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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 with boundary values on and at the puncture is a witness.
Facts & Assumptions
Given: The punctured disc , the boundary datum equal to on and at .
A bounded harmonic function on a punctured disc extends harmonically across the puncture (A bounded harmonic function near an isolated puncture extends harmonically).
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).
Perron lower functions satisfy a boundary limsup inequality, and is the upper-semicontinuous regularization of their pointwise supremum (The Perron lower family for continuous boundary data, The Perron envelope and its regularization).
A point is regular when every continuous datum has the correct Perron-envelope limit there (Barriers and regular boundary points).
A positive sum of subharmonic functions is subharmonic, and a subharmonic function with boundary limsup at most on a bounded domain is at most (Positive linear combinations and finite maxima preserve subharmonicity, A plane subharmonic function with an interior maximum is constant on its component).
Counterexample
Suppose were a continuous harmonic solution of this boundary-value problem on . Then is bounded on every punctured neighbourhood of because it extends continuously to the puncture with value . By [L1], extends to a harmonic function on the full unit disc.
We determine the Perron envelope directly. The constant is a Perron lower function for , so . Let be any other lower function and fix . Its boundary limsup at the puncture gives on all sufficiently small circles about . The boundary limsup at each point of the compact unit circle gives, by a finite subcover, some such that whenever . For any sufficiently small, compare on with the harmonic function [L3, L5, given] . The inner and outer boundary bounds give on both circles; [L5] and the maximum principle therefore give throughout the annulus. For a fixed with , letting yields . For any fixed , take above and sufficiently close to , then let . Thus every lower function is at most , so .
The extension still has boundary value on the unit circle, so [L2] forces on the closed unit disc. But then , contradicting the prescribed puncture value . Therefore no such harmonic solution exists.
In particular, as inside , whereas . By [L4] the puncture is irregular. Together with step 2.1, this proves both asserted failures for the given domain and datum.
Depends on
- A bounded harmonic function near an isolated puncture extends harmonically
- The bounded plane Dirichlet problem has at most one continuous harmonic solution
- The Perron lower family for continuous boundary data
- The Perron envelope and its regularization
- Barriers and regular boundary points
- Positive linear combinations and finite maxima preserve subharmonicity
- A plane subharmonic function with an interior maximum is constant on its component
Used by
Dependency tree · two levels
26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Boris Khoruzhenko, Potential Theory lecture notes (standard reference, not scraped)