Harnack's Inequality: Positive Solutions Can't Vary Too Wildly

What the inequality claims

Let Ω ⊂ ℝⁿ be open and let u ≥ 0 be harmonic on Ω, meaning Δu = 0 (equivalently ∑ᵢ ∂²u/∂xᵢ² = 0). The classical Harnack inequality says: for every compact set K ⋐ Ω (contained with positive distance to ∂Ω), there is a constant C = C(n, K, Ω), independent of u, such that

sup_{x∈K} u(x) ≤ C · inf_{x∈K} u(x).

The cleanest special case is a ball. If u ≥ 0 is harmonic on the ball B(x₀, R) ⊂ ℝⁿ, then for any x with |x − x₀| = r < R,

Rⁿ⁻² · (R−r)/(R+r)ⁿ⁻¹ · u(x₀) ≤ u(x) ≤ Rⁿ⁻² · (R+r)/(R−r)ⁿ⁻¹ · u(x₀).

Two features are essential and easy to miss. First, u must be nonnegative throughout Ω — the bound is false for sign-changing harmonic functions. Second, C is uniform: the same constant controls every nonnegative harmonic function on Ω at once. That single quantifier ordering — one C for all u — is the entire power of the theorem.

The picture: positivity is a straitjacket

Harmonic functions satisfy the mean-value property: u(x) equals its average over any sphere or ball centered at x. So the value at a point is a democratic vote of surrounding values. Now impose u ≥ 0. If u were tiny at one interior point p but huge at a nearby point q, the average over a ball around q would have to be huge — yet that ball overlaps the region near p where u is small, and no negative values are available to compensate. Positivity forbids the cancellation that would let a function dip low and spike high in a small neighborhood.

The vivid image: think of u as brightness of a glowing medium in equilibrium (Δu = 0 is steady-state heat/electrostatics). Harnack says a positive glow can't have a pinprick-dark spot right next to a blazing one. The medium's equilibrium smooths ratios, not just differences. As you approach the boundary ∂Ω the constant C blows up — near the edge, extra freedom (Poisson kernel concentration) returns, which is exactly why K must stay compactly inside.

Key idea of the proof

Classical (mean-value) proof. Fix x, y in a ball B(x₀,R) and use the mean-value property over balls. If B(x, r₁) ⊂ B(y, r₂) ⊂ Ω, then since u ≥ 0, u(x) = ⨍_{B(x,r₁)} u ≤ (|B(y,r₂)|/|B(x,r₁)|) ⨍_{B(y,r₂)} u = (r₂/r₁)ⁿ u(y). The nonnegativity lets us enlarge the averaging ball from x's ball to y's ball and only gain — dropping no mass. Chaining such comparisons across a compact K (a covering/connectedness argument) yields sup_K u ≤ C inf_K u with C depending on how many overlapping balls tile K.

Divergence-form proof (Moser, 1961). For Lu = div(A(x)∇u) = 0 with only measurable, uniformly elliptic coefficients (λI ≤ A ≤ ΛI), there is no mean-value property. Moser's insight: test the weak equation with powers uᵖ, derive reverse-Hölder / energy estimates, and run Moser iteration to bound sup u ≤ C(‖u‖_{Lᵖ}) and (via powers p → 0 and the John–Nirenberg lemma controlling log u ∈ BMO) inf u ≥ c(‖u‖_{Lᵖ}). Comparing gives the Harnack estimate. Positivity enters by allowing the substitution v = log u and negative test powers.

Worked example and the Liouville payoff

Example. Let u ≥ 0 be harmonic on all of ℝⁿ. Apply the ball form on B(0,R) at the center: for |x| = r, u(x) ≤ Rⁿ⁻²·(R+r)/(R−r)ⁿ⁻¹·u(0). Now fix x and let R → ∞. Every factor tends to 1, so u(x) ≤ u(0). The same argument with x and 0 swapped gives u(0) ≤ u(x). Hence u ≡ u(0): every nonnegative harmonic function on ℝⁿ is constant — the Liouville theorem, obtained for free from the scaling of the Harnack constant. (The usual Liouville theorem assumes boundedness; Harnack strengthens it to one-sided boundedness.)

