3
$\begingroup$

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 $\mathcal{L}\mathcal{G}(\cdot, t)=\delta_{t}$?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ I think the simple answer is No, simply because it's not so hard to find examples of integral kernels whose inverses are not differential operators. Also, see this earlier question. $\endgroup$
    – Igor Khavkine
    Commented Dec 6, 2014 at 13:07
  • 3
    $\begingroup$ For example, the Gaussian $G(x,y) = \exp(-|x-y|^2/\sigma^2)$ corresponds to a differential operator of infinite order. $\endgroup$
    – Peter Michor
    Commented Dec 6, 2014 at 14:28
  • $\begingroup$ Dear @PeterMichor. Thank you for your answer. Excuse my lack of knowledge, but why is the infinite series expansion of differential operators that have a Gaussian Green's function not a linear differential operator? $\endgroup$
    – Lars Lau Raket
    Commented Dec 7, 2014 at 9:47
  • $\begingroup$ Differential operators are usually constrained to be of finite (or at least locally finite order). Otherwise, the line gets blurred, e.g., you could consider the translation $f(x) \mapsto f(x+y)$ as a differential operator of infinite order as well. $\endgroup$
    – Igor Khavkine
    Commented Dec 8, 2014 at 16:28

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .