Stochastic differential equations with coefficients in Sobolev spaces

Texte intégral

(1)Journal of Functional Analysis 259 (2010) 1129–1168 www.elsevier.com/locate/jfa. Stochastic differential equations with coefficients in Sobolev spaces Shizan Fang a,∗ , Dejun Luo b,c , Anton Thalmaier b a I.M.B., BP 47870, Université de Bourgogne, Dijon, France b UR Mathématiques, Université du Luxembourg, 6, rue Richard Coudenhove-Kalergi, L-1359 Luxembourg c Key Laboratory of Random Complex Structures and Data Science, Academy of Mathematics and Systems Science,. Chinese Academy of Sciences, Beijing 100190, China Received 3 February 2010; accepted 24 February 2010 Available online 16 March 2010 Communicated by Paul Malliavin. Abstract. m. j d j =1 Aj (Xt ) dwt + A0 (Xt ) dt on R . The 1,p d diffusion coefficients A1 , . . . , Am are supposed to be in the Sobolev space Wloc (R ) with p > d, and to have linear growth. For the drift coefficient A0 , we distinguish two cases: (i) A0 is a continuous vector field whose distributional divergence δ(A0 ) with respect to the Gaussian measure γd exists, (ii) A0 has Sobolev 1,p 2 2 regularity Wloc for some p > 1. Assume Rd exp[λ0 (|δ(A0 )| + m j =1 (|δ(Aj )| + |∇Aj | ))] dγd < +∞ for some λ0 > 0. In case (i), if the pathwise uniqueness of solutions holds, then the push-forward (Xt )# γd admits a density with respect to γd . In particular, if the coefficients are bounded Lipschitz continuous, then Xt leaves the Lebesgue measure Lebd quasi-invariant. In case (ii), we develop a method used by G. Crippa. We consider the Itô stochastic differential equation dXt =. and C. De Lellis for ODE and implemented by X. Zhang for SDE, to establish existence and uniqueness of stochastic flow of maps. © 2010 Elsevier Inc. All rights reserved. Keywords: Stochastic flows; Sobolev space coefficients; Density; Density estimate; Pathwise uniqueness; Gaussian measure; Ornstein–Uhlenbeck semigroup. * Corresponding author.. E-mail address: fang@u-bourgogne.fr (S. Fang). 0022-1236/$ – see front matter © 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.jfa.2010.02.014.

(2) 1130. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1. Introduction Let A0 , A1 , . . . , Am : Rd → Rd be continuous vector fields on Rd . We consider the following Itô stochastic differential equation on Rd (abbreviated as SDE) dXt =. m . j. Aj (Xt ) dwt + A0 (Xt ) dt,. X0 = x,. (1.1). j =1. where wt = (wt1 , . . . , wtm ) is the standard Brownian motion on Rm . It is a classical fact in the theory of SDE (see [16,17,21,30]) that, if the coefficients Aj are globally Lipschitz continuous, then SDE (1.1) has a unique strong solution which defines a stochastic flow of homeomorphisms on Rd ; however contrary to ordinary differential equations (abbreviated as ODE), the regularity of the homeomorphisms is only Hölder continuity of order 0 < α < 1. Thus it is not clear whether the Lebesgue measure Lebd on Rd admits a density under the flow Xt . In the case where the vector fields Aj , j = 0, 1, . . . , m, are in Cb∞ (Rd , Rd ), the SDE (1.1) defines a flow of diffeomorphisms, and Kunita [21] showed that the measures on Rd which have a strictly positive smooth density with respect to Lebd are quasi-invariant under the flow. This result was recently generalized in [27] to the case where the drift A0 is allowed to be only log-Lipschitz continuous. Studies on SDE beyond the Lipschitz setting attracted great interest during the last years, see for instance [10,13,12,19,20,23,24,29,34,35]. In the context of ODE, existence of a flow of quasi-invariant measurable maps associated to a vector field A0 belonging to Sobolev spaces appeared first in [6]. In the seminal paper [7], Di Perna and Lions developed transport equations to solve ODE without involving exponential integrability of |∇A0 |. On the other hand, L. Ambrosio [1] took advantage of using continuity equations which allowed him to construct quasi-invariant flows associated to vector fields A0 with only BV regularity. In the framework for Gaussian measures, the Di Perna–Lions method was developed in [4], also in [2,11] on the Wiener space. The situation for SDE is quite different: even for vector fields A0 , A1 , . . . , Am in C ∞ with linear growth, if no conditions were imposed on the growth of the derivatives, the SDE (1.1) may not define a flow of diffeomorphisms (see [25,26]). More precisely, let τx be the life time of the solution to (1.1) starting from x. The SDE (1.1) is said to be complete if for each x ∈ Rd , P(τx = +∞) = 1; it is said to be strongly complete if P(τx = +∞, x ∈ Rd ) = 1. The goal in [26] is to construct examples for which the coefficients are smooth, but such that the SDE (1.1) is not strongly complete (see [13,25] for positive examples). Now consider Σ = (w, x) ∈ Ω × Rd : τx (w) = +∞ . Suppose that SDE (1.1) is complete, then for any probability measure μ on Rd , Rd. 1Σ (w, x) dP(w) dμ(x) = 1.. Ω. Thus, by Fubini’s theorem, Ω ( Rd 1Σ (w, x) dμ(x)) dP(w) = 1. It follows that there exists a full measure subset Ω0 ⊂ Ω such that for all w ∈ Ω0 , τx (w) = +∞ holds for μ-almost every x ∈ Rd ..

(3) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1131. Now under the existence of a complete unique strong solution to SDE (1.1), we have a flow of measurable maps x → Xt (w, x). Recently, inspired by previous work due to Ambrosio, Lecumberry and Maniglia [3], Crippa and De Lellis [5] obtained some new type of estimates of perturbation for ODE whose coefficients have Sobolev regularity. More precisely, the absence of Lipschitz condition was filled by 1,1 the following inequality: for f ∈ Wloc (Rd ),.

(4). f (x) − f (y) Cd |x − y| MR |∇f |(x) + MR |∇f |(y) holds for x, y ∈ N c and |x − y| R, where N is a negligible set of Rd and MR g is the maximal function defined by 1 0<rR Lebd (B(x, r)). . MR g(x) = sup. g(y) dy,. B(x,r). where B(x, r) = {y ∈ Rd : |y − x| r}; the classical moment estimate is replaced by estimating the quantity . |Xt (x) − X˜ t (x)| + 1 dx, log σ. B(0,r). where σ > 0 is a small parameter. This method has recently been successfully implemented to SDE by X. Zhang in [36]. The aim in this paper is two-fold: first we shall study absolute continuity of the push-forward measure (Xt )# Lebd with respect to Lebd , once the SDE (1.1) has a unique strong solution; secondly we shall construct strong solutions (for almost all initial values) using the approach mentioned above for SDE with coefficients in Sobolev space. The key point is to obtain an a priori Lp estimate for the density. To this end, we shall work with the standard Gaussian measure γd ; this will be done in Section 2. The main result in Section 3 is the following d growth. Assume Theorem 1.1. Let A0 , A1 , . . . , Am be continuous vector fields on R of linear q that the diffusion coefficients A1 , . . . , Am are in the Sobolev space q>1 D1 (γd ) and that δ(A0 ) exists; furthermore there exists a constant λ0 > 0 such that. . exp λ0. Rd. m. . 2.

(5) 2. δ(Aj ) + |∇Aj |. δ(A0 ) + dγd < +∞.. (1.2). j =1. Suppose that pathwise uniqueness holds for SDE (1.1). Then (Xt )# γd is absolutely continuous with respect to γd and the density is in L1 log L1 . A consequence of this theorem concerns the following classical situation..

(6) 1132. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Theorem 1.2. Let A0 , A1 , . . . , Am be globally Lipschitz continuous. Suppose that there exists a constant C > 0 such that m .

