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The harmonic affine extension minimises the Dirichlet energy
Example
Example. Assume the Axiom of Choice (The Axiom of Choice). Let , , be a bounded domain and let be affine on , so that and is constant. Then is the unique minimiser of the Dirichlet energy on the affine trace class (The trace operator on a bounded domain): for every , and with equality if and only if almost everywhere, and then almost everywhere by the Poincare inequality on (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).
Facts & Assumptions
Given: The Axiom of Choice; a bounded domain , an affine function on (so is a constant field and ), the Dirichlet energy , and the affine trace class .
is nonempty, convex and weakly closed, and for every by the kernel description (The kernel of the trace is the closure of the test functions, The trace operator on a bounded domain).
Poincare's inequality on : if almost everywhere for , then (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).
Under the Axiom of Choice, the ultrafilter lemma, Dependent Choice and Hahn--Banach, the Dirichlet principle also identifies an affine-class minimiser with the unique weak solution (The Dirichlet principle for the Poisson equation). In the zero-forcing case here, is weakly harmonic with trace , since is constant and for every (The notation and the reserved zero-boundary symbol).
Verification
Orthogonality of the cross term. For one has by [F1], and is a constant field, so : each component of has vanishing integral, because is the -limit of functions in and for those the integral of each partial derivative vanishes by integration by parts against the smooth constant field (Divergence on a bounded C1 Euclidean domain). Passage to the limit is valid since by Holder (Holder's inequality for integrals, including the endpoint cases).
The energy identity. Expanding the square, , and integrating with step 1.1 gives .
Equality case. Equality holds exactly when almost everywhere, which by [F2] forces , that is almost everywhere.
is the minimiser. By steps 2.1 and 3.1 every satisfies with equality only for ; since , it is the unique minimiser of on . This direct completion-of-the-square argument proves the example's claim. Under the additional choice hypotheses stated in [F3], the general Dirichlet principle also identifies this minimiser with the weak solution; the weak harmonicity of was checked in [F3].
Depends on
- The Dirichlet principle for the Poisson equation
- The kernel of the trace is the closure of the test functions
- The $L^p$ trace operator on a bounded $C^1$ domain
- The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction
- The notation $H^k$ and the reserved zero-boundary symbol
- The Axiom of Choice
- Zero-boundary Sobolev space as a norm closure
- Divergence on a bounded C1 Euclidean domain
- Holder's inequality for integrals, including the endpoint cases
Used by
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Sources
- Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)
- Riccardo Cristoferi, Calculus of Variations: Lecture Notes, Carnegie Mellon University 2016 (complete 133-page notes) (standard reference, not scraped)