# Brownian motion, reproducing kernel Hilbert space, and the Laplace operator

Consider the standard Brownian motion on $[0,1]$:

$$dB_t, \; B_0 = 0,$$

defined on the probability space $(\Omega, P)$. It covariance function is $K(s,t) = \min \{s , t\}$ on $[0,1] \times [0,1]$. The RKHS with reproducing kernel $K$ is the Sobolev space

$$\mathcal{H}_K = \{f \; {\tt absolutely \; continuous}, \; f(0) = 0, f'\in L^2[0,1] \}$$

with inner product

$$\langle f, g \rangle_{\mathcal{H}_K} = \int f'g'.$$

This can be seen by noting that $t \mapsto \min \{ s,t\}$ has weak derivative $1_{[0, s]}$.

Questions

1. $\mathcal{H}_K$ is isomorphic to the Hilbert space generated by $\{ B_t\}_{t \in [0,1]}$, with isomorphism $K_t(s) = K(t,s) \mapsto B_t \in L^2(\Omega, P)$. So it looks like one can define a stochastic integral against $dB_t$ with deterministic integrands. Is there a name for this integral? It looks a bit strange when compared to the Ito integral. For example, the increment $B_{s_2} - B_{s_1}$ is identified with

$$\min(s_2, t) - \min(s_1, t).$$

1. I've encountered the claim that "(the differential operator) $-\frac{d^2}{dx^2}$ is the reproducing kernel of $B_t$". How is $-\frac{d^2}{dx^2}$ related to $K$? Subject to boundary conditions, integrating by parts can recover the inner product on $\mathcal{H}_K$ but I don't see an identification with the Sobolev space $\mathcal{H}_K$:

$$- \int f''g = \int f'g'.$$

1. Related to 2.: the infinitesmal generator of $B_t$ as a Markov process happens to be the Laplacian $\frac{d^2}{dx^2}$. Is this a related to the above?

2. The Cameron-Martin space of $B_t$ also just happens to be $\mathcal{H}_K$.

Same objects, all related to the Brownian motion $B_t$, keep coming up via (apparently) different constructions...what's happening here? Is there a way they fit together?

• Can you explain your statement in #1 a bit further? Specifically: (a) What exactly do you mean by "the Hilbert space generated by $\{B_t\}_{t \in [0,1]}$"? What are the elements of this space (paths, random variables, ???) and what is the inner product? What is the isomorphism between this space and $\mathcal{H}_K$? And how exactly are you defining the integral you mention? There is definitely something to be said here, but at this point I don't understand where your idea is heading. – Nate Eldredge Nov 28 '14 at 8:50
• Edited a bit on Hilbert space, which just consists of random variables. By an "integral", I just mean this unitary operator from $\mathcal{H}_K$ to the said Hilbert space. Like the Ito isometry. E.g. we can map $K_s(s') - K_t(s')$ to $B_s -B_t$. Under the Ito isometry, it's $1_{[t,s]}$ that's mapped to $B_s -B_t$. So it does looks a bit strange compared to the Ito integral. I am wondering whether this "integral" has been considered. – Michael Nov 28 '14 at 9:09
• It's a general result that the Cameron-Martin space of a Gaussian r.v. in $\mathcal C(X)$ is the same as the RKHS associated to the covariance kernel induced over $X$. One reference for this is Chapter 2 of Mathematical Statistics by E. Giné and R. Nickl. – Suhas Vijaykumar Jan 17 '18 at 16:49

## 1 Answer

For #1: Let's call your isomorphism $T$, the isometry from $\mathcal{H}_K$ to $L^2(P)$ which maps $K_t$ to $B_t$. Yes, $T$ really is a stochastic integral: $Tf$ is the Itô integral, not of $f$, but of $f'$. So $$Tf = \int_0^1 f'(t)\,dB_t$$ or in other words, for deterministic $g \in L^2([0,1])$, $$\int_0^1 g(t)\,dB_t = T\left(\int_0^\cdot g(s)\,ds\right).$$ So this recovers the Itô integral for deterministic $L^2$ integrands. This special case of the Itô integral is sometimes called the Wiener integral.

Another way of thinking about this is that the map $f \mapsto f'$ is an isometric isomorphism from $\mathcal{H}_K$ to $L^2([0,1])$. So if you identify $\mathcal{H}_K$ with $L^2([0,1])$ under this map, then $T$ really is the stochastic integral.

I will think about your other questions some more.

• Sorry, somehow I missed this and just saw. – Michael Dec 6 '14 at 10:37