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TauCeti.Analysis.Sobolev.W1p.InteriorMorrey

Morrey's embedding for W^{1,p}(Ω) in the interior of a domain #

Let Ω be an open subset of a finite-dimensional real inner product space of dimension n, with an additive Haar measure μ, and let n < p ≤ ∞. This file proves that every u ∈ W^{1,p}(Ω) is locally Hölder continuous in Ω with exponent α = 1 - n / p, in the form of the local Morrey inequality

|u*(x) - u*(y)| ≤ C ‖∇u‖_{Lᵖ(B(x, 2|x - y|))} |x - y|^α whenever closedBall x (2|x - y|) ⊆ Ω,

where u* is the precise representative of u (the limit of its averages over shrinking balls, which agrees with u almost everywhere on Ω by TauCeti.W1p.ae_eq_preciseRepresentative) and C depends only on n, p and μ. No regularity of ∂Ω is needed, and the endpoint p = ∞ is included: there α = 1, so a function of W^{1,∞}(Ω) is locally Lipschitz in Ω.

Morrey's embedding on the whole space, TauCeti.W1p.exists_holderWith_ae_eq_value, is proved by approximation with test functions, which are dense in W^{1,p}(ℝⁿ) but not in W^{1,p}(Ω). Here the estimate is instead a consequence of Morrey's Dirichlet growth theorem (TauCeti.W1p.dist_preciseRepresentative_le_of_setIntegral_norm_gradient_le): by Hölder's inequality on a ball B(z, r),

∫_{B(z, r)} ‖∇u‖ ≤ μ(B(z, r))^(1 - 1/p) ‖∇u‖_{Lᵖ(B(z, r))} = ω^(1 - 1/p) r^(n - 1 + α) ‖∇u‖_{Lᵖ},

with ω = μ(B(0, 1)), which is exactly the growth condition of that theorem.

Main declarations #

References #

theorem TauCeti.W1p.setIntegral_ball_norm_gradient_le {E : Type u_1} [MeasurableSpace E] [NormedAddCommGroup E] [InnerProductSpace ℝ E] [FiniteDimensional ℝ E] [BorelSpace E] {mu : MeasureTheory.Measure E} [mu.IsAddHaarMeasure] {p : ENNReal} [Fact (1 ≤ p)] {Omega : TopologicalSpace.Opens E} (u : ↥(W1p mu Omega p)) {z : E} {r : ℝ} (hball : Metric.ball z r ⊆ ↑Omega) :
∫ (w : E) in Metric.ball z r, ‖↑↑(gradient u) w‖ ∂mu ≤ mu.real (Metric.ball 0 1) ^ (1 - 1 / p.toReal) * r ^ (↑(Module.finrank ℝ E) * (1 - 1 / p.toReal)) * (MeasureTheory.eLpNorm (↑↑(gradient u)) p (mu.restrict (Metric.ball z r))).toReal

The gradient of a W^{1,p} function grows like a power of the radius. If B(z, r) ⊆ Ω, then by Hölder's inequality

∫_{B(z, r)} ‖∇u‖ ≤ ω^(1 - 1/p) r^(n (1 - 1/p)) ‖∇u‖_{Lᵖ(B(z, r))},

where n is the dimension and ω is the measure of the unit ball. For p > n the exponent n (1 - 1/p) is n - 1 + α with α = 1 - n / p, which is the growth condition of Morrey's Dirichlet growth theorem.

theorem TauCeti.W1p.exists_dist_preciseRepresentative_le {E : Type u_1} [MeasurableSpace E] [NormedAddCommGroup E] [InnerProductSpace ℝ E] [FiniteDimensional ℝ E] [BorelSpace E] {mu : MeasureTheory.Measure E} [mu.IsAddHaarMeasure] {p : ENNReal} [Fact (1 ≤ p)] (hp : ↑(Module.finrank ℝ E) < p) :
∃ (C : ℝ), 0 < C ∧ ∀ {Omega : TopologicalSpace.Opens E} (u : ↥(W1p mu Omega p)) {x y : E}, Metric.closedBall x (2 * dist x y) ⊆ ↑Omega → dist (MeasureTheory.preciseRepresentative mu (↑↑(value u)) x) (MeasureTheory.preciseRepresentative mu (↑↑(value u)) y) ≤ C * (MeasureTheory.eLpNorm (↑↑(gradient u)) p (mu.restrict (Metric.ball x (2 * dist x y)))).toReal * dist x y ^ ↑(1 - ↑(Module.finrank ℝ E) / p).toNNReal

