dZ(t)
0
For the power-Harnack inequality we use Hölder’s inequality to find for all x, y ∈ Rd
and non-negative f ∈ Cb (Rd )
i
h
PT f (y) = E PTℓ,v f (y) ℓ=Z
v=V




ℓ(T )2−2H
2 ΘH |x − y| 1


p−1
≤ E  PTℓ,v f p (x) p exp  R T

2 
−K(t)
e
dℓ(t)
ℓ=Z
0
p
≤ (PT f (x)
p1

v=V

1− p1
pZ(T )2−2H
ΘH |x − y|2
2
E exp  (p−1)
RT

−K(t) dZ(t) 2
e
0
.
Step 2: For the general case, we use the approximation argument proposed in [19,
part (c) of proof of Theorem 2.1]. Let
b̃t (x) := bt (x) − k(t)x,
t ≥ 0, x ∈ Rd .
(k(t) is the constant appearing in (H).) Using (H), it is not difficult to see that the
mapping id −n−1 b̃t : Rd → Rd is injective for any n ∈ N and t ≥ 0. The maps
−1
(n)
−1
bt (x) := n id −n b̃t
(x) − x + k(t)x, n ∈ N, t ≥ 0, x ∈ Rd .
are, uniformly for t in compact intervals, globally Lipschitz continuous, see [4].
(n)
(n)
Denote by (Xt (x))t≥0 the solution of (4) with b replaced by b(n) , and define Pt
HARNACK INEQUALITIES FOR SDES
11
(n)
by (6) with Xt (x) replaced by Xt (x). Because of the first part of the proof, the
(n)
statements of Theorem 3.1 hold with PT replaced by PT .
On the other hand, we see as in [19, part (c) of proof of Theorem 2.1], that
(n)
lim XT (x) = XT (x) a.s., hence,
n→∞
(n)
lim PT f = PT f
n→∞
for all f ∈ Cb (Rd ).
Therefore, the claim follows if we let n → ∞.
3.3. Applications. Let Z = (Z(t))t≥0 be a subordinator, whose Laplace transform
is given by
E e−rZ(t) = e−tφ(r) ,
r > 0, t ≥ 0.
It is well-known that the Laplace exponent φ is a Bernstein function, i.e. a function
which has the following representation
Z
(18)
φ(r) = ϑr +
1 − e−rx ν(dx), r > 0,
(0,∞)
where ϑ ≥R0 is the drift parameter and ν is a Lévy measure, i.e. a measure on (0, ∞)
such that (0,∞) (1 ∧ x) ν(dx) < ∞.
For the constant k(t) from (H) and its primitive K(t), cf. (7), we set
"
#
K ∗ (T ) := exp 2 sup K(t) ,
T > 0.
t∈[0,T ]
Obviously, if k(t) ≤ 0 for all t ≥ 0, then K ∗ (T ) ≤ 1 for all T > 0.
Corollary 3.3. Let Z be a subordinator whose characteristic exponent is the Bernstein function φ and assume that (H) holds. We have for all T > 0, x, y ∈ Rd and
all bounded Borel functions f : Rd → [0, ∞) the following assertions:
i) If lim inf r→∞ φ(r)r −ρ > 0 for some ρ > 0, then
CH,ρ K ∗ (T )|x − y|2
PT log f (y) ≤ log PT f (x) +
, f ≥ 1,
(T ∧ 1)2H/ρ
n
2 o CH,ρ K ∗ (T )
2
2
.
|∇PT f | (x) ≤ PT f (x) − PT f (x)
(T ∧ 1)2H/ρ
If, in addition, lim inf r↓0 φ(r)r −ρ > 0, then we can replace T ∧ 1 by T and get
CH,ρ K ∗ (T )|x − y|2
, f ≥ 1,
PT log f (y) ≤ log PT f (x) +
T 2H/ρ
n
2 o CH,ρ K ∗ (T )
|∇PT f |2 (x) ≤ PT f 2 (x) − PT f (x)
.
T 2H/ρ
ii) If lim inf r→∞ φ(r)r −ρ > 0 for some ρ > 2H/(1 + 2H) and p > 1, then
p
PT f (y) ≤ PT f p (x)
#
"
ρ
1
(pK ∗ (T )|x − y|2) ρ−2H(1−ρ)
CH,ρ p ∗
2
K (T ) 1 + 2H/ρ |x − y| + CH,ρ
.
× exp
ρ+2H(1−ρ)
2H
p−1
T
(p − 1) ρ−2H(1−ρ) T ρ−2H(1−ρ)
12
C.-S. DENG AND R. L. SCHILLING
If, in addition, lim inf r↓0 φ(r)r −ρ > 0, then
p
PT f (y) ≤ PT f p (x)
#
"
ρ
(pK ∗ (T )|x − y|2) ρ−2H(1−ρ)
CH,ρ p K ∗ (T )|x − y|2
+ CH,ρ
× exp
.
