Consider a stationary Continuous-time AutoRegressive (CAR) process on
a bounded time-interval $(a, \, b)$. This article by Emmanuel
Parzen describes
the corresponding Reproducing Kernel Hilbert Space (RKHS) $\mathcal{K}$ and its inner product for
the first and second-order CARs. For instance consider the first
order
$$
\frac{\text{d}}{\text{d}t} X(t) + \beta X(t) = \varepsilon(t)
$$
where $\varepsilon(t)$ is a Gaussian white noise with unit variance
and $\beta >0$ - a more rigorous formulation would be $\text{d} X(t) +
\beta X(t)\text{d}t = \text{d}W(t)$ with $W(t)$ being a standard
Wiener process. Then $\mathcal{K}$ contains all differentiable
functions on $(a, \,b)$ and the inner product is given by formula
(4.16) p. 477
$$
\begin{aligned}
\langle h, \, g \rangle_{\mathcal{K}} &=
\frac{1}{2\beta} \,\int_{a}^b
\left[h' + \beta h\right]\left[g' + \beta g\right]
\text{d}t + h(a) g(a) \\
& = \frac{1}{2\beta} \,\int_{a}^b
\left[h'g' + \beta^2 hg \right] \text{d}t +
\frac{1}{2}\left\{ h(a) g(a) + h(b) g(b) \right\}
\end{aligned}.
$$
The second form is more attractive because of its symmetry w.r.t. to
the two end-points $a$ and $b$ which relates to the time-reversibilty
of the process; it is easily derived from the non-symmetric form using
integration by parts. Parzen gives the RKHS for the second-order CAR
but only with a non-symmetric form for the inner product. The same is true for
a $\text{CAR}(p)$ with order $p$ defined as the stationary solution
(assumed to exist) of
$$
D^{p} X(t) + a_{1} D^{p-1} X(t) + \dots + a_p X(t) = \varepsilon(t)
$$
where $D := \text{d}/ d\text{t}$ and again with a more rigorous form
involving $\text{d}W(t)$. See formulas (4.7), (4.8) and (4.9) in the
article.
For the $\text{CAR}(p)$ case, what is the expression of the RKHS inner product which is symmetric in the two end-points $a$ and $b$? I guess that this could involve the formal adjoint of the linear differential operator but did not find the answer with Google searches.
