Integration by Parts Formula and the Stein Lemma on Abstract Wiener Space Hui-Hsiung Kuo

Communications on Stochastic Analysis

Volume 5 | Number 2 Article 10

6-1-2011 Integration by parts formula and the Stein lemma on abstract Wiener space Hui-Hsiung Kuo

Yuh-Jia Lee

Follow this and additional works at: https://digitalcommons.lsu.edu/cosa Part of the Analysis Commons, and the Other Mathematics Commons

Recommended Citation Kuo, Hui-Hsiung and Lee, Yuh-Jia (2011) "Integration by parts formula and the Stein lemma on abstract Wiener space," Communications on Stochastic Analysis: Vol. 5 : No. 2 , Article 10. DOI: 10.31390/cosa.5.2.10 Available at: https://digitalcommons.lsu.edu/cosa/vol5/iss2/10 Communications on Stochastic Analysis Serials Publications Vol. 5, No. 2 (2011) 405-418 www.serialspublications.com

INTEGRATION BY PARTS FORMULA AND THE STEIN LEMMA ON ABSTRACT WIENER SPACE

HUI-HSIUNG KUO AND YUH-JIA LEE*

Abstract. We prove the integration by parts formula for abstract Wiener measures under much weaker conditions than those assumed in [11]. Thus the formula holds for a much larger class of functions. For applications, we characterize the abstract Wiener measures, extend the Stein lemma to inﬁnite dimensions, and evaluate some functional integrals.

1. Introduction The first infinite dimensional integration by parts formula was obtained by Cameron [2] in 1951 for the classical Wiener space. Later in 1964, Donsker [5] also proved this formula and applied it to study Fréchet-Volterra differential equations. In 1974 Kuo [11] generalized the formula to abstract Wiener space (Gross [8, 9], see also the book [12]) for abstract Wiener measures and applied the formula to study Fourier-Wiener transform and to evaluate some functional integrals. The purpose of this paper is to prove the integration by parts formula in [11] under much weaker conditions. In fact, we also simplify the proof. This will be done in Section 2. As an interesting application of this formula, we will extend the Stein lemma [1, 3, 14, 15] to infinite dimensions in Section 3. Then in Section 4 we will apply this integration by parts formula to evaluate some functional integrals on an abstract Wiener space.

2. Integration by Parts Formula Let (i, H, B) be an abstract Wiener space, i.e., H is a real Hilbert space with norm | · | and inner product h·, ·i, B is the completion of H with respect to a measurable norm k · k on H, and i : H,→ B is the inclusion map. The dual space B∗ of B is embedded in H by using the Riesz representation theorem to identify the dual space H∗ of H with H. It follows from this identiﬁcation that (y, x) = hy, xi, ∀ y ∈ B∗, x ∈ H, where (·, ·) is the bilinear pairing of B∗ and B. Let pt denote the Gaussian measure on (i, H, B) with mean 0 and variance t. If y ∈ B∗, then (y, x) is deﬁned for all x ∈ B and (y, ·) is a normal random

Received 2010-4-25; Communicated by the editors. 2000 Mathematics Subject Classiﬁcation. Primary 28C20, 46E25; Secondary 60G15, 26E20. Key words and phrases. Abstract Wiener space, probability measure, integration on inﬁnite dimensional spaces. * This research is supported by the NSC of Taiwan. 405 406 HUI-HSIUNG KUO AND YUH-JIA LEE

