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Cell averaging is the $L^{2}$ projection onto P0

Statement

For a finite conforming simplicial mesh of bounded $D$ , define $P_{0} : L^{2} (D) \to P_{0} (T_{h})$ cellwise by $(P_{0} g) ∣_{K} = ∣ K ∣^{- 1} \int_{K} g .$ Then $P_{0}$ is linear, equals the identity on piecewise constants, is the $L^{2}$ -orthogonal projection onto that space, satisfies $∥ P_{0} g ∥_{2} \leq ∥ g ∥_{2}$ , and preserves the integral $\int_{D} P_{0} g = \int_{D} g$ .

Facts & Assumptions

Given: A finite conforming mesh of bounded $D$ and $g \in L^{2} (D)$ .

[F1]

The mesh cells have positive finite measure, disjoint interiors, and cover $D$ up to their measure-zero affine faces. (Matching affine simplicial meshes and shape regularity, Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null)

[F2]

Cauchy–Schwarz gives $∣ \int_{K} g ∣ \leq ∣ K ∣^{1 / 2} ∥ g ∥_{L^{2} (K)}$ . (Cauchy-Schwarz inequality for $L^{2}$ )

Proof

1.1F1givenalgebra

The cell integral is linear, so the formula defines a linear piecewise-constant map. If $g$ is constant on each cell, its cell average is the same constant, so $P_{0} g = g$ . For any piecewise constant $q_{h}$ , $\int_{D} (g - P_{0} g) q_{h} = \sum_{K} q_{K} (\int_{K} g - ∣ K ∣ ∣ K ∣^{- 1} \int_{K} g) = 0.$ Thus the range is orthogonal to the residual and $P_{0}$ is the orthogonal projection onto $P_{0} (T_{h})$ .

2.1F1F2algebra∎

By [F2], $∣ K ∣ ∣ (P_{0} g) ∣_{K} ∣^{2} = ∣ K ∣^{- 1} ∣ \int_{K} g ∣^{2} \leq \int_{K} ∣ g ∣^{2}$ . Summing over cells gives $∥ P_{0} g ∥_{2}^{2} \leq ∥ g ∥_{2}^{2}$ . Summing the defining averages multiplied by cell measures gives $\int_{D} P_{0} g = \sum_{K} \int_{K} g = \int_{D} g$ .

Depends on

Used by

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Part III Chapter 14: H(div) finite elements (standard reference, not scraped)
Daniele Boffi, Franco Brezzi and Michel Fortin, Mixed Finite Element Methods and Applications (standard reference, not scraped)

Cell averaging is the L2L2 projection onto P0