Obtaining Classical Wiener Space from abstract Wiener measure

Question

The question

I'm working on understanding the Abstract Wiener Space construction and wanted to rederive the defining property of the classical counterpart, $$\require{cancel} \xcancel{\xi_{t+s} - \xi_t}\ B_{t+s} - B_t\sim \mathcal{N}(0, s) \quad(\text{for } s> 0), \tag{1}$$ from the Abstract Wiener Space as constructed on the Wikipedia page and here.

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The setup

That is, I assume to know that $\xi \in W^{2,1}_0[0,T] =: \mathcal{H}$, i.e., $\xi: [0,T] \to \mathbb{R}$ is an absolutely continuous path with square-integrable first derivative and $\xi(0) = 0$. The inner product on $\mathcal{H}$ shall be defined as $$(\xi, \zeta) = \int_0^T \dot{\xi}(t) \dot{\zeta}(t) \,\mathrm{d}t, \tag{a}$$ and from the construction we know there exists a measure $\mu$ which acts on the algebraic dual space $E^a$ and which has the properties (Defs. 20 and 25 of Velhinho) $$\forall\, \xi \in \mathcal{H}: \quad \chi(\xi) := \int_{E^a} e^{i\phi(\xi)}\, \mathrm{d}\mu(\phi) = e^{-\frac{1}{2}(\xi,\xi)} \tag{b}$$ and (Theorem 11, which I believe is a special case of the cylinder set measure property, correct me if I'm wrong) $$\forall\, \xi \in \mathcal{H}, A \in \mathcal{B}(\mathbb{R}): \quad \mu_\xi(A) := \mu(\{\phi \in E^a \mid \phi(\xi) \in A \}) = \frac{1}{\sqrt{2\pi (\xi,\xi)}} \int_A e^{-\frac{x^2}{2(\xi,\xi)}} \,\mathrm{d}x. \tag{c}$$

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My attempt at a solution (see edit below!)

I have the following idea, but I'm not sure if it's right:

What we want to show in (1) is equivalent to $$P(\{\xi(t+s) - \xi(t) \in A\}) = \frac{1}{\sqrt{2\pi s}}\int_A e^{-\frac{x^2}{2s}} \,\mathrm{d}x$$ for any Borel set $A \in \mathcal{B}(\mathbb{R})$. We may note that $$\xi(t+s) - \xi(t) = \int_t^{t+s} \dot{\xi}(\tau)\, \mathrm{d}\tau = \int_0^{T} \dot{f}(t) \dot{\xi}(\tau)\, \mathrm{d}\tau \quad \text{with } \dot{f}(\tau) = 1_{[t,t+s]}(\tau).$$ Now, since every $\xi \in \mathcal{H}$ will also have a dual element $\phi_\xi \in E^a$ and $f$ as implicitly defined above is a valid element of $\mathcal{H}$, we can swap the roles of $\xi$ and $f$ to define $$\xi(t+s) - \xi(t) =: \phi_\xi(f).$$ Then we can apply (c) to find that $$\mu_f(A) = \frac{1}{\sqrt{2\pi (f,f)}}\int_A e^{-\frac{a^2}{2(f,f)}}\,\mathrm{d}x = \frac{1}{\sqrt{2\pi s}}\int_A e^{-\frac{a^2}{2s}}\,\mathrm{d}x,$$ where we used $(f,f) = s$ via definition (a).

This looks like the right result, but the way there seems a bit odd, and it also glosses over the fact that the set that is measured in (c) includes $\phi$ that are not duals $\phi_\xi$ of some $\xi$. Is it obvious that these will contribute with measure zero?

Feel free to point me to a good introductory text that gives more context, if necessary!

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Edit: Revised attempt

I found part of the problem: In (1), $\xi$ was the Brownian motion itself, whereas in (a) through (c), $\xi$ was an element of the Cameron–Martin Hilbert space. However, the Brownian motion sample paths are not elements of the Hilbert space, but rather of the Banach space $E^a$. Thus, I should replace (1) with $$B_{t+s} - B_t \sim \mathcal{N}(0, s)$$ to make the distinction between $\xi$ and $B$ obvious.