ρ+2H(1−ρ)
2H
p−1
T 2H/ρ
(p − 1) ρ−2H(1−ρ) T ρ−2H(1−ρ)
Proof. Since we have
(19)
Z(T )
2−2H
Z
0
T
−K(t)
e
dZ(t)
−2
≤ K ∗ (T )Z(T )−2H ,
T > 0,
the assertion follows from Theorem 3.1 and [7, Theorem 3.8 (a) and (b)].
We will now assume that the subordinator S is strictly increasing, i.e. we have
ν(0, ∞) = ∞ or ϑ > 0. Define the (generalized, right-continuous) inverse of S
S −1 (t) := inf{s ≥ 0 : S(s) > t} = sup{s ≥ 0 : S(s) ≤ t},
t ≥ 0.
We will call S −1 = (S −1 (t))t≥0 an inverse subordinator associated with the Bernstein
function φ. Since we assume that the subordinator S is strictly increasing, we know
that almost all paths of S −1 are continuous and non-decreasing. We will frequently
need the following identity:
(20)
P (S(r) ≥ t) = P S −1 (t) ≤ r , r, t > 0.
Corollary 3.4. Let Z be an inverse subordinator associated with the Bernstein
function φ and assume that (H) holds.
If lim supr↓0 φ(r)r −σ < ∞ and lim supr→∞ φ(r)r −σ < ∞ for some σ > 0, then the
following assertions hold.
i) For any T > 0, x, y ∈ Rd and all bounded Borel functions f : Rd → [1, ∞)
PT log f (y) ≤ log PT f (x) +
CH,σ K ∗ (T )
|x − y|2.
T 2Hσ
ii) For any T > 0, x ∈ Rd and all bounded Borel functions f : Rd → [0, ∞)
n
2 o CH,σ K ∗ (T )
|∇PT f |2 (x) ≤ PT f 2 (x) − PT f (x)
.
T 2Hσ
Corollary 3.4 follows, if we combine (19) with Theorem 3.1 i), ii) and the next
lemma.
Lemma 3.5. Let S −1 be an inverse subordinator with Bernstein function φ satisfying
the conditions of Corollary 3.4. For any θ ∈ (0, 1),
h
−θ i
E S −1 (t)
≤ Cσ,θ t−σθ , t > 0.
Proof. By our assumption, there exists a constant c = c(σ) > 0 such that φ(r) ≤ c r σ
for all r > 0. Combining this with
Z ∞
x
2S(s)
,
=
1 − e−xr e−r dr, x > 0,
1[t,∞) (S(s)) ≤
t + S(s)
1+x
0
HARNACK INEQUALITIES FOR SDES
13
and Tonelli’s theorem, we get that for all s, t > 0
P (S(s) ≥ t) = E 1[t,∞) (S(s))
S(s)/t
≤ 2E
1 + S(s)/t
Z ∞
−r
−rS(s)/t
= 2E
1−e
e dr
0
Z ∞
=2
1 − e−sφ(r/t) e−r dr
Z0 ∞ r −r
≤ 2s
e dr
φ
t
0
Z ∞ σ
r
e−r dr
≤ 2c s
t
0
= 2c Γ(σ + 1)st−σ .
This yields for all t > 0
E S
−1
(t)
−θ =
≤
Z
∞
Z0 ∞
0
=
P S(s−1/θ ) ≥ t ds
1 ∧ 2c Γ(σ + 1)s−1/θ t−σ ds
1
[2c Γ(σ + 1)]θ t−σθ .
1−θ
Remark 3.6. Let α ∈ (0, 1). For an α-stable subordinator S we have σ = α.
Because of (20) and the well-known two-sided estimate
P (S(r) ≥ t) ≍ 1 ∧ rt−α , r, t > 0,
(f ≍ g means that c−1 f (t) ≤ g(t) ≤ cf (t) for some c ≥ 1 and all t) we have for any
t>0
Z ∞
−1 −θ −θ
E S (t)
=
P S −1 (t)
≥ s ds
Z0 ∞
=
P S(s−1/θ ) ≥ t ds
(21)
Z0 ∞
≍
1 ∧ s−1/θ t−α ds
0
=
1 −αθ
t .
1−θ
This shows that Lemma 3.5 is sharp for α-stable subordinators.
Remark 3.7. Let Z be an inverse α-stable subordinator, i.e. Z(t) = S −1 (t) for
t ≥ 0, where (S(t))t≥0 is an α-stable subordinator. For any t, θ, δ > 0 we have
δ
= ∞.
