Stein's method, Malliavin calculus and infinite-dimensional Gaussian

On the other hand, by noting h A 1(x) = h(A 1 x), one obtains by standard computations (recall
that A is symmetric) that Hess U 0 g(x) = Hess h A 1(x) = A 1 Hess h(A 1 x)A 1 ; yielding
sup kHess U 0 g(x)k H:S: = sup kA 1 Hess h(A 1 x)A 1 k H:S:
x2R d x2R d
= sup
x2R d kA 1 Hess h(x)A 1 k H:S:
kA 1 k 2 op sup
x2R d kHess h(x)k H:S: (8.23)
kA 1 k 2 op kAk op kgk L (8.24)
kC 1 k op kCk 1=2
op kgk L : (8.25)
The chain of inequalities appearing in formulae (8.23)–(8.25) are mainly a consequence of the
usual properties of the Hilbert-Schmidt and operator norms. Indeed, to prove inequality (8.23)
we used the relations
kA 1 Hess h(x)A 1 k H:S: kA 1 k op kHess h(x)A 1 k H:S:
kA 1 k op kHess h(x)k H:S: kA 1 k op ;
relation (8.24) is a consequence of (8.22); …nally, to show the inequality (8.25), one uses the fact
that
q
q
q q
kA 1 k op kA 1 A 1 k op = kC 1 k op and kAk op kAAk op = kCk op :
We are now left with the proof of Point (i) in the statement. The fact that a vector Y N d (0; C)
necessarily veri…es (8.19) can be proved by standard integration by parts. On the other hand,
suppose that Y veri…es (8.19). Then, according to Point (ii), for every g 2 C 2 (R d ) with bounded
…rst and second derivatives,
E(g(Y )) E(g(Z)) = R(hY; rU 0 g(Y )i R d hC; Hess U 0 g(Y )i H:S: ) = 0;
where Z N d (0; C). Since the collection of all such functions g generates the Borel -…eld on
R d , this implies that Y Law = Z, thus yielding the desired conclusion.
9 Explicit bounds using Malliavin operators
9.1 One-dimensional normal approximation
Consider a standard Gaussian random variable N N (0; 1), as well as a functional F of some
isonormal Gaussian process X = fX (h) : h 2 Hg. We are interested in assessing the distance
between the law of N and the law of F by using relations (8.6)–(8.8). As shown in the next
statement, which has been …rst proved in [57], this task is particularly easy if one assumes that
F is also Malliavin di¤erentiable.
Theorem 9.1 (See [57]) Let F 2 D 1;2 be such that E [F ] = 0. Then, one has that
d W (F; N) E 1 DF; DL 1 F H
(9.1)
E[(1 DF; DL 1 F H )2 ] 1=2
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