Then the rest of the argument continues as before, i.e. $$ B(t+s) - B(t) = \int_0^T \dot{B}(\tau) \dot{\xi}(\tau)\, \mathrm{d}\tau \quad \text{with } \dot{\xi}(\tau) = 1_{[t,t+s]}(\tau) \implies \mu_\xi(A) = \frac{1}{\sqrt{2\pi s}} \int_A e^{-x^2/(2s)} \,\mathrm{d} x, $$ but we now face a different problem, namely that $B(\tau)$ is a.s. not differentiable and that hence the integral that I just wrote does not exist for a.e. $B$. I've considered interpreting $\dot{B}$ in the distributional sense, but $\xi$ is not a test function, so that doesn't seem to work, I don't know if there's some closure/completion/limiting property that can be used instead.

Answer 1

I still welcome other answers from more knowledgeable people, but I believe I have found the solution to this problem:

In the standard formulation of abstract Wiener spaces, the measure $\mu$ is not on the algebraic dual space explicitly, but just on some Banach space $W$, and the abstract Wiener space itself is said to be the triple $i: \mathcal{H} \to W$. In the classical case, $W$ is the set of continuous functions $B(t) \in C[0,T]$ with $B(0) = 0$ and $i$ is the inclusion map from $W^{2,1}_0[0,T]$ to $W$. Let $j: W^\ast \to \mathcal{H}$ be the adjoint of $i$, i.e., $(j(\ell), h) = \ell(i(h))\ \forall \ell \in W^\ast, h \in \mathcal{H}$.

On this space, the evaluation functional $\delta_s : B \mapsto B(s)$ is a continuous linear functional and thus an element of the continuous dual space $W^\ast$. Then $\delta_{t+s} - \delta_t$ is also an element of $W^\ast$, and we may note that $j(\delta_{t+s} - \delta_t) = f$ as given in the question (i.e., the "stopped" ramp function fulfilling $\dot{f}(\tau) = 1_{[t,t+s]}(\tau)$ with $f(0) = 0$).

Now, finally, by definition/construction, the Wiener measure $\mu$ is the radonification of the canonical Gaussian cylinder set measure on $\mathcal{H}$. In particular, this means that any $\ell \in W^\ast$ is distributed as $\mathcal{N}(0, {\|\left(\ell \circ i\right)^\sharp\|_\mathcal{H}}^2) = \mathcal{N}(0, {\|j(\ell)\|_\mathcal{H}}^2)$, where $\left(\ell \circ i\right)^\sharp \in \mathcal{H}$ is the Riesz representative of $\ell \circ i \in \mathcal{H}^\ast$. Thus, since $\delta_{t+s} - \delta_t =: \ell \in W^\ast$ corresponds to the random variable I was looking for and $j(\ell) = f$ as given in the question, we find that $${\|j(\ell)\|_\mathcal{H}}^2 = \int_0^T \left(1_{[t,t+s]}(\tau) \right)^2 \,\mathrm{d}\tau = s$$ and therefore $B_{t+s} - B_{t} \sim \mathcal{N}(0,s)$ (for $s > 0$).

I wouldn't have figured this out without the help of Nathaniel Eldredge's and David Elworthy's lecture notes (in particular, sections 3 through 6 of the latter).

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However, I'm still not entirely sure how this can be translated to the algebraic-dual-space formulation of Velhinho without appealing to the same argument using $i$ and $j$ as I did here. I have the following idea, which doesn't seem quite satisfactory to me:

For any $s \in [0, T]$, the function $g_s: [0,T] \to \mathbb{R}$ defined by $$g_s(\tau) = \begin{cases} \tau, & \tau < s\\ s, & \tau \geq s\end{cases}$$ is an element of $\mathcal{H}$. We can somewhat arbitrarily define this to be the evaluation functional on $E^a$—motivated by $\int_0^T \dot{B}(\tau) \dot{g}_s(\tau)\ \mathrm{d}\tau$ when it exists—in the sense that $\delta_s(B) = B(g_s) := B_s$ for $B \in E^a$. Then $g_{t+s} - g_{t}$ is the $f$ from the question with ${\|g_{t+s} - g_t\|_\mathcal{H}}^2 = s$ and hence $$B(g_{t+s} - g_s) = B_{t+s} - B_s \sim \mathcal{N}(0, s).$$ I still wonder if there is some way to get around the apparent arbitrariness of this way of defining the evaluation functional.