---
tag: YNH832D50W
kind: statement
role: theorem
name: Harnack's inequality
msc: [31A05, 31B05]
uses: [0DEB0JMN0W, W19567AVYG]
about: [0DEB0JMN0W]
sources:
  - url: https://en.wikipedia.org/wiki/Harnack%27s_inequality
same_as:
  wikidata: Q868279
  wikipedia: Harnack's inequality
---

## Statement

Let $a \in \C$, let $R > 0$, and let $u$ be a [[harmonic function|0DEB0JMN0W]] on the disc $\set{z \in \C : \abs{z - a} < R}$ with $u \ge 0$.
Then for every $z$ with $r = \abs{z - a} < R$,
$$\frac{R - r}{R + r}\, u(a) \le u(z) \le \frac{R + r}{R - r}\, u(a).$$

## Remarks

For $\rho$ with $r < \rho < R$, the Poisson formula on the disc of radius $\rho$ gives $u(z)$ as an average of $u$ on the circle $\abs{w - a} = \rho$ against the [[Poisson kernel|W19567AVYG]] $(\rho^2 - r^2)/\abs{w - z}^2$.
Since $\rho - r \le \abs{w - z} \le \rho + r$, this kernel lies between $(\rho - r)/(\rho + r)$ and $(\rho + r)/(\rho - r)$, and the average of $u$ over the circle is $u(a)$.
Letting $\rho \to R$ gives the inequality.
The hypothesis $u \ge 0$ is necessary: $u(z) = \operatorname{Re} z$ on the unit disc has $u(0) = 0$ but takes nonzero values.
In $\R^n$ with $n \ge 2$, the same argument gives $\dfrac{R^{n-2}(R - r)}{(R + r)^{n-1}}\, u(a) \le u(x) \le \dfrac{R^{n-2}(R + r)}{(R - r)^{n-1}}\, u(a)$ for $\abs{x - a} = r < R$.

## Explanations

### L1 human

A nonnegative harmonic function cannot change too fast.
If you know its value at the centre of a disc, you know its values at other points up to a factor.
This factor depends only on how close the points are to the edge.
For example, at half the radius the value is between one third and three times the value at the centre.

### L2 human

A chain of discs gives the general form on a connected open set $\Omega$ with a compact subset $K$.
There is a constant $C$ such that $\sup_K u \le C \inf_K u$ for every nonnegative harmonic $u$ on $\Omega$.
This is the key estimate behind [[Harnack's principle|P15K8RKHDV]] on increasing sequences of harmonic functions.
It is used in Perron's method for the [[Dirichlet problem|8D56GQ4AG8]] and in the study of positive harmonic functions, such as the [[Herglotz representation theorem|9ZBAX4ZPX0]].
Moser extended Harnack's inequality to solutions of elliptic equations in divergence form with measurable coefficients.