The local Morrey inequality. Let n < p ≤ ∞, where n is the dimension, and let α = 1 - n / p. There is a constant C > 0, depending only on n, p and the Haar measure μ, such that the following holds. Let u ∈ W^{1,p}(Ω) and let x, y be points with closedBall x (2 |x - y|) ⊆ Ω. Then the precise representative u* of u satisfies

|u*(x) - u*(y)| ≤ C ‖∇u‖_{Lᵖ(B(x, 2 |x - y|))} |x - y|^α.

theorem TauCeti.W1p.exists_holderOnWith_preciseRepresentative {E : Type u_1} [MeasurableSpace E] [NormedAddCommGroup E] [InnerProductSpace ℝ E] [FiniteDimensional ℝ E] [BorelSpace E] {mu : MeasureTheory.Measure E} [mu.IsAddHaarMeasure] {p : ENNReal} [Fact (1 ≤ p)] (hp : ↑(Module.finrank ℝ E) < p) :
∃ (C : ℝ), 0 < C ∧ ∀ {Omega : TopologicalSpace.Opens E} (u : ↥(W1p mu Omega p)) {K : Set E} {R : ℝ}, (∀ x ∈ K, ∀ y ∈ K, dist x y ≤ R) → Metric.cthickening (2 * R) K ⊆ ↑Omega → HolderOnWith (C * ‖gradient u‖).toNNReal (1 - ↑(Module.finrank ℝ E) / p).toNNReal (MeasureTheory.preciseRepresentative mu ↑↑(value u)) K

Morrey's embedding in the interior of a domain. Let n < p ≤ ∞, where n is the dimension, and let α = 1 - n / p. There is a constant C > 0, depending only on n, p and the Haar measure μ, such that the following holds. Let u ∈ W^{1,p}(Ω), and let K be a set of diameter at most R whose closed 2R-thickening lies in Ω. Then the precise representative of u is α-Hölder continuous on K with constant C ‖∇u‖_{Lᵖ(Ω)}.

Continuity of W^{1,p} functions for p above the dimension. If n < p ≤ ∞, where n is the dimension, the precise representative of any u ∈ W^{1,p}(Ω), which agrees with u almost everywhere on Ω, is continuous on Ω.

The continuous precise representative of a W^{1,∞}(Ω) function is bounded pointwise on Ω by the essential supremum of its value.

The endpoint local Morrey estimate. A W^{1,∞}(Ω) precise representative satisfies an interior Lipschitz estimate with constant C ‖∇u‖, where C > 0 depends only on the space and Haar measure. The estimate applies whenever the closed ball of radius twice the distance of the two points lies in Ω.

theorem TauCeti.W1p.exists_holderOnWith_preciseRepresentative_of_isCompact {E : Type u_1} [MeasurableSpace E] [NormedAddCommGroup E] [InnerProductSpace ℝ E] [FiniteDimensional ℝ E] [BorelSpace E] {mu : MeasureTheory.Measure E} [mu.IsAddHaarMeasure] {p : ENNReal} [Fact (1 ≤ p)] (hp : ↑(Module.finrank ℝ E) < p) {Omega : TopologicalSpace.Opens E} (u : ↥(W1p mu Omega p)) {K : Set E} (hK : IsCompact K) (hKΩ : K ⊆ ↑Omega) :

Morrey's embedding on compact subsets. If n < p ≤ ∞, where n is the dimension, the precise representative of any u ∈ W^{1,p}(Ω) is Hölder continuous with exponent 1 - n / p on every compact subset of Ω.

W^{1,∞} functions are locally Lipschitz. The precise representative of any u ∈ W^{1,∞}(Ω) is Lipschitz on every compact subset of Ω.

On the whole space, for finite p above the dimension, the precise representative of u ∈ W^{1,p}(ℝⁿ) is the canonical Morrey representative TauCeti.W1p.morreyRepresentative: both are continuous and agree with u almost everywhere.