Question

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.

Answer 1

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.

Answer 2

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 $u=E\star f$.

Answer 3

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.