Here is what I think I know: Given a symmetric linear differential operator $\mathcal{L}$ that is positive definite on a function space(/space of distributions) $\mathcal{H}$, we can find its inverse, which we know will be an integral transform due to the Schwartz kernel theorem. The kernel of the transform will be the Green's function $\mathcal{G}$ corresponding to $\mathcal{L}$ on $\mathcal{H}$. Do you agree this far?

So what about the other way around? If I have a symmetric positive definite kernel $\mathcal{G}$ on $\mathcal{H}$, will there always exist a linear differential operator $\mathcal{L}$ for which $\mathcal{L}\mathcal{G}(\cdot, t)=\delta_{t}$?

Here is what I think I know: Given a symmetric linear differential operator $\mathcal{L}$ that is positive definite on a function space(/space of distributions) $\mathcal{H}$, we can find its inverse, which we know will be an integral transform due to the Schwartz kernel theorem. The kernel of the transform will be the Green's function $\mathcal{G}$ corresponding to $\mathcal{L}$ on $\mathcal{H}$.

So what about the other way around? If I have a symmetric positive definite kernel $\mathcal{G}$ on $\mathcal{H}$, will there always exist a linear differential operator $\mathcal{L}$ for which $\mathcal{L}\mathcal{G}(\cdot, t)=\delta_{t}$?