Theorem 1 (Weyl) Let ${f \in L^2(U)}$, where ${U}$ is the unit disk with Lebesgue measure. If $\displaystyle \int_U f \Delta \phi = 0$  for all ${\phi \in C^{\infty}(U)}$ with compact support, then ${f}$ 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 $\displaystyle \int_{\partial D} f \frac{dg}{dn} - g \frac{df}{dn} = \int_D f\Delta g - g \Delta f$

which is stated in a form that follows easily from Stokes theorem (I will discuss the above equation when I talk more about ${1}$-forms on Riemann surfaces). Here the ${dn}$ derivatives are with respect to the outward unit normal. This is valid for all sufficiently smooth ${f,g}$ on a reasonable region ${D}$

So if ${f}$ is smooth, then by choosing ${g}$ to be ${C^{\infty}}$ and compactly supported we find $\displaystyle \int_U g \Delta f = 0,$

which implies by the arbitrariness of ${g}$ that ${\Delta f \equiv 0}$

Approximations to the identity

The general case will be handled by the common procedure detailed below.

Let ${\phi \in L^1(\mathbb{R}^n)}$ be a function of norm ${C}$ and integral ${1}$, and define ${\phi_t(x) := t^{-n} \phi(x/t)}$. As ${t \rightarrow 0}$ the ${\phi_t}$ 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 ${1 \leq p < \infty}$. If ${g \in L^p}$, then ${f \ast g \in L^p, ||g \ast \phi_t||_p \leq C||g||_p}$ and $\displaystyle ||g - g \ast \phi_t||_p \rightarrow 0$

as ${t \rightarrow 0}$.

The first fact is Young’s inequality. The interesting part is the last statement. The difference ${g - g \ast \phi_t}$ can be written as $\displaystyle (g - g \ast \phi_t)(x) = \int_{\mathbb{R}^n} (g(x) - g(x-u) )\phi_t(u) du.$

Consider the family of linear operators ${F_t: L^p \rightarrow L^p}$, ${F_t(g) := g - g \ast \phi_t}$. Each is bounded with norm at most ${C+1 }$ by the first inequality. Because of this, it will be enough to prove that $\displaystyle F_tg \rightarrow 0$

in ${L^p}$ as ${t \rightarrow 0}$ for ${g}$ continuous with compact support. Now $\displaystyle F_t g(x) = \int_{\mathbb{R}^n} (g(t^{-1}x) - g(t^{-1}x-u) )\phi(u) du .$

As ${t \rightarrow 0}$, the integrand tends to zero almost everywhere and is bounded by a constant times ${|\phi|}$; therefore ${F_t g}$ tends to zero pointwise. Since ${F_tg}$ is supported in a fixed compact set for all ${t<1}$ and always uniformly bounded, the dominated convergence theorem implies ${||F_tg||_p\rightarrow 0}$ as ${t \rightarrow 0}$

Now if ${g_u}$ denotes the translation ${x \rightarrow g(x-u)}$ for ${u \in \mathbb{R}^n}$, then ${G(u) := ||g_u - g||_{L^p}}$ tends to zero as ${u \rightarrow 0}$; this is checked on the continuous functions with compact support, which are dense.When ${\phi}$ is smooth and, say, compactly supported or a Schwarz function, so is ${g \ast \phi_t}$; 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 ${L^p}$; by multiplying by a cutoff function, we find that the smooth functions with compact support are dense in ${L^p}$ as well.

Proof of Weyl’s lemma

I am now going to write ${||\cdot||}$ for ${||\cdot||_2}$

First, choose ${\phi}$ to be a radial smooth function supported in a small disk centered at the origin (say ${D_{1}(0)}$) and of total integral 1.

Consider ${f}$ as a function in ${L^2(\mathbb{R}^2)}$ by extending by zero. Then we can consider the convolutions ${f_t := f \ast \phi_t}$, which converge to ${f}$ in ${L^2}$.  Moreover, I claim that the ${f_t}$ are mostly harmonic.

Indeed, ${f_t}$ can be viewed as a weighted average of translates of ${f}$ by perturbations ${u}$ with ${|u| < t}$. These perturbations of ${f}$ satisfy the Weyl condition (1) in the smaller disk ${D_{1-t}}$, so the average ${f_t}$ satisfies (1) too—and by the smooth version of the result, already proved, we find that ${f_t}$ is harmonic in ${D_{1-t}(0)}$.

Next, I claim that if ${\tau>t}$, then $\displaystyle f_t = f_{\tau}$

in ${D_{1-\tau}(0)}$. Indeed, ${f_t = f_t \ast \phi_{\tau} = f \ast \phi_t \ast \phi_{\tau}}$ in ${D_{1-\tau}(0)}$; this is because ${f_t}$ is harmonic in that disk, and we have a mean-value property (which incidentally explains the radiality condition we imposed on ${\phi}$).

Similarly, ${f_{\tau} = f_{\tau} \ast \phi_t = f \ast \phi_{\tau} \ast \phi_t}$ in ${D_{1-\tau}(0)}$. Since convolution is commutative, this proves the claim.

So we define the function ${\tilde{f}}$ by gluing together all the ${f_t}$. Since the ${||f_t||}$ are bounded, ${\tilde{f} \in L^2(D)}$. Moreover, $\displaystyle || \tilde{f} - f || \leq || \tilde{f} - f_t|| + ||f_t - f||$ for any ${t}$. Now ${\tilde{f}-f_t}$ is uniformly bounded and pointwise tends to zero; hence if we let ${t \rightarrow 1}$ the first term disappears. So does the second, by the second section. Thus ${\tilde{f}}$ is the harmonic equivalent to ${f}$.