Harnack inequality for semilinear SPDE with multiplicative noise

Statistics and Probability Letters 83 (2013) 1184–1192

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Harnack inequality for semilinear SPDE with multiplicative noise✩ Shao-Qin Zhang ∗ School of Math. Sci. and Lab. Math. Com. Sys., Beijing Normal University, Beijing 100875, China

article

info

Article history: Received 12 September 2012 Received in revised form 7 January 2013 Accepted 8 January 2013 Available online 14 January 2013

abstract The dimensional free Harnack inequality is established for a class of semilinear stochastic partial differential equations in the Hilbert space with multiplicative noise by perturbing the linear term of the equation by a suitable linear operator. © 2013 Elsevier B.V. All rights reserved.

MSC: 60J60 Keywords: Harnack inequality Log-Harnack inequality Multiplicative noise Stochastic partial differential equation

1. Introduction Many kinds of dynamics with random influence can be modelled by stochastic partial differential equations, see e.g. Da Prato and Zabczyk (1992), and Prévôt and Röckner (2007). The main aim of this paper is to establish Wang’s Harnack inequality for the following type of stochastic partial differential equations (SPDE) in Hilbert spaces with multiplicative noise dxt = −Axt dt + F (t , xt )dt + B(t , xt )dWt .

(1.1)

The Harnack inequality we shall establish was introduced in Wang (1997) for the first time, and has become a powerful tool in infinite dimensional stochastic analysis. It has been applied to the study of the strong Feller property, heat kernel estimates and so on for diffusion semigroups. There are many papers establishing this type of inequality for SPDE with additive noise, see e.g. Wang (2007), Liu and Wang (2008), Da Prato et al. (2009), Liu (2009), Ouyang (2009, 2011), Ouyang and Röckner (2009), Wang et al. (2011), Wang and Xu (2011), Ouyang et al. (2012) and references therein. For the case of nonadditive noise, the log-Harnack inequality has been established for the same kind of equation in Röckner and Wang (2010) by gradient estimates and finite-dimension approximation. In Wang (2011), the coupling method was used to establish the Harnack inequality for stochastic differential equations (SDE) with multiplicative noise, and in Wang and Yuan (2011), Shao et al. (2012), and Shao (2012), this method was further developed to deal with stochastic functional differential equations and SDE with irregular coefficients. We develop this method to the case of semilinear SPDE. In infinite dimensional space, a new difficulty occurs, that is, the inverse of B(t , xt ) may be unbounded.

✩ This work was supported by the NSFC (11131003), SRFDP, 985-Project.

∗

Tel.: +86 15117924615. E-mail address: [email protected].

0167-7152/$ – see front matter © 2013 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2013.01.009

S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

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2. The model and results Let H be a real separable Hilbert space with inner product ⟨·, ·⟩ and norm ∥ · ∥. Consider the following SPDE on H: dxt = −Axt dt + F (t , xt )dt + B(t , xt )dWt ,

(2.1)

where W = {W (t )}t ≥0 is a cylindrical Brownian motion on H with covariance operator I defined on a complete filtered probability space (Ω , F , P, (Ft )t ≥0 ), and the coefficients satisfy some of the following hypotheses: (H1) A is a non-negative self-adjoint operator with discrete spectrum: 0 ≤ λ1 ≤ λ2 ≤ · · · ≤ λn → ∞,

(2.2)

where {λn , n ∈ N} are the eigenvalues of A, and are the corresponding eigenvectors. (H2) F : [0, ∞) × Ω × H → H and B : [0, ∞) × Ω × H → L(H ) are P∞ × B (H ) measurable, where P∞ is a predictable σ -algebra on [0, ∞) × Ω and L(H ) is the set of bounded linear operators on H. There exists an increasing function K1 : [0, ∞) → [0, ∞) such that

{en }+∞ n =1

∥F (t , x) − F (t , y)∥ + ∥B(t , x) − B(t , y)∥HS ≤ K1 (t )∥x − y∥,

(2.3)

for all t ≥ 0, x ∈ H , P-a.s., where ∥ · ∥HS denotes the Hilbert–Schmidt norm, and there exists r > 1 such that for all t > 0,

 E

t

∥F (s, 0)∥ ds

r

u

 + sup

u∈[0,t ]

0

0

  2r  1r E e−(u−s)A B(s, 0)HS ds < ∞.

