# Does infinite-dimensional Brownian motion live in hyperplanes?

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!

• Wild guess: the answer depends (only) on whether $W^*$ is separable. Jul 23, 2012 at 19:01
• It can't be hyperplanar, because any sequence of IID random variables will be dense in the support of the distribution. This just uses second countability so that the notion of support makes sense. Jul 23, 2012 at 19:03
• (Trying to save face) Okay, so it's because $W$ is separable... Jul 23, 2012 at 19:11

As suggested in my comment, here's a simple fact which applies to any probability measure $\mu$ on (the Borel σ-algebra of) a second countable topological space $X$. There is a unique minimal closed subset $S$ of $X$ with $\mu(S)=1$ -- the support of $\mu$ -- and, if $X_1,X_2,\ldots$ is an IID sequence of random variables, each with distribution $\mu$, then $\lbrace X_1,X_2,\ldots\rbrace$ is almost surely a dense subset of $S$.

Now, in the situation described in the question, choose a sequence $0\le t_0 < t_1 < t_2 < \cdots \le1$. Then, $X_k\equiv(t_k-t_{k-1})^{-1/2}(B_{t_k}-B_{t_{k-1}})$ is an IID sequence of random variables each with distribution $\mu$. Also, as it is non-degenerate, the support of $\mu$ is not contained in a closed proper subspace of the Banach space $W$.

• Thanks again, George. I just wanted to let you know we've acknowledged you in our paper, and cited this answer. arxiv.org/abs/1310.8010 Apr 21, 2014 at 16:28

Shortly after posting this, I discussed it with Clinton Conley and we came up with what is essentially the same as George Lowther's argument.

The point is that, by the Hahn-Banach theorem, $\omega$ is hyperplanar iff the linear span of $\{\omega(t) : t \in [0,1]\}$ is not dense in $W$. But with probability 1, this span is dense.

$W$ is a separable metric space, so it has a countable basis $\{U_i\}_{i=1}^\infty$. Choose an infinite sequence $0 < t_1 < t_2 < \dots < 1$ and set $\Delta_n = (t_{n+1}-t_n)^{-1/2}(B_{t_{n+1}}-B_{t_n})$ so that the $\Delta_i$ are iid with distribution $\mu$. For each $i$, $\mu(U_i) > 0$ by non-degeneracy, so almost surely, one of the $\Delta_n$ lies in $U_i$. In particular, the linear span of $\{B_t\}$ meets $U_i$. Taking a countable intersection, almost surely, the linear span of $\{B_t\}$ meets every $U_i$ and hence is dense.

Edit: This characterization also makes it clear that the set of hyperplanar paths is much better than just analytic. Indeed, for fixed $i$, the set of paths $\omega$ such that the linear span of $\{\omega(t) : 0 \le t \le 1\}$ meets $U_i$ is easily seen to be open. The set of non-hyperplanar paths is thus $G_\delta$. It is also dense (since a measure with full support gives it measure 1, or by the simpler argument that any hyperplanar path can be slightly perturbed to make it non-hyperplanar), and in particular comeager. So this also answers the Baire category analogue of my question.