Rigidity at an upper bound equal to the average
ProvedMeasureTheory.eqOn_of_le_setAveragemaximum-principlemean-value-propertymeasure-theory
Let μ be a measure that assigns positive measure to every nonempty open set. Let s be an open set of finite measure and let f be a continuous, integrable real function on s. If f(x) is at most its normalized average over s for every x in s, then f is identically equal to that average on s. This includes the empty set. The conclusion follows because the nonnegative continuous deficit from the average has integral zero.
Preamble
import Mathlib.MeasureTheory.Integral.Average import Mathlib.MeasureTheory.Measure.OpenPos open MeasureTheory MeasureTheory.Measure Set open scoped ENNReal set_option autoImplicit false
Formal statement
theorem MeasureTheory.eqOn_of_le_setAverage {X : Type*} [TopologicalSpace X] [MeasurableSpace X]
[OpensMeasurableSpace X] {μ : Measure X} [IsOpenPosMeasure μ]
{s : Set X} {f : X → ℝ} (hs : IsOpen s) (hfinite : μ s ≠ ∞)
(hc : ContinuousOn f s) (hi : IntegrableOn f s μ)
(hle : ∀ x ∈ s, f x ≤ ⨍ y in s, f y ∂μ) :
EqOn f (fun _ => ⨍ y in s, f y ∂μ) s := by sorrySource
Hunter, Notes on Partial Differential Equations, revised 6/18/2014, https://math.ucdavis.edu/~hunter/pdes/pde_notes.pdf, p. 26, proof of Theorem 2.13: equality in the mean-value bound forces constancy on a ball. This theorem isolates that measure-theoretic argument and generalizes it to finite-measure open sets and measures positive on nonempty open sets.