E exp
Z(t)θ
14
C.-S. DENG AND R. L. SCHILLING
This follows similar to (21):
Z ∞ h
−θ i
δ
−1
E exp
≥
≥
s
ds
P
exp
δ
S
(t)
Z(t)θ
1
Z ∞ =
P S (δ/ log s)1/θ ≥ t ds
Z1 ∞ i
h
≍
1 ∧ (δ/ log s)1/θ t−α ds
1
= ∞.
This means that we cannot expect, in the setting of Corollary 3.4, that we get a
power-Harnack inequality as we did in Corollary 3.3 iii).
4. SDEs with non-Lipschitz drift and anisotropic noise
Let W Hi ,i = (WtHi ,i )t≥0 , Z (i) = (Z (i) (t))t≥0 , 1 ≤ i ≤ d, V = (Vt )t≥0 be 2d + 1
independent stochastic processes such that
 Hi ,i
W
are fBMs on R with Hurst parameter Hi ∈ (0, 1/2);


(22)
Z (i) are non-decreasing processes on [0, ∞) with Z (i) (0) = 0;


V is a locally bounded measurable process on Rd with V0 = 0.
We consider the following stochastic equation on Rd :
Z t
1 ,1
d ,d
+ Vt ,
(23) Xt (x) = x +
, . . . , WZH(d)
bs Xs (x) ds + WZH(1)
(t)
(t)
t ≥ 0, x ∈ Rd ,
0
where b = (b(1) , . . . , b(d) ) : [0, ∞) × Rd → Rd , b = bt (x), is measurable, locally
bounded in the variable t ≥ 0 and continuous as a function of x. By U we denote
the family of functions u : (0, ∞) → (0, ∞)Rwhich are continuous, non-decreasing,
ds
= ∞. Typical examples of such
grow at most linearly as x → ∞ and satisfy 0+ u(s)
−1
functions are u(s) = s, u(s) = s log(e∨s ), u(s) = s·{log(e∨s−1 )}·log log(ee ∨s−1 ).
In this section, we will use the ℓ1 -norm on Rd which we denote by kxk1 := |x(1) | +
· · · + |x(d) |, x ∈ Rd , and we replace the one-sided Lipschitz condition (H) by the
following Yamada–Watanabe type condition

There exists some u ∈ U and a locally bounded


measurable function k : [0, ∞) → [0, ∞) such that
(A)


kbt (x) − bt (y)k1 ≤ k(t)u(kx − yk1), t ≥ 0, x, y ∈ Rd .
It is easy to see that (A) guarantees the existence, uniqueness and non-explosion of
the solution to (23). As before, we define for bounded Borel functions f : Rd → R
the operator
Pt f (x) := Ef (Xt (x))
t ≥ 0, x ∈ Rd .
Remark 4.1. As is well known, (A) is incomparable with the condition (H) used
in Section 3.
HARNACK INEQUALITIES FOR SDES
15
4.1. Statement of the
R t main result. Let k(t) be the constant appearing in (A),
and denote by K(t) = 0 k(s) ds its primitive as in (7). For i ∈ {1, . . . , d} we define
ΘHi by (8) with Hi instead of H. Finally, we set for u ∈ U and k(t)
( R1
ds
, if r ∈ [0, 1),
−
Gu (r) := R r rdsu(s)
,
if r ∈ [1, ∞),
1 u(s)
and
Φu,k (t, r) := r +
Z
0
t
k(s) u ◦ G−1
u Gu (r) + K(s) ds,
t, r ≥ 0;
G−1
u is the inverse function of Gu . Since u ∈ U , it is easy to see that Gu is strictly
increasing with Gu (0) = −∞ and limr↑∞ Gu (r) = ∞, so that Φu,k is well-defined.
If, in particular, u(s) = cs for some constant c > 0, then
Z t
cK(s)
Φu,k (t, r) = 1 + c
k(s) e
ds r, t, r ≥ 0.
0
1
Since we use the ℓ -norm in this section, the local Lipschitz constant of a function
f on Rd at the point x ∈ Rd is defined by
|f (y) − f (x)|
|∇f |(x) := lim sup
.
ky − xk1
y→x
Theorem 4.2. Denote by Xt (x) the unique solution of the SDE (23), and assume
that (22) and (A) hold.
i) For T > 0, x, y ∈ Rd , and any bounded Borel function f : Rd → [1, ∞)
d
X
Θ Hi
2
PT log f (y) ≤ log PT f (x) + Φu,k (T, kx − yk1)
E
.
(i) (T ))2Hi
(Z
i=1
ii) For T > 0, x, y ∈ Rd , p > 1 and any bounded Borel function f : Rd → [0, ∞)
#!p−1
"
d
X
p
Θ
p
H
i
Φ2 (T, kx − yk1 )
.