(2.4)

(H3) There exists a decreasing function ρ : [0, ∞) → (0, ∞) and a bounded self-adjoint operator B0 such that B(t , x)B(t , x)∗ ≥ ρ(t )2 B20 ,

x ∈ H , t ≥ 0, P-a.s.,

(2.5)

{ bn } ∞ n =1 ,

where B0 is given by a positive sequence i.e. B0 en = bn en . 1 (H4) Ran(B(t , x) − B(t , y)) ⊂ Ran(B− ) holds for all x , y ∈ H with x − y ∈ H0 , t ≥ 0, P-a.s., where Ran(B) means the range 0 1 of linear operator B, and H0 := D (B− ) . There exists an increasing function K2 : [0, ∞) → R such that 0 2 −1 2 2 2⟨F (t , x) − F (t , y), B− 0 (x − y)⟩ + ∥B0 (B(t , x) − B(t , y))∥HS ≤ K2 (t )∥x − y∥H0 2 2 −2 2 holds for all x, y ∈ D (B− 0 ), t ≥ 0, P-a.s., where ∥x∥H0 := n=0 bn ⟨x, en ⟩ . (H5) There exists an increasing function K3 : [0, ∞) → [0, ∞) such that

+∞

2 ∥(B(t , x)∗ − B(t , y)∗ )B− 0 (x − y)∥ ≤ K3 (t )∥x − y∥H0

(2.6)

holds for all x, y ∈ H , t ≥ 0 and x − y ∈ D (B0 ), P-a.s. −1

Remark 2.1. (1) These conditions are analogous to those used in Röckner and Wang (2010), and Wang (2011). Comparing with (A2) in Wang (2011), we use the self-adjoint operator B0 in (H3) instead of the identity operator, which is the same as (H2) in Röckner and Wang (2010). We use the term BB∗ in the left hand side of the inequality (H3) instead of B∗ B which is used both in (A2) in Röckner and Wang (2010) and (H2) in Wang (2011). (H3) is further equivalent to that 1 −1 Ran(B(t , x)) ⊃ RanB0 and ∥B(t , x)−1 z ∥ ≤ ρ(t )−1 ∥B− ), t ≥ 0, P-a.s. 0 z ∥, for all z ∈ D (B0  −2 (2) Let Hn = span{e1 , . . . , en }. Under (H1), we can replace D (B0 ) in (H4) by n Hn , which is the condition (H4) in Röckner 1 and Wang (2010). (H5) will be used to deduce the Harnack inequality. If (H4) holds, then B− 0 (B(t , x) − B(t , y)) is a −1 bounded operator for all x, y ∈ H , t ≥ 0, P-a.s. So in (H5) we only require x − y ∈ D (B0 ). A sufficient condition for 1 (H5) is that B− 0 (B(t , x) − B(t , y)) is uniformly bounded. Define PT f (x) = Ef (xT (x)) for all x ∈ H. Denote Bb (H )(Bb+ (H )) the set of bounded (non-negative) Borel measurable functions on H. Our main result of this paper is the following theorem Theorem 2.2. If (H1)–(H4) hold, then PT log f (y) ≤ log PT f (x) +

K2 (T )∥x − y∥2H0 2ρ(T )2 (1 − e−TK2 (T ) )



K (T )

3 If, in addition, (H5) holds, then for p > 1 + ρ( T)

2

,

f ∈ Bb (H ), f ≥ 1, x, y ∈ H , T > 0.

, δp,T = K3 (T ) ∨

ρ(T ) 2

√

p

K2 (T ) p

holds for all T > 0, x, y ∈ H and f ∈ Bb+ (H ).



p − 1 , the Harnack inequality

  √ √ p − 1 ∥x − y∥2H0 √   (PT f (y)) ≤ (PT f (x)) exp , 4δp,T p − 1 ρ(T ) − δp,T (1 − e−TK2 (T ) ) 

p

(2.7)

(2.8)

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S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

Firstly, we shall give a concise introduction to the method used in Wang (2011). Consider the following SDE on Rd dxt = σ (t , xt )dBt + b(t , xt )dt ,

(2.9)

where Bt is a d-dimensional Brownian motion on a complete filtered probability space (Ω , {Ft }t ≥0 , P). Consider the following coupling dxt = σ (t , xt )dBt + b(t , xt )dt , dyt = σ (t , yt )dBt + b(t , yt )dt +

x0 = x, 1

ξt

(2.10)

σ (t , yt )σ (t , xt )−1 (xt − yt )dt ,

y0 = y,

(2.11)

then xt − yt satisfies d(xt − yt ) = (σ (t , xt ) − σ (t , yt ))dBt + (b(t , xt ) − b(t , yt ))dt

−

1

ξt

σ (t , yt )σ (t , xt )−1 (xt − yt )dt ,

x0 − y0 = x − y,

(2.12)

where ξ is some continuous positive function on [0, T ] with ξ (T ) = 0. Under the assumptions of Wang (2011), it was proved that (2.11) has the strong solution up to the explosion time ζ . Let

