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The mean-zero Poincare inequality on bounded John domains

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 2$ , let $Ω \subseteq R^{n}$ be a bounded John domain with admissible constant $c_{J}$ , and let $1 \leq p < \infty$ . There is a constant $C (n, p, c_{J})$ with $∥ u - u_{Ω} ∥_{L^{p} (Ω)} \leq C (n, p, c_{J}) diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ for every $u \in W^{1, p} (Ω; K)$ , where $u_{Ω} = ∣ Ω ∣^{- 1} \int_{Ω} u$ . The factor $diam (Ω)$ is necessary: for the function $u (x) = x_{1}$ on a ball of radius $R$ with the centre as distinguished point, the ratio $∥ u - u_{B} ∥_{L^{p} (B)} / ∥ D u ∥_{L^{p} (B)}$ grows linearly in $R$ , while the John constant of that pair is $1$ for every $R$ .

Facts & Assumptions

Given: The Axiom of Choice, whose Countable-Choice consequence is used for the measure-theoretic interfaces; a bounded John domain $Ω$ with distinguished point $x_{0}$ and admissible constant $c_{J}$ ; $1 \leq p < \infty$ ; a field $K$ ; and a class $u \in W^{1, p} (Ω; K)$ .

[F1]

The John chain lemma: with $r_{0} = \frac{1}{4} dist (x_{0}, \partial Ω)$ and $B_{0} = B (x_{0}, r_{0})$ there is $M = M (n, c_{J})$ such that for every $x \in Ω$ there are balls $B_{i} = B (x_{i}, r_{i}) \subseteq Ω$ with $∣ B_{i} \cup B_{i + 1} ∣ \leq M ∣ B_{i} \cap B_{i + 1} ∣$ , $dist (x, B_{i}) \leq M r_{i}$ , $r_{i} \to 0$ , $x_{i} \to x$ , and multiplicity at most $M$ (Bounded-overlap ball chains in a bounded John domain; $x_{0}, c_{J}$ as in John domains and the John constant).

[F2]

Ball oscillation and ball Poincare: for $v \in W^{1, p} (B)$ , $∥ v - v_{B} ∥_{L^{p} (B)} \leq C_{1} (n, p) r ∥ D v ∥_{L^{p} (B)}$ and $\int_{B} ∣ v - v_{B} ∣ \leq C_{2} (n) r \int_{B} ∣ D v ∣$ for a ball $B = B (x, r)$ (Poincare inequality on a ball, Ball-mean oscillation bound by the Riesz potential of the gradient).

[F3]

At almost every Lebesgue point $x$ of $u$ , the centered averages $A_{r} u (x)$ converge to $u (x)$ as $r ↓ 0$ (Lebesgue differentiation theorem on $R^{n}$ , The average of a locally integrable function over a Euclidean ball).

[F4]

The truncated Riesz kernel bound: for measurable $Ω^{'} \subseteq B (z_{0}, ρ)$ and $f \in L^{p} (Ω^{'})$ , $∥ \int_{Ω^{'}} ∣ x - y ∣^{1 - n} ∣ f (y) ∣ d y ∥_{L^{p} (Ω^{'})} \leq C (n) ρ ∥ f ∥_{L^{p} (Ω^{'})}$ (The truncated Riesz kernel is bounded on $L^{p}$ of a bounded set).

[F5]

Holder's inequality and Tonelli's theorem for nonnegative functions on sigma-finite products (Holder's inequality for integrals, including the endpoint cases, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product); $W^{1, p}$ consists of $L^{p}$ classes with gradient in $L^{p}$ (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F6]

Linear substitution scales Lebesgue measure by the absolute determinant (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not); smooth classical derivatives are weak derivatives (Classical derivatives agree with weak derivatives), and balls have positive finite measure scaling with the nth power of the radius (Sphere and ball measures scale in Rn).

