Malliavin Calculus by David Nualart 1 1 One-dimensional Gaussian analysis Consider the probability space (R; B(R); γ), where • γ = N(0; 1) is the standard Gaussian probability on R with density 1 −x2=2 p(x) = p e ; x 2 R: 2π • The probability of any interval [a; b] is given by b 1 Z 2 γ([a; b]) = p e−x =2dx: 2π a 1 We are going to introduce two basic differential operators. For any f 2 C (R) we define: • Derivative operator: Df(x) = f 0(x): • Divergence operator: δf(x) = xf(x) − f 0(x): k m m Denote by Cp (R ) the space of functions f : R ! R, which are k times continuously differentiable, and such that for some N ≥ 1, jf (k)(x)j ≤ C(1 + jxjN ). Lemma 1.1. The operators D and δ are adjoint with respect to the measure γ. That means, 1 for any f; g 2 Cp (R), we have hDf; giL2(R,γ) = hf; δgiL2(R,γ) : Proof. Integrating by parts and using p0(x) = −xp(x) we get Z Z f 0(x)g(x)p(x)dx = − f(x)(g(x)p(x))0dx R R Z Z = − f(x)g0(x)p(x)dx + f(x)g(x)xp(x)dx R R Z = f(x)δg(x)p(x)dx: R 2 Lemma 1.2 (Heisenberg's commutation relation). Let f 2 C (R). Then (Dδ − δD)f = f Proof. We can write Dδf(x) = D(xf(x) − f 0(x)) = f(x) + xf 0(x) − f 00(x) and, on the other hand, δDf(x) = δf 0(x) = xf 0(x) − f 00(x): This completes the proof. 2 n More generally, if f 2 C (R) for n ≥ 2, we have (Dδn − δnD)f = nδn−1f Proof. Using induction on n, we can write Dδnf = Dδ(δn−1f) = δD(δn−1f) + δn−1f = δ δn−1Df + (n − 1)δn−2f + δn−1f = δnDf + nδn−1f: Next we will introduce the Hermite polynomials. Define H0(x) = 1, and for n ≥ 1 put n Hn(x) = δ 1. In particular, for n = 1; 2; 3, we have H1(x) = δ1 = x 2 H2(x) = δx = x − 1 2 3 H3(x) = δ(x − 1) = x − 3x: We have the following formula for the derivatives of the Hermite polynomials: 0 Hn = nHn−1 In fact, 0 n n n−1 Hn = Dδ 1 = δ D1 + nδ 1 = nHn−1: Proposition 1.1. The sequence of normalized Hermite polynomials f p1 H ; n ≥ 0g form a n! n 2 complete orthonormal system of functions in the Hilbert space L (R; γ). Proof. For n; m ≥ 0, we can write ( Z n! if n = m Hn(x)Hm(x)p(x)dx = R 0 if n 6= m Indeed, using the properties of Hermite polynomials, we obtain Z Z m Hn(x)Hm(x)p(x)dx = Hn(x)δ 1(x)p(x)dx R R Z 0 m−1 = Hn(x)δ 1(x)p(x)dx R Z = n Hn−1(x)Hm−1(x)p(x)dx: R 2 To show completeness, it suffices to prove that if f 2 L (R; γ) is orthogonal to all Hermite polynomials, then f = 0. Because the leading coefficient of Hn(x) is 1, we have that f is n orthogonal to all monomials x . As a consequence, for all t 2 R, 1 Z X (it)n Z f(x)eitxp(x)dx = f(x)xnp(x)dx = 0: n! R n=0 R 3 We can commute the integral and the series because 1 X Z jtxjn Z jf(x)jp(x)dx = ejtxjjf(x)jp(x)dx n! n=0 R R 1 Z Z 2 ≤ f 2(x)p(x)dx e2jtxjp(x)dx < 1: R R Therefore, the Fourier transform of fp is zero, so fp = 0, which implies f = 0. This completes the proof. For each a 2 R, we have the following series expansion, which will play an important role. 1 n 2 X a ax− a H (x) = e 2 : (1) n! n n=0 n n n Proof of (1): In fact, taking into account that Hn = δ 1 and that δ is the adjoint of D , we obtain 1 ax X 1 a· e = he ;H i 2 H (x) n! n L (R,γ) n n=0 1 X 1 a· n = he ; δ 1i 2 H (x) n! L (R,γ) n n=0 1 X 1 n a· = hD (e ); 1i 2 H (x) n! L (R,γ) n n=0 1 n X a a· = he ; 1i 2 H (x): n! L (R,γ) n n=0 Finally, 1 Z x2 a2 a· ax− 2 2 he ; 1iL2(R,γ) = p e dx = e : 2π R and (1) holds true. Let us now define the Ornstein Uhlenbeck operator, which is a second order differential 2 operator. For f 2 C (R) we set Lf(x) = −xf 0(x) + f 00(x): This operator has the following properties. 