Harnack inequalities and heat kernel estimates for SDEs with singular drifts

Harnack inequalities and heat kernel estimates for SDEs with singular drifts

Accepted Manuscript Harnack inequalities and heat kernel estimates for SDEs with singular drifts Jinghai Shao PII: DOI: Reference: S0007-4497(12)001...

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Accepted Manuscript Harnack inequalities and heat kernel estimates for SDEs with singular drifts Jinghai Shao

PII: DOI: Reference:

S0007-4497(12)00114-5 10.1016/j.bulsci.2012.12.003 BULSCI 2531

To appear in:

Bulletin des Sciences Mathématiques

Received date: 5 November 2012

Please cite this article in press as: J. Shao, Harnack inequalities and heat kernel estimates for SDEs with singular drifts, Bull. Sci. math. (2012), http://dx.doi.org/10.1016/j.bulsci.2012.12.003

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Harnack inequalities and heat kernel estimates for SDEs with singular drifts Jinghai Shao Laboratory of Mathematics and Complex Systems, Ministry of Education, School of Mathematical Sciences, Beijing Normal University, Beijing, 100875, China Email: [email protected]

Abstract In this work, we establish the Harnack-type inequalities for the solutions of SDEs with singular drifts and multiplicative noises. Overcoming the difficulties caused by the singularity of the drifts, we establish the Harnack inequalities using the coupling method. As an application, we obtain the estimates of the corresponding heat kernels.

AMS subject Classification: 60H10, 60J60. Keywords: Harnack inequalities, heat kernel estimates, singular drifts, stochastic differential equations

1

Introduction

The aim of this work is to establish the Harnack inequalities for stochastic differential equations (SDEs) with singular drifts and state-dependent diffusion coefficients. We are interested in a type of dimension-free Harnack inequality which was firstly introduced by Wang in [1]. This remarkable work has caused a lot of studies in this topic in recent years. For example, it has been applied to study functional inequalities in [2, 3, 4]; the short time behavior of infinite dimensional diffusions in [5, 6]; and the heat kernel estimates [7, 8]. Recently, this inequality was established in [9] for a class of stochastic porous media equations; in [10] for Ornstein-Uhlenbeck Semigroups; in [11] for stochastic 1

fast-diffusion equations; in [12] for stochastic Burgers equations; in [13] for solutions of stochastic functional differential equations. Harnack inequalities are able to be established by the curvature-dimension condition, but we emphasize here the approach proposed by Arnaudon-Thalmaier-Wang in [7]. There they constructed a successful coupling and applied the Girsanov theorem to establish the dimension-free Harnack inequality for diffusion semigroups on Riemannian manifolds with curvatures unbounded below. This approach overcomes the difficulty to check the curvature condition, which is usually very hard to be verified when the diffusion processes are with multiplicative noise. In the work [14], Wang showed a new technique to construct the coupling for a diffusion process with multiplicative noise. Precisely, consider the following SDE on Rd : dXt = b(t, Xt )dt + σ(t, Xt )dBt ,

(1.1)

where σ : [0, ∞) × Rd × Ω → Rd×d , b : [0, ∞) × Rd × Ω → Rd are progressively measurable. Under the condition σ(t, x) − σ(t, y)2HS + 2b(t, x) − b(t, y), x − y ≤ K|x − y|2 , x, y ∈ Rd , t ≥ 0,

(1.2)

for some K > 0, and the condition d  

σij (t, x)ξi

2

≥ λ2 |ξ|2 , ξ = (ξ1 , . . . , ξd ) ∈ Rd , t ≥ 0,

(1.3)

i,j=1

for some λ > 0, the Harnack inequalities are established. The most important discovery in [14] is that to deal with the multiplicative noise, one can construct a coupling process with an unbounded drift to guarantee the coupling to be successful before a fixed time. This is very similar to the representation of Brownian bridge in the differential equation form. Recall that the solution of SDE dUt = dBt −