2 variable with mean 0 and variance t|y| on the probability space (B, pt). Suppose ∗ h ∈ H. Take a sequence {yn} in B such that yn → h in H. Then {(yn, ·)} is a 2 Cauchy sequence in L (B, pt). The limit of this sequence is well defined, i.e., it is independent of the choice of the sequence {yn}. For simplicity, we will use hh, ·i to denote this limit. Then hh, ·i, defined almost everywhere on (B, pt), is a normal random variable with mean 0 an variance t|h|2. Let f be a function defined on an open subset U of B with values in a real Banach space W . The H-derivative DH f(x) of f at x ∈ U belongs to the Banach space L(H,W ) of bounded linear operators from H into W . A function f from 1 U into W is called a CH -function if DH f is a continuous function from U into L k ∈ (H,W ). The k-th H-derivative DH f(x) of f at x U is a bounded k-linear mapping from H × · · · × H (k factors) into W . In particular, when W = R, 2 the second H-derivative DH f(x) of f at x can be identified as a bounded linear 2 operator on H. When DH f(x) is a trace class operator on H, its trace is known as the Gross Laplacian of f: 2 ∆Gf(x) = trace DH f(x) . Now we can state the integration by parts formula from [11] on an abstract Wiener space (i, H, B). 1 Theorem 2.1 (Kuo [11]). Let f be a CH -function from B into a real Hilbert space K satisfying the following conditions: 2 (a) kf(·)kK ∈ L (pt) with k · kK being the norm on K; (b) There exist constants r > 0 and M > 0 such that Z

kf(x + h)kK dpt(x) < M, ∀ |h| < r; B (c) The function

sup kDH f(x + hkL(H,K), x ∈ B, |h|

is pt-integrable. Then for any h ∈ H, we have Z Z

hh, xif(x) dpt(x) = t [DH f(x)](h) dpt(x). (2.1) B B The conditions in Theorem 2.1 are too strong. We will prove in Theorem 2.6 (see below) that the integration by parts formula in Equation (2.1) holds under much weaker conditions. For the proof of Theorem 2.6, we need to prepare several lemmas. We start with a generalization of the Stein Lemma (see [1, 3, 14, 15].) Lemma 2.2. Let f and g be C1-functions on R with g(∞) = 0. Then the following equality holds: Z ∞ Z ∞ f 0(x)g(x) dx = − f(x)g0(x) dx, (2.2) −∞ −∞ where the existence of either improper integral implies that of the other. Assume further that f 0g and fg0 are Lebesgue integrable. Then Equation (2.2) also holds in the sense of Lebesgue integral. INTEGRATION BY PARTS FORMULA AND STEIN LEMMA 407 R ∞ 0 Proof. We may assume that −∞ f(x)g (x) dxR exists since the proof of the other ∞ 0 case is similar. Write the improper integral −∞ f (x)g(x) dx as Z Z Z ∞ 0 ∞ f 0(x)g(x) dx = f 0(x)g(x) dx + f 0(x)g(x) dx. (2.3) −∞ −∞ 0 For the ﬁrst improper integral in the right-hand side, we take the limit and change the order of integration of a continuous function on a ﬁnite region to get Z Z Z 0 0 x f 0(x)g(x) dx = f 0(x) g0(y) dy dx −∞ −∞ −∞ Z Z 0 x = lim f 0(x) g0(y) dy dx a→∞ − − Z a Z a 0 0 = lim f 0(x) dx g0(y) dy a→∞ − Z a y 0 = lim f(0) − f(y) g0(y) dy a→∞ − a Z 0 = lim f(0) g(0) − g(−a) − f(y)g0(y) dy a→∞ − Z a 0 = f(0)g(0) − f(x)g0(x) dx, (2.4) −∞ R 0 0 which shows that −∞ f (x)g(x) dx exists. Similarly, for the second improper integral, we have Z ∞ Z ∞ f 0(x)g(x) dx = −f(0)g(0) − f(x)g0(x) dx, (2.5) 0 R 0 ∞ 0 which impliesR that 0 f (x)g(x) dx exists. Hence by Equation (2.3) the improper ∞ 0 integral −∞ f (x)g(x) dx also exists. Then Equation (2.2) follows from Equations (2.4) and (2.5). To prove the second assertion, suppose f 0g and g0f are Lebesgue integrable. Then by the Lebesgue dominated covergence theorem, Z Z ∞ b f 0(x)g(x) dx = lim f 0(x)g(x) dx −∞ a,b→∞ − Z a Z b ∞ = − lim f(x)g0(x) dx = − f(x)g0(x) dx, →∞ a,b −a −∞ which shows that Equation (2.2) holds in the sense of Lebesgue integral. Lemma 2.3. Let f and g be Cn-functions on R with g(j)(∞) = 0 for all j = 0, 1, . . . , n − 1. Then the following equality holds: Z ∞ Z ∞ f (n)(x)g(x) dx = (−1)n f(x)g(n)(x) dx, (2.6) −∞ −∞ where the existence of either improper integral implies that of the other. Assume further that f (n)g and fg(n) are Lebesgue integrable. Then Equation (2.6) also holds in the sense of Lebesgue integral. 408 HUI-HSIUNG KUO AND YUH-JIA LEE