PT f (y) ≤ PT f p (x) · E exp
(i) (T ))2Hi
(p − 1)2 u,k
(Z
i=1
iii) If (A) holds with u(s) = cs for some constant c > 0, then we have for T > 0,
x ∈ Rd and any bounded Borel function f : Rd → R
n
2 o
|∇PT f |2(x) ≤ PT f 2 (x) − PT f (x)
2 X
Z T
d
Θ Hi
cK(s)
×2 1+c
.
k(s) e
ds
E
(Z (i) (T ))2Hi
0
i=1
4.2. Deterministic time-changes. The proof of Theorem 4.2 follows the same
strategy as the proof of Theorem 3.1. Because of independence of the random timechange and the driving processes, we consider first a deterministic time-change ℓ =
(ℓ(1) , . . . , ℓ(d) ) : [0, ∞) → [0, ∞)d such that for each i ∈ {1, . . . , d} the map t 7→ ℓ(i) (t)
is non-decreasing and càdlàg with ℓ(i) (0) = 0. Let v = (v (1) , . . . , v (d) ) : [0, ∞) → Rd
be a locally bounded measurable function such that v0 = 0. Under (A), the following
SDE has a unique non-explosive strong solution
Z t
ℓ,v
+ vt , t ≥ 0, x ∈ Rd ,
(24) Xt (x) = x +
, . . . , WℓH(d)d ,d
bs Xsℓ,v (x) ds + WℓH(1)1 ,1
(t)
(t)
0
16
C.-S. DENG AND R. L. SCHILLING
and, as before, we set for any bounded Borel function f : Rd → R
Ptℓ,v f (x) := Ef Xtℓ,v (x)
t ≥ 0, x ∈ Rd .
Proposition 4.3. Assume (A) and denote by X ℓ,v (x) the unique solution to the
SDE (24).
i) For T > 0, x, y ∈ Rd and any bounded Borel function f : Rd → [1, ∞)
PTℓ,v log f (y) ≤ log PTℓ,v f (x) + Φ2u,k (T, kx − yk1)
d
X
i=1
Θ Hi
(ℓ(i) (T ))
2Hi
.
ii) For T > 0, p > 1, x, y ∈ Rd and any bounded Borel function f : Rd → [0, ∞)
"
#
d
X
p
Θ
p
Hi
PTℓ,v f (y) ≤ PTℓ,v f p (x) · exp
.
Φ2 (T, kx − yk1)
2Hi
(i)
p − 1 u,k
(ℓ
(T
))
i=1
As in Section 3.2, we approximate ℓ(i) by strictly increasing, absolutely continuous
functions
Z
Z 1
1 t+ǫ (i)
ǫ,i
ℓ (t) :=
ℓ (s) ds + ǫt =
ℓ(i) (ǫs + t) ds + ǫt, ǫ ∈ (0, 1), 1 ≤ i ≤ d, t ≥ 0.
ǫ t
0
By construction, ℓǫ,i (t) ↓ ℓ(i) (t) as ǫ ↓ 0. Denote by γ ǫ,i the inverse function of ℓǫ,i .
We consider the following approximation of the SDE (24)
Z t
(i)
ℓǫ ,v,i
Hi ,i
(i)
ℓǫ ,v
(25) Xt (x) = x +
b(i)
i = 1, . . . , d,
s Xs (x) ds + Wℓǫ,i (t)−ℓǫ,i (0) + vt ,
0
ǫ
ǫ
ǫ
where t ≥ 0, x ∈ R and Xtℓ ,v (x) = Xtℓ ,v,1 (x), . . . , Xtℓ ,v,d (x) . Again, for all
bounded Borel functions f : Rd → R
ǫ
ǫ
Ptℓ ,v f (x) := Ef Xtℓ ,v (x) , t ≥ 0, x ∈ Rd .
d
ǫ
We will first prove the Harnack inequalities for Ptℓ ,v using a modification of the
arguments from Lemma 3.2, compare also [20].
ǫ
Lemma 4.4. Assume (A) and denote by Xtℓ ,v (x), ǫ ∈ (0, 1), the unique solution to
the SDE (25).
i) For T > 0, x, y ∈ Rd and any bounded Borel function f : Rd → [1, ∞)
ǫ
ǫ
PTℓ ,v log f (y) ≤ log PTℓ ,v f (x) + Φ2u,k (T, kx − yk1 )
d
X
i=1
Θ Hi
(ℓǫ,i (T ) − ℓǫ,i (0))2Hi
ii) For T > 0, p > 1, x, y ∈ Rd and any bounded Borel function f : Rd → [0, ∞)
#
"
d
X
ǫ
ǫ
Θ
p
p
Hi
.