Ut dt 1−t

provides us the Brownian bridge (Ut )t∈[0,1) which satisfies U0 = 0 and limt→1 Ut = 0 a.s.. In view of the importance of the dimension-free Harnack inequality, it will be useful to establish it for more general stochastic processes. In the present work, we shall establish Harnack-type inequalities for the solutions of SDEs with singular drifts. This further reveals the general applicability of the dimension-free Harnack inequalities. At the same time, we want to reveal the different role played by diffusion coefficient σ and drift coefficient b. In Arnaudon-Thalmaier-Wang’s approach, there are two key points. One is the 2

construction of successful coupling. Another is the application of the Girsanov theorem. By the present work, one will find that the diffusion coefficient σ, indeed, plays more important role in guaranteeing the coupling to be successful, and that the drift coefficient b affects mainly in the application of the Girsanov theorem. So for the drift coefficient b, it is the growth condition that plays important role in Arnaudon-Thalmaier-Wang’s approach to establish the Harnack-type inequalities. Now let’s recall some facts on the solutions of SDEs with singular drifts. Consider the following SDE in Rd :  t Xt = x + b(s, Xs )ds + Bt , t ≥ 0, (1.4) 0

where (Bt ) is a d-dimensional Brownian motion on a filtered probability space (Ω, Ft , P), x ∈ Rd and b : [0, ∞) × Rd → Rd is a measurable vector field belonging to Lqp (T ) :=   Lq [0, T ]; Lp (Rd ) for T > 0 and some p, q ∈ (1, ∞) satisfying the condition d 2 + < 1. p q

(1.5)

In the remarkable work [15], Krylov and R¨ockner proved the existence and uniqueness of the strong solutions to SDE (1.4). Their drift b could be nowhere continuous, which makes the classical methods to deal with the existence and uniqueness of strong solution for SDEs with Lipschitz coefficients (cf. Ikeda and Watanabe [16]) or with non-Lipschitz coefficients (cf. Fang and Zhang [17]) are not applicable either. Krylov and R¨ockner’s approach is based on the Yamada-Watanabe theorem, and following their work, Fedrizzi and Flandoli [18] provided a different argument to the same result. Moreover, [18] also proved the α-H¨older continuity of x → Xt (x) for α ∈ (0, 1). X.C. Zhang [19] extended the result to state-dependent diffusion coefficients under some regularity assumptions, whose method is based on the Zwonkin’s transform.

2

Set up and main results

In this work, we consider the following SDE in Rd : dXt = bt (Xt )dt + σt (Xt )dBt ,

X0 = x ∈ Rd ,

(2.1)

where b : [0, ∞) × Rd → Rd , σ : [0, ∞) × Rd → Rd×d are measurable, (Bt ) is a ddimensional Brownian motion on a filtered probability space (Ω, Ft , P). For T > 0, let 3

Lqp (T ) = Lq ([0, T ]; Lp (Rd )) for some p, q ∈ (1, ∞) satisfying the condition d 2 + < 1. p q

(2.2)

A measurable function f : [0, T ] × Rd → Rn , n ∈ N is called in Lqp (0, T ) = Lqp (T ) if f Lqp (T ) :=



T 0

 Rd

|f (r, y)|p dy

q/p

1/q dr

< ∞.

Assume that the drift coefficient b belongs to Lqp (T ) for every T > 0 with p, q satisfying (2.2). Suppose the diffusion coefficient satisfies the following conditions: (Hσ1 ) σt (x) is uniformly continuous w.r.t x ∈ Rd and locally uniformly continuous w.r.t. t ∈ [0, ∞), and there exist constants γ0 , γ1 > 0 such that for all (t, x) ∈ [0, ∞) × Rd ,  γ0 |ξ|2 ≤ (σi,j (t, x)ξi )2 ≤ γ1 |ξ|2 , ξ = (ξ1 , . . . , ξd ) ∈ Rd . i,j

(Hσ2 ) For each T > 0, |∇σt | ∈ Lqp (T ) with the same p, q as required on b, where ∇ denotes the generalized gradient w.r.t. x. According to [19, Theorem 1.1], for any T > 0 and x ∈ Rd , there exists a unique strong solution Xt (x) of SDE (2.1) such that 

P ω:



T 0

 |bs (Xs (ω))|ds +

T

0



2

|σs (Xs (ω))| ds < +∞ = 1,

which also implies that the solution is nonexplosive. Let Pt be the semigroup corresponding to the solution (Xt ) of SDE (2.1), that is, Pt f (x) = Ef (Xt ), with X0 = x, f ∈ Bb (Rd ). To establish the Harnack inequality for Pt , in addition to (Hσ1 ) and (Hσ2 ), we need some further conditions: (Hσ3 ) There exists λ > 0 such that σt (x) − σt (y)2HS ≤ λ|x − y|2 ,

4

x, y ∈ Rd .

