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 tobe subharmonic if whenever $D(z,r)\subset G$ (where $D(z,r)$ is a closeddisc 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 pointno 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 maximumin a region $G$ (it would achieve it at the boundary if it is continuous there). Similarly for a superharmonicfunction, but then a non-constant superharmonic function does not achieve itsminumum 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 $\phi$ wouldin fact be a harmonic function. That is, when $\phi$ is both subharmonic andsuperharmonic, 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 thesuperharmonic case. The integral will then however need to be theLebesgueintegral (http://planetmath.org/Integral2) rather than the Riemann integral which may not be defined for sucha 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 thereal 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 functionthen $\phi (z):=\log |f(z)|$ is a subharmonic function if wedefine the value of $\phi (z)$ at the zeros of $f$ as $-\mathrm{\infty }$.Again it is important to note that with thisgeneralization we again must use the Lebesgue integral.