Are the paths of the Brownian motion contained in a suitable RKHS?

Question

Let $H_B$ be the reproducing kernel Hilbert space (RKHS) of the Brownian Motion $(B_t)$ on $[0,1]$. It is well known that with probability 1 the paths of $(B_t)$ are not contained in $H_B$. But is there another (potentially larger) RKHS $H$ that does contain the paths of $(B_t)$ with probability 1?

Background information. An RKHS $H$ on $[0,1]$ is a Hilbert space of functions $f:[0,1]\to \mathbb R$ such that the point evaluations \begin{align*} \delta_t :H &\to \mathbb R \\ f &\mapsto f(t) \end{align*} are continuous for all $t\in [0,1]$. In this case, $k:[0,1]^2\to \mathbb R$ defined by \begin{align*} k(s,t) := \langle \delta_s,\delta_t\rangle_{H'} \end{align*} is the reproducing kernel of $H$, where $H'$ denotes the dual of $H$. Reproducing kernels are exactly those functions that are symmetric and positive semi-definite. It is well-known that there is a one-to-one relationship between reproducing kernels and RKHSs.

For a centered stochastic process $(X_t)$, the covariance function $(s,t) \mapsto \mathbb E X_s X_t$ is symmetric and positive semi-definite and thus a reproducing kernel.

In the case of the Brownian motion, we have $\mathbb E B_s B_t = \min\{s,t\}$. In my question, the corresponding, uniquely defined RKHS of this reproducing kernel is denoted by $H_B$.

Answer 1

(I will work with the Brownian bridge instead). This is to explain why there is no such RKHS of the form

$$ W=\left\{f(x)=\sum_{k=1}^\infty \hat{f}_k\sin(\pi k x):\sum_{k=1}^\infty w_k\hat{f}^2_k<\infty\right\}. $$

where $w_k$ is any sequence of positive numbers. This includes in particular fractional Sobolev spaces $W^{s,2}.$

For a Brownian bridge $B_t$, we have $\hat{f}_k=\frac{1}{k}Z_k,$ where $Z_k$ are i.i.d. Gaussians, and the condition $B_t\in W$ almost surely is equivalent to $\sum_{k=1}^\infty \frac{w_k}{k^2}Z^2_k<\infty$ almost surely, which happens if and only if $\sum_{k=1}^\infty \frac{w_k}{k^2}<\infty$, by Kolmogorov three series theorem.

On the other hand, the $W$-norm of the evaluation $f\mapsto f(x)=\sum_{k=1}^\infty \hat{f}_k\sin(\pi k x)$ is given by $\sum_{k=1}^\infty w^{-1}_k\sin^2(\pi k x)$, and for $W$ to be RKHS, we need that to be finite for all $x$.

However, by Cauchy-Schwarz, we have $$\sum_{k=1}^\infty\frac{1}{k}|\sin(\pi k x)|\leq\left(\sum_{k=1}^\infty \frac{w_k}{k^2}\right)^\frac12\left(\sum_{k=1}^\infty w^{-1}_k\sin^2(\pi k x)\right)^\frac12,$$ and the left-hand side diverges, say, for $x=\frac12.$