Consistency bounds for the Crouzeix–Raviart Poisson method

Apply [F1] on each cell and sum. The volume products are integrable by [F6] because $\mathrm{\nabla }u,{\mathrm{\nabla }}_h{v}_h,f,v_h\in L^2$ on each cell. For an interior facet $F=K^{+}\cap K^{-}$, [F2] identifies both traces of $\mathrm{\nabla }u$, and the two outward normals are opposite. Their normal traces and the jumps are in $L^2(F)$, so [F6] also makes each boundary product integrable. The boundary terms therefore combine to ${\mathcal{R}}_h(u;v_h)={\sum }_{F\in {\mathcal{F}}^{\circ }}{\int }_F({\gamma }_F\mathrm{\nabla }u\cdot n_{K^{+},F})[v_h{]}_{F}dS+{\sum }_{F\in {\mathcal{F}}^{\mathrm{\partial }}}{\int }_F({\gamma }_F\mathrm{\nabla }u\cdot n_{K,F})v_{h}dS\mathrm{.}$ The sign follows from $-\mathrm{\Delta }u=f$ in the Green identity.

Write $e_K=\zeta -I_K^{CR}\zeta$. On an interior facet the common trace of $\zeta$ gives $[I_h^{CR}\zeta {]}_F=-{\gamma }_{F,K^{+}}{e}_{K^{+}}+{\gamma }_{F,K^{-}}{e}_{K^{-}}$. On a boundary facet, ${\gamma }_{F,K}{I}_K^{CR}\zeta =-{\gamma }_{F,K}{e}_K$ because $\zeta$ has zero trace. Applying [F5], summing over the at most $d+1$ facets of each cell, and using nonnegativity of the local fractional seminorm integrals gives ${\sum }_F{h}_F^{-1}\mathrm{\parallel }[I_h^{CR}\zeta {]}_F{\mathrm{\parallel }}_{L^2(F)}^2\le C{h}^{2s}\mathrm{\mid }\zeta {\mathrm{\mid }}_{H^{1+s}(\mathrm{\Omega })}^2\mathrm{.}$ The face mean of each such jump is zero by [F5].

Each jump in step 1.1 has zero face mean by [F2], so subtract the mean of its normal-gradient factor. Regard the finitely many facets as a disjoint union with surface measure on each component; Holder [F6] with the component functions $h_F^{1\mathrm{/}2}(q_F-{\mathrm{avg}}_F{q}_F)$ and $h_F^{-1\mathrm{/}2}[v_h{]}_F$ gives $\mathrm{\mid }{\mathcal{R}}_h(u;v_h)\mathrm{\mid }\le {({\sum }_F{h}_F\mathrm{\parallel }{q}_F-{\mathrm{avg}}_F{q}_F{\mathrm{\parallel }}_{L^2(F)}^2)}^{1\mathrm{/}2}{({\sum }_F{h}_F^{-1}\mathrm{\parallel }[v_h{]}_F{\mathrm{\parallel }}_{L^2(F)}^2)}^{1\mathrm{/}2},$ where $q_F$ denotes the corresponding outward normal trace (and the boundary jump is $v_h$). By [F3], the first factor is at most $C{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}(\mathrm{\Omega })}$: each cell contributes at most $d+1$ facet terms, and the sum of the cell Gagliardo seminorms is bounded by the global seminorm because the integrands are nonnegative and $K\times K\subset \mathrm{\Omega }\times \mathrm{\Omega }$. The second factor is at most $C\mathrm{\parallel }{\mathrm{\nabla }}_h{v}_h{\mathrm{\parallel }}_2$ by [F4]. This proves the general bound.

Use the same face representation as step 1.1 with $v_h=I_h^{CR}\zeta$. Subtract the normal-trace means as in step 2.1. Holder [F6] on the disjoint union of facets bounds the product of the first weighted factor, at most $C{h}^r\mathrm{\mid }u{\mathrm{\mid }}_{H^{1+r}(\mathrm{\Omega })}$ by [F3], and the second, at most $C{h}^s\mathrm{\mid }\zeta {\mathrm{\mid }}_{H^{1+s}(\mathrm{\Omega })}$ by step 1.2. Their product is the sharper asserted bound.