(7) 2. x, Aj (x) C 1 + |x|2. for all x ∈ Rd .. (1.3). j =1. Then the stochastic flow of homeomorphisms Xt generated by SDE (1.1) leaves the Lebesgue measure Lebd quasi-invariant. Remark that condition (1.3) not only includes the case of bounded Lipschitz diffusion coefficients, but also, maybe more significant, indicates the role of dispersion: the vector fields A1 , . . . , Am should not go radially to infinity. The purpose of Section 4 is to find conditions that guarantee strict positivity of the density, in case where existence of the inverse flow is not known, see Theorem 4.4. The main result of Section 5 is Theorem 1.3. Assume that the diffusion coefficients A1 , . . . , Am belong to the Sobolev space. q q D (γ q>1 1 d ) and the drift A0 ∈ D1 (γd ) for some q > 1. Assume condition (1.2) and that the coefficients A0 , A1 , . . . , Am are of linear growth, then there is a unique stochastic flow of measurable maps X : [0, T ] × Ω × Rd → Rd , which solves (1.1) for almost all initial x ∈ Rd and the push-forward (Xt (w, ·))# γd admits a density with respect to γd , which is in L1 log L1 . When the diffusion coefficients satisfy uniform ellipticity, a classical result due to Stroock and Varadhan [32] says that if the diffusion coefficients A1 , . . . , Am are bounded continuous and the drift A0 is bounded Borel measurable, then weak uniqueness holds, that is uniqueness in law of the diffusion. This result was strengthened by Veretennikov [33], saying that in fact pathwise uniqueness holds. When A0 is not bounded, some conditions on the diffusion coefficients were needed. In the case where the diffusion matrix a = (aij ) is the identity, the drift A0 in (1.1) p can be quite singular: A0 ∈ Lloc (Rd ) with p > d + 2 implies that SDE (1.1) has the pathwise uniqueness (see Krylov and Röckner [20] for a more complete study); if the diffusion coeffi1,2(d+1) cients A1 , . . . , Am are bounded continuous, under a Sobolev condition, namely, Aj ∈ Wloc 2(d+1) for j = 1, . . . , m and A0 ∈ Lloc (Rd ), X. Zhang proved in [34] that SDE (1.1) admits a unique strong solution. Note that even in this uniformly non-degenerated case, if the diffusion coefficients lose the continuity, there are counterexamples for which weak uniqueness does not hold, see [19,31]. Finally we would like to mention that under weaker Sobolev type conditions, the connection between weak solutions and Fokker–Planck equations has been investigated in [14,22]; some notions of “generalized solutions”, as well as the phenomena of coalescence and splitting, have been explored in [23,24]. Stochastic transport equations are studied in [15,36]. 2. Lp estimate of the density The purpose of this section is to derive a priori estimates for the density of the push-forwards under the flow. We assume that the coefficients A0 , A1 , . . . , Am of SDE (1.1) are smooth with compact support in Rd . Then the solution Xt , i.e., x → Xt (x), is a stochastic flow of diffeomorphisms on Rd . Moreover SDE (1.1) is equivalent to the following Stratonovich SDE.

(8) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. dXt =. m . j Aj (Xt ) ◦ dwt + A˜ 0 (Xt ) dt,. X0 = x,. 1133. (2.1). j =1. where A˜ 0 = A0 − 12 m j =1 LAj Aj and LA denotes the Lie derivative with respect to A. Let γd be the standard Gaussian measure on Rd , and γt = (Xt )# γd , γ˜t = (Xt−1 )# γd the pushforwards of γd respectively by the flow Xt and its inverse flow Xt−1 . To fix ideas, we denote dγt by (Ω, F , P) the probability space on which the Brownian motion wt is defined. Let Kt = dγ d and K˜ t = dγ˜t be the densities with respect to γd . By Lemma 4.3.1 in [21], the Radon–Nikodym dγd. derivative K˜ t has the following explicit expression K˜ t (x) = exp −. m . t.

(9) j δ(Aj ) Xs (x) ◦ dws −. j =1 0. t. .

(10) δ(A˜ 0 ) Xs (x) ds ,. (2.2). 0. where δ(Aj ) denotes the divergence of Aj with respect to the Gaussian measure γd : . ∇ϕ, Aj

(11) dγd =. Rd. ϕδ(Aj ) dγd ,.

(12) ϕ ∈ Cc1 Rd .. Rd. It is easy to see that Kt and K˜ t are related to each other by the equality below:

(13) −1 Kt (x) = K˜ t Xt−1 (x) .. (2.3). In fact, for any ψ ∈ Cc∞ (Rd ), we have . ψ(x) dγd (x) =. Rd.

(14) ψ Xt Xt−1 (x) dγd (x). Rd. . =. ψ Xt (y) K˜ t (y) dγd (y). Rd. . =.

(15) ψ(x)K˜ t Xt−1 (x) Kt (x) dγd (x),. Rd. which leads to (2.3) due to the arbitrariness of ψ ∈ Cc∞ (Rd ). In the following we shall estimate the Lp (P × γd ) norm of Kt . We rewrite the density (2.2) with the Itô integral: K˜ t (x) = exp −. m j =1 0. t.

(16) j δ(Aj ) Xs (x) dws −. t m.

(17) 1 LAj δ(Aj ) + δ(A˜ 0 ) Xs (x) ds . 2 0. j =1. (2.4).

(18) 1134. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Lemma 2.1. We have 1 1 1 ∇Aj , (∇Aj )∗ , LAj δ(Aj ) + δ(A˜ 0 ) = δ(A0 ) + |Aj |2 + 2 2 2 m. m. m. j =1. j =1. j =1. (2.5). where ·,·

(19) denotes the inner product of Rd ⊗ Rd and (∇Aj )∗ the transpose of ∇Aj . Proof. Let A be a C 2 vector field on Rd . From the expression. δ(A) =. d . xk Ak −. k=1. ∂Ak , ∂xk. we get. LA δ(A) =. d . A Ak δk + A xk. ,k=1. ∂Ak ∂ 2 Ak . − A ∂x ∂x ∂xk. (2.6). Note that k ∂ ∂Ak ∂A ∂ 2 Ak ∂A A = + A . ∂xk ∂x ∂x ∂xk ∂xk ∂x Thus, by means of (2.6), we obtain LA δ(A) = |A|2 + δ(LA A) + ∇A, (∇A)∗ .. (2.7). Recall that δ(A˜ 0 ) = δ(A0 ) − 12 m j =1 δ(LAj Aj ). Hence, replacing A by Aj in (2.7) and summing over j , gives formula (2.5). 2 We can now prove the following key estimate. Theorem 2.2. For p > 1, Kt Lp (P×γd ) Rd. p−1 m p(2p−1). .

(20) 2 |Aj |2 + |∇Aj |2 + 2(p − 1) δ(Aj ). dγd exp pt 2 δ(A0 ) + . . . j =1. (2.8).

(21) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1135. Proof. Using relation (2.3), we have . p E Kt (x) dγd (x) = E. Rd. . −1

(22) −p K˜ t Xt (x) dγd (x). Rd. . =E . −p K˜ t (y) K˜ t (y) dγd (y). Rd. =.

(23) −p+1 dγd (x). E K˜ t (x). (2.9). Rd. To simplify the notation, denote the right-hand side of (2.5) by Φ. Then K˜ t (x) rewrites as m t t .

(24) j.

(25) ˜ Kt (x) = exp − δ(Aj ) Xs (x) dws − Φ Xs (x) ds . j =1 0. 0. Fixing an arbitrary r > 0, we get m t t .

(26) −r.

(27) j.

(28) K˜ t (x) = exp r δ(Aj ) Xs (x) dws + r Φ Xs (x) ds j =1 0. . 0. m t . = exp r. m .

(29) j δ(Aj ) Xs (x) dws − r 2. j =1 0. t.

(30). δ(Aj ) Xs (x) 2 ds. . j =1 0. t . m .

(31) 2. δ(Aj ) + rΦ Xs (x) ds . r2. × exp. j =1. 0. By Cauchy–Schwarz’s inequality, 1/2 m t m t .

(32) −r .

(33) j.

(34). 2. δ(Aj ) Xs (x) ds E K˜ t (x) E exp 2r δ(Aj ) Xs (x) dws − 2r 2 j =1 0. . j =1 0. t . 1/2 m . 2.

(35). δ(Aj ) + 2rΦ Xs (x) ds 2r. × E exp. 2. j =1. 0. . t . = E exp. 1/2 m . 2.

(36). δ(Aj ) + 2rΦ Xs (x) ds , 2r 2. 0. j =1. (2.10).

(37) 1136. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. since the first term on the right-hand side of the inequality in (2.10) is the expectation of a martingale. Let m . 2

(38). Φ˜ r = 2r δ(A0 ) + r |Aj |2 + |∇Aj |2 + 2r δ(Aj ) . j =1. Then by (2.10), along with the definition of Φ and Cauchy–Schwarz’s inequality, we obtain .

(39) −r dγd E K˜ t (x). Rd. . t E exp. Rd. 1/2.

(40) ˜ Φr Xs (x) ds dγd .. (2.11). 0. Following the idea of A.B. Cruzeiro ([6, Corollary 2.2], see also Theorem 7.3 in [8]) and by Jensen’s inequality, t exp. t t.

(41)

(42) ds 1 ˜ t Φ˜ r Xs (x) et Φr (Xs (x)) ds. Φ˜ r Xs (x) ds = exp t t. 0. 0. . Define I (t) = sup0st by Hölder’s inequality, . Rd. t E exp. Rd. 0. p. E[Kt (x)] dγd . Integrating on both sides of the above inequality and. t .

(43) 1 ˜ Φ˜ r Xs (x) ds dγd (x) E et Φr (Xs (x)) dγd (x) ds t. 0. 1 = t. 0. Rd. t. E. 0. 1 t. t. ˜. et Φr (y) Ks (y) dγd (y) ds. Rd. t Φ˜ e r . Lq (γd ). Ks Lp (P×γd ) ds. 0. ˜ et Φr . Lq (γd ). I (t)1/p ,. where q is the conjugate number of p. Thus it follows from (2.11) that . ˜ 1/2

(44) −r dγd (x) et Φr Lq (γ ) I (t)1/2p . E K˜ t (x) d. Rd. Taking r = p − 1 in the above estimate and by (2.9), we obtain . ˜ 1/2 p E Kt (x) dγd (x) et Φp−1 Lq (γ ) I (t)1/2p . d. Rd. (2.12).