Proof

technique · direct

1.1F1F2F3algebra

Lebesgue points and the near case. Extend $u$ by zero outside $Ω$ ; it is locally integrable. The cited differentiation theorem implies $A_{s} (∣ u - u (x) ∣) (x) \to 0$ at almost every $x$ : apply it simultaneously to $∣ u - a ∣$ for the countable dense set $a \in Q$ (or $Q + i Q$ ), and bound ${lim sup}_{s ↓ 0} A_{s} (∣ u - u (x) ∣) (x) \leq 2 ∣ a - u (x) ∣$ ; let $a \to u (x)$ . Write $B_{0} = B (x_{0}, r_{0})$ for the fixed central ball and $B_{*} : = B (x_{0}, 2 r_{0}) \subset Ω$ . For almost every $x \in B_{*}$ , [F2] gives $∣ u (x) - u_{B_{*}} ∣ \leq C (n) \int_{B_{*}} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} d y$ . Also $∣ u_{B_{*}} - u_{B_{0}} ∣ \leq ∣ B_{0} ∣^{- 1} \int_{B_{*}} ∣ u - u_{B_{*}} ∣ \leq C (n) r_{0}^{1 - n} \int_{B_{*}} ∣ D u ∣$ by the $L^{1}$ ball Poincare estimate. Since $∣ x - y ∣ \leq 4 r_{0}$ for $x, y \in B_{*}$ , this last bound is at most $C (n) \int_{B_{*}} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} d y$ . Thus $∣ u (x) - u_{B_{0}} ∣ \leq C (n) \int_{Ω} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} d y$ in the near case, with the same fixed $B_{0}$ for all $x$ .

1.2F1F2F3givenalgebra

Telescoping along the chain. Fix a Lebesgue point $x \notin B (x_{0}, 2 r_{0})$ of $u$ and a chain ${B_{i} = B (x_{i}, r_{i})}$ from [F1], with overlap and distance constant $M$ . Since $dist (x, B_{i}) \leq M r_{i}$ , we have $∣ x_{i} - x ∣ \leq (M + 1) r_{i}$ and hence $B_{i} \subseteq B (x, (M + 2) r_{i})$ . The volume ratio is $∣ B (x, (M + 2) r_{i}) ∣ / ∣ B_{i} ∣ = (M + 2)^{n}$ , so [F3] gives $∣ u_{B_{i}} - u (x) ∣ \leq \frac{1}{∣ B_{i} ∣} \int_{B_{i}} ∣ u (y) - u (x) ∣ d y \leq (M + 2)^{n} A_{(M + 2) r_{i}} (∣ u - u (x) ∣) (x) ⟶ 0.$ Thus $u_{B_{i}} \to u (x)$ , and $∣ u (x) - u_{B_{0}} ∣ \leq \sum_{i \geq 0} ∣ u_{B_{i}} - u_{B_{i + 1}} ∣$ . Writing each difference of means as an average over the intersection and using $∣ B_{i} ∣, ∣ B_{i + 1} ∣ \leq M ∣ B_{i} \cap B_{i + 1} ∣$ from [F1], $∣ u_{B_{i}} - u_{B_{i + 1}} ∣ \leq M (\frac{\int_{B_{i}} ∣ u - u_{B_{i}} ∣}{∣ B_{i} ∣} + \frac{\int_{B_{i + 1}} ∣ u - u_{B_{i + 1}} ∣}{∣ B_{i + 1} ∣}) .$ Applying the $L^{1}$ ball Poincare inequality [F2] on each ball and using the radius comparability supplied by [F1] gives $∣ u_{B_{i}} - u_{B_{i + 1}} ∣ \leq C_{3} (n, c_{J}) \sum_{j \in {i, i + 1}} r_{j} \frac{\int_{B_{j}} ∣ D u ∣}{∣ B_{j} ∣} .$

2.1F1F5step 1.1step 1.2givenalgebra

The potential bound. From step 1.2, $∣ u (x) - u_{B_{0}} ∣ \leq C_{3} \sum_{i} r_{i} \int_{B_{i}} ∣ D u ∣ / ∣ B_{i} ∣$ (each ball counted with a bounded number of neighbours with comparable radii, the constant absorbed into $C_{3}$ ). For $y \in B_{i}$ the chain property $dist (x, B_{i}) \leq M r_{i}$ gives $∣ x - y ∣ \leq (M + 2) r_{i}$ , hence $∣ B_{i} ∣ = ω_{n - 1} r_{i}^{n} / n$ and $r_{i} / ∣ B_{i} ∣ = c (n) r_{i}^{1 - n} \leq c (n, M) ∣ x - y ∣^{1 - n}$ ; summing over $i$ and using the multiplicity bound of [F1], $\sum_{i} r_{i} \int_{B_{i}} ∣ D u ∣ / ∣ B_{i} ∣ \leq c (n, M) \int_{Ω} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} N (x, y) d y \leq c (n, M) \int_{Ω} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} d y$ with $N (x, y) \leq M$ the number of balls containing $y$ ; the exchange of sum and integral is Tonelli [F5]. The index $i = 0$ needs the separate bound $r_{0} \int_{B_{0}} ∣ D u ∣ / ∣ B_{0} ∣ \leq c (n, c_{J}) \int_{B_{0}} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} d y$ : the John condition at $x$ gives $diam Ω \leq 8 c_{J} r_{0}$ , so $∣ x - y ∣ \leq 8 c_{J} r_{0}$ for $y \in B_{0}$ and $r_{0}^{1 - n} \leq (8 c_{J})^{n - 1} ∣ x - y ∣^{1 - n}$ . Together with step 1.1, this gives $∣ u (x) - u_{B_{0}} ∣ \leq C_{4} (n, c_{J}) \int_{Ω} ∣ D u (y) ∣ ∣ x - y ∣^{1 - n} d y$ for almost every $x \in Ω$ .

