# 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. – Mark Meckes Jul 23 '12 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. – George Lowther Jul 23 '12 at 19:03
• (Trying to save face) Okay, so it's because $W$ is separable... – Mark Meckes Jul 23 '12 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 – Nate Eldredge Apr 21 '14 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.