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. (more…)

The topic for the next few weeks will be Riemann surfaces.  First, however, I need to briefly review harmonic functions because I will be talking about harmonic forms.  I will have more to say about them later, and I actually won’t use most of today’s post even until then.  But it’s fun.

Some of this material has also been covered by hilbertthm90 at A Mind for Madness.

Definition

A ${C^2}$ function ${f}$ on an open subset of ${\mathbb{R}^n}$, ${n >1}$, is called harmonic if it satisfies the Laplace equation $\displaystyle \Delta f = \sum \frac{\partial^2f}{\partial x_i^2} = 0.$ For now, we are primarily interested in the case ${n=2}$, and we will identify ${\mathbb{R}^2}$ with ${\mathbb{C}}$.  In this case, as is well-known, harmonic functions are locally the real parts of holomorphic functions.

The Poisson Integral

The following fact is well-known: given a continuous function ${f}$ on the circle ${C_1(0)}$, there is a unique continuous function on the closed unit disk ${\overline{U}}$ which is harmonic in the interior and coincides with ${f}$ on the boundary.The idea of the proof is that ${f}$ can be represented as a Fourier series, $\displaystyle f(e^{it}) = \sum_{n \in \mathbb{Z}} c_n e^{int}$

where the ${c_n}$ are obtained through the orthogonality relations $\displaystyle c_n = ( f, e^{-int} )$

where the inner product is the ${L^2}$ product taken with respect to the Haar measure on the circle group. This convergence holds in ${L^2}$, because the exponentials form an orthonormal basis for that space. Indeed, orthonormality can be checked by integration, and the Stone-Weierstrass theorem implies their linear combinations are dense in the space of continuous functions on the circle. It is even the case that convergence holds uniformly if ${f}$ is well-behaved (say, ${C^2}$). But this is only for motivational purposes, and I refer anyone interested to, say, Zygmund’s book on trigonometric series for a whole lot fo such results.

Now, it is clear that the functions $\displaystyle z \rightarrow r^n e^{int}, \ z \rightarrow r^n e^{-int}$

are harmonic (where ${t = Arg(z), r = |z|}$) as the real parts of ${z^n, \bar{z}^n}$.

It thus makes sense to define the extended function ${\tilde{f}}$ as $\displaystyle \tilde{f}(re^{it}) = \sum_n c_n r^{|n|} e^{int}.$ (more…)