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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Face jumps are controlled by the broken gradient

Statement

Assume Countable Choice. For every matching shape-regular simplicial mesh and $v_{h} \in V_{h}^{C R, 0}$ , $\sum_{F \in F^{\circ} \cup F^{\partial}} h_{F}^{- 1} ∥ [v_{h}]_{F} ∥_{L^{2} (F)}^{2} \leq C ∥ \nabla_{h} v_{h} ∥_{L^{2} (Ω)}^{2} .$ On an interior facet, $[v_{h}]_{F} = v_{K^{+}} - v_{K^{-}}$ ; on a boundary facet, $[v_{h}]_{F} = v_{K}$ . The constant depends only on $d$ and the shape-regularity bound.

Facts & Assumptions

Given: Countable Choice, a finite matching shape-regular simplicial mesh, and $v_{h}$ in the global CR space of The global homogeneous Crouzeix–Raviart space.

[A1]

Countable Choice is required by the shape-regular affine simplex scaling interface used below. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The mean of every interior jump and every boundary trace jump is zero by the defining face-average constraints. (The global homogeneous Crouzeix–Raviart space)

[F2]

On a shape-regular simplex, $∣ K ∣ ≍ h_{K}^{d}$ , $∣ F ∣ ≍ h_{K}^{d - 1}$ , and adjacent cell/facet diameters are comparable. (Shape regularity controls affine simplex geometry)

Proof

1.1F1givenalgebra

Let $F$ be a facet of $K$ and let $b_{F}$ be its barycenter. For any affine $j_{F}$ on $F$ , $j_{F} (x) = j_{F} (b_{F}) + \nabla_{F} j_{F} \cdot (x - b_{F})$ and $\int_{F} (x - b_{F}) d S = 0$ by the definition $b_{F} = ∣ F ∣^{- 1} \int_{F} x d S$ ; hence its face average is exactly $j_{F} (b_{F})$ . On an interior facet, the jump $j_{F} = [v_{h}]_{F}$ is affine with zero mean by [F1], hence $j_{F} (b_{F}) = 0$ . On a boundary facet, $j_{F} = v_{K}$ has zero mean and likewise $j_{F} (b_{F}) = 0$ . Therefore for every $x \in F$ , $∣ j_{F} (x) ∣ = ∣ \nabla_{F} j_{F} \cdot (x - b_{F}) ∣ \leq h_{F} (∣ \nabla v_{K^{+}} ∣ + ∣ \nabla v_{K^{-}} ∣),$ with the absent cell term omitted on the boundary.

2.1A1F2step 1.1algebra∎

Squaring step 1.1 and integrating on $F$ gives $h_{F}^{- 1} ∥ j_{F} ∥_{L^{2} (F)}^{2} \leq C h_{F}^{d} (∣ \nabla v_{K^{+}} ∣^{2} + ∣ \nabla v_{K^{-}} ∣^{2})$ , where the factor $h_{F}^{d}$ is $h_{F}^{- 1} \cdot h_{F}^{2} \cdot ∣ F ∣$ . By [F2], $h_{F}^{d} ∣ \nabla v_{K} ∣^{2} \leq C ∣ K ∣ ∣ \nabla v_{K} ∣^{2} = C ∥ \nabla v_{K} ∥_{L^{2} (K)}^{2}$ for each incident cell. Summing over facets counts each cell at most $d + 1$ times, proving the estimate.

Depends on

Used by

Dependency tree · two levels

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Part VIII, Chapter 36, Crouzeix–Raviart approximation (standard reference, not scraped)
Joachim Schöberl, Numerical Methods for Partial Differential Equations, Chapter 25, Non-conforming Finite Element Methods (standard reference, not scraped)