Dividing a tempered distribution by a polynomial

distribution-theoryfunctional-analysisschwartz-space

Let $p=p(x_1,…,x_N)$ be a non-zero polynomial in $N$ variables (real coefficients). Let $\mathscr{S}$ be the Schwartz space on $\mathbb{R}^N$ and let $\mathscr{S}'$ be its topological dual (i.e. the space of tempered distributions). I know that the map

$\mathscr{S}'\to \mathscr{S}', \ \ \ T\mapsto pT$

(where $\langle pT,\psi \rangle:=\langle T, p\psi \rangle$) is linear and continous. My class notes affirm that this map has also a continous inverse given by

$\mathscr{S}'\to \mathscr{S}', \ \ \ T\mapsto T/p$

and I don't understand this. This works if $p$ hasn't any real zeroes, but otherwise $T/p$ shouldn't be well-defined. I understand that the zeroes of $p$ form a null measure set so maybe there is some way of solving this issue. But I don't know how. Can you help me?

Best Answer

The result you ask of is one of the cornerstones of the linear theory of PDEs, proved by Lars Hörmander at the beginning of the second half of the 20th century: the original paper is available and a link is given below.
It is not easy to describe the method of proof he used: the only thing I can say is that Hörmander relies on a deep theorem in logic proved by Seidenberg and Tarski a few years before. By using it, he was able to prove two estimates for the growth of the polynomial $p(\xi)$ as a function of the distance of its argument $\xi$ (assumed to be real) from the set of its real zeros $N$: precisely, if $N=\emptyset$ (i.e. $p$ is an elliptic polynomial) then $$ |p(\xi)|\ge c (1+|\xi|^2)^{-\mu^\prime}\tag{2.5} $$ while if $N\neq\emptyset$ $$ |p(\xi)|\ge c (1+|\xi|^2)^{-\mu^\prime}d(\xi,N)^{\mu^{\prime\prime}}\tag{2.6} $$ where $c>0$, $\mu^\prime$ and $\mu^{\prime\prime}$ are constants.

Reference

[1] Lars Hörmander, "On the division of distributions by polynomials" (English) Arkiv för Matematik 3, DOI:10.1007/BF02589517, 555-568 (1958), MR0124734, Zbl 0131.11903.