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A bounded-domain Dirichlet Green function is unique and positive

Statement

Assume Countable Choice and n≥2. Let Ω⊆Rn be bounded, nonempty, open and connected, and suppose a Dirichlet Green function as in Dirichlet Green function for minus Laplacian exists. Then it is unique and GΩ(x,y)>0 for every distinct x,y∈Ω. No boundary differentiability is needed, and existence is not asserted.

Facts & Assumptions

Given: The objects and hypotheses in the statement, the kernel Φ fixed by Fundamental solution for the positive operator minus Laplacian, and correctors supplied by the Green-function definition.

[A1]

Countable Choice, written ACω, is an assumption of the Green and kernel conventions used here (The Axiom of Countable Choice (ACω)).

[F1]

For each pole, the Green definition supplies a corrector Hy∈C2(Ω)∩C(Ω‾) with boundary values Hy(z)=Φ(z−y) and GΩ(x,y)=Φ(x−y)−Hy(x) away from the pole (Dirichlet Green function for minus Laplacian).

[F2]

The kernel is Φ(x)=∣x∣2−n/((n−2)ωn−1) for n≥3 and Φ(x)=−(2π)−1log⁡∣x∣ for n=2, with ωn−1>0 (Fundamental solution for the positive operator minus Laplacian).

[F3]

A real C2 function on a bounded open set that is continuous on its closure and has Δu≥0 has its closure maximum on the boundary (Weak maximum principle for the laplacian).

[F4]

If instead Δu≤0, its closure minimum is on the boundary (Weak minimum principle for the laplacian).

[F5]

For every η>0 there is an integer N≥1 with 1/N<η (For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[F6]

Positive reciprocals reverse strict order: 0<a<b implies 0<b−1<a−1 (Inverses of positives are positive, and reciprocation reverses order).

[F7]

The natural logarithm is strictly increasing and onto R, and log⁡(1/s)=−log⁡s for s>0 (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F9]

If n≥2, deleting one point from a nonempty connected open subset of Rn leaves a nonempty, open, connected, path-connected set (Puncturing a connected open subset of Rn preserves path-connectedness for n≥2).

[F10]

A harmonic function on a connected domain that attains a global maximum or minimum in the domain is constant (Strong maximum principle for harmonic functions).

[F11]

For fixed y, GΩ(⋅,y) is harmonic away from y, extends continuously to Ω‾∖{y}, and has zero boundary trace (Dirichlet Green function for minus Laplacian).

[F12]

Mathematical induction applies to properties of natural numbers (The principle of mathematical induction).

[F13]

The order on R makes it a totally ordered field (The reals form a totally ordered field).

[F14]

Positive elements of an ordered field are closed under multiplication; in particular, a product of nonnegative reals is nonnegative (Ordered field, The reals form a totally ordered field).

[F15]

The Laplacian of a C2 function is the sum of its pure second partial derivatives (The Laplacian of a C2 function and of a C2 vector field).

[F17]

A C2 function is harmonic exactly when its Laplacian is zero (The Laplacian of a C2 function and of a C2 vector field).

[F16]

Total derivatives obey sum and scalar rules, and their coordinate partials are obtained by applying them to standard basis vectors; applying these facts twice gives linearity of each second partial of C2 functions (Sums and scalar multiples of totally differentiable maps are totally differentiable with the expected derivatives, A total derivative computes every directional derivative, and its matrix is the Jacobian).

Proof

technique · direct
1.1F1F3F4F15F16F17givenalgebra

Suppose G and G~ are two Green functions with correctors Hy and H~y. For fixed y, w=Hy−H~y is C2, and [F15]–[F16] give Δw=ΔHy−ΔH~y=0, so it is harmonic by [F17]; it is continuous on Ω‾ and zero on ∂Ω. The weak maximum principle [F3] gives w≤0, and the weak minimum principle [F4] gives w≥0. Hence w=0 and G(x,y)=G~(x,y) for all x≠y. This holds for every pole, proving uniqueness.

1.2F1givenchoose

Fix y∈Ω. Continuity of Hy at y gives r0>0 and M=∣Hy(y)∣+1 such that ∣Hy(x)∣≤M whenever ∣x−y∣<r0; shrink r0 if needed so B‾2(y,r0)⊂Ω.

