Theorem 1 (Weyl) Let , where is the unit disk with Lebesgue measure. If for all with compact support, then is harmonic (in particular smooth).
I dropped out of the groove for a couple of days due to other activities; I’m back today to talk about Weyl’s lemma (for the Laplacian—it generalizes to elliptic operators), a tool we will need for the special case of the Hodge decomposition theorem on Riemann surfaces. The result states that a “weak” solution to the Laplace equation is actually a strong one.
The smooth case
First, recall Green’s theorem
which is stated in a form that follows easily from Stokes theorem (I will discuss the above equation when I talk more about -forms on Riemann surfaces). Here the derivatives are with respect to the outward unit normal. This is valid for all sufficiently smooth on a reasonable region .
So if is smooth, then by choosing to be and compactly supported we find
which implies by the arbitrariness of that .
Approximations to the identity
The general case will be handled by the common procedure detailed below.
Let be a function of norm and integral , and define . As the can be thought of as approaching the Dirac delta function (which is true in the distribution sense).
The following is the main theorem:
Theorem 2Let . If , then and
The first fact is Young’s inequality. The interesting part is the last statement. The difference can be written as
Consider the family of linear operators , . Each is bounded with norm at most by the first inequality. Because of this, it will be enough to prove that
in as for continuous with compact support. Now
As , the integrand tends to zero almost everywhere and is bounded by a constant times ; therefore tends to zero pointwise. Since is supported in a fixed compact set for all and always uniformly bounded, the dominated convergence theorem implies as .
Now if denotes the translation for , then tends to zero as ; this is checked on the continuous functions with compact support, which are dense.When is smooth and, say, compactly supported or a Schwarz function, so is ; in fact, we can differentiate under the integral with justification from the dominated convergence theorem.As a corollary, we find that the smooth functions are dense in ; by multiplying by a cutoff function, we find that the smooth functions with compact support are dense in as well.
Proof of Weyl’s lemma
I am now going to write for .
First, choose to be a radial smooth function supported in a small disk centered at the origin (say ) and of total integral 1.
Consider as a function in by extending by zero. Then we can consider the convolutions , which converge to in . Moreover, I claim that the are mostly harmonic.
Indeed, can be viewed as a weighted average of translates of by perturbations with . These perturbations of satisfy the Weyl condition (1) in the smaller disk , so the average satisfies (1) too—and by the smooth version of the result, already proved, we find that is harmonic in .
Next, I claim that if , then
in . Indeed, in ; this is because is harmonic in that disk, and we have a mean-value property (which incidentally explains the radiality condition we imposed on ).
Similarly, in . Since convolution is commutative, this proves the claim.
So we define the function by gluing together all the . Since the are bounded, . Moreover,
for any . Now is uniformly bounded and pointwise tends to zero; hence if we let the first term disappears. So does the second, by the second section. Thus is the harmonic equivalent to .