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A bounded-domain Poincare inequality for H1-zero fields

Statement

Assume Countable Choice. Let $D \subset R^{d}$ be bounded and open, and suppose its first coordinate projection is contained in an interval $(a, b)$ of width $L = b - a$ . For every real $v \in H_{0}^{1} (D)$ , $∥ v ∥_{L^{2} (D)} \leq L ∥ \partial_{1} v ∥_{L^{2} (D)} \leq L ∥ \nabla v ∥_{L^{2} (D)} .$

Facts & Assumptions

Given: A bounded open $D \subset (a, b) \times R^{d - 1}$ and $v \in H_{0}^{1} (D)$ , where $H_{0}^{1}$ is the $H^{1}$ -closure of $C_{c}^{\infty} (D)$ .

[A1]

Countable Choice permits selecting the approximating test functions from the H0 closure at each n. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$H_{0}^{1} (D)$ is the closure of compactly supported smooth functions in the $W^{1, 2}$ norm. (Zero-boundary Sobolev space as a norm closure)

[F2]

The $W^{1, 2}$ norm includes the $L^{2}$ norm and the $L^{2}$ norms of all first weak derivatives. (Integer-order Sobolev spaces and their norms)

[F3]

Fubini permits iterated integration of the nonnegative integrand on the bounded coordinate slab. (Fubini's theorem for L^1 functions on a sigma-finite product)

[F4]

The L2 Cauchy--Schwarz inequality gives $\int ∣ f g ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ . (Cauchy-Schwarz inequality for $L^{2}$ )

Proof

1.1givenF1F4

First take $v \in C_{c}^{\infty} (D)$ and extend it by zero to $(a, b) \times R^{d - 1}$ . For every $x^{'} \in R^{d - 1}$ and $x_{1} \in (a, b)$ , the fundamental theorem of calculus gives $v (x_{1}, x^{'}) = \int_{a}^{x_{1}} \partial_{1} v (t, x^{'}) d t$ . Cauchy--Schwarz yields $∣ v (x_{1}, x^{'}) ∣^{2} \leq L \int_{a}^{b} ∣ \partial_{1} v (t, x^{'}) ∣^{2} d t$ .

2.1F2F3step 1.1

Integrating the last inequality over $x_{1}$ gives a second factor $L$ ; integrating over $x^{'}$ and using the zero extension proves $∥ v ∥_{2}^{2} \leq L^{2} ∥ \partial_{1} v ∥_{2}^{2}$ .

3.1A1F1F2F3algebrastep 2.1∎

For general $v \in H_{0}^{1} (D)$ , [A1, F1] supplies a sequence $v_{n} \in C_{c}^{\infty} (D)$ converging to $v$ in $W^{1, 2}$ . The two norms in step 2.1 converge by [F2], so passing to the limit gives the same inequality for $v$ . Finally $∣ \partial_{1} v ∣ \leq ∣ \nabla v ∣$ almost everywhere.

Depends on

Used by

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Chapters 53–54 (complete course notes) (standard reference, not scraped)
Daniele Boffi, Franco Brezzi, Michel Fortin, Mixed Finite Element Methods and Applications (complete textbook) (standard reference, not scraped)