In this post, we prove a generalization of the result that any (clasically) harmonic function defined on a bounded, open set is analytic. If is *weakly harmonic*, i.e. for any compactly supported function (the space of such functions is denoted , then is strongly or classically harmonic and therefore equal a.e. to an analytic function. This result is due to Hermann Weyl and appears to be known in the literature as the Weyl Lemma.

The proof I will give is based on an outline in a problem set I found online.

Let be the radial mollifier defined by

where is the normalizing constant. Define an approximate identity . I leave it to the reader to verify that is indeed (this is a standard calculus problem) and that is an approximate identity.

For , define an open subset of by (i.e. the set of points in which are at least distance from the boundary of ). As is supported on , the convolution is well-defined. Furthermore, is . Iif , then , so is integrable on . By Young’s inequality, we have the estimate

I claim that is (classicaly) harmonic on . Indeed, as , we can use the Lebesgue dominated convergence theorem to differentiate inside the integral in the definition of (look at the Taylor expansion of ) to obtain

since and is weakly harmonic. Consequently, has the mean value property.

We now want to extract a sequence of the which converges to some continuous function and then argue that this function must be harmonic. To accomplish the first task, we use the Arzelà-Ascoli theorem. First, we have that is equicontinuous and uniformly bounded. Fix an , and set .

Lemma 1The family is uniformly bounded on for .

*Proof:* For , satisfies the mean value property. For ,

If , then , so we obtain the estimate

for all .

Lemma 2The family is equicontinuous on for .

*Proof:* For any and .

As is uniformly bounded on for all (with constant independent of ) and the Lebesgue measure of the two domains depends only on and goes to zero as , we obtain equicontinuity.

By the Arzelà-Ascoli theorem, there exists a sequence which converges uniformly to a continuous function . Replacing with a countable subset with indices tending to zero, we may assume that .

To prove that is harmonic, we need the following lemma.

Lemma 3If has the mean value property, then is (classically) harmonic.

*Proof:* Set . Observe that by the Lebesgue dominated convergence theorem is continuous. By the mean value property,

where we use the translation invariance of the Lebesgue measure to obtain the penultimate equality. Hence, .

We now show that is equal to a smooth function a.e. on . Set . Then

In particular, , so by the converse to the mean value property (see [Evans] Theorem 3, pp. 26), is harmonic.

As is an approximate identity, as . Hence, a.e. As we proved last time, is therefore equal to an analytic function a.e.

Next time, we will extend the Weyl lemma to distributions that satisfy the Laplace equation.

- L.C. Evans,
*Partial Differential Equations*(Second Edition), AMS, 2010.

Matt,

it seems to me that being smooth harmonic, coincides with by the mean value property, since we are doing averages on circles. As a result . So on growing (as goes to ) subsets of , the coincide. Since as goes to , converges to , this means is smooth in the interior of . Whether is mooth up to the boundary is another story.

Beatrice,

Thank you for your comment. We could have indeed proceeded more succinctly as you suggested. In fact, your outline generalizes nicely to the case where is an arbitrary distribution.

Regards,

Matt

What is the set V here and what is R?

I edited the post to include these details.