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Positive harmonic functions on a disc satisfy Harnack's inequality

Statement

Let $u$ be positive and harmonic on a neighbourhood of $\overline{D (a, R)}$ , and let $z$ satisfy $∣ z - a ∣ = ρ < R$ . Then

$\frac{R - ρ}{R + ρ} u (a) \leq u (z) \leq \frac{R + ρ}{R - ρ} u (a) .$

In particular, for every $r < R$ , the values of $u$ on $\overline{D (a, r)}$ are bounded above and below by fixed multiples of $u (a)$ .

Facts & Assumptions

Given: A positive harmonic function $u$ on a neighbourhood of $\overline{D (a, R)}$ and a point $z = a + ρ e^{i ϕ}$ with $0 \leq ρ < R$ .

[L1]

The Poisson representation on the radius- $R$ circle is $u (z) = \frac{1}{2 π} \int_{0}^{2 π} \frac{R^{2} - ρ^{2}}{R^{2} - 2 R ρ \cos (ϕ - t) + ρ^{2}} u (a + R e^{i t}) d t$ (A harmonic function is recovered from its values on any containing circle by the Poisson formula).

[L2]

The center value is the average on the radius- $R$ circle: $u (a) = \frac{1}{2 π} \int_{0}^{2 π} u (a + R e^{i t}) d t$ (Plane harmonic functions satisfy the mean-value property).

Proof

technique · direct

1.1

For every $t$ , the denominator in [L1] lies between $(R - ρ)^{2}$ and $(R + ρ)^{2}$ , so the Poisson kernel there satisfies $\frac{R - ρ}{R + ρ} \leq \frac{R^{2} - ρ^{2}}{R^{2} - 2 R ρ \cos (ϕ - t) + ρ^{2}} \leq \frac{R + ρ}{R - ρ} .$

givenalgebra

2.1

Multiplying the bounds of step 1.1 by the positive boundary values $u (a + R e^{i t})$ and integrating, [L1] and [L2] give $\frac{R - ρ}{R + ρ} u (a) \leq u (z) \leq \frac{R + ρ}{R - ρ} u (a) .$

step 1.1L1L2

3.1

The constants in step 2.1 depend only on $ρ / R$ , so the same bound holds for every $∣ z - a ∣ \leq r < R$ after replacing $ρ$ by $r$ .

step 2.1algebra∎

Depends on

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Sources

Sigurdur Helgason, MIT 18.112 Lecture 16: Harmonic Functions (standard reference, not scraped)