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.

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