(Hσ4 ) There exists a positive constant δ such that   (σt (x) − σt (y))(x − y) ≤ δ|x − y|,

x, y ∈ Rd .

(Hb ) b and |b|2 are in Lqp (T ) for each T > 0 with p, q satisfying (2.2). Theorem 2.1 Suppose the conditions (Hσ1 )-(Hσ3 ) and (Hb ) hold. Then (1)

for each T > 0, it holds PT log f (y) ≤ log PT f (x) +

4C 8λ|x − y|2 |b|2 Lqp (T ) + C˜ + , γ0 γ0 (1 − e−λT )

(2.3)

for every f ≥ 1, x, y ∈ Rd , where C, C˜ are positive constants depending on T, p, q, γ0 , γ1 and bLqp (T ) , |b|2 Lqp (T ) .  −1 γ2 (2) Suppose further (Hσ4 ) holds. Then for each T > 0, and p˜ > 1 + 4 1 + 8δ0 − 1 , it holds

p˜   p˜  p˜|x − y|2 PT f (y) ≤ PT f p˜(x) C41+α exp , (2.4) (10 + 12α)(1 − e−λT )   γ2 1 + 8δ0 − 1 and C4 is positive constant depending on T, p, q, γ0 , γ1 and where α = 14 bLqp (T ) , |b|2 Lqp (T ) . Remark 2.2 In comparison with the Harnack inequalities established in [20, Theorem 1.1], the additional terms in (2.3) and (2.4) on their right hand sides is not equal to 1 when x = y, which seems less optimal. But it is natural to get this term in applying the coupling method, as we merely assume the integrability condition on the drift b. If one posed some additional condition such as H¨older continuity to b, this defect could be deleted. Moreover, this kind of phenomenon has appeared in [7] to establish Harnack inequalities on Riemannian manifolds with curvature unbounded below. As a standard application of dimension-free Harnack inequalities, one can obtain estimates of heat kernels which are also independent of dimension. Indeed, applying our Harnack inequalities for SDEs with singular drifts, we can get the following heat kernel estimates. Corollary 2.3 Assume that the semigroup Pt is reversible w.r.t. μ and let pt (x, y) be the heat kernel of Pt w.r.t. μ. Then

5

(i) (2.3) implies pt (x, y) ≥ exp −

4C 8λ|x − y|2 2 ˜ −C − |b| Lqp ( 2t ) , x, y ∈ Rd , t > 0. γ0 (1 − e−λt/2 ) γ0

(ii) If p˜ ∈ (1, 2), then (2.4) implies

4˜ pt 2˜ p log C4 2t + + exp α(10+12α)(1−e

|x − y|2 1 −λϑt/2 ) α(1+α) (ϑ− 2−α )t , exp − pt (x, y) ≤ √ √ 2ϑt μ(Bx ( 2t))μ(By ( 2t))

(2.5)

(2.6)

for ϑ > 1/(2 − α), t > 0, and x, y ∈ Rd . The argument of Corollary 2.3 (i) is the same as that of [21, Corollary 1.3], and the argument of Corollary 2.3 (ii) is similar to [7, Corollary 3] or [22, Lemma 4.7.1]. We omit the proof to save space.

3

Proofs of main results

In this section we shall provide the proof of Theorem 2.1. Take T > 0 and fix it throughout this section. For θ ∈ (0, 1), let ξt =

 1 − θ 1 − eλ(t−T ) , λ

t ∈ [0, T ],

then ξt is a strictly positive function on [0, T ) such that 1 − λξt + ξt = θ

T and T −ε ξt−2 dt = ∞, for any ε > 0. For any x, y ∈ Rd , we construct the coupling processes as follows: dXt = bt (Xt )dt + σt (Xt )dBt , X0 = x, dYt = bt (Yt )dt + σt (Yt )dBt + ξt−1 σt (Yt )σt (Xt )−1 (Xt − Yt )dt,

Y0 = y.