Φ2 (T, kx − yk1 )
PTℓ ,v f (y) ≤ PTℓ ,v f p (x) · exp
ǫ,i
ǫ,i (0))2Hi
p − 1 u,k
(ℓ
(T
)
−
ℓ
i=1
Proof. Fix ǫ ∈ (0, 1), T > 0, x, y ∈ Rd and denote the coordinates by the superscript
(i). Let (Yt )t≥0 be a solution of the equation
Z t
(i)
(i)
(i)
Yt = y +
bs(i) (Ys ) ds + WℓHǫ,ii ,i(t)−ℓǫ,i (0) + vt
0
(26)
+ξ
(i)
Z
t
0
ǫ
(i)
X ℓ ,v,i (x) − Ys
1[0,τi ) (s) sℓǫ ,v,i
dℓǫ,i (s),
Xs (x) − Ys(i) HARNACK INEQUALITIES FOR SDES
17
where t ≥ 0, i = 1, . . . , d and
ξ (i) :=
Φu,k (T, kx − yk1)
,
ℓǫ,i (δT ) − ℓǫ,i (0)
ǫ
(i) τi := inf t ≥ 0 : Xtℓ ,v,i (x) = Yt .
δ ∈ (0, 1),
(i)
As in the proof of Lemma 3.2, there is a unique solution to (26) such that Yt =
ǫ
Xtℓ ,v,i (x) for t ≥ τi on the set {τi < ∞}, and Rwe use the differential versions of the
equations (26) and the observation that |η| = sgn(η) dη to get for i = 1, . . . , d and
t≥0
ℓǫ ,v,i
Xt (x) − Yt(i) − |x(i) − y (i) |
Z t
ǫ
(i)
Xsℓ ,v,i (x) − Ys
(i)
ℓǫ ,v
(i)
(i) ǫ,i
ǫ,i
b
X
(x)
−
b
(Y
)
ds
−
ξ
ℓ
(t
∧
τ
)
−
ℓ
(0)
=
ℓǫ ,v,i
s
i
s
s
s
(i)
(x) − Ys 0 Xs
Z t
(i) ℓǫ ,v bs Xs (x) − b(i)
ds − ξ (i) ℓǫ,i (t ∧ τi ) − ℓǫ,i (0) .
≤
(Y
)
s
s
0
Summing over i we obtain, using (A),
ℓǫ ,v
Xt (x) − Yt 1
Z t
d
X
ℓǫ ,v
≤ kx − yk1 +
bs Xs (x) − bs (Ys ) 1 ds −
ξ (i) ℓǫ,i (t ∧ τi ) − ℓǫ,i (0)
0
≤ kx − yk1 +
Z
0
i=1
t
d
X
ǫ
k(s) u Xsℓ ,v (x) − Ys 1 ds −
ξ (i) ℓǫ,i (t ∧ τi ) − ℓǫ,i (0) .
i=1
We can now apply Bihari’s inequality (cf. [3, Section 3]) to conclude
ℓǫ ,v
X (x) − Ys ≤ G−1 Gu (kx − yk1) + K(s) , s ≥ 0.
s
u
1
Inserting this into the previous inequality yields for any t ∈ [0, T ]
d
X
i=1
ξ (i) ℓǫ,i (t ∧ τi ) − ℓǫ,i (0)
≤ kx − yk1 +
Z
0
t
k(s) u ◦ G−1
u Gu (kx − yk1 ) + K(s) ds
≤ Φu,k (T, kx − yk1),
which means that we have for each n = 1, . . . , d
d
ℓǫ,n (t ∧ τn ) − ℓǫ,n (0) X ℓǫ,i (t ∧ τi ) − ℓǫ,i (0)
≤
≤ 1.
ǫ,i (δT ) − ℓǫ,i (0)
ℓǫ,n (δT ) − ℓǫ,n (0)
ℓ
i=1
Taking t = T implies ℓǫ,n (T ∧ τn ) ≤ ℓǫ,n (δT ) and this is only possible if τn < T as
δ ∈ (0, 1) and ℓǫ,n is strictly increasing.
Let W (i) = (W (i) )t≥0 , 1 ≤ i ≤ d, be independent one-dimensional standard
Brownian motions. Define
Z t
(i)
(i)
f
Wℓǫ,i (t)−ℓǫ,i (0) := Wℓǫ,i (t)−ℓǫ,i (0) +
ηs(i) dℓǫ,i (s), t ≥ 0, 1 ≤ i ≤ d,
0
18
C.-S. DENG AND R. L. SCHILLING
where
ηs(i)
gr(i)
:=
−1
1[0,τi ) (s)KH
i
Z
·
0
gr(i)
dr
ǫ
ℓǫ,i (t) − ℓǫ,i (0) ,
s ≥ 0,
(i)
Xγℓ ǫ,i,v,i
(x) − Yγ ǫ,i (r+ℓǫ,i (0))
(r+ℓǫ,i (0))
:= ξ 1[0,ℓǫ,i(τi )−ℓǫ,i (0)) (r) ℓǫ ,v,i
,
(i)
X ǫ,i
(x) − Yγ ǫ,i (r+ℓǫ,i (0)) γ (r+ℓǫ,i (0))
(i)
r ≥ 0.