Proof. We only prove the first assertion for the improper integral case since the second one for the Lebesgue integral case can be handled by the same arguments as those used in the proofR of Lemma 2.2. ∞ (n) We may assume that −∞ f(x)g (x) dx exists since the proof of the other − Rcase is similar. Apply Lemma 2.2 to the functions f and g(n 1) to conclude that ∞ 0 (n−1) −∞ f (x)g (x) dx exists and Z ∞ Z ∞ f 0(x)g(n−1)(x) dx = − f(x)g(n)(x) dx. −∞ −∞ 0 − ThenR again apply Lemma 2.2 to the functions f and g(n 2) to conclude that ∞ 00 (n−2) −∞ f (x)g (x) dx exists and Z ∞ Z ∞ f 00(x)g(n−2)(x) dx = − f 0(x)g(n−1)(x) dx. −∞ −∞ Repeat the argument until finallyR when we apply Lemma 2.2 to the functions (n−1) ∞ (n) f and g to conclude that −∞ f (x)g(x) dx exists and Z ∞ Z ∞ f (n)(x)g(x) dx = − f (n−1)(x)g0(x) dx. −∞ −∞ Putting the above equalities together, we immediately obtain Equation (2.6). For the next theorem, we recall the Hermite polynomial of degree n defined by − n x2/2 n −x2/2 Hn(x) = ( 1) e Dx e . Theorem 2.4. Let µ be the standard Gaussian measure on R, namely, dµ(x) = 2 √1 e−x /2 dx. Let f be a CN -function on R such that 2π Z ∞ f(x)Hn(x) dµ(x) < ∞, ∀ 1 ≤ n ≤ N. −∞ Then we have Z ∞ Z ∞ (n) f (x) dµ(x) = f(x)Hn(x) dµ(x), ∀ 1 ≤ n ≤ N. (2.7) −∞ −∞ Proof. For each n = 1, 2,...,N, apply Lemma 2.3 to the functions f(x) and 2 g(x) = √1 e−x /2. Note that 2π

(n) n 1 −x2/2 g = (−1) Hn(x)√ e . 2π Hence the resulting equality from Equation (2.6) becomes Equation (2.7). Now we go back to an abstract Wiener space (i, H, B). We ﬁrst prove a special case of Theorem 2.6 (see below) for K = R.

Theorem 2.5. Let pt be the Gaussian measure on an abstract Wiener space → R 1 | · | ∈ (i, H, B) with variance t. Let f : B be a CH -function such that DH f( ) L1(B, p ). Then for any h ∈ H, we have t Z Z

hh, xif(x) dpt(x) = t hDH f(x), hi dpt(x). (2.8) B B INTEGRATION BY PARTS FORMULA AND STEIN LEMMA 409

Proof. For h =6 0, put e = h/|h|. Let P denote the projection P x = hx, eie and − × 0 let Qx = x P x. Then pt can be decomposed as a product measure pt = µt µt. R 0 Here µt is the Gaussian measure on with mean 0 and variance t and µt is the Gaussian measure on QB with mean 0 and variance t. Then we have Z Z Z ∞ h i | | 0 h, x f(x) dpt(x) = h λ f(λe + v) dµt(λ)dµt(v) B QB −∞ Z Z ∞ n o | | d 0 = t h f(λe + v) dµt(λ)dµt(v) QB −∞ dλ Z Z ∞ h i 0 = t DH f(λe + v), h dµt(λ)dµt(v) ZQB −∞

