subharmonic and superharmonic functions

First let’s look at the most general definition.

Definition.

Let $G\subset {ℝ}^n$ and let $\phi :G\to ℝ\cup \{-\mathrm{\infty }\}$ be an upper semi-continuous function, then $\phi$ is subharmonic if for every $x\in G$ and $r>0$ such that $\overline{B(x,r)}\subset G$ (the closure of the open ball of radius $r$ around $x$ is still in $G$) and every real valued continuous function $h$ on $\overline{B(x,r)}$ that is harmonic in $B(x,r)$ and satisfies $\phi (x)\le h(x)$ for all $x\in \partial B(x,r)$ (boundary of $B(x,r)$) we have that $\phi (x)\le h(x)$ holds for all $x\in B(x,r)$.

Note that by the above, the function which is identically $-\mathrm{\infty }$ is subharmonic, but some authors exclude this function by definition. We can define superharmonic functions in a similar fashion to get that $\phi$ is superharmonic if and only if $-\phi$ is subharmonic.

If we restrict our domain to the complex plane we can get the following definition.

Definition.

Let $G\subset ℂ$ be a region and let $\phi :G\to ℝ$ be a continuous function. $\phi$ is said to be subharmonic if whenever $D(z,r)\subset G$ (where $D(z,r)$ is a closed disc around $z$ of radius $r$) we have

$$\phi (z)\le \frac{1}{2\pi }{\int }_0^{2\pi }\phi (z+r{e}^{i\theta })𝑑\theta ,$$

and $\phi$ is said to be superharmonic if whenever $D(z,r)\subset G$ we have

$$\phi (z)\ge \frac{1}{2\pi }{\int }_0^{2\pi }\phi (z+r{e}^{i\theta })𝑑\theta .$$

Intuitively what this means is that a subharmonic function is at any point no greater than the average of the values in a circle around that point. This implies that a non-constant subharmonic function does not achieve its maximum in a region $G$ (it would achieve it at the boundary if it is continuous there). Similarly for a superharmonic function, but then a non-constant superharmonic function does not achieve its minumum in $G$. It is also easy to see that $\phi$ is subharmonic if and only if $-\phi$ is superharmonic.

Note that when equality always holds in the above equation then $\phi$ would in fact be a harmonic function. That is, when $\phi$ is both subharmonic and superharmonic, then $\phi$ is harmonic.

It is possible to relax the continuity statement above to take $\phi$ only upper semi-continuous in the subharmonic case and lower semi-continuous in the superharmonic case. The integral will then however need to be the Lebesgue integral (http://planetmath.org/Integral2) rather than the Riemann integral which may not be defined for such a function. Another thing to note here is that we may take ${ℝ}^2$ instead of $ℂ$ since we never did use complex multiplication. In that case however we must rewrite the expression $z+r{e}^{i\theta }$ in of the real and imaginary parts to get an expression in ${ℝ}^2$.

It is also possible generalize the range of the functions as well. A subharmonic function could have a range of $ℝ\cup \{-\mathrm{\infty }\}$ and a superharmonic function could have a range of $ℝ\cup \{\mathrm{\infty }\}$. With this generalization, if $f$ is a holomorphic function then $\phi (z):=\log |f(z)|$ is a subharmonic function if we define the value of $\phi (z)$ at the zeros of $f$ as $-\mathrm{\infty }$. Again it is important to note that with this generalization we again must use the Lebesgue integral.

Title	subharmonic and superharmonic functions
Canonical name	SubharmonicAndSuperharmonicFunctions
Date of creation	2013-03-22 14:19:39
Last modified on	2013-03-22 14:19:39
Owner	jirka (4157)
Last modified by	jirka (4157)
Numerical id	12
Author	jirka (4157)
Entry type	Definition
Classification	msc 31C05
Classification	msc 31A05
Classification	msc 31B05
Related topic	HarmonicFunction
Defines	subharmonic
Defines	subharmonic function
Defines	superharmonic
Defines	superharmonic function