So, we’re given a constant-coefficient operator

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

Let be the polynomial ; then the notation suggests the substitution of differential operators in a polynomial, as it should, even though technically should be replaced with . This should not cause any confusion. We are interested in finding local solutions to an equation

where 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 is equivalent to

so formally we write

This, of course, is utter nonsense. The polynomial 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: is a compactly supported smooth function. Thus the Fourier transform

extends to and is an entire function of . 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:

by looking at the definition. Moreover, if remains bounded, then I claim rapidly as ; indeed,

and the family of functions for bounded is bounded in the topology of .

It turns out in fact that if the Fourier transform satisfies these properties, then 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 , we may be able to make sense of things by shifting the contours to avoid the singularities of .

So, let’s start by writing , where the lower terms are of lower degree in . We can do this by changing coordinates in if necessary. Then for , there are finitely many such that

It is possible to choose as a function of such that for has no zeros for and in such a way that 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

which is entirely kosher because of the rapid decrease of , and the way we’ve nimbly moved the contour up a step to dodge the roots of .

**Checking the solution **

This is well-defined and integrable because is bounded and measurable. Moreover, by the same justification is smooth, and we can differentiate with respect to . We can write

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

We can now state what we have proved:

Theorem 1If is a smooth, compactly supported function then there is a smooth solution to

This argument is due to Niremberg, incidentally.

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