The Ornstein Uhlenbeck semigroup

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1 Lecture 2 The Ornstein Uhlenbeck semigrou All of us know the imortance of the Lalacian oerator f(x) = d 2 2 (x), T(t)f(x) = i and of the heat semigrou, (4 t) d/2 e x y 2 /4t f(y)dy, t > 0, R d that serve as rototyes for ellitic di erential oerators and semigrous of oerators, resectively. Choosing = L 2 (R d, d), the Lalacian : D( )=W 2,2 (R d, d)! is the infinitesimal generator of T (t), namely given any f 2, thereexiststhelimit lim t!0 +(T (t)f f)/t if and only if f 2 W 2,2 (R d, d), and in this cashe limit is f. The W 2,2 norm is equivalent to the grah norm. Moreover, the realization of the Lalacian in is the oerator associated with the quadratic form Q(u, v) = ru rv dx, u,v2 W,2 (R d, d). R d This means that D( ) consists recisely of the elements u 2 W,2 (R d, d) such that the function W,2 (R d, d)! R, ' 7! R R ru r' dx has a linear bounded extension to the d whole, namely there exists g 2 L 2 (R d, d) such that ru r' dx = g ' dx, ' 2 W,2 (R d, d), R d R d and in this case g = u. Ifu 2 W 2,2 (R d, d), the above formula follows just integrating by arts, and it is the basic formula that relates the Lalacian and the gradient. Moreover, for u 2 W 2,2 (R d, d) wehave u =divru, wherhe divergence div is (minus) the adjoint of the gradient r : W,2 (R d, d)! L 2 (R d, d; R d ), and for a vector field v 2 W,2 (R d, d; R d ) it is given by P d i/@x i. In this lecture and in the next ones we introduche Ornstein Uhlenbeck oerator and the Ornstein Uhlenbeck semigrou, that lay the role of the Lalacian and of the heat 45

2 46 Lecture 2 semigrou if the Lebesgue measure is relaced by the standard Gaussian measure d, and that have natural extensions to our infinite dimensional setting (,,H). As before, is a searable Banach sace endowed with a centred nondegenerate Gaussian measure, and H is the relevant Cameron Martin sace. 2. The Ornstein Uhlenbeck semigrou In this section we definhe Ornstein Uhlenbeck semigrou; we start by defining it in the sace of the bounded continuous functions and then we extend it to L (, ), for every 2 [, +). The Ornstein Uhlenbeck semigrou in C b () is defined as follows: T (0) = I, and for t>0, T (t)f is defined by the Mehler formula T (t)f(x) := f( x + y) (dy). (2..) We list some roerties of the family of oerators {T (t) : t 0}. Proosition 2... {T (t) : t 0} is a contraction semigrou in C b (). Moreover, for every f 2 C b () we have T (t)f d = fd, t > 0. (2..2) Proof. First of all we notichat T (t)f(x) alekfk for every x 2 and t 0. The fact that T (t)f 2 C b () follows by Dominated Convergence Theorem. So, T (t) 2 L(C b ()) and kt (t)k L(Cb ()) ale. Taking f, we have T (t)f so that kt (t)k L(Cb ()) = for every t 0. Let us rovhat {T (t) : t 0} is a semigrou. For every f 2 C b () and t, s > 0we have (T (t)(t (s)f))(x) = (T (s)f)( x + y) (dy) = f(e s ( x + y)+ e 2s z) (dy) (dz). Setting now (y, z) =e s y + e 2s 2s e z, and using Proosition 2.2.7(iv), we 2t 2s get (T (t)(t (s)f))(x) = f(e s t x + 2s (y, z)) (dy) (dz) = f( s x + 2s )(( ) )(d ) = f( s x + 2s ) (d ) = T (t + s)f(x).

