8 1. THE DERIVATIVE OPERATOR
Proposition 1.14. Let ϕ :
Rm
R be a function such that
|ϕ(x) ϕ(y)| K|x y|
for any x, y Rm. Suppose that F = (F 1, . . . , F m) is a random vector whose
components belong to the space D1,2. Then ϕ(F ) D1,2, and there exists a random
vector G = (G1, . . . , Gm) bounded by K such that
(1.16) D(ϕ(F )) =
m
i=1
GiDF
i.
Proof. If the function ϕ is continuously differentiable, then the result reduces
to that of Proposition 1.10 with Gi =
∂ϕ
∂xi
(F ). Let αn(x) be a sequence of regular-
ization kernels of the form αn(x) = nmα(nx), where α is a nonnegative function
belonging to C∞(Rm) whose support is the unit ball and such that
Rm
α(x)dx = 1.
Set ϕn = ϕ αn. It is easy to check that limn ϕn(x) = ϕ(x) uniformly with respect
to x, and the functions ϕn are C∞ with |∇ϕn|≤ K. For each n we have
(1.17) D(ϕn(F )) =
m
i=1
∂ϕn
∂xi
(F )DF
i.
The sequence ϕn(F ) converges to ϕ(F ) in
L2(Ω)
as n tends to infinity. On the
other hand, the sequence {D(ϕn(F )), n 1} is bounded in L2(Ω; H). Hence, by
Lemma 1.13 ϕ(F ) D1,2 and {D(ϕn(F )), n 1} converges in the weak topol-
ogy of L2(Ω; H) to D(ϕ(F )). On the other hand, the sequence {∇ϕn(F ), n
1} is bounded by K. Hence, there exists a subsequence {∇ϕn(k)(F ), k 1}
that converges to some random vector G = (G1, . . . , Gm) in the weak topology
σ(L2(Ω; Rm)). Moreover, G is bounded by K. Then, taking the limit in (1.17), we
obtain Equation (1.16). The proof of the lemma is now complete.
The following proposition asserts that Bernoulli random variables cannot be
differentiable.
Proposition 1.15. Let A F. Then the indicator function of A belongs to
D1,p, p 1, if and only if P (A) is equal to zero or one.
Proof. By the chain rule (Proposition 1.10) applied to a function ϕ C0 ∞(R),
which is equal to x2 on [0, 1], we have
D1A =
D(1A)2
= 21AD1A
and, therefore, D1A = 0 because from the above equality we get that this derivative
is zero on
Ac
and equal to twice its value on A. So, we obtain 1A = P (A).
Here is an application of the properties of the derivative operator.
Proposition 1.16. Let F
D1,2
be such that E(|F
|−2
DF
2
H
) ∞. Then
the event {F 0} has probability zero or one.
Proof. For any ε 0, consider the function ϕε(x) =
x
−∞
ψε(y)dy, where
ψε(x) =
4x
ε2
1[0,
ε
2
]
(x) +
4(ε x)
ε2
1(
ε
2
,ε]
(x).
By Proposition 1.10, ϕε(F ) belongs to
D1,2,
and
Dϕε(F )
H
= ψε(F ) DF
H

2
ε
1{0≤F
≤ε}
DF
H
.
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