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}.