I'll begin this question with the finite-dimensional case, as a
warmup.

Let me say a continuous path $\omega : [0,1] \to \mathbb{R}^d$ is
**hyperplanar** if there exists a nonzero $x \in \mathbb{R}^d$ such
that $\omega(t) \cdot x = 0$ for all $t \in [0,1]$.

It is not hard to show:

> 
> **Proposition.** Let $B\_t$ be a standard Brownian motion in $\mathbb{R}^d$. Almost surely, $(B\_t : 0 \le t \le 1)$ is not hyperplanar.
> 
> 
> 

To avoid possible confusion, let me emphasize that the hyperplane is
allowed to be random. That is, I am claiming:
$$\mathbb{P}(\exists x \in \mathbb{R}^d\\, \forall t \in [0,1] : B\_t \cdot
x = 0) = 0$$
which should not be confused with the weaker statement
$$\forall x \in \mathbb{R}^d :\mathbb{P}(\forall t \in [0,1] : B\_t \cdot
x = 0) = 0.$$

One possible proof of the proposition is to choose $0 < t\_1 < \dots <
t\_d < 1$, and show by induction that, almost surely, $B\_{t\_1}, \dots,
B\_{t\_d}$ are linearly independent in $\mathbb{R}^d$. (By the
induction hypothesis, $B\_{t\_1}, \dots, B\_{t\_{k-1}}$ span a
$k-1$-dimensional subspace of $\mathbb{R}^d$; by the Markov property
and the absolute continuity of the Gaussian distribution, $B\_{t\_k}$ is
almost surely not in this subspace.)

---

I am interested in the analogous statement for infinite dimensions.
Let $W$ be a real separable Banach space, and say a continuous path $\omega :
[0,1] \to W$ is **hyperplanar** if there exists a nonzero continuous
linear functional $f \in W^\*$ such that $f(\omega(t)) = 0$ for all $t
\in [0,1]$. Let $\mu$ be a non-degenerate Gaussian measure on $W$
(so that $(W,\mu)$ is an abstract Wiener space), and let $B\_t$ be a standard Brownian motion in $W$. That is, the process
$B\_t$ has continuous sample paths and independent increments, starts
at 0, and the increments are distributed such that $(t-s)^{-1/2}(B\_t -
B\_s) \sim \mu$.

> 
> For infinite-dimensional $W$, what is the probability that $(B\_t : 0 \le t \le 1)$ is hyperplanar?
> 
> 
> 

As a start, I can show that the set of hyperplanar paths is analytic
in $C([0,1], W)$, and hence universally measurable, so the question
actually makes sense. We can also use a scaling argument and the
Blumenthal zero-one law to see that the probability must be either 0
or 1.

I am not sure which way my intuition points here. On the one hand, we
expect a Brownian motion to be pretty irregular and unconstrained,
suggesting the answer is 0 as in finite dimensions. On the other
hand, an infinite-dimensional space has a lot of hyperplanes. In principle, the answer could depend on the abstract Wiener space $(W,\mu)$.

Thanks!