8
$\begingroup$

What theories are there for generalized functions (distributions) in infinite dimensions?

In particular, suppose your "infinite dimensional manifold" is $\mathfrak{M}:=C^\infty(S^1)$. Is there a theory of generalized functions based on the test functions which would roughly be smooth functions on $\mathfrak{M}$? Perhaps using the convenient calculus of Kriegl/Michor.

$\endgroup$

1 Answer 1

6
$\begingroup$

The study of an "infinite-dimensional delta function" has been motivated to a large extent by applications in quantum field theory. Here is some relevant literature:

$\endgroup$
4
  • 2
    $\begingroup$ Nagaev's notes are about probability distributions, mainly on Banach spaces, whereas the question is about generalized functions on "infinite dimensional manifolds" (rather than the usual generalized functions on $\mathbb R^n$). Of course, any probability distribution on $\mathbb R^n$ is a generalized function, but not vice versa, and the subject of the theory of generalized functions is very different from that of the theory of probability distributions. $\endgroup$ Jan 17, 2023 at 20:16
  • 1
    $\begingroup$ my mistake, thank you for correcting me; I deleted Nagaev and added further references. $\endgroup$ Jan 17, 2023 at 20:38
  • $\begingroup$ I have two points of confusion, here. 1. Is it possible to see the usual distributions on $\mathbb{R}^n$ as a special case of this theory? When I try, the test functions seem to be analytic instead of smooth, but perhaps I'm missing something. $\endgroup$
    – SnowRabbit
    Jan 18, 2023 at 20:07
  • $\begingroup$ 2. The distributions seem to be functions of duals of nuclear spaces, so in my example above, this would be functions of $\mathscr{D}'(S^1)$, instead of $\mathscr{D}(S^1)$. So if one has a partial differential operator acting on smooth functions of $\mathscr{D}(S^1)$, it doesn't seem to necessarily extend to the "distributions". Is that correct? $\endgroup$
    – SnowRabbit
    Jan 18, 2023 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.