# Obtaining Classical Wiener Space from abstract Wiener measure

## 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.

## 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) $$\forall\, \xi \in \mathcal{H}, A \in \mathcal{B}(\mathbb{R}): \quad \mu_\xi(A) := \mu(\{\phi \in E^a \mid \phi(\xi) \in A \}) = \frac{1}{\sqrt{2\pi (\xi,\xi)}} \int_A e^{-\frac{x^2}{2(\xi,\xi)}} \,\mathrm{d}x. \tag{c}$$

## 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!

## 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.

• I am not an expert on abstract Wiener space. Just found the following paper which seems a friendly introduction hopefully bridging the gap you are struggling with. Sep 2, 2022 at 18:55

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).

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.