= t hDH f(x), hi dpt(x), B where the second equality follows from Lemma 2.2. Next we state the main theorem of this paper, which assumes much weaker conditions than those in Theorem 2.1. 1 Theorem 2.6. Let (i, H, B) be an abstract WienerR space and f a CH -function k k ∞ from B into a real Hilbert space K such that B DH f(x) L(H,K) dpt(x) < . Then for any h ∈ H, we have Z Z

hh, xif(x) dpt(x) = t [DH f(x)](h) dpt(x), (2.9) B B where the integrals are K-valued Bochner integrals. Proof. Without loss of generality, we may assume that the space K is separable by replacing K, if necessary, with the closure of f(B). Under this assumption, the function f can be taken to be either strongly measurable or weakly measurable and the integrals in Equation (2.9) exist as Bochner integrals. For Φ ∈ K∗, the composition Φ ◦ f is a C1 -function from B into R and H hDH (Φ ◦ f)(x), hi = Φ ◦ [DH f(x)] (h), h ∈ H. (2.10) Then we haveZ Z

DH (Φ ◦ f)(x) dpt(x) = Φ ◦ [DH f(x)] dpt(x) B ZB

≤ kΦkK∗ kDH f(x)kL(H,K) dpt(x) B Z

= kΦkK∗ kDH f(x)kL(H,K) dpt(x) B < ∞ (by assumption). Hence we can apply Theorem 2.5 and use Equation (2.10) to show that Z Z

hh, xiΦ(f(x)) dpt(x) = t hDH (Φ ◦ f)(x), hi dpt(x) B ZB ∗ = t Φ ◦ [DH f(x)] (h) dpt(x), ∀ Φ ∈ K , B 410 HUI-HSIUNG KUO AND YUH-JIA LEE or equivalently, Z n o ∗ Φ hh, xif(x) − t DH f(x) (h) dpt(x) = 0, ∀ Φ ∈ K . B Since K∗ separates points of K, we can conclude that Z hh, xif(x) − t DH f(x) (h) dpt(x) = 0, B which gives Equation (2.9).

The next theorem follows from Theorem 2.6. We will use kT kHS to denote the Hilbert–Schmidt norm of an operator T . Theorem 2.7. Let (i, H, B) be an abstract Wiener space and A a bounded operator 1 from B into a real Hilbert space K. Suppose f is a CH -function from B into K satisfying the following conditions: (a) D f(x) is a Hilbert–Schmidt operator from H into K and H Z k k ∞ DH f(x) HS dpt(x) < ; R B k k | | ∞ (b) B f(x) K Ax K dpt(x) < . Then we haveZ Z h i | ∗ f(x), Ax K dpt(x) = t traceK DH f(x)(A H ) dpt(x), B B h· ·i where , K denotes the inner product of K, traceK denotes the trace of a trace class operator of K, and A|H denotes the restriction of the operator A to H. Remark 2.8. We point out that if A is a bounded operator from B to K, then its restriction A|H to H is a Hilbert–Schmidt operator from H into K. This is a fact from Corollary 4.4 (page 85 [12]), which follows from Theorem 4.6 (page 83 [12]) due to V. Goodman. Moreover, we have the next theorem which shows that Theorem 2 in [11] is valid under a much weaker condition.

1 Theorem 2.9. Let u and v be CH -functions from B into real Banach spaces F and G, respectively. Assume that the following condition holds: Z n o

kDH u(x)kL(H,F ) |v(x)|G + |u(x)|F kDH v(x)kL(H,G) dpt(x) < ∞, B where | · |G and | · |F denote the norms on G and F , respectively. Then for any bounded bilinear map Φ from F × G into a real Hilbert space K, we have Z Φ u(x), [DH v(x)](h) dpt(x) B Z n o 1 = hh, xiΦ u(x), v(x) − Φ [DH u(x)](h), v(x) dpt(x). B t Proof. Just apply Theorem 2.5 to the function f(x) = Φ(u(x), v(x)). INTEGRATION BY PARTS FORMULA AND STEIN LEMMA 411

We can use Theorem 2.5 to generalize another formulation of integration by parts formula given in Theorem 3.7 [13].

