Back to elliptic regularity. We have a constant-coefficient partial differential operator ${P = \sum_{a: |a| \leq k} C_a D^a}$ which is elliptic, i.e. the polynomial

$\displaystyle Q(\xi) = \sum_{a: |a| \leq k} C_a \xi^a$

satisfies ${|Q(\xi)| \geq \epsilon |\xi|^k}$ for ${|\xi|}$ large. We used this last property to find a near-fundamental solution to ${P}$. That is, we chose ${E}$ such that ${\hat{E} = (1-\varphi) Q^{-1}}$, where ${\varphi}$ was our arbitrary cut-off function equal to one in some neighborhood of the origin. The point of all this was that

$\displaystyle P(E) = \delta - \hat{\varphi}.$

In other words, ${E}$ is near the fundamental solution. So given that ${Pf = g}$, we can use ${E}$ to “almost” obtain ${f}$ from ${g}$ by convolution ${E \ast g}$—if this were exact, we’d have the fundamental solution itself.

We now want to show that ${E}$ isn’t all that badly behaved.

The singular locus of the parametrix

We are going to show that ${\mathrm{sing} E = \{0\}}$. The basic lemma we need is the following. Fix ${m}$. Consider a smooth function ${\phi}$ such that, for each ${a}$, there is a constant ${M_a}$ with

$\displaystyle |D^a \phi(x)| \leq M_a (1+|x|)^{m-|a|};$

then this is a distribution, but it is not necessarily a Schwarz function. And ${\hat{\phi}}$ cannot be expected to be one, thus. Nevertheless:

Lemma 1 ${\hat{\phi}}$ is regular outside the origin.

Indeed, let’s consider

$\displaystyle x^a D^b \hat{\phi} = \frac{1}{(2\pi i)^{|b|} } \widehat{ D_a (x^b \phi) } ;$

when ${a}$ is large enough (relative to ${b}$), the distribution ${ D_a(x^b \phi )}$ is actually an integrable function. So ${x^a D^b \hat{\phi} \in L^2}$ for ${a}$ large. As a result ${\hat{\phi}}$ and its derivatives are rapidly decreasing at ${\infty}$, whence the lemma; things may blow up at ${0}$. So, given that ${\hat{E}}$ was just the inverse of a polynomial multiplied by one minus some cutoff, we find:

Corollary 2 $\displaystyle \mathrm{sing} E = \{ 0\}.$

Elliptic regularity

Now we can return to the parametrix ${E}$ as before. We can deduce the following result.

Theorem 3 (Elliptic Regularity) Let ${f,g}$ be distributions. Suppose ${Pf = g}$ and ${P}$ is elliptic. Then ${\mathrm{sing} f \subset \mathrm{sing} g}$. In particular, if ${g}$ is smooth, so is ${f}$.

Before proving it, here is a special case. When ${P}$ is the Laplacian, this is a stronger form of Weyl’s lemma. Indeed, we find that a weak (distribution) solution of the Laplace equation is necessarily a smooth function, which must therefore be harmonic. Indeed, we get a version of Weyl’s lemma in higher dimensions. Let’s first assume ${f,g}$ are compactly supported, so we can talk about convolutions.

The idea is that

$\displaystyle P(E \ast f) = PE \ast f = E \ast g.$

We look at the last two terms.

$\displaystyle f - \hat{\varphi} \ast f = E \ast g.$

So, since ${\hat{\varphi} \ast f}$ is smooth, we find ${\mathrm{sing} f \subset \mathrm{sing} (E \ast g) \subset \mathrm{sing} g}$. This establishes the claim under the compact support hypothesis. In general, just multiply ${f,g}$ by some cutoff function equal to 1 in a large neighborhood of the origin. The proof of this result is complete.