1.1 Classical Wiener Space

Proof. $(1)$: Let $R \ge 0$, then

where each $\bracs{\theta \in C([0, \infty); \real^d): \abs{\theta(t)}/(1+t) \le R}$ is a closed subset of $C([0, \infty); \real^{d})$, and their intersection is also closed. Thus $\norm{\cdot}_{\sps}$ is lower semicontinuous.

$(2)$: Let $K \subset [0, \infty)$ be compact, then $M = \sup_{t \in K}(1 + t) < \infty$. Thus $\norm{\theta|_K}_{u} \le M\norm{\theta}_{\sps}$.

$(3)$: Since $\theta(0) = 0$ and $\limv{t}\theta(t)/t = 0$ for all $\theta \in \sps$,

and $\norm{\Lambda \theta}_{u, (0, \infty)}= \norm{\theta}_{\sps}$. On the other hand, for any $f \in C_{0}((0, \infty); \real^{d})$, let $\Lambda^{-1}f$ be defined by $(\Lambda^{-1}f)(0) = 0$ and $(\Lambda^{-1}f)(t) = f(t)(1 + t)$ for $t > 0$. Since $\lim_{t \to 0}f(t) = \limv{t}f(t) = 0$,

so $\Lambda^{-1}f \in \sps$ with $\normn{\Lambda^{-1}f}_{\sps}= \norm{f}_{u, (0, \infty)}$.

As $C_{0}((0, \infty); \real^{d})$ is a separable Banach space, so is $\sps$.$\square$

Proof. Let $\Lambda: \sps \to C_{0}((0, \infty); \real^{d})$ be the isomorphism defined above, then $\phi \circ \Lambda^{-1}\in C_{0}((0, \infty); \real^{d})^{*}$. By the Riesz representation theorem, there exists a vector measure $\lambda_{0} \in M((0, \infty); \real^{d})$ such that

and $\norm{\lambda_0}_{\text{var}}= \normn{\phi \circ \lambda^{-1}}_{\sps}$. Thus for any $\theta \in \sps$,

and

is the desired measure.$\square$

Proof. Firstly, $\lim_{t \to \infty}\theta(t)/t = 0$ if and only if for every $n \in \nat$, there exists $k \in \nat$ such that $\abs{\theta(s)}/s \le 1/n$ for all $s \ge k$. Thus

and

are Borel. Thus $\sps = \Theta_{0} \cap \Theta_{\infty} \cap \Theta_{b}$ is Borel in $C([0, \infty); \real^{d})$.

Now, by lower semicontinuity of $\norm{\cdot}_{\sps}$, every closed set in $\sps$ is the intersection of $\sps$ and a Borel set in $C([0, \infty), \real^{d})$. Therefore $\cb(\sps)$ contains the induced $\sigma$-algebra from $\cb(C([0, \infty); \real^{d}))$. On the other hand, the inclusion $\iota: \sps \to C([0, \infty); \real^{d})$ is continuous. Thus $\cb(\sps)$ contains the induced $\sigma$-algebra.$\square$

Proof. $(1)$: By the Brownian Strong Law of Large Numbers, $B(\omega) \in \sps$ for almost every $\omega$.

$(2)$: By Doob’s maximal inequality, for any $r \ge 0$,

and for any $\theta \in \sps$,

Thus

$\square$

Proof. $(1)$: Dominated Convergence Theorem.

$(2)$: Let $\bracsn{x_j^*}_{1}^{n} \subset E^{*}$ and $\mu_{(x_1^*, \cdots, x_n^*)}$ be the distribution of $(x_{1}^{*}, \cdots, x_{n}^{*}): E \to \real^{n}$ under $\mu$, then the characteristic function $\mu_{(x_1^*, \cdots, x_n^*)}$ is

Thus $\wh \mu$ uniquely determines the joint characteristic function of $(x_{1}^{*}, \cdots, x_{n}^{*})$. So for any $\bracsn{x_j^*}_{1}^{n} \subset E^{*}$ and Borel sets $\seqf{B_j}\subset \cb(\real)$,