Theorem 2.10. Let pt be the Gaussian measure on an abstract Wiener space (i, H, B) with mean 0 ana variance t. Suppose f : B → B∗ is an H-diﬀerentiable function satisfying the following conditions: 1 (a) kDH f(·)ktr ∈ L (B, pt), (k · ktr: the traceR class norm); | |α ∞ | · | ∗ (b) There exists a constant α > 1 such that B f(x) B∗ dpt(x) < , ( B : the norm on B∗). Then the following equality holds: Z Z

f(x), x dpt(x) = t traceH DH f(x) dpt(x), (2.11) B B · · ∗ where ( , ) denotes the bilinear pairing of B and B and traceH denotes the trace of a trace class operator on H. Proof. We need a fact from the book [12] (page 66, Corollary 4.2). Let (i, H, B) be an abstract Wiener space. Then there exist another abstract Wiener space { }∞ (i0,H,B0) and an increasing sequence Pn n=1 of ﬁnite dimensional orthogonal projections converging strongly to the identity in H such that:

(1) The B0-norm is stronger than the B-norm, (hence B0 ⊂ B). e (2) Each Pn extends by continuity to a projection Pn of B0. e (3) Pn converges strongly to the identity in B0 with respect to B0-norm.

Moreover, let pt and pet denote the Gaussian measures on B and B0, respectively, with mean 0 and variance t. Then the equality Z Z

g(x) dpt(x) = ge(x) dpet(x) (2.12) B B0 holds for all nonnegative measurable or pt-integrable functions g on B. Here ge denotes the the restriction of g to B0. Now let f be a function given as in the theorem. Apply Equation (2.12) to the function g(x) = (f(x), x) to get Z Z e (f(x), x) dpt(x) = (f(x), x) dpet(x). (2.13) B B0 e e Since Pn is the extension of Pn by continuity, we see that Pn(B0) must be the closure of Pn(H) with respect to the B0-norm. On the other hand, note that e Pn(H) is ﬁnite dimensional. Hence we can conclude that Pn(B0) = Pn(H). { }∞ Let ej j=1 be an orthonormal basis for H. Then Z X∞ Z e e e (f(x), P˜nx) dp˜t(x) = hf(x), ejihPnx, eji dpet(x) B0 j=1 B0 X∞ Z e = hf(x), ejihx, Pneji dpet(x) j=1 B0 412 HUI-HSIUNG KUO AND YUH-JIA LEE

Apply Theorem 2.6 to show that Z X∞ Z e e (f(x), P˜nx) dp˜t(x) = t hDH f(x)(Pnej), eji dp˜t(x) B B 0 Zj=1 0 e ◦ = t traceH DH f(x) Pn dp˜t(x). (2.14) B0

Note that for all x ∈ B0,

e traceH [DH f(x) ◦ Pn] ≤ kDH f(x)ktr,

e e e e e (f(x), Pnx) ≤ |f(x)|B∗ kPnxk0 ≤ |f(x)|B∗ kxk0. By the Fernique theorem [7] and the condition (b), we have Z Z 1/α Z 1/α0 α α0 | e | ∗ k k e ≤ | e | e k k e f(x) B x 0 d pt(x) f(x) B∗ d pt(x) x 0 d pt(x) B0 B0 B0 < ∞, (2.15) where 1/α + 1/α0 = 1. With condition (a) and Equation (2.15), we can apply the Lebesgue dominated convergence theorem and the same arguments as those in [13] to let n → ∞ in Equation (2.14) to obtain Z Z e e (f(x), x) dp˜t(x) = t traceH DH f(x) dp˜t(x) B0 ZB0

= t traceH DH f(x) dpt(x). (2.16) B Thus Equation (2.11) follows from Equations (2.13) and (2.16).