3.1F1F4F5step 2.1givenalgebra

$L^{p}$ norms and the mean. Take $L^{p} (Ω)$ norms in step 2.1 and apply the truncated kernel bound [F4] with $Ω^{'} = Ω \subseteq B (x_{0}, diam Ω)$ and $f = ∣ D u ∣$ : $∥ u - u_{B_{0}} ∥_{L^{p} (Ω)} \leq C_{5} (n, c_{J}) diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ . Since $u_{B_{0}}$ is a constant, $u_{Ω} - u_{B_{0}} = ∣ Ω ∣^{- 1} \int_{Ω} (u - u_{B_{0}})$ , so by Holder [F5] $∣ u_{B_{0}} - u_{Ω} ∣ \leq ∣ Ω ∣^{- 1} \int_{Ω} ∣ u - u_{B_{0}} ∣ \leq ∣ Ω ∣^{- 1 / p} ∥ u - u_{B_{0}} ∥_{L^{p} (Ω)}$ . Hence $∥ u - u_{Ω} ∥_{L^{p} (Ω)} \leq ∥ u - u_{B_{0}} ∥_{L^{p} (Ω)} + ∣ Ω ∣^{1 / p} ∣ u_{B_{0}} - u_{Ω} ∣ \leq 2 ∥ u - u_{B_{0}} ∥_{L^{p} (Ω)} \leq C (n, p, c_{J}) diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ with $C (n, p, c_{J}) : = 2 C_{5} (n, c_{J})$ .

4.1F1F6algebra∎

Necessity of length scaling. On $B = B (0, R)$ take $u (x) = x_{1}$ . By [F6] its weak gradient is $e_{1}$ , and reflection in the first coordinate gives $u_{B} = 0$ . Substituting $x = R y$ yields $∥ u ∥_{L^{p} (B)} = R^{1 + n / p} (\int_{B (0, 1)} ∣ y_{1} ∣^{p} d y)^{1 / p}$ and $∥ D u ∥_{L^{p} (B)} = R^{n / p} ∣ B (0, 1) ∣^{1 / p}$ . The first integral is finite and positive, since the unit ball contains a ball on which $∣ y_{1} ∣$ is bounded below by a positive number. Their ratio is therefore $c (n, p) R$ with $c (n, p) > 0$ . The radial segment from any $x$ to $0$ satisfies $R - ∣ γ (t) ∣ \geq ∣ x ∣ - ∣ γ (t) ∣ = ∣ x - γ (t) ∣$ , so this distinguished pair admits John constant $1$ for every $R$ . Thus no dimension-and-John-constant bound can omit the length factor.

Source notes

Kinnunen proves the Sobolev-Poincare inequality on John domains (Theorem 5.33, printed pp. 141-143) by exactly this chaining: the telescoping over $∣ u_{B_{i}} - u_{B_{i + 1}} ∣$ , the ball Poincare inequality, the comparison $r_{i} \approx ∣ x - y ∣$ on $B_{i}$ , the multiplicity bound, and the truncated-kernel estimate. The present item states the $L^{p}$ (rather than $L^{p^{*}}$ ) mean-zero form, which is what the surrounding page promises; the chaining argument is the same, and no Sobolev exponent is used. The endpoint index $i = 0$ is absorbed with the John bound $diam Ω \leq 8 c_{J} r_{0}$ .

Depends on

Used by

Poincare-Wirtinger fails on disconnected bounded domains Counterexample
Domain classes covered by the mean-zero Poincare inequality Remark
The Poincare inequality with a positive-measure zero set Theorem

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Sources

Juha Kinnunen, Sobolev Spaces (Aalto University, 2026, complete graduate lecture notes) (standard reference, not scraped)
Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, complete 158-page graduate notes) (standard reference, not scraped)