(45) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168 ˜. 1137. 1/2. Thus we have I (t) et Φp−1 Lq (γd ) I (t)1/2p . Solving this inequality for I (t) gives . p E Kt (x) dγd (x) I (t) . Rd. Rd. . p−1 2p−1 pt ˜ Φp−1 (x) dγd (x) exp . p−1. Now the desired estimate follows from the definition of Φ˜ p−1 .. 2. Corollary 2.3. For any p > 1, K˜ t Lp (P×γd ) . 1 m 2p+1. . 2

(46). 2 2. |Aj | + |∇Aj | + 2p δ(Aj ) dγd exp (p + 1)t 2 δ(A0 ) + .. . . j =1. Rd. (2.13) Proof. Similar to (2.12), we have for r > 0, ˜ 1/2

(47) r E K˜ t (x) dγd (x) et Φr Lq (γ ) I (t)1/2p ,. (2.14). d. Rd p/(2p−1) ˜ where Φ˜ r and I (t) are defined as above. Since I (t) et Φp−1 Lq (γd ) , by taking r = p − 1, we get

(48) p−1 dγd (x) E K˜ t (x) Rd. ˜ p/(2p−1) et Φp−1 Lq (γ ) d. = Rd. p−1 m 2p−1. .

(49) 2 |Aj |2 + |∇Aj |2 + 2(p − 1) δ(Aj ). dγd exp pt 2 δ(A0 ) + . . . j =1. Replacing p by p + 1 in the last inequality gives the claimed estimate.. 2. 3. Absolute continuity under flows generated by SDEs Now assume that the coefficients Aj in SDE (1.1) are continuous and of linear growth. Then it is well known that SDE (1.1) has a weak solution of infinite life time. In order to apply the results of the preceding section, we shall regularize the vector fields using the Ornstein–Uhlenbeck semigroup {Pε }ε>0 on Rd : .

(50) Pε A(x) = A e−ε x + 1 − e−2ε y dγd (y). Rd. We have the following simple properties..

(51) 1138. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Lemma 3.1. Assume that A is continuous and |A(x)| C(1 + |x|q ) for some q 0. Then (i) there is Cq > 0 independent of ε, such that.

(52). Pε A(x) Cq 1 + |x|q ,. for all x ∈ Rd ;. (ii) Pε A converges uniformly to A on any compact subset as ε → 0. √ Proof. (i) Note that |e−ε x + 1 − e−2ε y| |x| + |y| and that there exists a constant C > 0 such that (|x| + |y|)q C(|x|q + |y|q ). Using the growth condition on A, we have for some constant C > 0 (depending on q), . Pε A(x) . . −ε

(53). A e x + 1 − e−2ε y dγd (y). Rd. . C.

(54)

(55). 1 + |x|q + |y|q dγd (y) C 1 + |x|q + Mq. Rd. where Mq = Rd |y|q dγd (y). Changing the constant yields (i). (ii) Fix R > 0 and x in the closed ball B(R) of radius R, centered at 0. Let R1 > R be arbitrary. We have. Pε A(x) − A(x) . . . −ε

(56). A e x + 1 − e−2ε y − A(x) dγd (y). Rd. . +. = B(R1 ). . −ε

(57). A e x + 1 − e−2ε y − A(x) dγd (y). B(R1 )c. =: I1 + I2 .. (3.1). By the growth condition on A, for some constant Cq > 0, independent of ε, we have . .

(58). −ε

(59). A e x + 1 − e−2ε y + A(x) dγd (y). I2 B(R1 )c. . Cq.

(60) 1 + R q + |y|q dγd (y),. B(R1 )c. where the last term tends to 0 as R1 → +∞. For given η > 0, we may take R1 large enough such that I2 < η. Then there exists εR1 > 0 such that for ε < εR1 and |y| R1 , . −ε. e x + 1 − e−2ε y e−ε R + 1 − e−2ε R1 R1 ..

(61) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1139. Note that √. −ε. e x + 1 − e−2ε y − x εR + 2εR1 ,. for |x| R, |y| R1 .. Since A is uniformly continuous on B(R1 ), there exits ε0 εR1 such that . −ε

(62). A e x + 1 − e−2ε y − A(x) η. for all y ∈ B(R1 ), ε ε0 .. As a result, the term I1 η. Therefore by (3.1), for any ε ε0 ,. sup Pε A(x) − A(x) 2η.. |x|R. The result follows from the arbitrariness of η > 0.. 2. The vector field Pε A is smooth on Rd but does not have compact support. We introduce cut-off functions ϕε ∈ Cc∞ (Rd , [0, 1]) satisfying 1 ϕε (x) = 1 if |x| , ε. ϕε (x) = 0 if |x| . 1 +2 ε. and ∇ϕε ∞ 1.. Set Aεj = ϕε Pε Aj ,. j = 0, 1, . . . , m.. Now consider the Itô SDE (1.1) with Aj being replaced by Aεj (j = 0, 1, . . . , m), and denote the corresponding terms by adding the superscript ε, e.g. Xtε , Ktε , etc. In the sequel, we shall give a uniform estimate to Ktε . To this end, we need some preparations in the spirit of Malliavin calculus [28]. For a vector field A on Rd and p > 1, we say that p A ∈ D1 (γd ) if A ∈ Lp (γd ) and if there exists ∇A : Rd → Rd ⊗ Rd in Lp (γd ) such that for any d v∈R , A(x + ηv) − A(x) η→0 η. ∇A(x)(v) = ∂v A := lim. . holds in Lp (γd ) for any p < p.. p. For such A ∈ D1 (γd ), the divergence δ(A) ∈ Lp (γd ) exists and the following relations hold: ∇Pε A = e−ε Pε (∇A),.

(63) δ(Pε A) = eε Pε δ(A) .. (3.2). Note that the second term in (3.2) holds once the divergence δ(A) ∈ Lp exists for some p > 1. If p A ∈ Lp (γd ), then Pε A ∈ D1 (γd ) and limε→0 Pε A − ALp = 0. Lemma 3.2. Assume the vector field A ∈ Lp (γd ) admits the divergence δ(A) ∈ Lp (γd ) for p > 1, and denote by Aε = ϕε Pε A. Then for ε ∈ ]0, 1],.

(64) 1140. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168.

(65). ε

(66). δ A Pε |A| + e δ(A) ,. ε 2.

(67). A Pε |A|2 ,.

(68) . ε

(69) 2. . δ A Pε 2 |A|2 + e2 δ(A) 2 . p. If furthermore A ∈ D1 (γd ), then.

(70) . ∇Aε 2 Pε 2 |A|2 + |∇A|2 . Proof. Note that according to (3.2), δ(Aε ) = δ(ϕε Pε A) = ϕε eε Pε δ(A) − ∇ϕε , Pε A

(71) , from where the first inequality follows. In the same way, the other results are obtained. 2 Applying Theorem 2.2 to Ktε with p = 2, we have . ε K t. L2 (P×γd ). . 1/6 m. ε

(72) . ε 2.

(73).

(74) 2 2. A + ∇Aε + 2 δ Aε. dγd exp 2t 2 δ A0 + . j j j . j =1. Rd. (3.3) By Lemma 3.2, m.

(75) .

(76)

(77). ε 2. A + ∇Aε 2 + 2 δ Aε 2 2 δ Aε0 + j j j. Pε. j =1. m. . 2

(78). 2 2 2. 7|Aj | + 2|∇Aj | + 4e δ(Aj ) . 2|A0 | + 2e δ(A0 ) + j =1. We deduce from Jensen’s inequality and the invariance of γd under the action of the semigroup Pε that ε K t. L2 (P×γd ). . 1/6 m. .

(79) 2 4|Aj |2 + |∇Aj |2 + 2e2 δ(Aj ). dγd exp 4t |A0 | + e δ(A0 ) + . Rd. j =1. (3.4) for any ε 1. According to (3.4), we consider the following conditions. Assumptions (H).. q (A1) For j = 1, . . . , m, Aj ∈ q1 D1 (γd ), A0 is continuous and δ(A0 ) exists. (A2) The vector fields A0 , A1 , . . . , Am have linear growth..

(80) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1141. (A3) There exists λ0 > 0 such that . m. . 2. δ(Aj ). exp λ0 δ(A0 ) + dγd < +∞. j =1. Rd. (A4) There exists λ0 > 0 such that . . exp λ0. m . |∇Aj |. 2. dγd < +∞.. j =1. Rd. Note that by Sobolev’s embedding theorem, the diffusion coefficients A1 , . . . , Am admit Hölder continuous versions. In what follows, we consider these continuous versions. It is clear that under the conditions (A2)–(A4), there exists T0 > 0 small enough, such that ΛT0 :=. . . exp 4T0 |A0 | + e δ(A0 ). Rd. 1/6 m .