3. Infinite Dimensional Stein Lemma We can apply Theorem 2.5 to obtain a generalization of the Stein lemma [1, 3, 14, 15] to infinite dimensional spaces. Let B be a real separable Banach space. A probability measure µ on B is called a Gaussian measure with mean 0 if for each ξ in the dual space B∗ of B, the random variable (ξ, ·) on (B, µ) is Gaussian with mean 0. Theorem 3.1. Let X be a random variable with values in a real separable Banach space B. The distribution µ of X on B is a Gaussian measure with mean 0 if and only if there exists a real separable Hilbert space H such that (i, H, B) is an abstract Wiener space and the equality Z n o hh, xif(x) − hDH f(x), hi dµ(x) = 0, ∀ h ∈ H, (3.1) B 1 | · | ∈ 1 holds for all CH -functions f such that DH f( ) L (B, µ). Proof. We first prove the necessity part. Suppose the distribution µ of X is a Gaussian measure on B with mean 0. By the Kuelbs Theorem [10], there exists a real separable Hilbert space H such that (i, H, B) is an abstract Wiener space INTEGRATION BY PARTS FORMULA AND STEIN LEMMA 413 and µ = p1, the Gaussian measure on B with mean 0 and variance 1. Then we can apply Theorem 2.5 with t = 1 to obtain Equation (3.1) from Equation (2.8). Next we prove the sufficiency part. Assume that (i, H, B) is an abstract Wiener space and the distribution µ of X satisfies Equation (3.1). Let ξ ∈ B∗ be fixed. The characteristic function of the random variable (ξ, X) is given by Z θ(λ) = Eeiλ(ξ,X) = eiλ(ξ,x) dµ(x), λ ∈ R. B The assumption on µ allows us to differentiate θ to get Z θ0(λ) = i (ξ, x)eiλ(ξ,x) dµ(x). (3.2) B Use Equation (3.1) for h = ξ and f(x) = cos(λ(ξ, x)) to get Z Z (ξ, x) cos(λ(ξ, x)) dµ(x) = − λ|ξ|2 sin(λ(ξ, x)) dµ(x). (3.3) B B Similarly, with h = ξ and f(x) = sin(λ(ξ, x)) in Equation (3.1), we have Z Z (ξ, x) sin(λ(ξ, x)) dµ(x) = λ|ξ|2 cos(λ(ξ, x)) dµ(x). (3.4) B B It follows from Equations (3.2), (3.3), and (3.4) that θ0(λ) = −λ|ξ|2θ(λ), which together with the condition θ(0) = 1 yields the equality − 1 λ2|ξ|2 θ(λ) = e 2 . Hence we have proved that Z iλ(ξ,x) − 1 λ2|ξ|2 ∗ e dµ(x) = e 2 , ∀ λ ∈ R, ξ ∈ B . B This equality implies that µ is a Gaussian measure on B with mean 0. Thus the theorem is proved. More generally, we have the next theorem, which can be easily proved by using Theorem 2.7 and similar arguments as those in the proof of Theorem 3.1. Theorem 3.2. Let X be a random variable taking values in a real separable Ba- nach space B. Then the distribution µ of X on B is a Gaussian measure with mean 0 if and only if there exists a real separable Hilbert space H such that (i, H, B) is an abstract Wiener space and the equality Z n o h i − | ∗ f(x), Ax K traceK DH f(x)(A H ) dµ(x) = 0 B holds for all bounded operators A from B into a real separable Hilbert space K and 1 for all CH -functions f from B into K satisfying the following conditions: (a) D f(x) is a Hilbert–Schmidt operator from H into K and H Z k k ∞ DH f(x) HS dµ(x) < ; R B k k | | ∞ (b) B f(x) K Ax K dµ(x) < . 414 HUI-HSIUNG KUO AND YUH-JIA LEE

On the other hand, if we apply Theorem 2.10 instead of Theorem 2.5 or 2.7, then the inﬁnite dimensional Stein lemma can be reformulated as given by the next theorem. Theorem 3.3. Let X be a random variable with values in a real separable Banach space B. The distribution µ of X on B is a Gaussian measure with mean 0 if and only if there exists a real separable Hilbert space H such that (i, H, B) is an abstract Wiener space and the equality Z n o − f(x), x traceH DH f(x) dµ(x) = 0 B 1 → ∗ holds for all CH -functions f : B B satisfying the following conditions: 1 (a) kDH f(·)ktr ∈ L (B, µ); R | |α ∞ (b) There exists a constant α > 1 such that B f(x) B∗ dµ(x) < . 4. Evaluation of Function Integrals on Wiener Space Let (i, H, B) be an abstract Wiener space and let µ denote the Gaussian measure on B with mean 0 and variance 1. The following formula was derived in [4]: Z h i z exp (Ax, x) + (h, x) dµ(x) B 2 D E −1/2 1 = det(I − zA| ) exp (I − zA )−1h, h , (4.1) H 2 H

