# What's the use of Malgrange preparation theorem?

The Malgrange preparation theorem,which is the $C^{\infty}$ version of the classical Weierstrass preparation theorem,says that if $f(t,x)$ is a $C^{\infty}$ function of $(t,x)\in\mathbb{R}^{n+1}$ near $(0,0)$ which satisfies $$f=\frac{\partial{f}}{\partial{t}}=\cdots =\frac{\partial^{k-1}{f}}{\partial{t^{k-1}}}=0\quad \frac{\partial^{k}{f}}{\partial{t^{k}}}\ne 0\quad\text{at}(0,0)$$ Then there exists a factorization $$f(t,x)=c(t,x)(t^{k}+a_{k-1}(x)t^{k-1}+\cdots +a_{0}(x))$$ where $a_j$ and $c$ are $C^{\infty}$ functions near $0$ and $(0,0)$ respectively,$c(0,0)\ne 0$ and $a_{j}(0)=0$.As a corollary, there is a division thereom just like the Weierstrass formula.However, unlike the analytic case,this factorization is not unique.The result is said to be highly non-trival even when $k=1$,the difficulty is then the zeros may be lost,For example,$t^{2}+x$ has two real zeros when $x<0$ but none when $x>0$.The proof can be seen in Theorem 7.5.6 in Hormander's The Analysis of linear partial differential operators.

My question is What's the use of Malgrange preparation theorem in mathematics?Is this a verey useful formula in analysis ? Can anyone take some examples to apply this theorem?(In hormander's book,this is used in the method of Stationary Phase).

A quick google search shows that there is also a algebraic version which can be restated as a theorem about modules over rings of smooth, real-valued germs.

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The origin of the differential preparation theorem as well as its relation with distribution division problem (solved independently by Lars H\"ormander and Stanis\l aw \L ojasiewicz; a version is mentioned in Denis Serre's answer) is nicely explained by Malgrange himself in the paper MR2065138 (2006i:46039) Malgrange, Bernard(F-GREN-F) Idéaux de fonctions différentiables et division des distributions. (French) [Ideals of differentiable functions and division of distributions] With an Appendix: "Stanisław Łojasiewicz (1926–2002)''. Distributions, 1–21, Ed. Éc. Polytech., Palaiseau, 2003 – Margaret Friedland Apr 25 at 18:30

The Malgrange preparation theorem is extremely useful in singularity theory, where it can be used to prove Mather's fundamental theorem on the equivalence of infinitesimal stability and stability for $C^\infty$ mappings, and can also be used to derive the normal forms of some stable singularities. The book "Stable mappings and their singularities" by Golubitsky and Guillemin is a good reference.

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Malgrange elaborated his preparation theorem in order to prove that if $L$ is a linear differential operator with constant coefficients, then it admits a fundamental solution, that is a distribution $E$ such that $LE=\delta_{x=0}$. Then if $f$ is a smooth function with compact support, the equation $Lu=f$ admits the solution $u=E\star f$.

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Thanks a lot. I didn't know that before,by the way,the proof of this theorem(thm 7.3.10) in Hormander's book is by constructing some particular measure.Maybe that's originally from Ehrenpreis who also proved the result. – user23078 Oct 8 '12 at 13:14
@Denis Serre Are you sure of this ? After all constant coefficients PDE are described by a polynomial with $n$ variables, a very simple type of analytic function. Why the preparation theorem, usually devised to dealing with smooth non-analytic functions satisfying a finite type assumption could be useful in that context ? – Bazin Oct 11 '12 at 20:20
This answer refers to the Ehrenpreis-Malgrange Theorem which is very different from the Malgrange Preparation Theorem. – Sönke Hansen Apr 19 '13 at 15:08
@SönkeHansen Actually, the Malgrange preparation theorem and the Malgrange-Ehrenpreis theorem are not unrelated. The former came as a spinoff of of Malgrange's work on the latter, as he tells us himself at the end of his book "Ideals of Differentiable Functions" (OUP, 1966). To quote, "In other words, every linear differential operator with constant coefficients has a temperate fundamental solution (i.e. one in $\mathscr{S}'$). This is mainly of historical interest (...). We have, however, given this here because it was the origin of a large part of the results contained in this book." – Pedro Lauridsen Ribeiro Apr 25 at 19:21

Take $k=1$ in your statement. There are two easy cases: the first one is when $f$ is real-valued, then you have only to use the implicit function theorem to get a normal form $t+a(x)$, up to a unit (a non-vanishing factor). The next easy case is when $f$ is analytic: then again implicit function theorem for holomorphic functions, same normal form, of course with $a$ complex-valued and analytic.

What if $f$ is only $C^\infty$ and complex-valued? Malgrange preparation theorem answers precisely to this question. Same normal form with $a$ complex-valued $C^\infty$. Caveats: no uniqueness, very hard proof. A example of application: a $C^\infty$ complex valued principal-type symbol (smooth function on the cotangent bundle of a smooth manifold) can be reduced to $$\xi_1+iq(x_1,x',\xi').$$ It means that this type of equation can be reduced to an evolution equation (time variable $x_1$). Without Malgrange preparation theorem, no thorough analysis of scalar complex-valued pseudodifferential equations of principal type in the $C^\infty$ framework.

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@Bazin Thanks very much,this is very useful to me – user23078 Oct 9 '12 at 0:36

One use is to prove that the sheaf $\mathscr{O}_\infty$ of complex smooth functions on a smooth algebraic variety $X/\mathbb{C}$ is flat over $\mathscr{O}_X$ (the sheaf of holomorphic functions). This was for example used by Deligne [(3.2.3) b), 1] to construct filtered resolutions of the logarithmic de Rham complex.

[1] P. Deligne Thêorie de Hodge, II. Pub. Mat. IHES 40 (1971), pp 5-57

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