For t ≥ 0 we set
Mt := −
d Z
X
i=1
=−
d Z
X
i=1
t
0
(i)
1[0,τi ) (s)ηs(i) dWℓǫ,i (s)−ℓǫ,i (0)
ℓǫ,i (t∧τi )−ℓǫ,i (0)
0
(i)
ηγ ǫ,i (s+ℓǫ,i (0)) dWs(i) .
Using (3) we find for s ∈ [0, ℓǫ,i (τi ) − ℓǫ,i (0))
Z s
(i)
1
1
1
1
Hi − 2
−Hi (i)
−Hi − 2
η ǫ,i ǫ,i = 2
s
g
(s
−
r)
dr
r
r
γ (s+ℓ (0))
1
Γ 2 − Hi
0
Z s
1
1
1
1
ξ (i) sHi − 2
r 2 −Hi (s − r)−Hi− 2 dr
≤
1
Γ 2 − Hi
0
3
1
B 2 − Hi , 2 − Hi (i) 1 −Hi
ξ s2
.
=
Γ 12 − Hi
Therefore, the compensator of the martingale satisfies
hMi∞ =
d Z
X
i=1
≤
d
X
ℓǫ,i (τi )−ℓǫ,i (0)
0
B
i=1
3
2
(i)
η ǫ,i ǫ,i 2 ds
γ (s+ℓ (0))
− Hi , 21 − Hi
Γ 21 − Hi
= 2Φ2u,k (T, kx − yk1)
d
X
i=1
Set
!2
Θ Hi
(i) 2
(ξ )
Z
ℓǫ,i (T )−ℓǫ,i (0)
s1−2Hi ds
0
2−2H
i
[ℓǫ,i (T ) − ℓǫ,i (0)]
.
[ℓǫ,i (δT ) − ℓǫ,i (0)]2
1
R := exp M∞ − hMi∞ .
2
Using Novikov’s criterion shows that ER = 1, and by Girsanov’s theorem,
fℓǫ (t)−ℓǫ (0) := W
f (1)
f (d)
W
,
.
.
.
,
W
,
ǫ,1
ǫ,1
ǫ,d
ǫ,d
ℓ (t)−ℓ (0)
ℓ (t)−ℓ (0)
ǫ
ǫ
t≥0
is a d-dimensional (Ftℓ )-martingale under RP, where Ftℓ is the σ-algebra generated
(i)
by {Wℓǫ,i (s)−ℓǫ,i (0) : 0 ≤ s ≤ t, 1 ≤ i ≤ d}. For 0 ≤ s ≤ t and θ = (θ(1) , . . . , θ(d) ) ∈
HARNACK INEQUALITIES FOR SDES
19
Rd , it is easy to see that
fℓǫ (t)−ℓǫ (0) − W
fℓǫ (s)−ℓǫ (0) F ℓǫ
ERP exp i θ, W
s
#
" d
1 X (i) 2 ǫ,i
(θ ) ℓ (t) − ℓǫ,i (s)
= exp
2 i=1
ǫ = E exp i θ, Wℓǫ (t)−ℓǫ (0) − Wℓǫ (s)−ℓǫ (0) Fsℓ ,
where
(1)
(d)
Wℓǫ (t)−ℓǫ (0) := Wℓǫ,1 (t)−ℓǫ,1 (0) , . . . , Wℓǫ,d (t)−ℓǫ,d (0) ,
t ≥ 0.
fℓǫ (t)−ℓǫ (0) )t≥0 under RP coincides with the law
This shows that the distribution of (W
of (Wℓǫ (t)−ℓǫ (0) )t≥0 under P. If we rewrite (25) and (26) for t ≥ 0 and i = 1, . . . , d as
Z t
(i)
ℓǫ ,v,i
(i)
ℓǫ ,v
Xt (x) = x +
b(i)
s Xs (x) ds + vt
0
Z t
(i)
+
KHi ℓǫ,i (t) − ℓǫ,i (0), ℓǫ,i (s) − ℓǫ,i (0) dWℓǫ,i (s)−ℓǫ,i (0) ,
0
and
(i)
Yt
=y
(i)
+
Z
t
0
+
Z
(i)
b(i)
s (Ys ) ds + vt
t
0
f (i)
KHi ℓǫ,i (t) − ℓǫ,i (0), ℓǫ,i(s) − ℓǫ,i (0) dW
,
ℓǫ,i (s)−ℓǫ,i (0)
ǫ
respectively, we see that the distribution of (Xtℓ ,v (y))t≥0 under P coincides with the
distribution of (Yt )t≥0 under RP.