(81) 2 + dγd 4|Aj |2 + |∇Aj |2 + 2e2 δ(Aj ). < ∞.. (3.5). j =1. In this case, for t ∈ [0, T0 ], sup Ktε L2 (P×γ ) ΛT0 .. (3.6). d. 0<ε1. Theorem 3.3. Let T > 0 be given. Under (A1)–(A4) in Assumptions (H), there are two positive constants C1 and C2 , independent of ε, such that sup E 0<ε1. Ktε log Ktε dγd 2(C1 T )1/2 ΛT0 + C2 T Λ2T0 ,. for all t ∈ [0, T ].. Rd. Proof. We follow the arguments of Proposition 4.4 in [11]. By (2.3) and (2.4), we have m t t .

(82) .

(83) .

(84)

(85) −1 j Ktε Xtε (x) = K˜ tε (x) = exp δ Aεj Xsε (x) dws + Φε Xsε (x) ds , j =1 0. 0. where m m. ε 2 1

(86) 1.

(87) ∗ . A + Φε = δ Aε0 + ∇Aεj , ∇Aεj . j 2 2 j =1. Thus. j =1.

(88) 1142. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. . . Ktε log Ktε dγd. E. =E. Rd.

(89). log K ε X ε (x) dγd (x) t. t. Rd. . m t ε

(90) ε

(91) j. δ Aj Xs (x) dws dγd (x) E. Rd. j =1 0. . t ε

(92) . + E Φε Xs (x) ds dγd (x). Rd 0. =: I1 + I2 .. (3.7). Using Burkholder’s inequality, we get. m t. m t 1/2 .

(93) ε

(94) ε

(95) j.

(96) 2. ε ε. δ A X (x) ds E. δ Aj Xs (x) dws 2E . s j. j =1 0. j =1 0. For the sake of simplifying the notations, write Ψε = t I1 2. E. m. ε 2 j =1 |δ(Aj )| .. ε

(97). Ψε X (x) dγd (x) ds. By Cauchy’s inequality,. 1/2 .. s. (3.8). Rd. 0. α Now we are going to estimate E Rd |Ψε (Xsε (x))|2 dγd (x) for α ∈ Z+ which will be done inductively. First if s ∈ [0, T0 ], then by (3.4) and (3.6), along with Cauchy’s inequality, E. ε

(98) 2α. Ψε X (x) dγd (x) = E s. Rd. . α. Ψε (y) 2 K ε (y) dγd (y) s. Rd α. Ψε 2 2α+1 L. (γd ). ε K . s L2 (P×γd ). α. ΛT0 Ψε 2 2α+1 L. (γd ). .. (3.9). Now for s ∈ ]T0 , 2T0 ], we shall use the flow property of Xsε : let (θT0 w)t := wT0 +t − wT0 and ε,T Xt 0 be the solution of the Itô SDE driven by the new Brownian motion (θT0 w)t , then ε,T0 . XTε 0 +t (x, w) = Xt ε,T0. and Xt.

(99) XTε 0 (x, w), θT0 w ,. enjoys the same properties as Xtε . Therefore,. for all t 0,.

(100) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. E. ε

(101) 2α. Ψε X (x) dγd (x) = E. . s. Rd. 1143. ε,T0 ε

(102)

(103) 2α. dγd (x). Ψε X s−T0 XT0 (x). Rd. . =E. ε,T0

(104) 2α ε. K (y) dγd (y). Ψε X T0 s−T0 (y). Rd. which is dominated, using Cauchy–Schwarz inequality 1/2. ε,T ε

(105) 2α+1 K 2 E Ψε Xs−T00 (y). dγd (y) T0 L (P×γ. d). Rd. α+1 ΛT0 Ψε 2 2α+2 L.

(106) 1/2 (γd ). −1. α. ΛT0 = Λ1+2 Ψε 2 2α+2 T0 L. (γd ). .. Repeating this procedure, we finally obtain, for all s ∈ [0, T ], E. ε

(107) 2α. Ψε X (x) dγd (x) Λ1+2−1 +···+2−N+1 Ψε 2α α+N s. T0. L2. (γd ). ,. Rd. where N ∈ Z+ is the unique integer such that (N − 1)T0 < T N T0 . In particular, taking α = 0 gives E. ε

(108). Ψε X (x) dγd (x) Λ2 Ψε 2N . s T0 L (γ ). (3.10). d. Rd. By Lemma 3.2, m . 2

(109) . 2 2. |Aj | + e δ(Aj ) sup Ψε L2N (γ ) 2 d 0<ε1 j =1. =: C1 N. L2 (γd ). whose right-hand side is finite under the assumptions (A2)–(A4). This along with (3.8) and (3.10) leads to I1 2(C1 T )1/2 ΛT0 .. (3.11). The same manipulation works for the term I2 and we get I2 C2 T Λ2T0 ,. (3.12). where m m . 3 2 2 C2 = |A0 | + e δ(A0 ) + |Aj | + |∇Aj | 2 j =1. N L2 (γd ). j =1. Now we draw the conclusion from (3.7), (3.11) and (3.12).. 2. < ∞..

(110) 1144. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. It follows from Theorem 3.3 that the family {K.ε }0<ε1 is weakly compact in L1 ([0, T ] × Ω × Rd ). Along a subsequence, K.ε converges weakly to some K. ∈ L1 ([0, T ] × Ω × Rd ) as ε → 0. Let .

(111) 1 d 1/2 2 E[ut log ut ] dγd 2(C1 T ) ΛT0 + C2 T ΛT0 . C = u ∈ L [0, T ] × Ω × R : ut 0, Rd. By convexity of the function s → s log s, it is clear that C is a convex subset of L1 ([0, T ] × Ω × Rd ). Since the weak closure of C coincides with the strong one, there exists a sequence of functions u(n) ∈ C which converges to K in L1 ([0, T ] × Ω × Rd ). Along a subsequence, u(n) converges to K almost everywhere. Hence by Fatou’s lemma, we get for almost all t ∈ [0, T ], E(Kt log Kt ) dγd 2(C1 T )1/2 ΛT0 + C2 T Λ2T0 .. (3.13). Rd. Theorem 3.4. Assume conditions (A1)–(A4) and that pathwise uniqueness holds for SDE (1.1). Then for each t > 0, there is a full subset Ωt ⊂ Ω such that for all w ∈ Ωt , the density Kˆ t of (Xt )# γd with respect to γd exists and Kˆ t ∈ L1 log L1 . Proof. Under these assumptions, we can use Theorem A in [18]. For the convenience of the reader, we include the statement: Theorem 3.5. (See [18].) Let σn (x) and bn (x) be continuous, taking values respectively in the space of (d × m)-matrices and Rd . Suppose that .

(112).

(113) sup σn (x) + bn (x) C 1 + |x| , n. and for any R > 0, lim. .

(114) sup σn (x) − σ (x) + bn (x) − b(x) = 0.. n→+∞ |x|R. Suppose further that for the same Brownian motion B(t), Xn (x, t) solves the SDE.

(115)

(116) dXn (t) = σn Xn (t) dB(t) + bn Xn (t) dt,. Xn (0) = x.. If pathwise uniqueness holds for.

(117).

(118) dX(t) = σ X(t) dB(t) + b X(t) dt,. X(0) = x,. then for any R > 0, T > 0, lim. sup E. n→+∞ |x|R. . 2 . sup Xn (t, x) − X(t, x) = 0. 0tT. (3.14).

(119) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1145. We continue the proof of Theorem 3.4. By means of Lemma 3.1 and Theorem 3.5, for any T , R > 0, we get lim sup E. . ε→0 |x|R. 2 . sup Xtε (x) − Xt (x) = 0.. (3.15). 0tT. Now fixing arbitrary ξ ∈ L∞ (Ω) and ψ ∈ Cc∞ (Rd ), we have E. .

(120).

(121). ξ(·). ψ X ε (x) − ψ Xt (x) dγd (x) t. Rd. . +. ξ ∞.

(122).

(123). E ψ Xtε (x) − ψ Xt (x) dγd (x). B(R)c. B(R). =: J1 + J2 .. (3.16). By (3.15), . E Xtε (x) − Xt (x) dγd (x). J1 ξ ∞ ∇ψ∞ B(R). ξ ∞ ∇ψ∞. . sup E. |x|R. . 2 1/2. sup Xtε (x) − Xt (x). → 0,. (3.17). 0tT. as ε tends to 0. It is obvious that.

(124) J2 2ξ ∞ ψ∞ γd B(R)c .. (3.18). Combining (3.16), (3.17) and (3.18), we obtain . .

(125)

(126).

(127). lim sup E |ξ | ψ X ε (x) − ψ Xt (x) dγd (x) 2ξ ∞ ψ∞ γd B(R)c → 0 ε→0. t. Rd. as R ↑ ∞. Therefore lim E. ε→0.

(128) ξ ψ Xtε (x) dγd (x) = E. Rd. .

