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The Poincare inequality with a positive-measure zero set

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 2$ , let $Ω$ be a bounded John domain with admissible constant $c_{J}$ , let $1 \leq p < \infty$ , and let $A \subseteq Ω$ be measurable with $∣ A ∣ \geq γ ∣ Ω ∣$ for some $γ \in (0, 1]$ . If $u \in W^{1, p} (Ω; K)$ vanishes almost everywhere on $A$ , then $∥ u ∥_{L^{p} (Ω)} \leq C (n, p, c_{J}, γ) diam (Ω) ∥ D u ∥_{L^{p} (Ω)} .$

Facts & Assumptions

Given: The Axiom of Choice; a bounded John domain $Ω$ with constant $c_{J}$ (John domains and the John constant); $1 \leq p < \infty$ ; a measurable $A \subseteq Ω$ with $∣ A ∣ \geq γ ∣ Ω ∣$ , $γ \in (0, 1]$ ; and a class $u \in W^{1, p} (Ω; K)$ vanishing almost everywhere on $A$ .

[F1]

The mean-zero Poincare inequality on the John domain: $∥ w - w_{Ω} ∥_{L^{p} (Ω)} \leq C_{1} (n, p, c_{J}) diam (Ω) ∥ D w ∥_{L^{p} (Ω)}$ for every $w \in W^{1, p} (Ω; K)$ , where $w_{Ω} = ∣ Ω ∣^{- 1} \int_{Ω} w$ (The mean-zero Poincare inequality on bounded John domains, The average of a locally integrable function over a Euclidean ball).

[F2]

Weak derivatives are linear: $D (w - c) = D w$ for every constant $c$ (Linearity, locality, and commutation of weak derivatives); $W^{1, p}$ consists of $L^{p}$ classes with weak gradient in $L^{p}$ , and $L^{p}$ is a space of almost-everywhere classes (Integer-order Sobolev spaces and their norms, The space $L^{p} (μ)$ as the quotient by null functions).

[F3]

Holder's inequality: $\int_{A} ∣ v ∣ \leq ∣ A ∣^{1 - 1 / p} ∥ v ∥_{L^{p} (Ω)}$ for the finite measure set $A$ (Holder's inequality for integrals, including the endpoint cases).

Proof

technique · direct

1.1F1F2givenalgebra

The mean-zero part. Since constants have zero weak derivative, $v : = u - u_{Ω}$ satisfies $v \in W^{1, p} (Ω; K)$ and $D v = D u$ by [F2]; by [F1], $∥ v ∥_{L^{p} (Ω)} \leq C_{1} (n, p, c_{J}) diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ . Also $v = u - u_{Ω}$ almost everywhere on $A$ , so $u$ vanishes there if and only if $v = - u_{Ω}$ there.

2.1F3step 1.1givenalgebra

Recovering the mean from the zero set. Because $u = 0$ almost everywhere on $A$ and $∣ A ∣ \leq ∣ Ω ∣ < \infty$ , $\int_{A} v = \int_{A} (u - u_{Ω}) = - ∣ A ∣ u_{Ω}$ ; hence $∣ A ∣ ∣ u_{Ω} ∣ \leq \int_{A} ∣ v ∣ \leq ∣ A ∣^{1 - 1 / p} ∥ v ∥_{L^{p} (Ω)} \leq ∣ A ∣^{1 - 1 / p} C_{1} diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ by [F3] and step 1.1, and therefore $∣ u_{Ω} ∣ \leq ∣ A ∣^{- 1 / p} C_{1} diam (Ω) ∥ D u ∥_{L^{p} (Ω)} \leq (γ ∣ Ω ∣)^{- 1 / p} C_{1} diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ .

3.1step 1.1step 2.1givenalgebra∎

Conclusion. By the triangle inequality and steps 1.1 and 2.1, $∥ u ∥_{L^{p} (Ω)} \leq ∥ u - u_{Ω} ∥_{L^{p} (Ω)} + ∣ Ω ∣^{1 / p} ∣ u_{Ω} ∣ \leq C_{1} diam (Ω) ∥ D u ∥_{L^{p} (Ω)} + C_{1} γ^{- 1 / p} diam (Ω) ∥ D u ∥_{L^{p} (Ω)} = C (n, p, c_{J}, γ) diam (Ω) ∥ D u ∥_{L^{p} (Ω)}$ with $C (n, p, c_{J}, γ) : = C_{1} (1 + γ^{- 1 / p})$ , which is the asserted inequality.

Source notes

Kinnunen's Remark 3.20 records the zero-set variant: a function vanishing on a set of positive measure can be normalised without the mean, and the mean itself is controlled by the amount of mass on the complement. The proof above implements that normalisation: the mean-zero inequality controls $u - u_{Ω}$ , and the value $u_{Ω}$ is recovered from the zero set by integrating $u - u_{Ω}$ over $A$ . The exponent $- 1 / p$ in the mean bound is what produces the constant $γ^{- 1 / p}$ .

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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)