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Uniqueness for harmonic functions on a punctured half-space with zero trace, decaying at the pole and at infinity

Statement

Let H:={x∈R3:x3>0} be the open upper half-space, let a>0 and p:=ae3∈H, where e3:=(0,0,1). Let w:H‾∖{p}→R be a function (closure as in Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) with:

(i) w is continuous (Continuity of a map between metric spaces, at a point and globally, in the ε-δ form) on H‾∖{p} and of class C2 with Δw=0 on H∖{p} (The Laplacian of a C2 function and of a C2 vector field, Ck Euclidean maps and diffeomorphisms);

(ii) w=0 on ∂H={x3=0};

(iii) ∣x−p∣w(x)→0 as x→p, x≠p;

(iv) w(x)→0 as ∣x∣→∞ within H: for every ε>0 there is R>0 with ∣w(x)∣≤ε for all x∈H∖B‾R(0).

Then w=0 on H∖{p}.

The conclusion is a uniqueness statement for the stated class only: no removable-singularity theorem, no pole condition beyond the sub-Coulomb growth (iii), no boundary regularity of ∂H beyond the flat graph, and no claim about w outside H are asserted.

Facts & Assumptions

Given: The open upper half-space H={x3>0}, the point p=ae3 with a>0, and a function w:H‾∖{p}→R satisfying (i)-(iv) of the statement. Write Br(q) for the open ball and use the Euclidean norm and metric on R3.

[F1]

The weak maximum principle: if Ω⊆Rn, n≥2, is bounded, nonempty and open and u∈C2(Ω)∩C(Ω‾) satisfies Δu≥0 on Ω, then max⁡Ω‾u=max⁡∂Ωu (Weak maximum principle for the laplacian).

[F2]

A C2 function is harmonic where Δu=0; Δ is defined as the sum of the pure second partial derivatives, is linear on C2 functions and annihilates constants (The Laplacian of a C2 function and of a C2 vector field, Ck Euclidean maps and diffeomorphisms).

[F3]

For n=3 the fundamental solution is Φ(x)=∣x∣−1/ω2 with ω2>0, and for every pole q the function x↦Φ(x−q) is smooth and harmonic on R3∖{q}; the function bq(x):=∣x−q∣−1=ω2 Φ(x−q) is therefore a positive constant multiple of a harmonic function, hence C∞ and harmonic off q, and it is continuous wherever it is defined (Fundamental solution for the positive operator minus Laplacian, The Laplace fundamental solution is harmonic off its pole, The Laplacian of a C2 function and of a C2 vector field).

[F5]
[F6]

The boundary of the open set U is ∂U=U‾∖U; for an open set the interior is the set itself (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[F7]

Unwinding hypothesis (iii): for every ε>0 there is δ>0 with ∣w(x)∣≤ε/∣x−p∣ for all x in H‾∖{p} with 0<∣x−p∣<δ (Continuity of a map between metric spaces, at a point and globally, in the ε-δ form).

[F8]

Hypothesis (iv) says exactly that for every ε>0 there is ρ>0 with ∣w(x)∣≤ε for all x∈H with ∣x∣>ρ (Open ball, closed ball and sphere in a metric space).

[F9]

Countable Choice holds in the ambient measure conventions inherited from the kernel supplier; the argument below is pointwise and uses no choice principle (The Axiom of Countable Choice (ACω)).

Proof

technique · direct
1.1givenF3F7F8

Fix y∈H∖{p} and η>0. By [F8] there is ρ>0 with ∣w∣≤η on H∩{∣x∣>ρ}; put R:=max⁡{ρ,2a,∣y∣}+1, so that R>∣p∣+a, R>∣y∣ and ∣w∣≤η on H∩{∣x∣≥R}. By [F7] there is δ0>0 with ∣w(x)∣≤η/∣x−p∣ whenever 0<∣x−p∣<δ0; put δ:=12min⁡{δ0,a/2,∣y−p∣,R−∣p∣}, so 0<δ<a/2, δ<∣y−p∣, δ<R−∣p∣ and δ<δ0. Define the open set U:=H∩BR(0)∩{x:∣x−p∣>δ}, the barrier b(x):=∣x−p∣−1, and v±:=±w−η(1+b) on U.

2.1F4F5F8step 1.1

U is bounded, nonempty and open, and y∈U; explicitly, R>∣y∣ and δ<∣y−p∣ put y in U. The closure satisfies U‾⊆H‾∩BR(0)‾∩{x:∣x−p∣≥δ}: each of H‾, BR(0)‾ and {x:∣x−p∣≥δ} is closed (the last is the complement of the open ball Bδ(p)) and contains U, so [F5] applies factor by factor. In particular ∣x−p∣≥δ>0 on U‾, so p∉U‾ and U‾⊆H‾∖{p}.

3.1F2F3F4step 1.1step 2.1

Each v± lies in C2(U)∩C(U‾) and satisfies Δv±=0 on U. Indeed U⊆H∖{p} by construction, where w is C2 and Δw=0, while b=ω2Φ(⋅−p) is C∞ and harmonic off p [F3]; constants are C∞; Δ is linear and annihilates constants [F2], so Δv±=±Δw−η(0+Δb)=0. Continuity on U‾ holds because w is continuous there by (i) and step 2.1, and b is continuous on U‾ by [F3] and [F4], the bound ∣x−p∣≥δ giving ∣b(x)−b(y)∣≤∣x−y∣/δ2; sums and scalar multiples of continuous functions are continuous [F4].

3.2F6givenstep 1.1step 2.1

Each v± is nonpositive on ∂U. Let x∈∂U=U‾∖U [F6]. Since x∈U‾ and x∉U, at least one of x3=0, ∣x∣=R, ∣x−p∣=δ holds, by the description of U‾ in step 2.1 and the definition of U. If x3=0 then w(x)=0 by (ii), so v±(x)=−η(1+b(x))<0. Otherwise x3>0, and since at least one of the three conditions holds, ∣x∣=R or ∣x−p∣=δ (or both); if ∣x∣=R then ∣w(x)∣≤η by the choice of R in step 1.1, so v±(x)≤η−η(1+b(x))=−ηb(x)<0, and if ∣x−p∣=δ then ∣w(x)∣≤η/δ=η/∣x−p∣ by the choice of δ in step 1.1 (and x∈H‾∖{p} by step 2.1), so v±(x)≤ηb(x)−η(1+b(x))=−η<0.

4.1F1step 3.1step 3.2

The weak maximum principle applies to each v± on the bounded nonempty open set U [step 2.1, step 3.1] and gives max⁡U‾v±=max⁡∂Uv±≤0 by step 3.2, so v±≤0 on U‾. In particular at y∈U⊆U‾, ±w(y)≤η(1+∣y−p∣−1).

5.1givenF9step 1.1step 4.1∎

Since η>0 was arbitrary in step 1.1 and the bound of step 4.1 holds for every η>0, the real number t:=∣w(y)∣ satisfies t≤η(1+∣y−p∣−1) for every η>0; if t>0 then η:=t/(2(1+∣y−p∣−1))>0 gives t≤t/2<t, a contradiction, so t=0, that is w(y)=0. As y∈H∖{p} was arbitrary, w=0 on H∖{p}. This proves uniqueness in the stated sub-Coulomb class; no removable-singularity theorem, no boundedness or growth assumption beyond (iii)-(iv), and no boundary regularity beyond the flat trace (ii) was used, and the only ambient choice convention is the one inherited from the kernel interfaces [F9].

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