(129) ξ ψ Xt (x) dγd .. (3.19). Rd. On the other hand, by Theorem 3.3, for each fixed t ∈ [0, T ], up to a subsequence, Ktε converges weakly in L1 (Ω × Rd ) to some Kˆ t , hence ε

(130) E ξ ψ Xt (x) dγd (x) = E ξ ψ(y)Ktε (y) dγd (y) Rd. Rd. . →E Rd. ξ ψ(y)Kˆ t (y) dγd (y).. (3.20).

(131) 1146. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. This together with (3.19) leads to E.

(132) ξ ψ Xt (x) dγd (x) = E. Rd. . ξ ψ(y)Kˆ t (y) dγd (y).. Rd. By the arbitrariness of ξ ∈ L∞ (Ω), there exists a full measure subset Ωψ of Ω such that .

(133) ψ Xt (x) dγd (x) =. Rd. . ψ(y)Kˆ t (y) dγd (y),. for any ω ∈ Ωψ .. Rd. Now by the separability of Cc∞ (Rd ), there exists a full subset Ωt such that the above equality holds for any ψ ∈ Cc∞ (Rd ). Hence (Xt )# γd = Kˆ t γd . 2 Remark 3.6. The Kt (w, x) appearing in (3.13) is defined almost everywhere. It is easy to see that Kt (w, x) is a measurable modification of {Kˆ t (w, x); t ∈ [0, T ]}. Remark 3.7. Beyond the Lipschitz condition, several sufficient conditions guaranteeing pathwise uniqueness for SDE (1.1) can be found in the literature. For example in [12], the authors give the condition m .

(134). Aj (x) − Aj (y) 2 C|x − y|2 r |x − y|2 , j =1.

(135). A0 (x) − A0 (y) C|x − y|r |x − y|2 ,. for |x − y| c0 small enough, where r : ]0, c0 ] → ]0, +∞[ is C 1 satisfying (i) lims→0 r(s) = +∞, (s) (ii) lims→0 srr(s) = 0, and c0 ds (iii) 0 sr(s) = +∞. Corollary 3.8. Suppose that the vector fields A0 , A1 , . . . , Am are globally Lipschitz continuous and there exists a constant C > 0, such that m .

(136) 2. x, Aj (x) C 1 + |x|2. for all x ∈ Rd .. (3.21). j =1. Then (Xt )# Lebd Lebd for any t ∈ [0, T ]. Proof. It is obvious that hypotheses (A1), (A2) and (A4) are satisfied, and that for some constant C > 0,.

(137). δ(A0 ) (x) C 1 + |x|2 ..

(138) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Hence there exists λ0 > 0 such that. Rd. 1147. exp(λ0 |δ(A0 )|) dγd < +∞. Finally we have. m m m . 2. δ(Aj ) 2 (x) 2 x, Aj (x) + 2 Lip(Aj )2 . j =1. j =1. j =1. Therefore, under condition (3.21), there exists λ0 > 0 such that . . m . 2. δ(Aj ) exp λ0 dγd < +∞. j =1. Rd. ˆ Hence, hypothesis (A3) is satisfied as well. By Theorem 3.4, we have (X t )# γd = Kt γd . Let A be a d Borel subset of R such that Lebd (A) = 0, then γd (A) = 0; therefore Rd 1{Xt (x)∈A} dγd (x) = 0. It follows that 1{Xt (x)∈A} = 0 for Lebd almost every x, which implies Lebd (Xt ∈ A) = 0; this means that (Xt )# Lebd is absolutely continuous with respect to Lebd . 2 In the next section, we shall prove that under the conditions of Corollary 3.8, the density of (Xt )# Lebd with respect to Lebd is strictly positive, in other words, Lebd is quasi-invariant under Xt . Corollary 3.9. Assume that conditions (A1)–(A4) hold. Let σ = (Aij ) and suppose that for some C > 0, σ (x)σ (x)∗ C Id,. for all x ∈ Rd .. Then (Xt )# γd is absolutely continuous with respect to γd . Proof. The conditions (A1)–(A4) are stronger than those in Theorem 1.1 of [34] given by X. Zhang, so the pathwise uniqueness holds. Hence Theorem 3.4 applies to this case. 2 4. Quasi-invariance under stochastic flow In the sequel, by quasi-invariance we mean that the Radon–Nikodym derivative of the corresponding push-forward measure is strictly positive. First we prove that in the situation of Corollary 3.8, the Lebesgue measure is in fact quasi-invariant under the stochastic flow of homeomorphisms. To this end, we need some preparations. In what follows, T0 > 0 is chosen small enough such that (3.5) holds. Proposition 4.1. Let q 2. Then lim. ε→0 Rd. q . m t. ε

(139) ε

(140) j. δ Aj Xs − δ(Aj )(Xs ) dws dγd = 0. E sup. 0tT0. j =1 0. (4.1).

(141) 1148. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Proof. By Burkholder’s inequality,. m t. q .

(142)

(143). . j E sup. δ Aεj Xsε − δ(Aj )(Xs ) dws. 0tT0 . j =1 0. T0 CE. m . ε

(144) ε

(145). δ A X − δ(Aj )(Xs ) 2 ds s j. q/2 . 0 j =1 m . T0. q/2−1 CT0. q

(146).

(147)

(148) E δ Aεj Xsε − δ(Aj )(Xs ) ds.. j =1 0. Again by the inequality (a + b)q Cq (a q + bq ), there exists a constant Cq,T0 > 0 such that the above quantity is dominated by T0 T0 m . q

(149). ε

(150) ε

(151) ε

(152) q

(153). ε

(154) Cq,T0 E δ Aj Xs − δ(Aj ) Xs ds + E δ(Aj ) Xs − δ(Aj )(Xs ) ds . j =1. 0. 0. (4.2) Let I1ε and I2ε be the two terms in the squared bracket of (4.2). Note that .

(155)

(156)

(157) q

(158) E δ Aεj Xsε − δ(Aj ) Xsε dγd = E. Rd. . ε

(159). δ A − δ(Aj ) q K ε dγd s. j. Rd.

(160) q δ Aεj − δ(Aj )L2q (γ ) Ksε L2 (P×γ ) . d. According to (3.5), for s T0 , we have Ksε L2 (P×γd ) ΛT0 . Remark that

(161) δ Aεj = δ(ϕε Pε Aj ) = ϕε eε Pε δ(Aj ) − ∇ϕε , Pε Aj

(162) , which converges to δ(Aj ) in L2q (γd ). By (4.3), T0 . I1ε dγd Rd. = 0. .

(163)

(164)

(165) q

(166) E δ Aεj Xsε − δ(Aj ) Xsε dγd ds. Rd. q

(167) T0 ΛT0 δ Aεj − δ(Aj )L2q (γ. d). which tends to 0 as ε → 0.. d. (4.3).

(168) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. For the estimate of I2ε , we remark that exists ψ ∈ Cc (Rd ) such that . Rd. 1149. |δ(Aj )|2q dγd < +∞. Let η > 0 be given. There. δ(Aj ) − ψ 2q dγd η2 .. Rd. We have, for some constant Cq > 0, . q

(169).

(170) E δ(Aj ) Xsε − δ(Aj )(Xs ) dγd. Rd. Cq .

(171) q

(172).

(173) E δ(Aj ) Xsε − ψ Xsε dγd +. Rd. +. . q

(174).

(175) E ψ Xsε − ψ(Xs ) dγd. Rd. . q

(176). E ψ(Xs ) − δ(Aj )(Xs ) dγd .. (4.4). Rd. Again by (3.6), we find E. .

(177).

(178). δ(Aj ) X ε − ψ X ε q dγd = E. δ(Aj ) − ψ q K ε dγd s s s. Rd. Rd. q δ(Aj ) − ψ 2q. L (γd ). ΛT0 ΛT0 η.. In the same way, E. . δ(Aj )(Xs ) − ψ(Xs ) q dγd ΛT η. 0. Rd. To estimate the second term on the right-hand side of (4.4), we use Theorem 3.5: from (3.14), we see that up to a subsequence, Xsε (w, x) converges to Xs (w, x), for each s T0 and almost all (w, x) ∈ Ω × Rd . By Lebesgue’s dominated convergence theorem, lim. ε→0 Rd. In conclusion, limε→0. . ε Rd I2 dγd. q

(179).

(180) E ψ Xsε − ψ(Xs ) dγd = 0.. = 0. According to (4.2), the proof of (4.1) is complete.. 2. Proposition 4.2. Let Φ be defined by 1 1 Φ = δ(A0 ) + ∇Aj , (∇Aj )∗ , |Aj |2 + 2 2 m. m. j =1. j =1. (4.5).

(181) 1150. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. and analogously Φε where Aj is replaced by Aεj . Then T0 lim. ε→0 Rd 0. q

(182).

(183) E Φε Xsε − Φ(Xs ) ds dγd = 0.. (4.6). Proof. Along the lines of the proof of Proposition 4.1, it is sufficient to remark that lim Φε − ΦL2q (γd ) = 0.. (4.7). ε→0. To see this, let us check convergence for the last term in the definition of Φε . We have.

