Subharmonic Functions and Perron's Method

Subharmonic functions generalize harmonic functions and provide the framework for Perron's method, which solves the Dirichlet problem on very general domains.

Subharmonic Functions

Definition8.4Subharmonic function

An upper semicontinuous function $u: D \to [-\infty, \infty)$ is subharmonic if for every closed disk $\overline{D}(z_0, r) \subset D$ :

$u(z_0) \leq \frac{1}{2\pi}\int_0^{2\pi} u(z_0 + re^{i\theta})\,d\theta.$

That is, the value at the center is at most the average over any circle.

RemarkRelation to holomorphic functions

If $f$ is holomorphic and nonvanishing on $D$ , then $\log|f|$ is harmonic. More generally, $\log|f|$ is subharmonic on all of $D$ (allowing $f$ to have zeros, where $\log|f| = -\infty$ ). Also $|f|^p$ is subharmonic for any $p > 0$ .

ExampleExamples of subharmonic functions

$u(z) = |z|^2 = x^2 + y^2$ is subharmonic (since $\Delta u = 4 > 0$ ).
$u(z) = \text{Re}(z)$ is harmonic (hence subharmonic).
$u(z) = \max(\text{Re}(z), 0)$ is subharmonic but not harmonic.
$u(z) = \log|f(z)|$ for holomorphic $f$ is subharmonic.

Properties

Theorem8.8Maximum principle for subharmonic functions

If $u$ is subharmonic and non-constant on a domain $D$ , then $u$ has no local maximum in $D$ . If $D$ is bounded and $u$ extends upper semicontinuously to $\overline{D}$ , then

$u(z) \leq \max_{\partial D} u \quad \text{for all } z \in D.$

Theorem8.9Characterization via Laplacian

A $C^2$ function $u$ is subharmonic on $D$ if and only if $\Delta u \geq 0$ on $D$ . It is harmonic if and only if $\Delta u = 0$ .

Perron's Method

Definition8.5Perron family

For the Dirichlet problem with boundary data $g$ on $\partial D$ , the Perron family is

$\mathcal{S}_g = \{v : D \to \mathbb{R} \mid v \text{ subharmonic}, \limsup_{z \to \zeta} v(z) \leq g(\zeta) \text{ for all } \zeta \in \partial D\}.$

The Perron solution is $u(z) = \sup\{v(z) : v \in \mathcal{S}_g\}$ .

Theorem8.10Perron's theorem

The Perron solution $u$ is harmonic on $D$ . If $\partial D$ satisfies the exterior cone condition (or more generally, if every boundary point is regular), then $u$ is continuous on $\overline{D}$ and $u = g$ on $\partial D$ .

ExampleRegular and irregular boundary points

A boundary point $\zeta \in \partial D$ is regular if the Dirichlet problem with any continuous boundary data has the correct limit at $\zeta$ . Examples:

Every point of a smooth boundary is regular.
An isolated boundary point is irregular (the Perron solution ignores it).
The tip of a cusp can be irregular, depending on the cusp's sharpness.

The classic example of an irregular point: if $D = D(0,1) \setminus \{0\}$ , then $z = 0$ is irregular. The harmonic function $u$ with $u = 1$ on $|z| = 1$ and $u(0) = 0$ would need $u \equiv 1$ by removable singularity, contradicting $u(0) = 0$ .

Harnack's Inequality and Principle

Theorem8.11Harnack's inequality

If $u$ is harmonic and non-negative on $D(z_0, R)$ , then for $|z - z_0| = r < R$ :

$\frac{R-r}{R+r}\,u(z_0) \leq u(z) \leq \frac{R+r}{R-r}\,u(z_0).$

RemarkHarnack's principle

If $\{u_n\}$ is an increasing sequence of harmonic functions on $D$ , then either $u_n \to \infty$ uniformly on compact subsets, or $u_n$ converges to a harmonic function uniformly on compact subsets. This is an immediate consequence of Harnack's inequality.