8
$\begingroup$

(This question was previously posted on MSE and I decided to post it here too.)

I am studying the proof of the Malgrange preparation theorem given in the book "Stable mappings and their singularities" written by Golubitsky and Guillemin (see Chapter IV). The statement is the following.

Let $F\in C^\infty(\mathbb{R}\times\mathbb{R}^n; \mathbb{R})$ be defined on a neighborhood of $0\in\mathbb{R}\times\mathbb{R}^n$ and such that $F(t, 0) = g(t)t^k $ where $g(0)\neq0$ and $g$ is $C^\infty$ and defined on a neighborhood of $0\in\mathbb{R}$. Then there exist $C^\infty$ functions $q$, $\lambda_0$, ..., $\lambda_{k-1}$ such that $(qF)(t, x) = t^k + \sum_{i=0}^{k-1}\lambda_i(x)t^i$ on a neighborhood of $0\in\mathbb{R}\times\mathbb{R}^n$ and $q(0)\neq0$.

After a careful reading of the proof, it seemed to me that the result is preserved even if the assumption "$F\in C^\infty(\mathbb{R}\times\mathbb{R}^n; \mathbb{R})$" is replaced by "$F(t, \cdot)$ is $C^1$ for all $t$ and $F(\cdot, x)$ is $C^\infty$ for all $x$". (Obviously, the regularity of $q$, $\lambda_0$, ..., $\lambda_{k-1}$ must also be weakened in the same manner.) Indeed, all complex analysis arguments (inspired by Weierstrass preparation theorem) are used on functions depending on $t$, while only the implicit function theorem is used on a function depending on $x$.

My question is: Does anyone know a reference where this less regular version of the Malgrange preparation theorem is stated/proved/discussed?

Thank you.

$\endgroup$
11
$\begingroup$

Edit:

Here's a better reference, which actually states and proves more clearly the version that you are looking for. This one is due to Peter Michor (who is also on MO and may drop by?)

  • P. MICHOR, The division theorem on Banach spaces, Österrich. Akad. Wiss. Math.- Natur. Kl. Sitzungsber II,189 (1980), pp. 1–18. (link from Michor's website)

I think if you apply his theorem with $E$ trivial, $F = C^1(\mathbb{R}^n)$, you should get what you are looking for.


See

End of first paragraph reads:

In this paper we give a short, rather elementary proof of the Malgrange-Mather Division Theorem, which applies to analytic, $C^\infty$ or $C^m$ functions.

$\endgroup$
6
  • 1
    $\begingroup$ (Note: as the paper itself mentions, the $C^m$ version was obtained earlier by Lassalle in the same journal; but that proof is both longer and in French [though the two may be correlated].) $\endgroup$ – Willie Wong Apr 13 at 13:57
  • $\begingroup$ Thank you for your answer. However I do not think that the result of Lassalle answers the question. Indeed, his assumption is (with my notations) that $F$ is $C^m$ with respect to both $t$ and $x$, while I would like to have $F$ only $C^1$ with respect to $x$ and $C^\infty$ with respect to $t$. Moreover, the order $k$ at which $F$ vanishes in $t$ at $x=0$ is bounded by $m$ in the version of Lassalle, so applying his result for $m=1$ is not sufficient to obtain the result I am looking for (with $k\in\mathbb{N}$). $\endgroup$ – preparation_theorem Apr 13 at 15:16
  • $\begingroup$ Since it is equivalent to Lassalle's result, the problem is the same with Millman's result. $\endgroup$ – preparation_theorem Apr 13 at 15:32
  • 1
    $\begingroup$ @preparation_theorem: you are correct, however, take a look at Milman's paper. His approach to proving Theorem 1 is to think of the mapping as $C^\infty(V^d; E)$ where $E$ is some TVS, and the precise structure of $E$ doesn't matter. In particular, I am pretty sure, based on the proof given without any modifications, that Theorem 1 holds with $C^\infty_\pi(V^d\times \mathbb{R}^n)$ replaced by $C^\infty_\pi(V^d; E)$ for any Banach space $E$ and similarly for the codomain. $\endgroup$ – Willie Wong Apr 13 at 15:33
  • $\begingroup$ ... and I think the same is true for the proof in Section 2 of the preparation theorem from Theorem 1; but I haven't checked all the details. (What I mean is that Milman seems to have proved a more general theorem than he stated.) $\endgroup$ – Willie Wong Apr 13 at 15:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.