The next application I want to talk about here of Fourier analysis is to (a basic case of) ellipic regularity. Later we will use refinements of these techniques to obtain all kinds of estimates. Anyway, for now, a partial differential operator

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

is called elliptic if the homogeneous polynomial

$\displaystyle \sum_{a: |a| = k} C_a \xi^a, \quad \xi = (\xi_1, \dots, \xi_n)$

has no zeros outside the origin. For instance, the Laplace operator is elliptic. Later I will discuss how this generalizes to other PDEs, and how this polynomial becomes the symbol of the operator. For the moment, though let’s define ${Q(\xi) = \sum_{a: |a| \leq k} C_a (2 \pi i \xi)^a}$. The definition of ${Q}$ such that

$\displaystyle \widehat{ Pf } = Q \hat{f},$

and we know that ${|Q(\xi)| \geq \epsilon |\xi|^k}$ for ${|\xi|}$ large enough. This is a very important fact, because it shows that the Fourier transform of ${Pf}$ exerts control on that of ${f}$. However, we cannot quite solve for ${\hat{f}}$ by dividing ${\widehat{Pf}}$ by ${Q}$ because ${Q}$ is going to have zeros. So define a smoothing function ${\varphi}$ which vanishes outside a large disk ${D_r(0)}$. Outside this disk, an estimate ${|Q(\xi)| \geq \epsilon |\xi|^k}$ will be assumed to hold.

A parametrix for ${P}$

We’re going to start by finding a parametrix for the operator ${P}$; this is not quite a fundamental solution, but it is close enough. We could formally get a fundamental solution (since ${\hat{\delta} \equiv 1}$) by considering ${\widehat{1/Q}}$, but this is nonsense. Rather, consider the distribution ${E}$ with

$\displaystyle \hat{E} = (1- \varphi) Q^{-1},$

which is indeed well-defined as a tempered distribution by the hypothesis on ${\varphi}$. It is then clear that ${\widehat{PE} = 1 - \varphi}$, so

$\displaystyle PE = \delta - \hat{\varphi},$

where the ${\hat{\varphi}}$ is at least suitably controlled (e.g. in ${\mathcal{S}}$). So we have something close to a fundamental solution, and ${E}$ is called the parametrix. Since we convolve things with a Moreover, I claim—and this is crucial—that ${E}$ comes very close to being a function of ${\mathcal{S}}$ as well; the only problem occurs because of the not-that-fast decrease at infinity. In particular, the singular locus of ${E}$ is the origin. This will require some discussion of intermediate concepts.

How to convolve two distributions

We already know how to convolve a distribution ${\phi}$ and a function ${f}$. Take ${\breve{f}_x}$ defined by ${\breve{f}_x(y) = f(x-y)}$ and set

$\displaystyle (\phi \ast f)(x) := \phi(\breve{f}_x).$

It is easy to check that this coincides with the old definition of convolution when ${\phi \in L^1}$. This is always ${C^{\infty}}$. When ${\phi}$ is compactly supported, which is to say that ${\phi(f)=0}$ when ${f}$ vanishes inside a sufficiently large compact set ${K}$, then we have

$\displaystyle |\phi(f)| \leq M \sum_{a:|a| \leq N} \sup_{K} |D^a f|,$

for some ${M,N}$. In this case it follows that ${\phi \ast f \in \mathcal{S}}$ in fact. So given another ${\phi'}$, not necessarily of compact support, we can heuristically use the properties of convolution to “write”

$\displaystyle (\phi' \ast \phi) \ast f = \phi \ast (\phi \ast f).$

Make this a definition. Then in particular, we have a way of talking about ${\phi' \ast \phi}$ as a distribution in itself—just evaluate the convolution at ${\breve{f}_0,0}$.

Let’s go back to what I just said about compact support. We can generalize this: say that a distribution ${\phi}$ vanishes on an open set ${\Omega}$ if ${\phi(f)=0}$ for ${f}$ supported in ${\Omega}$. Then a partition of unity argument shows that there is a largest open set ${\Omega}$ on which ${\phi}$ vanishes; the complement is called the support ${\mathrm{supp} \phi}$. This is a generalization of the notion for functions, as is easily seen. Anyway, it is a well-known fact about functions that ${\mathrm{supp} f_1 \ast f_2 \subset \mathrm{supp} f_1 + \mathrm{supp} f_2}$. That this is true for distributions when one is compactly supported follows by regularizing each: given ${\phi_1, \phi_2}$, we convolve each with an approximation to the identity to get smooth functions, one of which is compactly supported, that approxiamte ${\phi_1, \phi_2}$ arbitrarily closely (in the weak* topology).

The singular locus of a distribution

Say that a distribution ${\phi }$ is regular in an open set ${\Omega}$ if for ${f}$ smooth and supported in ${\Omega}$, we have ${\phi(f) = \int gf}$ for ${g: \Omega \rightarrow \mathbb{R}^n}$ smooth. Basically, this means that when restricted to ${\Omega}$, ${\phi}$ behaves just like a smooth function.

Using a partition of unity, we see that if ${\phi}$ is regular in ${\Omega_1}$ and ${\Omega_2}$, then it is regular in ${\Omega_1 \cup \Omega_2}$, and moreover for infinite unions. In particular, there is a maximal open set on which ${\Omega}$ is regular. The complement of this set is written ${\mathrm{sing} \phi}$. For instance, ${\mathrm{sing} \delta = \{0\}}$. This behaves well with respect to convolution, if one is compactly supported :

Lemma 1 $\displaystyle \mathrm{sing} \phi_1 \ast \phi_2 \subset \mathrm{sing} \phi_1 + \mathrm{sing} \phi_2.$

The reason is that we write ${\phi_1 = \phi_1^a + \phi_1^b, \phi_2 = \phi_2^a + \phi_2^b}$ where ${\phi_1^a,\phi_2^a}$ have supports barely outside the singular loci of ${\phi_1, \phi_2}$ and ${\phi_1^b, \phi_2^b}$ are smooth. Then

$\displaystyle \phi_1 \ast \phi_2 = \phi_1^a \ast \phi_2^a + \phi_1^a \ast \phi_2^b+\phi_1^b \ast \phi_2^a+\phi_1^b \ast \phi_2^b,$

and all but the first term are smooth. The first term is supported in a small neighborhood of ${\mathrm{sing} \phi_1 + \mathrm{sing} \phi_2}$, which we can make arbitrarily small.