(3.1) (3.2)

Since (Xt )t≥0 is a well-defined nonexplosive process, the locally Lipschitzian property of y → ξt−1 σt (Xt )σt (Yt )−1 (Xt − Yt ) yields that it is bounded before T ∧ ζn for each n ∈ N, which are defined below. So the process (Yt ) is well-defined up to T ∧ ζ, where ζ is the explosion time of Yt ; namely, ζ = limn→∞ ζn and ζn := inf{t ∈ [0, T ) : |Yt | ≥ n}, 6

where inf ∅ := T . Let

˜t = dBt + ξ −1 σt (Xt )−1 (Xt − Yt )dt, t < T ∧ ζ. dB t

We shall prove ζ = T and 

 s  σ (X )−1 (X − Y )  1 s  σt (Xt )−1 (Xt − Yt ) 2 t t t t   dt , dBt − Rs := exp − ξ 2 ξ t t 0 0

(3.3)

(3.4)

is a uniformly integrable martingale for s ∈ [0, T ), then by the martingale convergence theorem, RT := limt↑T Rt exists and {Rt }t∈[0,T ] is a martingale. In this case, by the ˜t }t∈[0,T ) is a d-dimensional Brownian motion under the probability Girsanov theorem {B RT P. Note that the coupling process is constructed in the same way as in [14]. But since, in our case, the drift coefficient b does not satisfy any Lipschitz or monotone condition, we have to overcome this technical difficulty to prove the uniform integrability of {Rt }t∈[0,T ) and the fact (Xt ) and (Yt ) meet together before time T . ˜t ) as Rewrite (3.1) (3.2) in terms of (B Xt −Yt dt, ξt ˜t , Y0 = y. dYt = bt (Yt )dt + σt (Yt )dB

˜t − dXt = bt (Xt )dt+σt (Xt )dB

X0 = x,

(3.1 ) (3.2 )

 For  notational simplicity, set Zt = Xt − Yt . Letσ τn =σ inf t ∈ b[0, T ) : |Xt | + |Yt | ≥ n . According to [19], under the conditions (H1 ), (H2 ) and (H ), the process (Xt ) is nonexplosive, so limn→∞ τn = ζ. To show {Rs } is uniformly integrable, we introduce the following lemma. Lemma 3.1 Assume (Hσ1 )-(Hσ3 ) and (Hb ). Then (1)

  4C1 8λ|x − y|2 E Rs∧τn log Rs∧τn ≤ |b|2 Lqp (T ) + C4 + γ0 γ0 (1 − e−λT ) n≥1,s∈[0,T ) sup

(3.5)

where C1 , C4 are positive constants related to T, p, q, γ0 , γ1 and bLqp (T ) , |b|2 Lqp (T ) . Hence, Rs∧ζ := lim Rs∧τn ∧(T −1/n) , s ∈ [0, T ], RT ∧ζ := lim Rs∧ζ n→∞

s↑T

exist so that {Rs∧ζ }s∈[0,T ] is a uniformly integrable martingale. (2)

Let Q = RT ∧ζ P. Then Q(ζ = T ) = 1 so that Q = RT P.

7

Proof. By (3.4) and (3.3), for s ∈ (0, T ),     E Rs∧τn log Rs∧τn = Es,n log Rs∧τn 

 s∧τn 2 1 s∧τn  −1 −1 −1 ˜ ξt σt (Xt )−1 Zt  dt ξt σt (Xt ) Zt , dBt  + = Es,n − 2 0 0  s∧τn

2 |Zt | 1 ≤ Es,n dt , 2γ0 ξt2 0

(3.6)

where we have used the condition (Hσ1 ), and Es,n denotes taking the expectation w.r.t. Rs∧τn P. By (Hσ1 ) and (Hσ3 ), and the Itˆo formula, we get d

2     |Zt |2 2 ˜t  − 1 − λξt + ξt |Zt | dt + |bt (Xt ) − bt (Yt )|2 dt ≤ Zt , σt (Xt ) − σt (Yt ) dB 2 ξt ξt ξt 2   2 ˜t  − θ |Zt | dt + |bt (Xt ) − bt (Yt )|2 dt. ≤ Zt , σt (Xt ) − σt (Yt ) dB ξt ξt2

(3.7)

This yields Es,n



s∧τn 0

|Zt |2 2 dt ≤ Es,n ξt2 θ



2    bt (Xt )2 + bt (Yt )2 dt + |x − y| . θξ0

s∧τn 0

(3.8)