As in the proof of Lemma 3.2 we get for any bounded Borel function f : Rd →
[1, ∞)
ǫ
ǫ
PTℓ ,v log f (y) = E R log f XTℓ ,v (x)
ǫ
≤ log Ef XTℓ ,v (x) + E [R log R]
ǫ
= log PTℓ ,v f (x) + ERP log R
ǫ
≤ log PTℓ ,v f (x) + Φ2u,k (T, kx − yk1 )
d
X
i=1
2−2H
Θ Hi
i
[ℓǫ,i (T ) − ℓǫ,i (0)]
,
[ℓǫ,i (δT ) − ℓǫ,i (0)]2
and for any non-negative bounded Borel function f : Rd → [0, ∞)
p
ǫ
PTℓ ,v f (y)
p
ǫ
= E Rf XTℓ ,v (x)
p−1
ǫ
≤ Ef p XTℓ ,v (x)
E Rp/(p−1)
"
#
d
2−2Hi
ǫ,i
ǫ,i
X
ǫ
[ℓ
(T
)
−
ℓ
(0)]
p
Φ2 (T, kx − yk1 )
Θ Hi
≤ PTℓ ,v f p (x) · exp
.
ǫ,i (δT ) − ℓǫ,i (0)]2
p − 1 u,k
[ℓ
i=1
Letting δ ↑ 1 finishes the proof.
The following result is easy; for the sake of completeness, we include its simple
proof.
20
C.-S. DENG AND R. L. SCHILLING
Lemma 4.5. Assume (A). Then for any x ∈ Rd and t ≥ 0,
ǫ
lim Xtℓ ,v (x) = Xtℓ,v (x).
ǫ↓0
Proof. Fix T > 0, ǫ ∈ (0, 1), x ∈ Rd and observe that for t ∈ [0, T ]
ℓǫ ,v
Xt (x) − Xtℓ,v (x)
1
Z t
d
X
Hi ,i
Hi ,i ℓǫ ,v
ℓ,v
W ǫ,i
≤
bs Xs (x) − bs Xs (x) 1 ds +
ℓ (t)−ℓǫ,i (0) − Wℓ(i) (t)
0
≤
Z
i=1
t
0
d
X
ǫ
Hi ,i
Hi ,i W ǫ,i
k(s) u Xsℓ ,v (x) − Xsℓ,v (x)1 ds +
ℓ (t)−ℓǫ,i (0) − Wℓ(i) (t) .
i=1
ǫ
Since the processes Xtℓ ,v (x) and Xtℓ,v (x) are non-explosive, the
last integral expres-
P sion is finite. Applying Bihari’s lemma with g(ǫ, t) := di=1 WℓHǫ,ii ,i(t)−ℓǫ,i (0) − WℓH(i)i ,i(t) yields that for any t ∈ [0, T ]
ℓǫ ,v
Xt (x) − Xtℓ,v (x) ≤ G−1 Gu (g(ǫ, t)) + K(t) .
u
1
Since ℓǫ,i (t) → ℓ(i) (t), one has g(ǫ, t) → 0 as ǫ ↓ 0. Combining this with Gu (0+) =
−∞, we find
ǫ
lim Xtℓ ,v (x) − Xtℓ,v (x)1 = 0.
ǫ↓0
Hence,
ǫ
lim Xtℓ ,v (x) = Xtℓ,v (x) holds for all t ∈ [0, T ].
ǫ↓0
The claim follows since T > 0 is arbitrary.
4.3. Proof of Theorem 4.2. The proof parallels the argument which we have used
for Theorem 3.1; in particular, Lemma 4.4 plays now the same role as Lemma 3.2
for the proof of Theorem 3.1.
The first step is to prove the log- and power-Harnack inequalities stated in i)
and ii) for deterministic time-changes and for continuous functions f ∈ Cb (Rd ).
Lemma 3.1 has these inequalities for absolutely continuous time-changes and the
ǫ
operators P ℓ ,v ; letting ǫ ↓ 0, we get them for general time-changes and the operators
P ℓ,v .
Since the processes Z and V are independent of (W H1 ,1 , . . . , W Hd ,d ), we can indeed
treat them like deterministic processes Z = ℓ and V = v, i.e. just as in Theorem 3.1
the deterministically time-changed inequalities combined with the Jensen and Hölder
inequality prove Theorem 4.2 i) and ii).
Finally, the gradient estimate follows immediately from i) and [1, Proposition 2.3].
4.4. Two examples. As in Section 3.3, we apply our results to two typical examples
of stochastic time-changes Z (i) : subordinators and inverse subordinators.
