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 2
Let. If
, then
and
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as
.
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
.
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