In the following, we estimate the first two terms separately. ˜t is a Brownian motion w.r.t. Rs∧τn P for (a) Since Yt satisfies the SDE (3.2 ) and B 0 ≤ t ≤ s ∧ τn , by [19, Theorem 2.1], it holds

 s∧τn 2 Es,n |bt (Yt )| dt ≤ C1 |b|2 Lqp (T ) < ∞, s ∈ (0, T ), (3.9) 0

where C1 is a constant related to T , but independent of n.  ), the existence (b) As Xt satisfies the SDE (3.1   s∧τ  of the term (Xt − Yt )/ξt makes it n 2 |bt (Xt )| dt by the classical SDEs’ method. We difficult to estimate the term Es,n 0 overcome this difficulty by using a small trick. By the inequality ab ≤ b ln b − b + ea , a, b > 0, we have for β > 0,  s∧τn



 s∧τn 2 Es,n |bt (Xt )| dt = E βRs∧τn β −1 |bt (Xt )|2 dt 0 0 (3.10)  s∧τn 

   −1 2 |bt (Xt )| dt . ≤ βE Rs∧τn log Rs∧τn + β log β + log E exp β 0

8

Since Xt satisfies the SDE (3.1) with |b|2 in Lqp (T ), by [19, Theorem 2.2], we have E



t2 ∧τn t1 ∧τn

  |br (Xr )|2 drFt1 ≤ C2 |b|2 Lqp (t1 ,t2 ) ,

for 0 ≤ t1 < t2 ≤ T , where C2 is a positive constant depending on γ0 , γ1 , d, p, q, T and |b|2 Lqp (T ) . Then applying [19, Lemma 5.3], we get for every β > 0



E exp β

−1



s∧τn 0

|bt (Xt )|2 dt



≤ C3 (β, T ) < ∞,

(3.11)

where C3 (β, T ) is a constant depending on β, C2 and T and independent of n. See [15, Corollary 3.4] for the same estimate in the case σ ≡ 1. Combining with (3.10), we get

 s∧τn   (3.12) |bt (Xt )|2 dt ≤ βE Rs∧τn log Rs∧τn + β log β + log C3 . Es,n 0

Substituting (3.8), (3.9) and (3.12) into (3.6), we obtain 

 β   E Rs∧τn log Rs∧τn θγ0  |x − y|2 1  ≤ . C1 |b|2 Lqp (T ) + β log β + log C3 (β, T ) + θγ0 2θγ0 ξ0 1−

Hence, for 0 < β < θγ0 , it follows   E Rs∧τn log Rs∧τn |x − y|2 1

2 q ≤ C1 |b| Lp (T ) + β log β + log C3 (β, T ) + θγ0 − β 2ξ0 (θγ0 − β) Here the explicit form of C3 (β, T ) is complicated, but it satisfies limβ→0+ C3 (β, T ) = ∞. Thus we don’t try to optimize the choice of β, and just put β = γ0 /4, and θ = 1/2. Then   E Rs∧τn log Rs∧τn γ0 8λ|x − y|2 4

γ0 ≤ C1 |b|2 Lqp (T ) + log + log C3 (T ) + , s ∈ [0, T ), n ≥ 1. γ0 4 4 γ0 (1 − e−λT ) By the martingale convergence theorem and the Fatou lemma, {Rs∧ζ : s ∈ [0, T ]} is a well-defined martingale with   4C1 8λ|x − y|2 , s ∈ [0, T ], E Rs∧ζ log Rs∧ζ ≤ |b|2 Lqp (T ) + C4 + γ0 γ0 (1 − e−λT ) 9

where C4 = log γ0 + log C3 (T ). (2) Let σn = inf{t ≥ 0 : |Xt | ≥ n}. By conditions (Hσ1 ), (Hσ2 ) and (Hb ), σn ↑ ∞, ˜t ) is a Brownian motion up to T ∧ ζ, it follows from (3.7) P-a.s., hence Q-a.s.. Since (B that

|X 2 (n − m)2 t∧σm ∧ζn − Yt∧σm ∧ζn | Q(σm > t ≥ ζn ) ≤ EQ ξ0 ξt∧σm ∧ζn  t∧σm ∧ζn

2 |x − y| + EQ |br (Xr )|2 + |br (Yr )|2 dr . ≤ ξ0 0 By (3.5), (3.9) and (3.12), the last term of the previous equation is bounded for all n > m > 0 and t ∈ [0, T ). Therefore, letting first n ↑ ∞ then m ↑ ∞, we obtain Q(ζ < T ) = 0 for all t ∈ [0, T ), which yields Q(ζ = T ) = 1. Lemma 3.2 Assume (Hσ1 )-(Hσ4 ) and (Hb ). Then, for any p > 1, it holds

