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.

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