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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-27
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The punctured disc has an irregular boundary point and a continuous boundary datum with no harmonic solution

Statement refuted

A bounded plane domain need not solve the Dirichlet problem for every continuous boundary datum. The punctured disc Ω={z:0<∣z∣<1} with boundary values 0 on ∣z∣=1 and 1 at the puncture is a witness.

Facts & Assumptions

Given: The punctured disc Ω={0<∣z∣<1}, the boundary datum φ equal to 0 on ∣z∣=1 and 1 at 0.

[L1]

A bounded harmonic function on a punctured disc extends harmonically across the puncture (A bounded harmonic function near an isolated puncture extends harmonically).

[L2]

On the unit disc, the only continuous harmonic function with zero boundary values is the zero function (The bounded plane Dirichlet problem has at most one continuous harmonic solution).

[L3]

Perron lower functions satisfy a boundary limsup inequality, and Hφ is the upper-semicontinuous regularization of their pointwise supremum (The Perron lower family for continuous boundary data, The Perron envelope and its regularization).

[L4]

A point is regular when every continuous datum has the correct Perron-envelope limit there (Barriers and regular boundary points).

[L5]

A positive sum of subharmonic functions is subharmonic, and a subharmonic function with boundary limsup at most 0 on a bounded domain is at most 0 (Positive linear combinations and finite maxima preserve subharmonicity, A plane subharmonic function with an interior maximum is constant on its component).

Counterexample

technique · direct
1.1assume-contraL1

Suppose u were a continuous harmonic solution of this boundary-value problem on Ω. Then u is bounded on every punctured neighbourhood of 0 because it extends continuously to the puncture with value 1. By [L1], u extends to a harmonic function U on the full unit disc.

1.2L3L5given

We determine the Perron envelope directly. The constant 0 is a Perron lower function for φ, so Uφ≥0. Let v be any other lower function and fix ε>0. Its boundary limsup at the puncture gives v≤1+ε on all sufficiently small circles about 0. The boundary limsup at each point of the compact unit circle gives, by a finite subcover, some R<1 such that v≤ε whenever R≤∣z∣<1. For any 0<δ<R sufficiently small, compare v on δ<∣z∣<R with the harmonic function [L3, L5, given] hδ,R(z)=ε+log⁡(R/∣z∣)/log⁡(R/δ). The inner and outer boundary bounds give v−hδ,R≤0 on both circles; [L5] and the maximum principle therefore give v≤hδ,R throughout the annulus. For a fixed z with ∣z∣<R, letting δ↓0 yields v(z)≤ε. For any fixed z∈Ω, take R above ∣z∣ and sufficiently close to 1, then let ε↓0. Thus every lower function is at most 0, so Uφ=Hφ=0.

2.1L2step 1.1discharge-contradiction

The extension U still has boundary value 0 on the unit circle, so [L2] forces U≡0 on the closed unit disc. But then U(0)=0, contradicting the prescribed puncture value 1. Therefore no such harmonic solution exists.

3.1L4step 2.1step 1.2∎

In particular, Hφ(z)→0 as z→0 inside Ω, whereas φ(0)=1. By [L4] the puncture is irregular. Together with step 2.1, this proves both asserted failures for the given domain and datum.

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