 1+α  exp ≤ C E Rs∧τ 4 n 





q  (1 + α) θ|x − y|2 · p2 (q  p + p )(1 + α) − p2 (1 − θ)(1 − e−λT )

where q = p /(p − 1), C4 = C3



p2 δ ,T θq 



1+α (q  p +p )(1+α)−1

· C3



p2 δα ,T q  θ(1+α)



α (q  p +p )(1+α)−1

 p2 (p − 1)2 + 4(p − 1)η θ2 γ02 α= , with η = . 2p2 2δ   γ2 1 + 8δ0 − 1 , and Moreover, set p = 2, θ = 12 , then α = 14 −p (p − 1) +

1+α   |x − y|2 exp ≤ C , E RT1+α · 5 ∧ζ 10 + 12α 1 − e−λT 1+α α   5+6α   5+6α 4δα C3 1+α ,T . where C5 = C3 4δ, T Proof. By (3.3), (3.4) and (Hσ1 ),  1+α   α  E Rs∧τ = Es,n Rs∧τ n  s∧τnn 

   α s∧τn  −1 −1 −1 ˜ ξt σt (Xt )−1 Zt 2 dt ξt σt (Xt ) Zt , dBt + = Es,n exp − α 2 0 0 

q  α(q  α + 1)  s∧τn |Z |2 (q −1)/q t ≤ Es,n exp dt , for q  > 1, 2(q  − 1)γ02 0 ξt2 10

(3.13)

and (3.14)

(3.15)

(3.16)

where in the last step we have used the inequality     Es,n exp rMt + rM t /2 = Es,n exp rMt − r2 q  M t /2 + r(q  r + 1)M t /2 

rq  (rq  + 1) (q −1)/q ≤ Es,n exp , M  t 2(q  − 1) for any exponential integrable martingale Mt w.r.t. Rs∧τn P. To estimate the right hand side of (3.16), let’s invoke the inequality (3.7) and use (Hσ4 ), which yields, for r > 0,

 s∧τn |Z |2 t Es,n exp r dt 2 ξ 0 t  

r|x − y|2 

2rp  s∧τn  Z  1/p t ˜t ≤ exp , (σt (Xt ) − σt (Yt ))dB Es,n exp θξ0 θ 0 ξt 

rq   s∧τn 1/q 2 · Es,n exp |bt (Xt ) − bt (Yt )| dt θ 0

r|x − y|2 

4r2 p2 δ  s∧τn |Z |2  2p1  t ≤ exp dt Es,n exp 2 θξ0 θ2 ξ 0 t 

rq   s∧τn 1/q 2 · Es,n exp |bt (Xt ) − bt (Yt )| dt , θ 0 where p =

q . q  −1



Take r = r0 :=

θ2 4p2 δ

such that r =

4r2 p2 δ , θ2

then we get

|Zt |2 dt Es,n exp r0 ξt2 0  (3.17) 

r q   s∧τn  q (2p2p −1)

2p r |x − y|2 0 0 2 |bt (Xt ) − bt (Yt )| dt . · Es,n exp · ≤ exp 2p − 1 θξ0 θ 0 s∧τn

By H¨older’s inequality,

r q   s∧τn 0 |bt (Xt ) − bt (Yt )|2 dt Es,n exp θ 0 

4r q   s∧τn  12 

4r q   s∧τn  12 0 0 2 2 Es,n exp |bt (Yt )| dt |bt (Xt )| dt . ≤ Es,n exp θ θ 0 0

(3.18)

˜t )t≤s∧τn is a Brownian motion w.r.t. Rs∧τn P, by the same argument as that Noting that (B of (3.11), we have 

4r q   s∧τn 1/2 1/2  θ 0 2 Es,n exp |bt (Yt )| dt ≤ C3 , T . θ 4r0 q  0 11

The main difficulty lies in estimating the second term of (3.18). We use again the trick used in the proof of Lemma 3.1.