As these cylinder sets form a $\pi$ system that generates $\cb(E)$, $\mu = \nu$ by Dynkin’s Uniqueness Theorem.$\square$

Proof. $(1)$, Finite: First suppose that there exists $0 = t_{0} \le t_{1} < \cdots < t_{n} < \infty$ and $\seqf{\xi_j}\subset \real^{d}$ such that

for all $\theta \in \sps$. This allows decomposing

as a sum of independent centred Gaussian random variables. Thus $\theta \mapsto \angles{\theta, \lambda}_{\real^d}$ is a centred Gaussian random variable.

$(1)$, with Riemann-Stieltjes: Let $\mu: \cb((0, \infty)) \to \real$ be a finite signed measure, and $F(t) = \mu((0, t])$ be the distribution function of $\mu$, then $\norm{F}_{\text{var}}= \norm{\mu}_{\text{var}}< \infty$.

For any indicator function of the form $\one_{(a, b]}$ with $0 \le a < b < \infty$,

where the integral on the left is a Riemann-Stieltjes integral. As simple functions that vanish at $0$ are dense in $C_{0}((0, \infty); \real)$, and

for any simple function $\phi: (0, \infty) \to \real$,

for all $\phi \in C_{0}((0, \infty); \real)$.

As seen in the proof of Lemma 1.1.2, there exists a $\real^{d}$-valued finite Borel measure $\mu: \cb((0, \infty)) \to \real^{d}$ such that for any $\theta \in \sps$,

Let $F(t) = \mu((0, t])$, then since $\Lambda \theta \in C_{0}((0, \infty); \real)$,

can be expressed as a Riemann-Stieltjes integral. Thus

As the above holds for all $\theta \in \sps$, $\lambda$ viewed as a random variable is a^[2] sequential limit of Gaussian random variables, so it is also Gaussian.

$(1)$, with Approximations: As seen in Lemma 1.1.2, there exists a $\real^{d}$-valued Borel measure $\mu: \cb((0, \infty)) \to \real^{d}$ such that

Since there exists $\real^{d}$-valued Borel measures $\seq{\mu_k}$ with finite support such that $\mu_{k} \Rightarrow \mu$, each $\theta \mapsto \int \frac{\theta(t)}{1 + t}d\mu_{k}(t)$ is Gaussian random variable and $\theta \mapsto \angles{\theta, \lambda}_{\sps}$ is a pointwise limit of Gaussian random variables, thus it is also Gaussian.

$(2)$: Firstly, for any $0 \le s, t < \infty$ and $\xi, \eta \in \real^{d}$,

Let $F: [0, \infty) \to \real^{d}$ be measurable such that $d\lambda = Fd\abs{\lambda}$, then

By Fubini’s theorem, the expectation can be pushed inside the integral:

$(3)$: $\lambda$ is a Gaussian random variable with variance $\iint_{[0, \infty)^2}s \wedge t d\lambda(s)d\lambda(t)$.$\square$

Proof. $(1)$: Let $\theta \in \ccm$, then for each $t \ge 0$,

Thus

$(2)$: Follows from the density of $C_{c}^{\infty}(\real^{d})$ in $C_{0}((0, \infty); \real^{d})$ and thus in $\sps$.$\square$

Proof. Firstly, $\lambda$ is bounded, so $h_{\lambda}$ is absolutely continuous and $Dh_{\lambda}(s) = \lambda((s, \infty))$ by the Lebesgue differentiation theorem. From here,

so $h_{\lambda} \in \ccm$ and

Now let $F(t) = \lambda((0, t])$, then $Dh_{\lambda}(t) = \lambda((0, \infty)) - F(t)$ and

can be expressed as improper Riemann-Stieltjes integrals. Using the by parts formula for Riemann-Stieltjes integrals,

where since $\theta(T)/T \to 0$ as $T \to \infty$ and $(1 + t) \in L^{1}(\abs{\lambda})$,

Therefore

$\square$