(184) . ∇Aε , ∇Aε ∗ − ∇Aj , (∇Aj )∗. j j ∇Aεj − ∇Aj ∇Aεj + ∇Aj ∇Aεj − ∇Aj . Note that Aεj = ϕε Pε Aj . Thus ∇Aεj = ∇ϕε ⊗ Pε Aj + e−ε ϕε Pε (∇Aj ), which converges to ∇Aj in L2q (γd ) as ε → 0.. 2. Now we can prove Proposition 4.3. Under the conditions of Corollary 3.8, the Lebesgue measure Lebd is quasiinvariant under the stochastic flow. Proof. Let kt be the density of (Xt )# Lebd with respect to Lebd . We shall prove that kt is strictly positive. Set K˜ tε (x) = exp. −. m . t.

(185)

(186) j δ Aεj Xsε (x) dws −. j =1 0. t. ε

(187) Φε Xs (x) ds ,. (4.8). 0. where Φε is defined in Proposition 4.2. By (2.3) we have .

(188) ψ Xtε K˜ tε dγd =. Rd. ψ dγd ,.

(189) ψ ∈ Cc1 Rd .. (4.9). Rd. Applying Propositions 4.1 and 4.2, up to a subsequence, for each t T0 and almost every (w, x), the term K˜ tε (w, x) defined in (4.8) converges to K˜ t (x) = exp −. m j =1 0. t.

(190) j δ(Aj ) Xs (x) dws −. t 0. .

(191) Φ Xs (x) ds .. (4.10).

(192) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1151. By Corollary 2.3 and Lemma 3.2, we may assume that T0 is small enough so that for any t T0 , the family {K˜ tε : ε 1} is also bounded in L2 (P × γd ). Therefore, by the uniform integrability, letting ε → 0 in (4.9), we get P-almost surely, . ψ(Xt )K˜ t dγd =. Rd. ψ dγd ,.

(193) ψ ∈ Cc1 Rd .. (4.11). Rd. Now taking a Borel version of x → K˜ t (w, x). Under the assumptions, the solution Xt is a stochastic flow of homeomorphisms, hence the inverse flow Xt−1 exists. Consequently, if t T0 , we deduce from (4.11) that the density Kt (w, x) of (Xt )# γd with respect to γd admits the expression Kt (w, x) = [K˜ t (w, Xt−1 (w, x))]−1 which is strictly positive. For Xt+T0 with t T0 , we use the flow property: Xt+T0 (w, x) = Xt (θT0 w, XT0 (w, x)). Thus, for any ψ ∈ Cc1 (Rd ), . ψ(Xt+T0 ) dγd =. Rd.

(194) ψ Xt (XT0 ) dγd. Rd. . =. ψ(Xt )KT0 dγd Rd. . =.

(195) ψKT0 Xt−1 Kt dγd .. Rd. That is to say, the density Kt+T0 = KT0 (Xt−1 )Kt is strictly positive. Continuing in this way, we obtain that Kt is strictly positive for any t 0. Now if ρ(x) denotes the density of γd with respect to Lebd , then.

(196) −1 kt (w, x) = ρ Xt−1 (w, x) Kt (w, x)ρ(x) > 0 which concludes the proof.. 2. In what follows, we will give examples for which existence of the inverse flow is not known. Theorem 4.4. Let A1 , . . . , Am be bounded C 1 vector fields on Rd such that their derivatives are of linear growth; furthermore let A0 be continuous of linear growth such that δ(A0 ) exists. Define Aˆ 0 = A0 −. m . LAj Aj .. (4.12). j =1. Suppose that δ(Aˆ 0 ) exists and that Rd.

(197)

(198) . exp λ0 δ(A0 ) + δ(Aˆ 0 ) dγd < +∞,. for some λ0 > 0.. (4.13).

(199) 1152. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. If pathwise uniqueness holds both for SDE (1.1) and for dYt =. m . Aj (Yt ) dwt − Aˆ 0 (Yt ) dt, j. (4.14). j =1. then the solution Xt to SDE (1.1) leaves the Gaussian measure γd quasi-invariant. Proof. Obviously the conditions in Theorem 3.4 are satisfied; hence (Xt )# γd = Kt γd . Let t > 0 be given, we consider the dual SDE to (1.1): dYst =. m .

(200) t,j

(201) Aj Yst dws − Aˆ 0 Yst ds. j =1. for which pathwise uniqueness holds; here wst = wt−s − wt with s ∈ [0, t]. Let Aεj , j = 0, 1, . . . , m, be the vector fields defined as above. Consider dYst,ε =. m .

(202) t,j.

(203) Aεj Yst,ε dws − Aˆ ε0 Yst,ε ds,. j =1. ε −1 = Y t,ε . It is easy to check that for ε ε where Aˆ ε0 = Aε0 − m t j =1 LAj Aj . Then it is known that (Xt ) some constant C > 0 independent of ε,. ε.

(204). Aˆ (x) C 1 + |x| . 0. (4.15). Moreover,. LAεj Aεj. =. d .

(205) k Aεj. . k=1. ∂Aj ∂ϕε −ε Pε Aj + ϕε e Pε ∂xk ∂xk. which converges locally uniformly to LAj Aj . Therefore Aˆ ε0 converges uniformly over any compact subset to Aˆ 0 . By Theorem 3.5, lim sup E. ε→0 |x|R. . 2 sup Yst,ε − Yst = 0.. 0st. It follows that, along a sequence, Ytt,ε converges to Ytt for almost every (w, x). Now let ψ1 , ψ2 ∈ Cb (Rd ), we have for t T0 , Rd.

(206) ψ1 · ψ2 Xtε K˜ tε dγd =. Rd.

(207) ψ1 Ytt,ε · ψ2 dγd ..

(208) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1153. Letting ε → 0 leads to . . ψ1 · ψ2 (Xt )K˜ t dγd =. Rd.

(209) ψ1 Ytt · ψ2 dγd .. (4.16). Rd. Taking ψ1 and ψ2 positive in (4.16) and using a monotone class argument, we see that Eq. (4.16) holds for any positive Borel functions ψ1 and ψ2 . Hence taking a Borel version of K˜ t and setting ψ1 = 1/K˜ t in (4.16), we get . ψ2 (Xt ) dγd = Rd. t

(210) −1 K˜ t Yt ψ2 dγd .. (4.17). Rd. It follows that Kt = [K˜ t (Ytt )]−1 > 0 for t T0 . For Xt+T0 with t T0 , we shall use repeatedly (4.16). By the flow property, Xt+T0 (w, x) = Xt (θT0 w, XT0 (w, x)) where (θT0 w)t = wt+T0 − wT0 . Letting t = T0 and replacing ψ2 by ψ2 (Xt ) we get . ψ1 · ψ2 (Xt+T0 )K˜ T0 dγd =. Rd. . T

(211) ψ1 YT00 ψ2 (Xt ) dγd .. Rd. Taking ψ1 = 1/K˜ T0 in the above equality, we get . ψ2 (Xt+T0 ) dγd =. Rd. T

(212) −1 K˜ T0 Y 0 ψ2 (Xt ) dγd T0. Rd. . =. T

(213) −1 K˜ T0 YT00 ψ2 (Xt )K˜ t−1 K˜ t dγd. Rd. . =. T

(214)

(215) −1 t

(216) −1 K˜ T0 Y 0 Ytt K˜ t Yt ψ2 dγd , T0. Rd T where in the last equality we have used (4.16) with ψ1 = [K˜ T0 (YT00 )]−1 K˜ t−1 . It follows that the density Kt+T0 of (Xt+T0 )# γd with respect to γd is strictly positive, and so on. 2. Corollary 4.5. Let A1 , . . . , Am be bounded C 2 vector fields such that their derivatives up to order 2 grow at most linearly, and let A0 be a continuous vector field of linear growth. Suppose that. A0 (x) − A0 (y) CR |x − y| log. k. 1 |x − y|. for |x| R, |y| R, |x − y| c0 small enough, where logk s = (log s)(log log s) . . . (log . . . log s). Suppose further that. (4.18).

(217) 1154. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. div(A0 ) =. j d ∂A. 0. j =1. ∂xj. exists and is bounded. Then the stochastic flow Xt defined by SDE (1.1) leaves the Lebesgue measure quasi-invariant. Proof. It is obvious that Aˆ 0 defined in (4.12) satisfies condition (4.18); therefore by [12], pathwise uniqueness holds for SDE (1.1) and (4.14). Note that δ(A0 ) = x, A0

(218) − div(A0 ). Then condition (4.13) is satisfied; thus Theorem 4.4 yields the result. 2 5. The case A0 in a Sobolev space From now on, A0 is no longer supposed to be continuous, but to lie in some Sobolev space, that is, condition (A1) in (H) is replaced by (A1 ) For i = 1, . . . , m, Ai ∈. q q1 D1 (γd ),. q. A0 ∈ D1 (γd ) for some q > 1.. First we establish the following a priori estimate on perturbations, using the method developed in [36]. Let {A0 , A1 , . . . , Am } be a family of measurable vector fields on Rd . We first give a precise meaning of solution to the following SDE dXt =. m . Ai (Xt ) dwti + A0 (Xt ) dt,. X0 = x.. (5.1). i=1. Definition 5.1. We say that a measurable map X : Ω × Rd → C([0, T ], Rd ) is a solution to the Itô SDE (5.1) if (i) for each t ∈ [0, T ] and almost all x ∈ Rd , w → Xt (w, x) is measurable with respect to Ft , i.e., the natural filtration generated by the Brownian motion {ws : s t}; (ii) for each t ∈ [0, T ], there exists Kt ∈ L1 (P × Rd ) such that (Xt (w, ·))# γd admits Kt as the density with respect to γd ; (iii) almost surely m . .