∗ ∗ where z is a complex number, <(z) < 1, h ∈ B , A ∈ L(B,B ), and A is a H self-adjoint operator of H. Note that A H is a trace class operator of H in view − of a theorem (page 83 [12]) due to V. Goodman. Hence det(I zA H ) exists. We now show that the integration by parts formula can be used to derive the formula in Equation (4.1). Define a function U on H by Z h i z U(h) = exp (Ax, x) + (h, x) dµ(x), h ∈ H. (4.2) B 2 The Fréchet derivative U 0(h) of U at h is easily checked to be given by Z h i z hU 0(h), ki = (k, x) exp (Ax, x) + (h, x) dµ(x), k ∈ H. (4.3) B 2 Now let k ∈ B∗. We can check that the following function h i z f(x) = k exp (Ax, x) + (h, x) 2 satisfies the conditions in Theorem 2.10. Hence we can apply the integration by parts formula in Equation (2.11) to the function f. From the resulting equality and Equation (4.3), we get Z Z h 0 i U (h), k = f(x), x dµ(x) = traceH DH f(x) dµ(x). (4.4) B B Direct computation shows that the H-derivative of f is given by h i h i z D f(x)(·) = k z(Ax, ·) + hh, ·i exp (Ax, x) + (h, x) , H 2 INTEGRATION BY PARTS FORMULA AND STEIN LEMMA 415 which yields the following trace: h i h i z trace D f(x) = z(Ak, x) + hh, ki exp (Ax, x) + (h, x) . (4.5) H H 2 It follows from Equations (4.4) and (4.5) that Z h i h i z hU 0(h), ki = z(Ak, x) + hh, ki exp (Ax, x) + (h, x) dµ(x) B 2

= z U 0(h), Ak + U(h)hh, ki. Thus the function U satisﬁes the diﬀerential equation

U 0(h), (I − zA)k = U(h)hh, ki. (4.6)

e ∗ By replacing A with A = [A + (A H ) ]/2, if necessary, we may assume that the −1 operator A|H is self-adjoint. Then we can replace k with (I − zA) k in Equation (4.6) and use the assumption that A|H is self-adjoint to get

0 − −1 ∈ ∗ U (h), k = U(h) (I zA H ) h, k , k B . Therefore, we have the diﬀerential equation

0 − −1 ∈ U (h) = U(h)(I zA H ) h, h H. The solution of this differential equation is given by 1 U(h) = U(0) exp (I − zA )−1h, h . 2 H But from Equation (4.2) we see that Z h i z − −1/2 U(0) = exp (Ax, x) dµ(x) = det(I zA H ) . B 2 Therefore, 1 U(h) = det(I − zA )−1/2 exp (I − zA )−1h, h . (4.7) H 2 H In view of Equations (4.2) and (4.7) we have derived the formula in Equation (4.1). Example 4.1. Let C = C[0, 1] be the classical Wiener space with the Wiener measure W . Let z ∈ C and <(z) < π2. It is shown in [4] that " # Z Z Z 2 q z 1 1 √ √ exp x(t) − x(s)ds dt dW (x) = z csc z. (4.8) C 2 0 0 We give a different method to derive this formula by using Equation (4.1). Let C0 denote the Cameron–Matin space, namely, the space consisting of all functions f ∈ C which are absolutely continuous with f 0 ∈ L2[0, 1]. Then (i, C0, C) is an abstract Wiener space with standard Gaussian measure W . It is well known that C∗ can be identified as the space n o C∗ = h ∈ C0 : h0 is right continuous and h0(1) = 0 416 HUI-HSIUNG KUO AND YUH-JIA LEE