4r q   s∧τn 0 |bt (Xt )|2 dt Es,n exp θ 0

4r q   s∧τn   0 |bt (Xt )|2 dt = E Rs∧τn exp θ 0 α  1  1+α  1+α 

4r q  (1 + α)  s∧τn 0 1+α 2 |bt (Xt )| dt ≤ ERs∧τn E exp θα 0 1  1+α  α  1+α  θα 1+α , T ≤ ERs∧τn C3 . 4r0 q  (1 + α) Therefore,

r q   s∧τn 0 Es,n exp |bt (Xt ) − bt (Yt )|2 dt θ 0 1   2(1+α) α 1/2   θ  2(1+α) θα 1+α ,T C3 . ≤ C3 ERs∧τn ,T 4r0 q  4r0 q  (1 + α)

(3.19)

Substituting (3.19) into (3.17), we obtain

 Es,n exp r0

s∧τn 0

2p r   p |x − y|2  θ |Zt |2 q (2p −1) 0 C3 · dt ≤ exp ,T ξt2 2p − 1 θξ0 4r0 q      p     p α q (2p −1)(1+α) θα q (2p −1)(1+α) 1+α · C3 . ER , T s∧τ n 4r0 q  (1+α)

By (3.14), it holds

(3.20)

q  α(q  α + 1) θ2 . = r = 0 2(q  − 1)γ02 4p2 δ

Then combining (3.20) with (3.16), we finally get

 1+α  E Rs∧τ ≤ exp n

q  (1 + α) θ|x − y|2 · p2 (q  p + p )(1 + α) − p2 (1 − θ)(1 − e−δT )  p2 δ    1+α  p2 δα    α q p (1+α)−1 (q p +p )(1+α)−1 , T ,T C3  . · C3 θq  q θ(1 + α)

(3.21)

By the Fatou’s lemma, (3.15) follows immediately as s ↑ T and n → ∞, and we complete the proof. 12

After the preparation of Lemma 3.1 and Lemma 3.2, the argument of Theorem 2.1 is completely similar to the argument of [14, Theorem 1.1]. But for the convenience of readers, we give it out. Proof of Theorem 2.1: ˜t ) is a d(1) By Lemma 3.1, {Rs∧ζ }s∈[0,T ] is a uniformly integrable martingale and (B dimensional Brownian motion under the probability measure Q. Thus, by [19, Theorem 1.1], Yt can be solved up to time T . Let τ = inf{t > 0 : Xt = Yt }, and set inf ∅ = ∞. If for some ω ∈ Ω, τ (ω) > T , then inf {|Xt − Yt |2 (ω)} > 0.

t∈[0,T ]

So



 T |Xt − Yt |2 (ω) 1 dt = ∞ as dt = ∞. 2 2 ξt 0 0 ξt On the other hand, due to (3.8), (3.9), (3.12), and (3.5),

 T |X − Y |2 t t dt < ∞. EQ 2 ξ 0 t T

We conclude Q(τ > T ) = 0, and XT = YT , Q-a.s.. In view of Lemma 3.1, by the Young inequality, we have     PT log f (y) = EQ log f (XT ) = E RT ∧ζ log f (XT )  ≤ log E[f (XT )] + E RT ∧ζ log RT ∧ζ ] 4C1 8λ|x − y|2 ≤ log PT f (x) + |b|2 Lqp (T ) + C4 + γ0 γ0 (1 − e−λT ) for every measurable f ≥ 1. ˜t )t≤T is a Brownian motion w.r.t. Q = RT ∧ζ P, we have (2) Since XT = YT and (B  p˜   p˜   p˜ PT f (y) = EQ f (YT ) = E RT ∧ζ f (XT )    p˜−1  p˜/(1+α) ≤ PT f p˜(x) ERTq˜ ∧ζ ≤ PT f p˜(x) ERT1+α , ∧ζ where q˜ = p˜/(˜ p −1) ≤ 1+α as p˜ ≥ (1+α)/α. Then the desired result follows immediately from (3.15). Acknowledgment: This research is supported by FANEDD (No. 200917), 985-project and Fundamental Research Funds for the Central Universities. 13

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