(219).

(220). Ai Xs (w, x) 2 ds + A0 Xs (w, x) ds < +∞; T. T. i=1 0. 0. (iv) for almost all x ∈ Rd , Xt (w, x) = x +. m i=1 0. t.

(221) Ai Xs (w, x) dwsi +. t.

(222) A0 Xs (w, x) ds;. 0. (v) the flow property holds.

(223) Xt+s (w, x) = Xt θs w, Xs (w, x) ..

(224) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1155. Now let {Aˆ 0 , Aˆ 1 , . . . , Aˆ m } be another family of measurable vector fields on Rd , and denote by Xˆ t the solution to the SDE dXˆ t =. m . Aˆ i (Xˆ t ) dwti + Aˆ 0 (Xˆ t ) dt,. Xˆ 0 = x.. (5.2).

(225) Λp,T = sup Kt Lp (P×γd ) ∨ Kˆ t Lp (P×γd ) .. (5.3). i=1. Let Kˆ t be the density of (Xˆ t )# γd with respect to γd and define. 0tT. 2q Theorem 5.2. Let q > 1. Suppose that A1 , . . . , Am as well as Aˆ 1 , . . . , Aˆ m are in D1 (γd ) and q A0 , Aˆ 0 ∈ D1 (γd ). Then, for any T > 0 and R > 0, there exist constants Cd,q,R > 0 and CT > 0 such that for any σ > 0,. . . E. log. sup0tT |Xt − Xˆ t |2 σ2. GR. . + 1 dγd . . CT Λp,T. Cd,q,R ∇A0 Lq + . m . 1/2 ∇Ai 2L2q. i=1. m 1 1 + 2 Ai − Aˆ i 2L2q + A0 − Aˆ 0 Lq + σ σ. . i=1. +. m . ∇Ai 2L2q. i=1 m . 1/2 !. Ai − Aˆ i 2L2q. ,. i=1. where p is the conjugate number of q: 1/p + 1/q = 1, and # ". GR (w) = x ∈ Rd : sup Xt (w, x) ∨ Xˆ t (w, x) R .. (5.4). 0tT. Proof. Denote ξt = Xt − Xˆ t , then ξ0 = 0. By Itô’s formula, d|ξt |2 = 2. m . ξt , Ai (Xt ) − Aˆ i (Xˆ t ) dwti + 2 ξt , A0 (Xt ) − Aˆ 0 (Xˆ t ) dt. i=1. +. m . Ai (Xt ) − Aˆ i (Xˆ t ) 2 dt.. (5.5). i=1. For σ > 0, log(|ξt |2 /σ 2 + 1) = log(|ξt |2 + σ 2 ) − log σ 2 . Again by Itô’s formula,.

(226) d log |ξt |2 + σ 2 = Using (5.5), we obtain. d|ξt |2 14 − 2 2 2 |ξt | + σ. m. ˆ ˆ 2 i=1 ξt , Ai (Xt ) − Ai (Xt )

(227) (|ξt |2 + σ 2 )2. dt..

(228) 1156. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168 m

(229). ξt , Ai (Xt ) − Aˆ i (Xˆ t )

(230) ξt , A0 (Xt ) − Aˆ 0 (Xˆ t )

(231) i d log |ξt |2 + σ 2 = 2 dw + 2 dt t |ξt |2 + σ 2 |ξt |2 + σ 2 i=1. +. m |Ai (Xt ) − Aˆ i (Xˆ t )|2. |ξt |2 + σ 2. i=1. dt − 2. m ξt , Ai (Xt ) − Aˆ i (Xˆ t )

(232) 2. (|ξt |2 + σ 2 )2. i=1. =: dI1 (t) + dI2 (t) + dI3 (t) + dI4 (t).. dt (5.6). Let τR (x) = inf{t 0: |Xt (x)| ∨ |Xˆ t (x)| > R}. Remark that almost surely, GR ⊂ {x: τR (x) > T } and for any t 0, {τR > t} ⊂ B(R). Therefore . . sup I1 (t) dγd E. E GR. 0tT. sup. B(R). 0tT ∧τR. . I1 (t) dγd .. By Burkholder’s inequality,. E. sup. 0tT ∧τR. . I1 (t) 2 4E. T∧τR m i=1. 0. ξt , Ai (Xt ) − Aˆ i (Xˆ t )

(233) 2 dt , (|ξt |2 + σ 2 )2. which is obviously less than. 4E. T∧τR m 0. i=1. |Ai (Xt ) − Aˆ i (Xˆ t )|2 dt . |ξt |2 + σ 2. Hence E. sup. B(R). 0tT ∧τR. T 4. . E 0. I1 (t) dγd. . m |Ai (Xt ) − Aˆ i (Xˆ t )|2. {τR >t} i=1. |ξt |2 + σ 2. dγd dt. 1/2 .. (5.7). We have Ai (Xt ) − Aˆ i (Xˆ t ) = Ai (Xt ) − Ai (Xˆ t ) + Ai (Xˆ t ) − Aˆ i (Xˆ t ). Using the density Kˆ t , it is clear that E {τR >t}. |Ai (Xˆ t ) − Aˆ i (Xˆ t )|2 1 dγd 2 E 2 2 |ξt | + σ σ. . Ai (Xˆ t ) − Aˆ i (Xˆ t ) 2 dγd. Rd. =. 1 E σ2. . Rd. |Ai − Aˆ i |2 Kˆ t dγd ..

(234) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1157. Thus by Hölder’s inequality and according to (5.3), we have E {τR >t}. Λp,T |Ai (Xˆ t ) − Aˆ i (Xˆ t )|2 dγd Ai − Aˆ i 2L2q . 2 2 |ξt | + σ σ2. (5.8). Now we shall use Theorem A.1 in Appendix A to estimate the other term. Note that on the set {τR > t}, Xt , Xˆ t ∈ B(R), thus |Xt − Xˆ t | 2R. Since (Xt )# γd γd and (Xˆ t )# γd γd , we can apply (A.2) so that.

(235). Ai (Xt ) − Ai (Xˆ t ) Cd |Xt − Xˆ t | M2R |∇Ai |(Xt ) + M2R |∇Ai |(Xˆ t ) . Then E {τR >t}. |Ai (Xt ) − Ai (Xˆ t )|2 2 dγd Cd E |ξt |2 + σ 2.

(236) 2 M2R |∇Ai |(Xt ) + M2R |∇Ai |(Xˆ t ) dγd .. {τR >t}. Notice again that on {τR (x) > t}, Xt (x) and Xˆ t (x) are in B(R), therefore E {τR >t}. .

(237) 2 |Ai (Xt ) − Ai (Xˆ t )|2 2 2C M2R |∇Ai | (Kt + Kˆ t ) dγd dγ E d d 2 2 |ξt | + σ B(R). .

(238) 2q M2R |∇Ai | dγd. 4Cd2 Λp,T. 1/q .. (5.9). B(R). Remark that the maximal function inequality does not hold for the Gaussian measure γd on the whole space Rd . However, on each ball B(R), γd |B(R) . 1 2 Lebd |B(R) eR /2 γd |B(R) . (2π)d/2. Thus, according to (A.3), .

(239) 2q M2R |∇Ai | dγd . 1 (2π)d/2. B(R). .

(240) 2q M2R |∇Ai | dx. B(R). . Cd,q (2π)d/2. . |∇Ai |2q dx B(3R). Cd,q e. 9R 2 /2. |∇Ai |2q dγd. B(3R). Cd,q e9R. 2 /2. 2q. ∇Ai L2q ..

(241) 1158. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Therefore by (5.9), there exists a constant Cd,q,R > 0 such that E {τR >t}. |Ai (Xt ) − Ai (Xˆ t )|2 dγd Cd,q,R Λp,T ∇Ai 2L2q . |ξt |2 + σ 2. Combining this estimate with (5.7) and (5.8), we get . sup I1 (t) dγd. E GR. . 0tT. CT. 1/2. 1/2 Λp,T. Cd,q,R. m . ∇Ai 2L2q. i=1. m 1 + 2 Ai − Aˆ i 2L2q σ. 1/2 .. (5.10). i=1. Now we turn to deal with I2 (t) in (5.6). We have E GR. T . |A0 (Xt ) − Aˆ 0 (Xˆ t )|. sup I2 (t) dγd 2 dγd dt. E (|ξt |2 + σ 2 )1/2 0tT GR. 0. Note that for x ∈ GR , Xˆ t (x) ∈ B(R) for each t ∈ [0, T ], thus Λp,T 1 |A0 (Xˆ t ) − Aˆ 0 (Xˆ t )| E A0 − Aˆ 0 Lq . dγ |A0 − Aˆ 0 |Kˆ t dγd d 2 2 1/2 σ σ (|ξt | + σ ). E GR. B(R). Again using (A.2), E GR. .