3 The Ornstein Uhlenbeck semigrou 47 Let us rovhat (2..2) holds. For any f 2 C b () wehave T (t)f (dx) = f( x + y) (dy) (dx). Setting (x, y) = x + y, we aly Proosition 2.2.7(iv) with any 2 R such that = cos, =sin, and we get T (t)f d = f( )( ) (d ) = f( ) (d ). We oint out that the semigrou {T (t) : t 0} is not strongly continuous in C b (), and not even in the subsace BUC() of the bounded and uniformly continuous functions. In fact, we havhe following characterisation. Lemma Let f 2 BUC(). Then lim kt (t)f fk =0() lim kf( ) fk =0. t!0 + t!0 + Proof. For every t>0 and x 2 we have (T (t)f f( ))(x) = (f( x + y) f( x)) (dy) and the right hand side goes to 0, uniformly in, as t! 0 +. Indeed, given " > 0fixR>0 such that ( \ B(0,R)) ale ", and fix > 0 such that f(u) f(v) ale " for ku vk ale. Then, for every t such that R ale and for every x 2 we have (f( x + y) f( x)) (dy) = + B(0,R) ale " +2kfk ". \B(0,R) (f( x + y) f( x)) (dy) The simlest counterexamlo strong continuity in BUC() is one dimensional: see Exercise 2.. However, for every f 2 C b () the function (t, x) 7! T (t)f(x) is continuous in [0, +) by the Dominated Convergence Theorem. In articular, lim T (t)f(x) =f(x), 8x 2, t!0 + which is enough for many uroses. The semigrou {T (t) : t 0} enjoys imortant smoothing and summability imroving roerties. The first smoothing roerty is in the next roosition.

4 48 Lecture 2 Proosition For every f 2 C b () and t>0, T (t)f is H-di erentiable at every x 2, and we have [r H T (t)f(x),h] H = f( x + e 2t y)ĥ(y) (dy). (2..3) Therefore, Proof. Set r H T (t)f(x) H ale h Hkfk, x 2. (2..4) c(t) :=. For every h 2 H we have T (t)f(x + h) = f( x + = f( x + (c(t)h + y)) (dy) n z)ex c(t)ĥ(z) c(t)2 h 2 o H 2 (dz), by the Cameron Martin formula. Therefore, denoting by l t,x (h) the right hand side of (2..3), T (t)f(x + h) T (t)f(x) l t,x (h) ale h H ale f( x + n e h 2t y) ex c(t)ĥ(y) c(t)2 h 2 o H H 2 c(t)ĥ(y) ale kfk n ex c(t) c(t) 2 h 2 o H c(t) N (0, h 2 h H R 2 H)(d ) n = kfk ex c(t) h H c(t) 2 h 2 o H c(t) h H N (0, )(d ) 2 R (dy) wherhe right hand side vanishes as h H! 0, by the Dominated Convergence Theorem. This roves (2..3). In its turn, (2..3) yields [r H T (t)f(x),h] H ale c(t)kfk kĥk L (, ) ale c(t)kfk kĥk L 2 (, ) = c(t)kfk h H for every h 2 H, and (2..4) follows. as Notichat in the case = R d, = d,wehaver H = r and formula (2..3) reads D i T (t)f(x) = Let us consider now more regular functions f. f( x + y)y i d (dy), i =,...,d. (2..5) R d

5 The Ornstein Uhlenbeck semigrou 49 Proosition For every f 2 Cb (), T (t)f 2 C b () for any t 0, and its derivative at x is (T (t)f) 0 (x)(v) = f 0 ( x + y)(v) (dy). (2..6) T (t)f(x) T f (x), v 2, x 2. (2..7) For every f 2 Cb 2(), T (t)f 2 C2 b () for any t 0, and its second order derivative at x is (T (t)f) 00 (x)(u)(v) = f 00 ( x + y)(u)(v) (dy), (2..8) so 2 (x) = 2 f T (t) (x), u, v 2, x 2. Proof. Fix t>0 and set (x, y) = x + y. For every x, v 2 we have (T (t)f)(x + v) (T (t)f)(x) f 0 ( x + y)(v) (dy) ale f( (x, y)+ v) f( (x, y)) f 0 ( (x, y))( v) (dy) kvk. On the other hand, for every y 2 we have and (see Exercise 2.3) f( (x, y) v) f( (x, y)) f 0 ( (x, y))( v) lim =0, v!0 kvk kvk f( (x, y)+ v) f( (x, y)) f 0 ( (x, y))( v) kvk ale 2 su kf 0 (z)k, z2 and (2..6) follows by the Dominated Convergence Theorem. Formula (2..7) is an immediate consequence. The derivative (T (t)f) 0 is continuous still by the Dominated Convergence Theorem. The verification of (2..8) and (2..9) for f 2 Cb 2 () follow iterating this rocedure (see Exercise 2.4). Let us comare Proosition 2..3 and Proosition Proosition 2..3 describes a smoothing roerty of T (t), while Proosition 2..4 says that T (t) reserves the saces C b () and C2 b (). In general, T (t) regularises only along H and it does not ma C b() into C (). If = R d and = d we have H = R d,thisdi culty does not arise, Proosition 2..3 says that T (t) mas C b (R d )intoc b (Rd ) and in fact one can check that T (t) mas C b (R d )intoc k b (Rd ) for every k 2 N, as we shall see in the next lecture.

