Back to elliptic regularity. We have a constant-coefficient partial differential operator which is elliptic, i.e. the polynomial

satisfies for large. We used this last property to find a near-fundamental solution to . That is, we chose such that , where was our arbitrary cut-off function equal to one in some neighborhood of the origin. The point of all this was that

In other words, is near the fundamental solution. So given that , we can use to “almost” obtain from by convolution —if this were exact, we’d have the fundamental solution itself.

We now want to show that isn’t all that badly behaved.

**The singular locus of the parametrix **

We are going to show that . The basic lemma we need is the following. Fix . Consider a smooth function such that, for each , there is a constant with

then this is a distribution, but it is not necessarily a Schwarz function. And cannot be expected to be one, thus. Nevertheless:

**Lemma 1** * is regular outside the origin. *

Indeed, let’s consider

when is large enough (relative to ), the distribution is actually an integrable function. So for large. As a result and its derivatives are rapidly decreasing at , whence the lemma; things may blow up at . So, given that was just the inverse of a polynomial multiplied by one minus some cutoff, we find:

**Corollary 2**

*
*

**Elliptic regularity **

Now we can return to the parametrix as before. We can deduce the following result.

**Theorem 3 (Elliptic Regularity)** *Let be distributions. Suppose and is elliptic. Then . In particular, if is smooth, so is . *

Before proving it, here is a special case. When 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 are compactly supported, so we can talk about convolutions.

The idea is that

We look at the last two terms.

So, since is smooth, we find . This establishes the claim under the compact support hypothesis. In general, just multiply 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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