1. Lf = −δDf. Proof: δDf(x) = δf 0(x) = xf 0(x) − f 00(x): 2. LHn = −nHn, that is, Hn is an eigenvector of L with eigenvalue −n. Proof: 0 LHn = −δDHn = −δHn = −nδHn−1 = −nHn: 4 The operator L is the infinitesimal generator of the Ornstein-Uhlenbeck semigroup. Con- 2 −nt sider the semigroup of operators fPt; t ≥ 0g on L (R; γ), defined by PtHn = e Hn, that is, 1 X 1 −nt P f = hf; H i 2 e H : t n! n L (R,γ) n n=0 dPt Then, L is the generator of Pt, that is, dt = LPt. 2 Proposition 1.2 (Mehler's formula). For any function f 2 L (R; γ), we have the following formula for the Ornstein-Uhlenbeck semigroup: Z −t p −2t −t p −2t Ptf(x) = f(e x + 1 − e y)p(y)dy = E[f(e x + 1 − e Y )]; R where Y is a N(0; 1) random variable. p R −t −2t Proof. Set Ptf(x) = f(e x + 1 − e y)p(y)dy. e R 2 (i) We will first show that Pt and Pet are contraction operators on L (R; γ). Indeed, 1 X 1 kP fk2 = hf; H i2 e−2nt ≤ kfk2 ; t L2(R,γ) n! n L2(R,γ) L2(R,γ) n=0 and Z Z p 2 kP fk2 = f(e−tx + 1 − e−2ty)p(y)dy p(x)dx et L2(R,γ) R R Z p ≤ f 2(e−tx + 1 − e−2ty)p(y)p(x)dydx R2 p = E[f 2(e−tX + 1 − e−2tY )] = kfk2 ; L2(R,γ) where X and Y are independent N(0; 1)-random variables. ax 2 (ii) The functions fe ; a 2 Rg form a total system in L (R; γ). So, it suffices to show that a· a· Pte = Pete for each a 2 R. We have p 2 a· h axe−t+aY 1−e−2t i axe−t a (1−e−2t) (Pete )(x) = E e = e e 2 2 2 1 n −nt a axe−t− 1 a2e−2t a X a e = e 2 e 2 = e 2 H (x) n! n n=0 1 ! 2 n 2 2 a X a a a·− a = e 2 P H (x) = e 2 P e 2 (x) t n! n t n=0 a· = (Pte )(x): This completes the proof of the proposition. 5 The Ornstein-Uhlenbeck semigroup has the following propreties: 1. kPtfkLp(R,γ) ≤ kfkLp(R,γ) for any p ≥ 2. Proof: Using Mehler's formula and H¨older'sinequality, we can write Z Z p p −t p −2t kPtfk p = f(e x + 1 − e y)p(y)dy p(x)dx L (R,γ) R R Z p ≤ jf(e−tx + 1 − e−2ty)jpp(y)p(x)dydx R2 p = E[jf(e−tX + 1 − e−2tY )jp] = kfkp : Lp(R,γ) R 2. P0f = f and P1f = limt!1 Ptf = f(y)p(y)dy. R −t 3. DPtf = e PtDf. 4. f ≥ 0 implies Ptf ≥ 0. 2 5. For any f 2 L (R; γ) we have Z Z 1 f(x) − fdγ = − LPtf(x)dt: R 0 Proof: 1 Z X 1 f(x) − fdγ = hf; H i 2 H (x) n! n L (R,γ) n R n=1 1 Z 1 X 1 −nt = hf; H i 2 ne H (x)dt n! n L (R,γ) n n=1 0 1 ! Z 1 X 1 = hf; H i 2 (−LP H )(x) dt n! n L (R,γ) t n 0 n=1 Z 1 = − LPtf(x)dt: 0 1 Proposition 1.3 (First Poincar´einequality). For any f 2 Cp (R), Var(f) ≤ kf 0k2 L2(R,γ) Proof. Set f¯ = R fdγ. We can write R Z Var(f) = f(x)(f(x) − f¯)p(x)dx R Z 1 Z = − f(x)LPtf(x)p(x)dxdt 0 R Z 1 Z = f(x)δDPtf(x)p(x)dxdt 0 R Z 1 Z −t 0 0 = e f (x)Ptf (x)p(x)dxdt 0 R Z 1 −t 0 0 ≤ e kf kL2(R,γ)kPtf kL2(R,γ)dt 0 ≤ kf 0k2 : L2(R,γ) 6 This result as the following interpretation. If f 0 is small, f is concentrated around its mean value f¯ = R f(x)p(x)dx because R Z Var(f) = (f(x) − f¯)2p(x)dx: R The result can be extended to the Sobolev space 1;2 0 2 D = ff : f; f 2 L (R; γ)g defined as the completion of C1( ) by the norm kfk2 = kfk2 + kf 0k2 .