Deﬁne an operator A : C → C∗ by Z Z Z t s 1 (Ax)(t) = − x(v)dv − s x(v)dv ds, x ∈ C. 0 0 0 Then we have Z Z 1 1 2 hAx, xi = x(t) − x(s)ds dt. 0 0 0 0 ∗ Note that (Ax) (·) is a continuous function and (Ax) (1) = 0. Hence Ax ∈ C ∈ L C C∗ C0 and A ( , ). It is easy to see that A C0 is a self-adjoint operator of . Moreover, we can use the same arguments as those on page 50 of the book [12] to compute the eigenvalues of A C0 as given by 1 , n = 1, 2, 3,... n2π2

− Thus the determinant of I zA C0 is given by ∞ Y z 1 √ det(I − zA ) = 1 − = √ sin z, <(z) < π2. C0 n2π2 z n=1 Putting this value into Equation (4.1) with h = 0, we immediately obtain the formula in Equation (4.8). Example 4.2. For z ∈ C with <(z) < π2/4 and h ∈ C0, we have Z Z Z 1 z 1 exp i h(1) − h(t) dy(t) + y(t)2 dt dW (y) C 2 q 0 Z Z 0 √ 1 1 1 √ = sec z exp − K(s, t; z)h0(t)h0(s) dsdt , (4.9) 2 0 0 where the kernel function K is deﬁned by n o sec z K(s, t; z) = sin z(1 − |s − t|) − sin z(1 − |s + t|) . (4.10) 2z To derive the formula in Equation (4.9), put Z t eh(t) = (h(1) − h(s)) ds, 0 Z 1 (Ay)(t) = (t ∧ s)y(s) ds. 0 Then eh ∈ C∗, A ∈ L(C, C∗), and we have Z Z Z 1 z 1 exp i (h(1) − h(t)) dy(t) + y(t)2 dt dW (y) C 2 Z n0 o 0 z = exp i(eh, y) + (Ay, y) dW (y). C 2 INTEGRATION BY PARTS FORMULA AND STEIN LEMMA 417

Then apply the formula in Equation (4.1) to get Z Z Z 1 z 1 exp i (h(1) − h(t)) dy(t) + y(t)2 dt dW (y) 2 C 0 0

−1/2 1 = det(I − zA ) exp − (I − z A )−1eh, eh . (4.11) C0 2 C0

From page 50 of the book [12], the eigenvalues of A C0 are given by 4 , n = 1, 2, 3,... [(2n − 1)π)]2 Therefore, we have ∞ Y 4z √ det(I − zA ) = 1 − = cos z, <(z) < π2/4. (4.12) C0 [(2n − 1)π)]2 n=1

h − −1e ei Next we need to evaluate (I z A C0 ) h, h in Equation (4.11). Put − −1e u(t) = (I z A C0 ) h (t). − e Then we have (I z A C0 )u (t) = h(t), namely, Z Z 1 t u(t) − z (t ∧ s)u(s) ds = (h(1) − h(s)) ds, 0 0 which can be diﬀerentiated twice to produce the following diﬀerential equation with a boundary condition:

u00(t) + zu(t) = −h0(t), u(0) = u0(1) = 0.

By using the variation of parameters method we can derive the solution Z 1 √ √ 1 √ u(t) = √ sec z sin( z t) cos( z(1 − s))h0(s) ds z Z 0 t √ − sin( z(t − s))h0(s) ds . 0 Therefore, Z 1 − −1e e 0 e0 (I z A C0 ) h, h = u (t)h (t) dt Z0 Z 1 1 √ = K(s, t; z)h0(s)h0(t) dsdt, (4.13) 0 0 where the kernel function K is given by Equation (4.10). Upon putting Equations (4.12) and (4.13) into the right-hand side of Equation (4.11), we obtain the formula in Equation (4.9). 418 HUI-HSIUNG KUO AND YUH-JIA LEE

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Hui-Hsiung Kuo: Department of Mathematics, Louisiana State University, Baton Rouge, LA 70803-4918, USA E-mail address: [email protected] Yuh-Jia Lee: Department of Applied Mathematics, National University of Kaohsi- ung, Kaohsiung, Taiwan 811 E-mail address: [email protected]