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 $H_B$.

  • $\begingroup$ What exactly do you mean by RKHS here? Please don't just link the Wiki page... $\endgroup$
    – user479223
    Commented Jan 5 at 13:04
  • $\begingroup$ Hi, I added some background information. I hope this helps. $\endgroup$
    – Mueller
    Commented Jan 5 at 16:08
  • 3
    $\begingroup$ Perhaps it is worth mentioning that fractional Sobolev spaces $W^{s,2}$ do not work: $B_t\in W^{s,2}$ almost surely if and only if $s<\frac{1}{2}$, while the point evaluation is continuous if and only if $s>\frac{1}{2}$ $\endgroup$
    – Kostya_I
    Commented Jan 5 at 18:06
  • $\begingroup$ I think Theorem 4.12 in arxiv.org/abs/1807.02582 does construct such a RKHS that contains paths of the Brownian motion. (I did not check the conditions in detail.) $\endgroup$ Commented Jan 5 at 18:33
  • $\begingroup$ @MarkusLange-Hegermann, unfortunately, it looks like in the BM case, in order to apply their theorem, we would need to find a $\theta$ that satisfies $\theta>\frac12$ and $\theta<\frac12$ simultaneously (unsurprisingly, since their RKHS is just $W^{\theta,2}$ in this case...). $\endgroup$
    – Kostya_I
    Commented Jan 5 at 18:58

1 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 $x=\frac12.$

  • $\begingroup$ Thank you very much for your effort and response. Unfortunately, this only partially answers my question as not every RKHS needs to be of this specific form. But in any case, thanks a lot. $\endgroup$
    – Mueller
    Commented Jan 8 at 17:18

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