1.3F2F5F6F8F12F13F14algebrachoose

Suppose n≥3, put m=n−2≥1 and C=1/((n−2)ωn−1)>0. For 0<r<1, induction on k∈N for the property 0<rk+1≤r starts with equality at k=0; if it holds at k, then rk+2=rk+1r>0 and rk+1−rk+2=rk+1(1−r)≥0 by [F13, F14], so it holds at k+1. Thus [F12] gives rm≤r. By [F8], r2−n=r−m=1/rm≥1/r. Given L>0, [F5] with η=C/L gives N≥1 with 1/N<C/L; [F6] then gives N>L/C. Thus 0<r<min⁡{1,1/N} implies Φ(r)=Cr−m≥C/r>CN>L. This proves Φ(r)→+∞ as r↓0 for every n≥3.

1.4F2F5F6F7algebrachoose

Suppose n=2 and put C=1/(2π)>0. For any L>0, surjectivity in [F7] gives s0>0 with log⁡s0>L/C. Apply [F5] to 1/s0 and then [F6] to obtain an integer N>s0. Whenever 0<r<1/N, [F6] gives 1/r>N>s0, so [F7] yields Φ(r)=Clog⁡(1/r)>Clog⁡s0>L. Hence Φ(r)→+∞ as r↓0 in dimension two as well.

2.1F1step 1.2step 1.3step 1.4algebrachoose

By steps 1.3 and 1.4, choose 0<ry<r0 so that Φ(x−y)>M whenever 0<∣x−y∣≤ry. Then GΩ(x,y)=Φ(x−y)−Hy(x)>0 on that punctured closed ball. This is the local strict positivity near the pole.

3.1F4F11step 2.1givencases

Let x∈Ω with ∣x−y∣>ry and choose 0<ε<min⁡{ry,∣x−y∣}. The set D=Ω∖B‾2(y,ε) is bounded, open, and contains x. A point outside ∂Ω∪S2(y,ε) has a neighborhood either contained in D or disjoint from D, so ∂D⊆∂Ω∪S2(y,ε). By [F11], GΩ(⋅,y) is harmonic on D and continuous on D‾. Its boundary values are zero on ∂Ω and positive on S2(y,ε) by step 2.1. The weak minimum principle [F4] therefore gives GΩ(x,y)≥0. Points with 0<∣x−y∣≤ry already have strict positivity by step 2.1, so GΩ(⋅,y)≥0 throughout Ω∖{y}.

4.1F9F10F11F15F16F17step 2.1step 3.1assume-contracontradictiondischarge-contradictionalgebra

By [F9], Ω∖{y} is a connected open set. The Green function is harmonic there by [F11], so its negative is harmonic by [F15]–[F17]. Step 3.1 gives GΩ≥0 there. Assume it vanishes at some x≠y [assume-contra]. Then −GΩ(⋅,y) attains its global maximum 0 at that interior point. By [F10] it is constant on the punctured domain, contradicting the strict positivity near y from step 2.1. Therefore GΩ(x,y)>0 whenever x≠y [contradiction, discharge-contradiction].

5.1A1F1F2F5step 1.3step 1.4cases∎

The argument treats n=2 and all n≥3 separately, excludes n=1 and dimension zero by the hypothesis, and makes no boundary smoothness or existence claim. Countable Choice is retained exactly because the preceding Green and kernel conventions assume it; the maximum principles and puncture argument add no choice principle, and the pointwise thresholds use only the Archimedean property. There is no iff assertion.

Source notes

Teschl §5.4 Theorem 5.21, printed p.124, establishes uniqueness for the classical Dirichlet problem. Lemma 5.23, printed pp.126–127, assumes a bounded connected domain, proves Green positivity by using the blow-up at the pole and a strong minimum principle, and notes that connectedness is required for positivity. The proof here derives the power/log blow-up and punctured-domain connectedness explicitly; it uses weak minimum on the bounded punctured domains and the strong maximum principle on Ω∖{y}.

Schmidt §2.8 remarks (2)–(3), printed pp.44–45, derives uniqueness from uniqueness of the harmonic corrector and the weak maximum principle, then derives nonpositivity under ΔF=δ0. Here Φ=−F, so that sign comparison supports nonnegativity only; strict positivity is proved above.

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