Show that $G(y^*)(h) = \int \limits _X {y^* \circ F(t) \Bbb d \nu _h (t)}$ for some measure $\nu$ on $X$

Question

The following image is taken from this paper, page $129$.

Questions:

1. Why do we have $$G(y^*)(h) = \int \limits _X {y^* \circ F(t) \Bbb d \nu _h (t)} ?$$

(I couldn't get my hands on the book 'Gaussian Measures' by Bogachev, so I don't know what theorem is being used here.)

2. What is the meaning of $H(\gamma)$?

3. Why do we use Lebesgue's Dominated Convergence theorem to conclude $G$ is weak$^*$-to weak$^*$ continuous? I thought it can be done without the theorem?

(Suppose $y_n^* \to y^*$ in the weak$^*$ topology of $Y$. I want to show that $G(y_n^*)(x) \to G(y^*)(x) \forall x \in X$. By our assumption, clearly we have $G(y_n^*)(x) \to G(y^*)(x)$. Where do I need to apply the theorem?)

4. Why is $R$ a weak Gâteaux derivative of $F * \gamma$ at $x=0$?

Answer 1

The answer for (2)

In what follows, $D(F, x, h)$ will be the Gâteaux derivative of $F$ in $x$ along $h$.

The definition of $H_\gamma$ is given in 2.2.7 (page 44) of Bogachev's "Gaussian Measures" (1998 edition). Let us assume that $X^* \hookrightarrow L^2 (\gamma)$. Then for $\omega \in X^*$ we may define

$$a_\gamma (\omega) = \int \limits _X \omega (x) \Bbb d \gamma (x)$$

(this is the "mean", or "expectation", of $\omega$ with respect to $\gamma$).

Next, we may define

$$R_\gamma (\omega) (\eta) = \int \limits _X \big( \omega (x) - a_\gamma (\omega) \big) \big( \eta (x) - a_\gamma (\eta) \big) \Bbb d \gamma (x)$$

(this is the "covariance operator").

Define

$$|h| _\gamma = \sup \{ \omega (h) : \omega \in X^*, R_\gamma (\omega) (\omega) \le 1\} .$$

Then

$$H(\gamma) = \{ h \in X : |h|_\gamma < \infty \} ,$$

is called the Cameron-Martin space (or "reproducing kernel Hilbert space").

The whole sub-chapter 2.4 is devoted to the exploration of $H(\gamma)$. A notation that we shall need is introduced in lemma 2.4.1. For $h \in X*$, define $\hat h$ by

$$\omega h = \int \limits _X \big( \omega (x) - a_\gamma (\omega) \big) \hat h (x) \Bbb d \gamma (x) \ \forall \omega \in X^*$$

(so that $\hat h \in X^*$). This allows to define an inner product on $H(\gamma)$ by $(x, y) = \langle \hat x, \hat y \rangle _{L^2 (\gamma)}$ (remeber that we worked under the assumption that $X^* \hookrightarrow L^2 (\gamma)$.)

One would like now to derive measures with respect to vectors. Definition 5.1.5 (page 207) says: if $h \in X$, we say that $\gamma$ is derivable along $h$ if and only if $\lim \limits _{t \to 0} \dfrac {\gamma (A + th) - \gamma (A)} t$ exists for all Borel subsets $A$.

Bogachev shows that this is equivalent to the following more technical (but also more useful) definition: $\gamma$ is called derivable along $h \in X$ if there exist a function $\beta _h \in L^1 (\gamma)$ such that for all smooth cylindrical functions $f$, the integration by parts

$$\int \limits _X D(f,x,h) \Bbb d \gamma (x) = - \int \limits _X f(x) \beta _h (x) \Bbb d \gamma (x)$$

is true. Intuitively, $\beta _h \gamma$ is the derivative of $\gamma$ along $h$. (A smooth cylindrical function is a function of the form $\varphi (\omega_1 (x), \dots, \omega_n (x))$ for some $n \in \Bbb N ^*$, $\varphi$ bounded and smooth on $\Bbb R ^n$ and $\omega_1, \dots, \omega_n \in X^*$.)

In general, measures are not derivable along evey vector. Nevertheless, proposition 5.1.6 show that for $\gamma$ a Gaussian probability, the set of vectors along which $\gamma$ is derivable coincides with $H(\gamma)$. Thus, you may even forget everything written above and think of the Cameron-Martin space as the directions along you may derive $\gamma$.