(242) |A0 (Xt ) − A0 (Xˆ t )| C M2R |∇A0 |(Xt ) + M2R |∇A0 |(Xˆ t ) dγd , dγ E d d 2 2 1/2 (|ξt | + σ ) GR. which is dominated by Cd E. .

(243) ˆ M2R |∇A0 | (Kt + Kt ) dγd Cd,q,R ∇A0 Lq Λp,T .. B(R). Therefore we arrive at the following estimate for I2 : E GR. . 1 sup I2 (t) dγd 2T Λp,T Cd,q,R ∇A0 Lq + A0 − Aˆ 0 Lq . σ 0tT. (5.11).

(244) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1159. In the same way we get . sup I3 (t) dγd. E GR. . 0tT. . CT Λp,T Cd,q,R. m . ∇Ai 2L2q. i=1. m 1 2 + 2 Ai − Aˆ i L2q . σ. (5.12). i=1. The term I4 (t) is negative and hence omitted. Combining (5.6) and (5.10)–(5.12), the proof is completed. 2 Now we shall construct a solution to SDE (5.1). To this end, we take ε = 1/n and write Anj 1/n. instead of Aj introduced in Section 3. Then by assumption (A2) and Lemma 3.1, there is a constant C > 0 independent of n and i, such that. n.

(245). A (x) C 1 + |x| .. (5.13). i. Let Xtn be the solution to Itô SDE (5.1) with coefficients Ani (i = 0, 1, . . . , m). Then for any α 1 and T > 0, there exists Cα,T > 0 independent of n such that E. . α .

(246) sup Xtn Cα,T 1 + |x|α ,. for all x ∈ Rd .. (5.14). 0tT. Let Ktn be the density of (Xtn )# γd with respect to γd . Under the hypotheses (A2)–(A4), there exists T0 > 0 such that (recall that p is the conjugate number of q > 1): Λp,T0 :=. . . exp 2pT0 |A0 | + e δ(A0 ). Rd. p−1 m p(2p−1) .

(247) 2 + 2p|Aj |2 + |∇Aj |2 + 2(p − 1)e2 δ(Aj ). dγd < ∞. j =1. (5.15) Similar to (3.6), we have sup sup Ktn Lp (γ. t∈[0,T0 ] n1. d ×P). Λp,T0 < +∞.. Next we shall prove that the family {X.n : n 1} converges to some stochastic field.. (5.16).

(248) 1160. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. Theorem 5.3. Let T0 be given in (5.15). Then, under the assumptions (A1 ) and (A2)–(A4), there exists X : Ω × Rd → C([0, T0 ], Rd ) such that for any α 1, lim E. . α sup Xtn − Xt dγd = 0.. . n→∞. (5.17). 0tT0. Rd. Proof. We shall prove that {X n : n 1} is a Cauchy sequence in Lα (Ω × Rd ; C([0, T0 ], Rd )). Denote by · ∞,T0 the uniform norm on C([0, T0 ], Rd ). We have to prove that . n X − X k α. E. lim. n,k→+∞. ∞,T0. dγd = 0.. (5.18). Rd. First by (5.14), the quantity . n 2α X . Jα,T0 := sup E n1. ∞,T0. .

(249) 1 + |x|2α dγd dγd Cα,T0. Rd. (5.19). Rd. is obviously finite. Let R > 0 and set Gn,R (w) = x ∈ Rd : X n (w, x)∞,T R . 0. Using (5.19), for any α 1 and R > 0, we have

(250) Jα,T sup E γd Gcn,R 2α0 . R n1 Now by Cauchy–Schwarz inequality . . n X − X k α. E. ∞,T0. dγd. Gcn,R ∪Gck,R. 1/2 n c

(251)

(252) 1/2 c k 2α X − X ∞,T dγd E γd Gn,R ∪ Gk,R · E 0. . 2Jα,T0 R 2α. 1/2. Rd.

(253) 1/2. · 22α Jα,T0 .. (5.20). Given ε > 0, we may choose R > 1 sufficiently large such that the last quantity in inequality (5.20) is less than ε. Then, for any n, k 1, . . E. n X − X k α. ∞,T0. Gcn,R ∪Gck,R. dγd ε.. (5.21).

(254) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1161. Let . σn,k = An − Ak 0. 0 Lq. +. m n A − Ak 2 2q i. 1/2 ,. i L. i=1. which tends to 0 as n, k → +∞ since An0 converges to A0 in Lq (γd ) and Ani converges to Ai in L2q (γd ) for i = 1, . . . , m. Now applying Theorem 5.2 with Ai and Aˆ i being replaced respectively by Ani and Aki , we get . . In,k := E Gn,R ∩Gk,R. n. X − X k 2∞,T0 log + 1 dγd 2 σn,k . CT0 Λp,T0 Cd,q,R. ∇An . 0 Lq. +. m ∇An 2 2q i. 1/2. L. i=1. ! n 2 ∇An 2q + 2 . + i L i=1. Recall that Ani = ϕ1/n P1/n Ai . Thus ∇Ani = ∇ϕ1/n ⊗ P1/n Ai + ϕ1/n e−1/n P1/n ∇Ai and. n.

(255). ∇A P1/n |Ai | + |∇Ai | . i We obtain the following uniform estimates ∇An . 0 Lq. ∇An . A0 Dq ,. i. 1. L2q. Ai D2q . 1. Hence the quantity In,k is uniformly bounded with respect to n, k. Let Πˆ be the measure on Ω × Rd defined by . ˆ ψ(w, x) dΠ(w, x) = E. . . ψ(w, x) dγd (x) .. Gn,R ∩Gk,R. Ω×Rd. ˆ Obviously we have Π(Ω × Rd ) 1. Let η > 0 and consider Σn,k = (w, x) ∈ Ω × Rd : X n (w, x) − X k (w, x)∞,T η 0. which equals n. 2 X − X k 2∞,T0 η (w, x) ∈ Ω × Rd : log + 1 log + 1 . 2 2 σn,k σn,k It follows that as n, k → +∞, ˆ n,k ) Π(Σ. In,k → 0, 2 + 1) log(η2 /σn,k. (5.22).

(256) 1162. S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. since σn,k → 0 and the family {In,k : n, k 1} is bounded. Now . . n X − X k α. E. ∞,T0. dγd =. Gn,R ∩Gk,R. n X − X k α. ∞,T0. dΠˆ. Ω×Rd. . n X − X k α. =. ∞,T0. c Σn,k. . +. n X − X k α. dΠˆ. ∞,T0. ˆ dΠ.. (5.23). Σn,k. The first term on the right side of (5.23) is bounded by ηα , while the second one, due to (5.19) and (5.22), is dominated by $. % $ & α n − X k 2α ˆ n,k ) · & ˆ n,k ) → 0 as n, k → +∞. X Π(Σ E dγ 2 Jα,T0 Π(Σ ' d ∞,T0 Rd. Now taking η = ε 1/α and combining (5.21) and (5.23), we see that lim sup E n,k→+∞. n X − X k α. ∞,T0. dγd 2ε,. Rd. which implies (5.18). Let X ∈ Lα (Ω × Rd ; C([0, T0 ], Rd )) be the limit of X n in this space. We see that for each t ∈ [0, T ] and almost all x ∈ Rd , w → Xt (w, x) is Ft -measurable. 2 Proposition 5.4. There exists a family {Kˆ t : t ∈ [0, T0 ]} of density functions on Rd such that (Xt )# γd = Kˆ t γd for each t ∈ [0, T0 ]. Moreover, sup Kˆ t Lp (P×γd ) Λp,T0. 0tT0. where Λp,T0 is given by (5.16). Proof. It is the same as the proof of Theorem 3.4.. 2. The same arguments as in the proof of Propositions 4.1 and 4.2 yield the following Proposition 5.5. For any α 2, up to a subsequence, lim. n→∞ Rd. m t. α .

(257). . Ani Xsn − Ai (Xs ) dwsi dγd = 0, E sup. 0tT0 . i=1 0.

(258) S. Fang et al. / Journal of Functional Analysis 259 (2010) 1129–1168. 1163. and T0. α. n n

(259). E A0 Xs − A0 (Xs ) ds dγd = 0. lim. n→∞ Rd. 0. Now for regularized vector fields Ani , i = 0, 1, . . . , m, we have. Xtn (x) = x. +. m . t. Ani. n

(260) i Xs dws +. t. i=1 0.