RT ∧ζ

1

σ (t , xt )−1 (xt − yt )dt , t < T ∧ ζ , ξt      T ∧ζ 1 1 T ∧ζ = exp − σ (t , xt )−1 (xt − yt ), dB(t ) − ξt 2 0 0

dB˜ t = dBt +

2   1   σ (t , xt )−1 (xt − yt ) dt . ξ  t

By the Girsanov theorem, RT ∧ζ P is a probability measure, and under the new probability, ζ = T almost surely and {B˜ t }0≤t ≤T is a Brownian motion. Then one can rewrite the above equations in terms of B˜ t . The estimates such as ERT ∧ζ log RT ∧ζ will get easier under the new probability measure RT ∧ζ P. Next, we shall give a brief explanation of the difficulty which occurred in infinite dimensional space and our method to overcome it. On the one hand, the term ‘‘ ξ1 σ (t , xt )−1 (xt − yt )dt’’ involves the inverse of σ (t , xt ) which may be an unbounded t operator on H, so if we did as in Wang (2011) step by step, then we have to prove that xt − yt belongs to the domain 1 −1 1 of σ (t , xt ) . On the other hand, though B−1 is controlled by B− in (H3), there is no relation between {λn } and {b− n }. 0 1 −1 Therefore, it seems not easy to prove that ‘‘ ξ σ (t , xt ) (xt − yt )dt’’ make sense in infinite dimensional space. To overcome t

2 this difficulty, we perturb A by B− 0 , and get a new SPDE (3.2) below, then prove that the solution of (3.2) converges to the solution of (2.1) when ϵ goes to 0. Moreover we prove that Eq. (3.18) below admits a variational strong solution zt in the sense of Prévôt and Röckner (2007), which plays the same role as xt − yt mentioned above. Then the method used in Wang (2011) goes on well for Eq. (3.2) and we get the Harnack inequality for PT by the approximation argument. Finally, we explain the reason why we abandon the finite dimension approximation. One may find that the lower dimensional projection of an invertible operator may be degenerate. For example, an operator in R2 has the matrix form  

0 1

1 0

under the orthonormal basis {e1 , e2 }. It is degenerate after projecting to the subspace generated by e1 . But in

Wang (2011), it requires that the diffusion coefficients cannot be degenerate, seeing (A2) there. One may replace B by √ ∗ in Eq. (2.1) which admits the same solution as that of (2.1) in law, and the finite dimensional BB its symmetrization √ √ projections of BB∗ remain invertible. But it seems not easy to get similar estimates for BB∗ as in (H4) and (H5) from the original assumptions on B, and some estimates become worse, see e.g. the remark after (Araki and Yamagami, 1981, Theorem 1). 3. Proof of Theorem 2.2 Fix T > 0, we restrict our discussion on the interval [0, T ]. We denote Ki (T ) by Ki , i = 1, 2, 3, for the sake of simplicity. The next lemma provides the existence and uniqueness of the mild solution of Eq. (2.1). The proof of the lemma follows completely from (Da Prato and Zabczyk, 1992, Theorem 7.6). Lemma 3.1. Let x0 be a F0 -measurable H value random variable. Under the conditions (H1) and (H2), Eq. (2.1) has a pathwise unique mild solution and there is a positive constant C (r , T ) which is dependent on r , T (recall that r is given in (H2)) such that sup E∥xt ∥r ≤ C (r , T )(1 + E∥x0 ∥r ).

t ∈[0,T ]

(3.1)

S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

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For ϵ > 0, we define a new self-adjoint operator (Aϵ , D (Aϵ )): D (Aϵ ) = D (A)



D (B0 ), −2

2 Aϵ x = Ax + ϵ B− 0 x.

To see that Aϵ is a self-adjoint operator, it is sufficient to check that {en |n ∈ N} ⊂ D (Aϵ ), and x ∈ D (Aϵ ) if and only ∞ −2 2 2 −2 if n=1 (λn + ϵ bn ) ⟨x, en ⟩ < ∞. The eigenvalues of Aϵ are {λn,ϵ := λn + ϵ bn |n ∈ N} and the eigenvectors remain −2 {en |n ∈ N}. We shall perturb (2.1) by B0 . The following lemma gives us the relation between the original equation and the perturbed one. Lemma 3.2. Let xt be the mild solution of Eq. (2.1) with initial value x, and xϵt be the mild solution of the following equation dxϵt = −Aϵ xϵt dt + F (t , xϵt )dt + B(t , xϵt )dWt ,

xϵ0 = x.

(3.2)

Then lim E∥xt − xϵt ∥2 = 0,

t ∈ [0, T ].