Finally, we have theorem 5.1.8 that is used in the article that you are reading. Fix $h \in H$ and let $F : X \to \Bbb R$ be $\gamma$-measurable, such that the function $t \mapsto F(x + th)$ is locally absolutely continuous for $\gamma$-almost every $x$. If $x \mapsto D(F, x, h)$ and $x \mapsto F(x) \hat h (x)$ are integrable with respect to $\gamma$, then

$$\int \limits _X D(F, x, h) \Bbb d \gamma (x) = \int \limits _X F (x) \hat h (x) \Bbb d \gamma (x) .$$

The answer for (1)

Start with the equality

$$[(y^* \circ F)*\gamma] (x) = \int \limits _X (y^* \circ F) (x + u) \Bbb d \gamma (u) .$$

Then

$$G (y^*)(h) = D((y^* \circ F)*\gamma, 0, h) = \lim \limits _{t \to 0} \frac {[(y^* \circ F)*\gamma] (0 + th) - [(y^* \circ F)*\gamma] (0)} t = \\ \lim \limits _{t \to 0} \int \limits _X \frac {(y^* \circ F) (0 + th + u) - (y^* \circ F) (0 + u)} t \Bbb d \gamma (u) = \\ \int \limits _X \lim \limits _{t \to 0} \frac {(y^* \circ F) (0 + th + u) - (y^* \circ F) (0 + u)} t \Bbb d \gamma (u) = \int \limits _X D(y^* \circ F, u, h) \Bbb d \gamma (u) = \int \limits _X (y^* \circ F) (u) \hat h (u) \Bbb d \gamma (u) ,$$

where the last equality is obtained from an application of theorem 5.1.8 mentioned above. If we construct a measure by $\nu (A) = \int \limits _A \hat h (u) \Bbb d u$, then we may write

$$\int \limits _X (y^* \circ F) (u) \hat h (u) \Bbb d \gamma (u) = \int \limits _X (y^* \circ F) (u) \Bbb d \nu (u) ,$$

so that we may finally write

$$G (y^*)(h) = \int \limits _X (y^* \circ F) (u) \Bbb d \nu (u) .$$

There are three questions here: why have we been able to introduce $\lim \limits _{t \to 0}$ inside the integral, why is $y^* \circ F$ Gâteaux derivable and why is it a smooth cylindrical function? The answer to the first question can be easily obtained by noting that $y^* \circ F$ is a Lipschitz function and applying Lebesgue's dominated convergence theorem. I do not yet know the answer to the second and third, in fact I suspect the authors to be wrong (a smooth cylindrical function is, in particular, bounded, and the authors make no assumption of boundedness for $F$).

The answer for (3)

First, I do not understand your argument, I suspect that it is wrong.

Second, it seems that the authors construct $G : Y^* \to X^*$ by

$$G (y^*) = \left(h \mapsto \int \limits _X (y^* \circ F) (u) \hat h (u) \Bbb d \gamma (u) \right)$$

and would like to show that is is weak-$*$ to weak-$*$ continuous (it is obviously linear). In order to do this, they claim to use Lebesgue's dominated convergence theorem. I see a problem here: $X^*$ endowed with the weak-$*$ topology is not first-countable, unless $X$ is finite-dimensional, which means that one should use nets, not sequences; the problem is that Lebesgue's theorem is not always true for nets. Ignoring this, I do not see how they apply this theorem, unless again $F$ is bounded, in which case $y^* \circ F$ will also be bounded and since $\gamma$ is a finite measure, the rest follows easily.

The answer to (4)

With $G : Y^* \to X^*$ linear and weak-$*$ continuous, construct $R : X \to Y$ by $y^* (Rh) = G(y^*) (h)$. $R$ is clearly linear. To show continuity, it suffices to show it at $0$, therefore choose $h_n \to 0$. Then

$$y^* (R h_n) = G(y^*) (h_n) \to G(y^*) (0) = 0 .$$

Since this is true for every $y^* \in Y^*$, this shows that $R h_n \to 0$ weakly, i.e. that $R$ is continuous from the norm topology on $X$ to the weak topology on $Y$.