# subharmonic and superharmonic functions

First let’s look at the most general definition.

###### Definition.

Let $G\subset{\mathbb{R}}^{n}$ and let $\varphi\colon G\to{\mathbb{R}}\cup\{-\infty\}$ be an upper semi-continuous function, then $\varphi$ 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 $\varphi(x)\leq h(x)$ for all $x\in\partial B(x,r)$ (boundary of $B(x,r)$) we have that $\varphi(x)\leq h(x)$ holds for all $x\in B(x,r)$.

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

###### Definition.

Let $G\subset{\mathbb{C}}$ be a region and let $\varphi\colon G\to{\mathbb{R}}$ be a continuous function. $\varphi$ 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

 $\varphi(z)\leq\frac{1}{2\pi}\int_{0}^{2\pi}\varphi(z+re^{i\theta})d\theta,$

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

 $\varphi(z)\geq\frac{1}{2\pi}\int_{0}^{2\pi}\varphi(z+re^{i\theta})d\theta.$

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 $\varphi$ is subharmonic if and only if $-\varphi$ is superharmonic.

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

It is possible to relax the continuity statement above to take $\varphi$ 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 ${\mathbb{R}}^{2}$ instead of ${\mathbb{C}}$ since we never did use complex multiplication. In that case however we must rewrite the expression $z+re^{i\theta}$ in of the real and imaginary parts to get an expression in ${\mathbb{R}}^{2}$.

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

## References

• 1 John B. Conway. . Springer-Verlag, New York, New York, 1978.
• 2 Steven G. Krantz. , AMS Chelsea Publishing, Providence, Rhode Island, 1992.
 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