ϵ→0+

(3.3)

Consequently, for all f ∈ Cb (H ) and x ∈ H , limϵ→0+ Ef (xϵT (x)) = Ef (xT (x)), where Cb (H ) is the set of bounded continuous functions on H. Proof. Noting that F (t , x) and B(t , x) do not change, one can verify that (H1) and (H2) hold for A replaced by Aϵ , so (3.2) has a unique mild solution as in Lemma 3.1. Since xt = e−tA x +

t



e−(t −s)A F (s, xs )ds +

0 −2 xϵt = e−t (A+ϵ B0 ) x +

t



e−(t −s)A B(s, xs )dWs ,

(3.4)

0 t



−2 e−(t −s)(A+ϵ B0 ) F (s, xϵs )ds +

0



t

−2 e−(t −s)(A+ϵ B0 ) B(s, xϵs )dWs ,

(3.5)

0

we obtain that

 t 2   −2 2 −(t −s)A −(t −s)(A+ϵ B− ) ϵ  0 ∥xt − xϵt ∥2 ≤ 3∥(e−t ϵ B0 − 1)e−tA x∥2 + 3  ( e F ( s , x ) − e F ( s , x )) ds s s   0

 t 2   2 −(t −s)A −(t −s)(A+ϵ B− ) ϵ  0 + 3  (e B(s, xs ) − e B(s, xs ))dWs   0

=: I1 + I2 + I3 .

(3.6)

It is clear that limϵ→0+ I1 = 0, and for I2 , we have t

 I2 ≤ 6T

−2

∥(e−(t −s)A − e−(t −s)(A+ϵ B0 ) )F (s, xs )∥2 ds + 6T

0

t



−2

∥e−(t −s)(A+ϵ B0 ) (F (s, xs ) − F (s, xϵs ))∥2 ds

0

=: I2,1 + I2,2 .

(3.7)

Since −2

∥(e−(t −s)A − e−(t −s)(A+ϵ B0 ) )F (s, xs )∥ ≤ C (1 + ∥xs ∥),

(3.8)

−2

lim ∥(e−(t −s)A − e−(t −s)(A+ϵ B0 ) )F (s, xs )∥ = 0,

(3.9)

ϵ→0+

we obtain limϵ→0+ EI2,1 = 0 by the dominated convergence theorem. For I2,2 , we have t

 I2,2 ≤ 6T

−2

∥e−(t −s)(A+ϵ B0 ) (F (s, xs ) − F (s, xϵs ))∥2 ds

0 t



ϵ

∥F (s, xs ) − F (s, xs )∥ ds ≤

≤ 6T 0

2

6TK12

t



∥xs − xϵs ∥2 ds.

(3.10)

0

Next for I3 ,

 t 2  t 2     2 2 −(t −s)A −(t −s)(A+ϵ B− ) −(t −s)(A+ϵ B− ) ϵ    0 0 EI3 ≤ 6E  (e −e )B(s, xs )dWs  + 6E  e (B(s, xs ) − B(s, xs ))dWs   0

=: I3,1 + I3,2 ,

0

(3.11)

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S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

and

2  −2 (I − e−(t −s)ϵ B0 )(e−(t −s)A B(s, 0))dWs  

t

 

EI3,1 ≤ 12T E  

0

 t 2   2 −(t −s)A −(t −s)(A+ϵ B− )  0 + 12T E  ( e − e )( B ( s , x ) − B ( s , 0 )) dW s s  0  t −2 ∥(I − e−(t −s)ϵ B0 )(e−(t −s)A B(s, 0))∥2HS ds ≤ 12T E 0  t −2 ∥(I − e−(t −s)ϵ B0 )(e−(t −s)A (B(s, xs ) − B(s, 0)))∥2HS ds + 12T E 0

=: I3,1,1 + I3,1,2 .

(3.12)

Since −2

∥(I − e−(t −s)ϵ B0 )e−(t −s)A B(s, 0)∥2 =

+∞ 

−2

∥(e−(t −s)ϵ B0 − I )e−(t −s)A B(s, 0)en ∥2 ,

(3.13)

n=1 −2

∥(e−(t −s)ϵ B0 − I )e−(t −s)A B(s, 0)en ∥ ≤ ∥e−(t −s)A B(s, 0)en ∥,  t  t +∞ E ∥e−(t −s)A B(s, 0)en ∥2 ds = E ∥e−(t −s)A B(s, 0)∥2HS ds < ∞, 0

(3.14) (3.15)