Simplest concrete check. On ℝ, harmonic means u″ = 0, so u(x) = ax + b. Nonnegativity on all of ℝ forces a = 0 (a line eventually goes negative), giving u = b constant — consistent. On a bounded interval [−1,1] a positive line like u(x) = x + 2 satisfies Harnack with sup/inf = 3/1 = 3, a finite ratio, exactly as promised.

Why the hypotheses matter — and what breaks

Positivity is not optional. Drop u ≥ 0 and the inequality collapses. On ℝ², u(x,y) = x is harmonic and vanishes on the y-axis while being positive to the right; sup_K u / inf_K u = +∞ for any K straddling the axis where inf ≤ 0. Even keeping u > 0 but only on part of the domain fails — you need nonnegativity on all of Ω, not just on K, because the averaging balls reach outside K.

Compact containment K ⋐ Ω is essential. The half-space harmonic function u(x,y) = y > 0 on the upper half-plane has inf → 0 as you approach the boundary while sup stays positive, so no finite C works up to ∂Ω. That is why C = C(n,K,Ω) degenerates as dist(K, ∂Ω) → 0.

Ellipticity / connectedness. For general Lu = 0, the ellipticity ratio Λ/λ must be finite (Moser's C depends on it); degenerate coefficients can break it. And Ω must be connected — on two disjoint balls u can be 1 on one and 10⁶ on the other. Connections: Harnack is the engine of the De Giorgi–Nash–Moser theorem (Hölder continuity of solutions with merely measurable coefficients) and, via Ricci lower bounds, of geometric comparison theory.

Applications and significance

Harnack's inequality is a foundational tool, not just a curiosity. (1) Regularity. Oscillation control osc_K u ≤ (1 − 1/C)·osc_{K'} u iterates across dyadic scales to give Hölder continuity — this is the heart of the De Giorgi (1957) and Nash (1958) solutions of Hilbert's 19th problem, later unified by Moser. Merely measurable coefficients still yield C^{0,α} solutions, purely because positive solutions can't oscillate too fast.

(2) Liouville and rigidity. As shown, Harnack instantly yields Liouville theorems on ℝⁿ and, via Yau (1975) and Li–Yau (1986) gradient estimates, on manifolds with Ricci ≥ −K, controlling positive solutions of Δu = 0 and the heat equation. (3) Probability. Harmonic functions are expectations of Brownian motion; the parabolic Harnack inequality is equivalent to two-sided Gaussian heat-kernel bounds and to the conjunction of a Poincaré inequality with volume doubling (Grigor'yan, Saloff-Coste). (4) Free boundaries and geometric flows use Harnack to compare solutions and extract convergence. Wherever positivity meets a second-order equation, Harnack quantifies how gently the solution must vary.

Harnack's inequality across settings: hypotheses, the operator, and who proved it

Setting	Equation / operator	Key hypotheses	Constant C depends on	First proof
Harmonic, ℝⁿ	Δu = 0	u ≥ 0, domain Ω	n and geometry (K, Ω)	Harnack 1887 (n=2); classical for n≥2
Divergence-form elliptic	div(A(x)∇u) = 0	u ≥ 0, A measurable, uniformly elliptic λI ≤ A ≤ ΛI	n, ellipticity ratio Λ/λ, geometry	Moser 1961
Parabolic	∂ₜu = div(A(x,t)∇u)	u ≥ 0, uniform parabolicity	n, ellipticity ratio, time-lag between sup/inf	Moser 1964
Riemannian manifold	Δ_g u = 0	u ≥ 0, Ric ≥ −K (lower Ricci bound)	n, K, radius R	Yau 1975 / Li–Yau 1986
Non-example	Δu = 0 with sign change	u NOT ≥ 0	— (inequality fails)	Counterexample (see hypotheses)