Throughout this section we assume that (Xt (x))t≥0 is the unique non-explosive
solution to the SDE (23) and Pt f (x) = Ef (Xt (x)). Combining Theorem 4.2 and [7,
Theorem 3.8 (a) and (b)], we obtain the following result.
HARNACK INEQUALITIES FOR SDES
21
Corollary 4.6. Assume that (A) holds and that for each i = 1, . . . , d, Z (i) is a
subordinator with Bernstein function φi such that lim inf r→∞ φi (r)r −αi > 0 for some
αi > 0. Let
Hi
Hi
and κ2 := 2 max
.
κ1 := 2 min
1≤i≤d αi
1≤i≤d αi
Then there exists some constant C = Cα1 ,...,αd ,H1 ,...,Hd > 0 such that the following
assertions i)–iii) hold.
i) For T > 0, x, y ∈ Rd and all bounded Borel functions f : Rd → [1, ∞)
PT log f (y) ≤ log PT f (x) +
Cd
Φ2 (T, kx − yk1) .
(T ∧ 1)κ2 u,k
If, in addition, lim inf r↓0 φi (r)r −αi > 0 for each i, then
PT log f (y) ≤ log PT f (x) +
Cd
Φ2u,k (T, kx − yk1 ) .
κ
κ
1
2
T ∧T
ii) Assume that αi > 2Hi /(1 + 2Hi ) for each i = 1, . . . , d. For any T > 0, x, y ∈ Rd ,
p > 1 and all bounded Borel functions f : Rd → [0, ∞)
"
p
Cp 2
d
p
PT f (y) ≤PT f (x) · exp
Φ (T, kx − yk1) 1 + κ1
p − 1 u,k
T ∧ T κ2
#
αi
d X
2Hi
pΦ2u,k (T, kx − yk1) αi −2Hi (1−αi ) −
T αi −2Hi (1−αi ) .
+ C(p − 1)
2
(p
−
1)
i=1
If, in addition, lim inf r↓0 φi (r)r −αi > 0 for each i, then
"
p
Cdp Φ2u,k (T, kx − yk1)
p
PT f (y) ≤PT f (x) · exp
p−1
T κ1 ∧ T κ2
#
αi
d X
2Hi
pΦ2u,k (T, kx − yk1) αi −2Hi (1−αi ) −
+ C(p − 1)
T αi −2Hi (1−αi ) .
2
(p
−
1)
i=1
iii) If (A) holds for u(s) = cs and some constant c > 0, then for T > 0, x ∈ Rd and
all bounded Borel functions f : Rd → R
2
Z T
n
2 o
Cd
2
2
cK(s)
|∇PT f | (x) ≤ PT f (x) − PT f (x)
1+c
k(s) e
ds .
(T ∧ 1)κ2
0
If, in addition, lim inf r↓0 φi (r)r −αi > 0 for each i, then
2
Z T
n
2 o
Cd
cK(s)
2
2
1+c
k(s) e
ds .
|∇PT f | (x) ≤ PT f (x) − PT f (x)
T κ1 ∧ T κ2
0
If the Z (i) are inverse subordinators, we cannot expect that a power-Harnack
inequality will hold, see Remark 3.7. Combining Lemma 3.5 with Theorem 4.2 i) &
iii), we still have the following corollary.
Corollary 4.7. Assume that (A) holds and that Z (i) is for each i = 1, . . . , d an
inverse subordinator with Bernstein function φi .
22
C.-S. DENG AND R. L. SCHILLING
Moreover, assume that lim supr↓0 φi (r)r −αi < ∞ and lim supr→∞ φi (r)r −αi < ∞
for some αi > 0. Let
κ3 := 2 min Hi αi
1≤i≤d
and κ4 := 2 max Hi αi .
1≤i≤d
Then there exists some constant C = Cα1 ,...,αd ,H1 ,...,Hd > 0 such that the following
assertions i), ii) hold.
i) For T > 0, x, y ∈ Rd and all bounded Borel functions f : Rd → [1, ∞)
PT log f (y) ≤ log PT f (x) +
Cd
Φ2u,k (T, kx − yk1 ) .
κ
4
∧T
T κ3
ii) If (A) holds with u(s) = cs for some constant c > 0, then for T > 0, x ∈ Rd and
all bounded Borel functions f : Rd → R
2
Z T
n
2 o
Cd
cK(s)
2
2
1+c
k(s) e
ds .
|∇PT f | (x) ≤ PT f (x) − PT f (x)
T κ3 ∧ T κ4
0
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(C.-S. Deng) School of Mathematics and Statistics, Wuhan University, Wuhan
430072, China
E-mail address: dengcs@whu.edu.cn
(R. L. Schilling) TU Dresden, Fachrichtung Mathematik, Institut für Mathematische Stochastik, 01062 Dresden, Germany
E-mail address: rene.schilling@tu-dresden.de