0

n =1

we have limϵ→0 I3,1,1 = 0. Noting that B(s, xs ) − B(s, 0) ∈ LHS (H ) and −2

∥(I − e−(t −s)ϵ B0 )e−(t −s)A (B(s, xs ) − B(s, 0))∥2HS =

+∞ 

−2

∥(I − e−(t −s)ϵ B0 )e−(t −s)A (B(s, xs ) − B(s, 0))en ∥2 ,

n=1 2 −(t −s)ϵ B− 0

∥(I − e  t +∞

E

0

)(e

−(t −s)A

(B(s, xs ) − B(s, 0)))en ∥ ≤ ∥(B(s, xs ) − B(s, 0))en ∥2 ,  t ∥(B(s, xs ) − B(s, 0))en ∥2 ds ≤ K12 E ∥xs ∥2 ds < ∞, 2

0

n=1

we get that limϵ→0 EI3,1 = 0 according to the dominated convergence theorem. In addition,

 EI3,2 ≤ 6T E

t

∥B(s, xs ) − B(s, xϵs )∥2HS ds ≤ 6TK12 E



0

t

∥xs − xϵs ∥2 ds.

(3.16)

0

Now, combining (3.6)–(3.16), we finally obtain the following inequality ϵ 2

E∥xt − xt ∥ ≤ ψϵ (t ) + C (T ,

K12

)E

t



∥xs − xϵs ∥2 ds,

(3.17)

0

where ψϵ (t ) is a function satisfying limϵ→0 ψϵ (t ) = 0 for all t ∈ [0, T ]. Then by the Gronwall inequality, the proof is complete.  By Lemma 3.2, we only need to deal with Eq. (3.2). Let ξt = 2K−θ (1 − eK2 (t −T ) ). We denote 2

ϵ

ϵ

F (t , xt − v1 ) − F (t , xt − v2 ),

ϵ

ϵ

B(t , xt ) − B(t , xt − zt )

by F (t , v2 , v1 ), Bˆ (t , zt ) respectively. Firstly, we shall consider the following equation dzt = −Aϵ zt dt + F (t , zt , 0)dt + Bˆ (t , zt )dWt +

1

ξt

Bˆ (t , zt )B(t , xϵt )−1 zt dt −

1

ξt

zt dt ,

z 0 = z ∈ H0 .

(3.18)

According to (H2)–(H4), F (t , ·, 0) : H → H ,

Bˆ (t , ·) : H0 → LHS (H , H0 ),

Bˆ (t , ·)B(t , xϵt )−1 : H0 → L(H0 ),

(3.19)

where LHS (H , H0 ) denotes the set of Hilbert–Schmidt operators from H to H0 , and L(H0 ) denotes the set of bounded operators on H0 . It is natural to solve Eq. (3.2) in H0 in the variational framework (see e.g. Prévôt and Röckner, 2007), if Aϵ is an extension of a self-adjoint operator on H0 . In fact, we have

S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

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Lemma 3.3. Define A0,ϵ as follows D (A0,ϵ ) = B0 (D (Aϵ )),

A0,ϵ x = Aϵ x,

∀x ∈ B0 (D (Aϵ )).

(3.20)

1 Then A0,ϵ is a well defined self-adjoint operator on H0 and Aϵ is an extension of A0,ϵ , moreover (A0,ϵ , B0 (D (Aϵ ))) = (B0 Aϵ B− 0 , B0 (D (Aϵ ))).

Proof. It is well defined. In fact for all x ∈ B0 (D (Aϵ )), +∞ 

λ2n,ϵ ⟨x, en ⟩2 =

n =1

+∞ 

1 2 2 λ2n,ϵ b2n ⟨B− 0 x, en ⟩ ≤ ∥B∥H

n=0

+∞ 

1 2 λ2n,ϵ ⟨B− 0 x, en ⟩ < +∞,

(3.21)

n=1

then x ∈ D (Aϵ ). Since +∞ 

2 2 b− n ⟨Aϵ x, en ⟩ =

n =1

+∞ 

1 2 λ2n,ϵ ⟨B− 0 x, en ⟩ ,

(3.22)

n =1

1 Aϵ x ∈ D (B− 0 ) if and only if x ∈ B0 (D (Aϵ )). Then it is easy to see that A0,ϵ is a self-adjoint operator on H0 . Finally, for all x ∈ B0 (D (Aϵ )) we have 1 −1 B0 Aϵ B− 0 x = Aϵ B0 B0 x = Aϵ x = A0,ϵ x. 

