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.


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}.

 Thus, writing {F(r,t) := \tilde{f(re^{it})}}, we find

\displaystyle F(r,t) = \frac{1}{2\pi} \int_0^{2\pi} \sum_n f(e^{ix}) e^{-inx} e^{int} r^{|n|} dx which implies

\displaystyle \boxed{ F(r,t) = \frac{1}{2\pi} \int f(e^{ix}) P_r(t-x) }


\displaystyle P_r(y) = \sum_n r^{|n|} e^{iny} = \frac{1-r^2}{1-2r\cos y + r^2}

Theorem 1 The function {\tilde{f}} is continuous in {\bar{U}}, harmonic in the interior {U}, and equal to {f} on the boundary.

I’m not going to actually fully prove the theorem; the basic idea is that the {P_r} are all of norm {1} (because they are nonnegative and of integral 1), so we have {||\tilde{f}||_{\infty} \leq ||f||_{\infty}}. Consequently the result follows from its counterpart on trigonometric polynomials, which is evident since the Fourier series is finite! The approximation result proves useful again.

The Poisson integral shows that it is possible to solve the Dirichlet problem for the disk: that is, one can extend a continuous function on the boundary to a harmonic function. This does not work if {D} is the deleted disk, because there is no harmonic function {f} vanishing on {\{z: |z|=1\}} with {f(0)=1}. (Cf. the maximum principle below.)

It is in fact the case that {\tilde{f}} is the only such harmonic function, satisfying the conclusions of the theorem. This follows from the maximum principle—a nonconstant harmonic function has no local maxima, which in turn follows from the Laplace equation and the second derivative test as follows. If {f} is harmonic and has a local maximum at {0}, so does {h:=f + \epsilon(x^2+y^2)} for {\epsilon>0} small; however, {\Delta h > 0}, a contradiction.

In particular, in view of the expression for the Poisson kernel, a harmonic function is necessarily smooth on its domain; apparently this is more generally true for solutions to elliptic PDEs, but I haven”t learned about them yet.   This is probably one of the most important facts for us in the next few posts.

The mean value property

A harmonic function {f} must satisfy on any disk {D_r(a)} in its domain

\displaystyle \boxed{f(a) = \frac{1}{2\pi} \int_0^{2\pi} f(a + re^{it}) dt.}

 This is in fact a corollary of the Poisson formula and uniqueness above. The mean value property is actually a sufficient condition for harmonicity (together with, say, continuity), but that is not necessary for us, and I refer anyone interested to Rudin’s Real and Complex Analysis.

The Harnack principle

Positive harmonic functions take comparable values on compact subsets of their domains, according to the next result:

Theorem 2 (Harnack)

Let {D_0} be a compact subset of the open region {D}. Then there is a positive constant {c := c(D_0,D)} such that for any positive harmonic function on {D} and {z,z' \in D_0}\displaystyle c^{-1} \leq \frac{f(z)}{f(z')} \leq c . 


If {r<1} is sufficiently small, there is a positive constant {c_0} such that\displaystyle c_0^{-1} \leq P_{r'}(t) \leq c_0, \ \mathrm{when} \ r' \leq r , \forall t. This is evident, e.g., from the power series expression. (We actually could have said somewhat more.)

So let {U} be the unit disk and {U'} a small proper subdisk. If {f} is a positive harmonic function on {U}, the above observation implies that the values of {f} on {U'} satisfy\displaystyle d^{-1} \leq \frac{f(z)}{f(z')} \leq d  for some {d>0}. In particular, we get Harnack’s theorem locally. By covering {D_0} with a finite number of overlapping disks, we get the theorem globally.