6 50 Lecture 2 If f 2 C b () we can write r HT (t)f(x) intwodi erent ways, using (2..3) and (2..6): for every h 2 H we have [r H T (t)f(x),h] H = = f 0 ( x + We now extend T (t) tol (, ), ale <. f( x + e y)(h) (dy). 2t y)ĥ(y) (dy) Proosition Let t 0. For every f 2 C b () and 2 [, +) we have kt (t)fk L (, ) alekfk L (, ). (2..0) Hence {T (t) : t 0} is uniquely extendablo a contraction semigrou {T (t) : t 0} in L (, ). Moreover (i) {T (t) : t 0} is strongly continuous in L (, ), for every 2 [, +); (ii) T 2 (t) is self adjoint and nonnegative in L 2 (, ) for every t>0; (iii) is an invariant measure for {T (t) : t 0}. Proof. For every f 2 C b (), t>0 and x 2 the Hölder inequality yields T (t)f(x) ale Integrating over and using (2..2) we obtain f( x + y) d = T (t)( f )(x). T (t)f d ale T (t)( f ) d = f d. Since C b () isdenseinl (, ), T (t) has a unique bounded extension T (t) to the whole L (, ), such that kt (t)k L(L (, )) ale. In fact, taking f, T (t)f so that kt (t)k L(L (, )) =. Let us rovhat {T (t) : t 0} is strongly continuous. We already know that for f 2 C b () wehavelim t!0 + T (t)f(x) =f(x) for every x 2, and moreover T (t)f(x) ale kfk for every x. By the Dominated Convergence Theorem, lim t!0 + T (t)f = f in L (, ). Since C b () isdenseinl (, ) and kt (t)k L(L (, )) = for every t, then lim t!0 + T (t)f = f for every f 2 L (, ). Let us rove statement (ii). Let f, g 2 C b (), t>0. Then ht (t)f,gi L 2 (, ) = f( x + y)g(x) (dy) (dx)

7 The Ornstein Uhlenbeck semigrou 5 and setting u = x + y, v = x + y,(u, v) =:R(x, y), using Proosition 2.2.7(iii) and the fact that is centred we get ht (t)f,gi L 2 (, ) = f(u)g( u v)(( ) R )(d(u, v)) = f(u)g( u v) (dv) (du) = f(u)g( u + v) (dv) (du) = hf,t(t)gi L 2 (, ). Aroximating any f, g 2 L 2 (, ) by elements of C b (), we obtain ht 2 (t)f,gi L 2 (, ) = hf,t 2 (t)gi L 2 (, ). Still for every f 2 L 2 (, ) and t>0wehave ht 2 (t)f,fi L 2 (, ) = ht 2 (t/2)t 2 (t/2)f,fi L 2 (, ) = ht 2 (t/2)f,t 2 (t/2)fi L 2 (, ) 0. Statement (iii) is an immediate consequence of (2..2). To simlify some statements we extend the Ornstein Uhlenbeck semigrou T (t) to vector valued functions. For v 2 C b (; H) and t>0 we set T (t)v(x) = v( x + y) (dy), t > 0, x2. By Remark 9.2.6, for every orthonormal basis {h i : i 2 N} of H we have T (t)v(x) = T (t)([v( ),h i ] H )(x)h i. i= Using Proosition 9.2.5(i) we get the estimate T (t)v(x) H ale v( x + y) H (dy) =(T (t)( v H ))(x), x 2. (2..) Notichat the right hand side concerns the scalar valued function v H. Raising to the ower, integrating over and recalling (2..2), we obtain kt (t)vk L (, ;H) alekvk L (, ;H), t 0. (2..2) As in the case of real valued functions, since C b (; H) isdenseinl (, ; H), estimate (2..2) allows to extend T (t) to a bounded (contraction) oerator in L (, ; H), called T (t). We will not develo thheory for vector valued functions, but we shall ushis notion to write some formulae in a concise way, see e.g. (2..3). L gradient estimates for T (t)f are rovided by the next roosition. Proosition Let ale <.