(3.23)

Now, we can construct the Gelfand triple. Let

 

  1    2   (V , ∥ · ∥V ) = D A0,ϵ ,  . A0,ϵ · 1 2

(3.24)

H0

    − 12   . Then V ∗ ⊃ H0 ⊃ V is the triple we need. Noting that x ∈ V if and (V , ∥ · ∥V ∗ ) denotes the completion of H0 , A0,ϵ ·  H0 ∞ ∞ 2 −2 2 −2 2 2 only if n=1 (λn + ϵ b− n )bn ⟨x, en ⟩ < ∞, x ∈ D (Aϵ ) if and only if n=1 (λn + ϵ bn ) ⟨x, en ⟩ < ∞, we arrive at 

∗

2 V ∗ ⊃ H ⊃ H0 ⊃ D (B− 0 ) ⊃ V ⊃ D (Aϵ ) ⊃ D (A0,ϵ ).

(3.25) 1 2

By this relation, one can find that Aϵ and A0,ϵ can be extended to the same operator, A0,ϵ , from V to V ∗ , then there is no harm in replacing Aϵ by A0,ϵ when we discuss the variational solution of Eq. (3.18). Lemma 3.4. If (H1)–(H4) hold, Eq. (3.18) has a unique variational strong solution in H0 up to the explosion time τ := limn→∞ inf{t ∈ [0, T ) | ∥z ∥H0 ≥ n}. Proof. The proof is essentially due to Brzezniak (1997). Let Gn (t , v) =

 B(t , xϵt )−1 v, ϵ −1

B(t , xt )

nv

∥v∥H0

∥v∥H0 ≤ n, ,

(3.26)

∥v∥H0 > n.

Denote Gn (t , v1 ) − Gn (t , v2 ) by Gn (t , v1 , v2 ). We consider the following equation firstly, dzt = −A0,ϵ zt dt + F (t , zt , 0)dt −

1

ξt

zt dt +

1

ξt

Bˆ (t , zt )Gn (t , zt )dt + Bˆ (t , zt )dWt ,

z 0 = z ∈ H0 .

(3.27)

Though 1ˆ 2 2 ∥Bˆ (t , v)∥2LHS = ∥B− 0 B(t , v)∥HS ≤ K2 ∥v∥H0 +

2K1

ϵ

∥B0 ∥3 ∥v∥V ∥v∥H0

(3.28)

does not satisfy the first inequality of the condition (1.2) in Liu and Röckner (2010), one can check that the proof of Liu and Röckner (2010, Lemma 2.2) goes on well if we apply the basic inequality a·b≤

a2 2δ

δ + b2 , 2

δ>0

and the inequality (3.28) instead of the first inequality of condition (1.2) in Liu and Röckner (2010) to the proof of Liu and Röckner (2010, Lemma 2.2). Noting that Gn (t , ·) is a Lipschitz map from H0 to H, one can prove that Eq. (3.27) has a unique variational strong solution (ztn )t ∈[0,T0 ] for all T0 < T by Liu and Röckner (2010, Theorem 1.1). Furthermore, the solution can

1190

S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

be extended to the interval [0, T ) by the pathwise uniqueness and continuity. Next we shall let n go to infinity to get the solution of Eq. (3.18). Let m > n, τmn = inf{t ∈ [0, T ) | ∥ztm ∥H0 > n}, and set inf ∅ = T . Then for t < τmn , ztm = z0 +

 t

1

−A0,ϵ zsm + F (s, zsm , 0) −

ξs

0 t



1

−

ξs

0

Bˆ (s, zsm )B(s, xϵs )−1 zsm ds +

zsm

 ds

t



Bˆ (s, zsm )dWs .

(3.29)

0

By Itô’s formula, for t < τnn ∧ τmn we have d∥ztn − ztm ∥2H0 − 2⟨(Bˆ (t , ztn ) − Bˆ (t , ztm ))dWt , ztn − ztm ⟩H0 ≤ Ψs ∥ztn − ztm ∥2H0 , where

Ψs = K 2 +



2

 

ξs

K2 zsn 2H0

∥ ∥



+ n K1 +



2K1

2

zsn 2V

∥B0 ∥ ∥ ∥

ϵ

+

n2 K1 ∥B0 ∥2

ϵ 2 ξs2 δ 2

.

