So, we’re given a constant-coefficient operator

$\displaystyle P(D) = \sum_{a: |a| \leq k} C_a D^a.$

(Note that I’m using slightly different notations than before.)

Let ${P}$ be the polynomial ${P(\xi) := \sum_{a: |a| \leq k} C_a (2\pi i \xi)^a}$; then the notation ${P(D)}$ suggests the substitution of differential operators in a polynomial, as it should, even though technically ${D}$ should be replaced with ${D/2 \pi i}$. This should not cause any confusion. We are interested in finding local solutions to an equation

$\displaystyle P(D) g = f,$

where ${f}$ is smooth. This will follow when we prove the Malgrange-Ehrenpreis theorem and get a fundamental solution. For now, however, we’ll avoid that.

Informal motivation

Anyway, we’ll first proceed informally: the equation ${P(D)g = f}$ is equivalent to

$\displaystyle P \hat{g} = \hat{f},$

so formally we write

$\displaystyle \hat{g} = \frac{\hat{f}}{P}, \ i.e. \ g(x) = \int_{\mathbb{R}^n} e^{2 \pi i x.t} \frac{ \hat{f}(t) }{ P(t) } dt \ (*) .$

This, of course, is utter nonsense. The polynomial ${P}$ will generally have roots, in which case we need to find a way to make sense of this.

Analyticity of the Fourier transform

First, let’s clear things up by making the hypothesis: ${f}$ is a compactly supported smooth function. Thus the Fourier transform

$\displaystyle \hat{f}(z) = \int_{\mathbb{R}^n} e^{-2\pi i z.t} f(t) dt$

extends to ${z \in \mathbb{C}^n}$ and is an entire function of ${z}$. Note incidentally that “entire” here can be taken as holomorphic in each variable separately and smooth; at some point, if and when I talk about several complex variables, this will be explained properly. Finally, we can get an estimate:

$\displaystyle | \hat{f}(z)| \leq C \exp( |\Im(z) |)$

by looking at the definition. Moreover, if ${\Im(z)}$ remains bounded, then I claim ${\hat{f}(z) \rightarrow 0}$ rapidly as ${z \rightarrow \infty}$; indeed,

$\displaystyle \hat{f}(x+iy) = \widehat{ e^{2\pi y t} f(t) }(x) ,$

and the family of functions ${\widehat{ e^{2\pi y t} f(t) }(x)}$ for ${y}$ bounded is bounded in the topology of ${\mathcal{S}}$.

It turns out in fact that if the Fourier transform satisfies these properties, then ${f}$ satisfies the corresponding ones, in a certain sense. This opposite (and more difficult) statement is the Paley-Wiener theorem.

Shifting the contours

Anyway, the reason for all this talk about entireness is that in the “expression” for ${g}$, we may be able to make sense of things by shifting the contours to avoid the singularities of ${P}$.

So, let’s start by writing ${P(t) = c_k t_n^k + \mathrm{lower terms}}$, where the lower terms are of lower degree in ${t_n}$. We can do this by changing coordinates in ${\mathbb{R}^n}$ if necessary. Then for ${t' = (t_1, \dots, t_{n-1}) \in \mathbb{R}^{n-1}}$, there are finitely many ${t_n}$ such that

$\displaystyle P(t', t_n) = 0.$

It is possible to choose ${\xi(t')}$ as a function of ${t' \in \mathbb{R}^n}$ such that ${P(t', t_n + i x)}$ for ${|x-\xi(t')| \leq 1}$ has no zeros for ${t_n \in \mathbb{R}^n}$ and in such a way that ${\xi}$ is measurable and bounded. This is a technical fact, which is pretty believable. I refer the interested reader to Folland’s book on PDE, my source here.

With this in mind, let us set

$\displaystyle g(x) = \int_{\mathbb{R}^n} e^{2 \pi i x.(t, t_n + i \xi(t') } \frac{ \hat{f}(t', t_n + i \xi(t')) }{ P(t', t_n + i \xi(t')) } dt' dt_n$

which is entirely kosher because of the rapid decrease of ${\hat{f}}$, and the way we’ve nimbly moved the contour up a step to dodge the roots of ${P}$.

Checking the solution

This is well-defined and integrable because ${\xi}$ is bounded and measurable. Moreover, by the same justification ${g}$ is smooth, and we can differentiate with respect to ${x}$. We can write

$\displaystyle Pg (x) = \int_{\mathbb{R}^n} e^{2 \pi i x.(t', t_n + i \xi(t') } \hat{f}(t', t_n + i \xi(t')) dt' dt_n.$

Now, by Cauchy’s theorem(!), we can shift the contours to conclude this equals

$\displaystyle \int_{\mathbb{R}^n} e^{2 \pi i x.(t', t_n } \hat{f}(t', t_n ) dt' dt_n = f(x).$

We can now state what we have proved:

Theorem 1 If ${f}$ is a smooth, compactly supported function then there is a smooth solution ${g}$ to$\displaystyle P(D)g = f.$

This argument is due to Niremberg, incidentally.