8 52 Lecture 2 (i) For every f 2 W, (, ) and t>0, T (t)f 2 W, (, ) and r H T (t)f = T (t)(r H f), (2..3) kt (t)fk W, (, ) alekfk W, (, ). (2..4) (ii) If >, for every f 2 L (, ) and t>0, T (t)f 2 W, (, ) and r H T (t)f(x) H d ale c(t, ) R / 0 with c(t, ) = R 0 N (0, )(d ) /. f d, (2..5) Proof. (i). Let f 2 C b (). By Proosition 2..4, T (t)f 2 C b () and (@ ht (t)f)(x) = (T (t)(@ h f))(x) for every h 2 H, namely [(r H T (t)f)(x),h] H = T (t)([r H f,h] H )(x). Therefore, (r H T (t)f)(x) H ale T (t)( r H f H )(x), for every x 2. Consequently, r H T (t)f(x) H ale (T (t)( r H f H )(x)) ale (T (t)( r H f H )(x)) and integrating we obtain r H T (t)f(x) H d ale T (t)( r H f H ) d = r H f H d, so that kt (t)fk W, (, ) = kt (t)fk L (, ) + k r H T (t)f H k L (, ) alekfk L (, ) + k r H f H k L (, ) = kfk W, (, ). Since C b () isdenseinw, (, ), (2..4) follows. (ii) Let f 2 C b (). By Proosition 2..3, T (t)f is H-di erentiable at every x, and we have r H T (t)f(x) H = su h2h, h H = [r H T (t)f(x),h] H. Let us estimate [r H T (t)f(x),h] H = l t,x (h) (where l t,x (h) is as in the roof of Proosition 2..3), for h H =, using formula (2..3). We have l t,x (h) ale ale = f( x + e 2t y) ĥ(y) (dy) f( x + / y) (dy) (T (t)( f )(x)) / R 0 N (0, )(d ) / 0. ĥ 0 d / 0

9 The Ornstein Uhlenbeck semigrou 53 By the invariance roerty (2..2) of, r H T (t)f(x) H (dx) ale = (T (t)( f ) d f d 0 N (0, )(d ) R 0 N (0, )(d ) R / 0. / 0 Therefore, T (t)f 2 W, (, ) and estimate (2..5) holds for every f 2 C b (). Since C b () isdenseinl (, ), the statement follows. Nothat the roof of (ii) fails for =, since for every h 2 H the function ĥ does not belong to L, and the constant c(t, ) in estimate (2..5) blows u as!. Indeed, T (t) does not ma L (, )intow, (, ) continuously for t>0, even in the simlest case = R, = (see for instance [MPP, Corollary 5.]). 2.2 Exercises Exercise 2.. Let = R and set f(x) =sinx. Provhat T (t)f does not converge uniformly to f in R as t! 0. Exercise 2.2. Show that the argument used in Proosition 2..5 to rovhat T (t) is self adjoint in L 2 (, ) imlies that T 0(t) =T (t) for 2 (, +) with/ +/ 0 =. Exercise 2.3. Provhat for every f 2 C () and for every x, y 2 we have so that, if f 2 C b (), f(y) f(x) = 0 f 0 ( y +( )x)(y x) d, f(y) f(x) ale su kf 0 (z)k 0ky z2 xk. Exercise 2.4. Provhat for every f 2 Cb 2() and t>0, T (t)f 2 C2 b () and (2..8), (2..9) hold. Bibliograhy [B] V. I. Bogachev: Gaussian Measures. American Mathematical Society, 998. [MPP] G. Metafune, D. Pallara, E. Priola: Sectrum of Ornstein Uhlenbeck oerators in L saces with resect to invariant measures, J. Funct. Anal. (96) (2002)