(3.30)

Then



  E exp −

n t ∧τnn ∧τm





Ψs ds ∥

0

ztn∧τ n ∧τ n n m

−

= 0.

ztm∧τ n ∧τ n 2H0 n m

∥

(3.31)

t

t ∥zsn ∥2V ds < ∞ for all t < T , we have 0 ∥zsn ∥2V ds < ∞ for all t ∈ [0, T ), P-a.s. Then ztn∧τ n ∧τ n = ztm∧τ n ∧τ n , t ∈ n m n m [0, T ), P-a.s. Letting t ↑ T and by the continuity of the path, we have zτnn ∧τ n = zτmn ∧τ n , P-a.s. If τnn < τmn , then Since E

0

n

H

m

n

m

zτnn = zτmn ∈ ∂ Bn 0 (0), which is contradictory to the definition of τmn . Thus τnn ≥ τmn . Similarly, τnn ≤ τmn . So τnn = τmn and n n zτnn = zτmn , P-a.s. We can define zt = ztn for t < τnn . Then τ = supn τnn , and (z , τ ) is a variational strong solution of Eq. (3.18). n m By the same method, one can prove the uniqueness.  Proof of Theorem 2.2. Fix x, y ∈ H, and z = x − y ∈ H0 . Let 1

˜ s = dWs + dW

ξs

  Rs = exp −

s

B(s, xϵs )−1 zs ds,

s < T ∧ τ,

ξt−1 ⟨B(t , xϵt )−1 zt , dWt ⟩ −

0

τn = inf{t ∈ [0, T ) | ∥zt ∥H0 > n},

1 2



s

0

 ∥B(t , xϵt )−1 zt ∥2 dt , ξt

s < T ∧ τ,

Q = RT ∧τ P.

˜ t: Rewrite Eq. (3.18) in terms of W ˜t − dzt = −A0,ϵ zt dt + F (t , zt , 0)dt + Bˆ (t , zt )dW

1

ξt

zt dt ,

z0 = x − y.

(3.32)

By Itô’s formula and (H4), we have that for s ∈ [0, T ) and t < τn ∧ s, d∥zt ∥2H0 = −2∥zt ∥2V dt + 2V ∗ ⟨F (t , zt , 0), zt ⟩V dt −

2∥zt ∥2H0

ξt

dt

˜ t , zt ⟩H0 + ∥Bˆ (t , zt )∥2LHS (H ,H0 ) dt + 2⟨Bˆ (t , zt )dW ≤−

2∥zt ∥2H0

ξt

˜ t , zt ⟩H0 + K2 ∥zt ∥2H dt , dt + 2⟨Bˆ (t , zt )dW 0

(3.33)

and then d

∥zt ∥2H0 ξt

≤− ≤−

2∥zt ∥2H0

ξ

2 t

dt +

K2

ξt

∥zt ∥2H0 dt −

ξt′ 2 ˜ t , zt ⟩H0 ∥zt ∥2H0 dt + ⟨Bˆ (t , zt )dW ξt ξt2

θ 2 ˜ t , zt ⟩H0 ∥zt ∥2H0 dt + ⟨Bˆ (t , zt )dW ξt ξt2

(3.34)

˜ t }0≤t ≤s∧τn is a Wiener process under the probability measure according to the definition of ξ . By the Girsanov theorem, {W Qs,n := Rs∧τn P. By (3.34), s∧τn

 0

∥z0 ∥2H0 ∥zt ∥2 dt ≤ + θ ξ0 ξt2

s∧τn

 0

2

θ ξt

˜ t , zt ⟩H0 . ⟨Bˆ (t , zt )dW

(3.35)

S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

1191

Thus

∥z0 ∥2H0 ∥zt ∥2 . dt ≤ θ ξ0 ξt2

s∧τn

 EQs,n

0

(3.36)

By condition (H3), we have u

 log Ru ≤ −

˜ t⟩ + ξt−1 ⟨B(t , xϵt )−1 zt , dW

0

u



1 2ρ(T )2

∥zt ∥2H0

dt ,

ξt

0

u ≤ s ∧ τn .

(3.37)

Combining (3.36) with (3.37), we obtain

ERs∧τn log Rs∧τn ≤

∥z0 ∥2H0 2θ ξ0 ρ(T )2

,

s ∈ [0, T ), n ≥ 1.

(3.38)

As in Wang (2011), one can prove that {Rs∧τ |s ∈ [0, T ]} is a martingale. Since nQ(τn ≤ t )

ξ0

≤ EQ 1[τn ≤t ]

∥zt ∧τn ∥2H0 ξt ∧τn

≤ EQ

∥zt ∧τn ∥2H0

≤

ξt ∧τn

∥z0 ∥2H0 ξ0

,

(3.39)

letting n go to infinity, we have Q(τn ≤ t ) = 0 for all t ∈ [0, T ). Then Q(τ = T ) = 1, which further implies that ˜ t }t ∈[0,T ] is a Wiener process under Q. Let ζ = inf{t ∈ [0, T ] | ∥zt ∥H0 = 0}, and set Eq. (3.32) can be solved up to time T , and {W inf ∅ = +∞. We shall prove that ζ ≤ T . Otherwise, there exists a set Ω0 such that Q(Ω0 ) > 0, and for all ω ∈ Ω0 , ζ (ω) > T . Then by the continuity of the path, we have inft ∈[0,T ] ∥zt (ω)∥H0 > 0, and T



∥zt ∥2H0 ξt2

0

dt = +∞.

