3
$\begingroup$

The space $C^{\infty}(\mathbb{R}^n)$ can be turned into a topological ring using the Whitney topology. Whitney's Spectral Theorem says that the closure of an ideal in this ring is the ideal of all functions whose Taylor expansion up to order $r \in \mathbb{N}$ at any point equals the Taylor expansion up to order $r$ of some function in the ideal. I'm interested in a characterization of the closed ideals of this ring.

By Neil Strickland's answer here--Ideals of the ring of smooth functions, using Whitney's Spectral Theorem, one may describe the closed ideals in $C^{\infty}(\mathbb{R})$ as having the form $$\{f : f^{(i)}=0 \, \, \text{on} \, \, X \, \, \text{for all} \, \, i \leq n\}$$ for some closed set $X \subset \mathbb{R}$. This is probably very simple, but could someone quickly explain in more detail how Whitney's Spectral Theorem implies this result?

More generally, I'm interested in whether or not the question Martin Brandenburg hinted at in the comment section of that post is true--do the closed ideals arise as intersections of sets of the following form

$$\{f : f(p)=0 \,\, \text{and}\,\, \forall \,\, i\leq \alpha_i,\, j \leq n, \,\, (\partial_1^{\alpha_1}\dots \partial_j^i)f(p)=0 \}$$

for some $p \in \mathbb{R}^n$ and some multi-index $\alpha = (\alpha_1, \dots, \alpha_n)$?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ The standard references to this subject are Malgrange's book Ideals of differentiable functions and the one of Tougeron Ideaux de fonctions differentiables. In higher dimensions things become quite complicated: The function $f(x,y)=y^2 - \exp(-1/x^2)$ generates a closed ideal in $\mathscr E(\mathbb R^2)=C^\infty(\mathbb R^2)$ (i.e., $\lbrace fg: g\in\mathscr E(\mathbb R^2)\rbrace$ is closed) whereas $g(x,y)=y^2 + \exp(-1/x^2)$ does not (this is example 4.8 in Tougeron's book). $\endgroup$
    – Jochen Wengenroth
    Commented Feb 18, 2015 at 9:39

0

Reset to default

You must log in to answer this question.