Harnack's Inequality and Principle

Harnack's inequality provides quantitative control on harmonic functions, bounding the ratio of values at different points. Harnack's principle extends this to monotone sequences.

Statement

Theorem8.13Harnack's inequality

Let $u$ be harmonic and non-negative on the disk $D(z_0, R)$ . Then for $z$ with $|z - z_0| = r < R$ :

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

Proof

By the Poisson integral formula:

$u(z) = \frac{1}{2\pi}\int_0^{2\pi} P_r(\theta)\,u(z_0 + Re^{i\theta})\,d\theta$

where $P_r(\theta) = \frac{R^2 - r^2}{R^2 - 2Rr\cos\theta + r^2}$ .

Since $R^2 - 2Rr\cos\theta + r^2 \leq (R+r)^2$ and $\geq (R-r)^2$ :

$\frac{R-r}{R+r} \leq P_r(\theta) \cdot \frac{1}{1} \leq \frac{R+r}{R-r}.$

More precisely: $\frac{R^2-r^2}{(R+r)^2} \leq P_r(\theta) \leq \frac{R^2-r^2}{(R-r)^2}$ , i.e., $\frac{R-r}{R+r} \leq P_r(\theta) \leq \frac{R+r}{R-r}$ .

Since $u \geq 0$ on the boundary:

$u(z) \leq \frac{R+r}{R-r}\cdot\frac{1}{2\pi}\int_0^{2\pi} u(z_0+Re^{i\theta})\,d\theta = \frac{R+r}{R-r}\,u(z_0)$

by the mean value property. Similarly for the lower bound. $\blacksquare$

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Harnack's Principle

Theorem8.14Harnack's principle

Let $\{u_n\}$ be a monotone increasing sequence of harmonic functions on a connected domain $D$ . Then either:

$u_n(z) \to +\infty$ for all $z \in D$ (uniformly on compact subsets), or
$u_n \to u$ uniformly on compact subsets, where $u$ is harmonic on $D$ .

Proof

Let $S = \{z \in D : \lim u_n(z) < \infty\}$ . If $z_0 \in S$ , choose a disk $D(z_0, R) \subset D$ . For $|z - z_0| \leq R/2$ , Harnack's inequality gives:

$0 \leq u_m(z) - u_n(z) \leq 3(u_m(z_0) - u_n(z_0)) \to 0$

as $m, n \to \infty$ . So $\{u_n\}$ converges uniformly on $D(z_0, R/2)$ , and all points of $D(z_0, R/2)$ belong to $S$ . Thus $S$ is open.

Similarly, $D \setminus S$ is open (if $u_n(z_0) \to \infty$ , Harnack gives $u_n(z) \geq \frac{1}{3}u_n(z_0) \to \infty$ on $D(z_0, R/2)$ ).

By connectedness, $S = \emptyset$ or $S = D$ . In the latter case, the limit is harmonic by the theorem that uniform limits of harmonic functions are harmonic. $\blacksquare$

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Applications

ExampleHarnack distance and metric

The Harnack distance on a domain $D$ is defined by: $d_H(z_1, z_2) = \log\sup\{u(z_1)/u(z_2)\}$ where the supremum is over all positive harmonic functions $u$ on $D$ . Harnack's inequality gives explicit bounds on this distance in terms of the Euclidean distance and the distance to the boundary.

ExampleConvergence of harmonic series

Consider $u(z) = \sum_{n=1}^\infty v_n(z)$ where each $v_n$ is harmonic and non-negative. The partial sums form an increasing sequence of harmonic functions. By Harnack's principle, either $u \equiv \infty$ or $u$ is harmonic. If the series converges at one point, it converges everywhere and the sum is harmonic.

RemarkHarnack inequality in higher dimensions

Harnack's inequality generalizes to harmonic functions on $\mathbb{R}^n$ : on a ball $B(x_0, R)$ , $\frac{(R-r)R^{n-2}}{(R+r)^{n-1}} u(x_0) \leq u(x) \leq \frac{(R+r)R^{n-2}}{(R-r)^{n-1}} u(x_0)$ . The principle and its consequences extend verbatim.