(3.40)

Combining this fact with the Fatou lemma and (3.36), it leads to a contradiction. Hence, ζ ≤ T , Q-a.s. By the uniqueness of the solution of Eq. (3.32), we have zt ≡ 0, t > ζ , Q-a.s. Therefore, zT = 0, Q-a.s. Next, we shall construct the coupling. Let yt be the unique mild solution of the following equation

˜ t, dyt = −Aϵ yt dt + F (t , yt )dt + B(t , yt )dW

y0 = y.

(3.41)

ϵ

xt is also the solution of the following equation dxϵt = −Aϵ xϵt dt + F (t , xϵt )dt −

zt

ξt

˜ t, dt + B(t , xϵt )dW

xϵ0 = x,

(3.42)

then the process xϵt − yt is the mild solution of the following equation in H

˜t − dut = −Aϵ ut dt + F (t , ut , 0)dt + Bˆ (t , ut )dW Since EQ

T 0

∥zt ∥2H

0

ξt2

zt

ξt

dt ,

z0 = x − y.

(3.43)

dt < ∞ and Bˆ (t , h) is a Hilbert–Schmidt operator on H for all h ∈ H, it is easy to prove that Eq. (3.43)

has a unique variational strong solution in H by Liu and Röckner (2010, Theorem 1.1) (see also Prévôt and Röckner, 2007), which is just the process xϵt − yt . Furthermore, zt is also a variational solution of (3.43) in H0 according to (3.18), (3.32) and Lemma 3.4. By the relation (3.25), zt is an analytic weak solution of (3.43) in H, and also a mild solution of Eq. (3.43) (see Prévôt and Röckner, 2007, pp. 133–135). Then by uniqueness, zt = xϵt − yt , t ∈ [0, T ], Q-a.s. Just as in Wang (2011), we establish the log-Harnack inequality for PTϵ . Then by Lemma 3.2, letting ϵ ↓ 0 and choosing θ = 1, we establish the log-Harnack inequality for Eq. (2.1): PT log f (y) ≤ log PT f (x) +

K2 ∥x − y∥2H0 2ρ(T )2 (1 − e−K2 T )

,

f ∈ Cb (H ), f ≥ 1, x, y ∈ H , x − y ∈ H0 .

(3.44)

Since PT (x, ·)+PT (y, ·) is a regular measure on H, where PT (z , dz˜ ) means the transition probability measure for PT , f ∈ Bb (H ) can be approximated by bounded continuous functions in L1 (PT (x, ·) + PT (y, ·)). Then the inequality (3.44) holds for f ∈ Bb (H ), f ≥ 1. If (H5) holds in addition, we have

 

s∧τn

Es,n exp h 0

∥zt ∥2H0 ξt2

 dt

 ≤ exp



h∥x − y∥2H0

θ ξ0

 Es,n exp

θ ξ0 

≤ exp

h∥x − y∥2H0



 Es,n

exp

2h

θ 

s∧τn

 0

8h2 K32

θ2

1

ξt

 ˆ ˜ ⟨B(t , zt )dWt , zt ⟩

s∧τn

 0

∥zt ∥2H0 ξt2

 12 dt

.

(3.45)

1192

S.-Q. Zhang / Statistics and Probability Letters 83 (2013) 1184–1192

Choosing h =

θ2

8K32

 Es,n exp

, we have

θ2 8K32

s∧τn



∥zt ∥2H0 ξt2

0

 dt

θ K2 ∥x − y∥2H0

 ≤ exp



4K32 (2 − θ )(1 − e−K2 T )

.

(3.46)

Similar to Wang (2011), we get that

θ K2 (2K3 + θ ρ(T ))∥x − y∥2H0

 sup

s∈[0,T ]

+r ERs1∧τ

≤ exp

8K32 (2 − θ )(K3 + θ ρ(T ))(1 − e−K2 T )

For p > (1 + K3 )2 , δp,T := K3 ∨

ρ(T ) 2

√

 .

2K ρ(T ) p − 1 , choosing θ = √3p−1 , and by Lemma 3.2, we finally obtain



  √ √ p − 1 ∥x − y∥2H0 √   , (PT f (y)) ≤ (PT f (x)) exp 4δp,T p − 1 ρ(T ) − δp,T (1 − e−K2 T ) 

p

p

(3.47)

K2 ( T ) p

for f ∈ Bb+ (H ), x, y ∈ H and x − y ∈ H0 .

(3.48)



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