Global boundary-flux estimate on a matching mesh

Sum [F1] over all boundary facets. Each cell contributes at most $N_{\mathrm{\partial }}$ copies of its squared gradient norm, so ${\sum }_F{h}_F\mathrm{\parallel }{\mathrm{\partial }}_n{v}_h{\mathrm{\parallel }}_F^2\le N_{\mathrm{\partial }}{C}_{\mathrm{t}\mathrm{r}}^2{\sum }_K\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_K^2=N_{\mathrm{\partial }}{C}_{\mathrm{t}\mathrm{r}}^2\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{\mathrm{\Omega }}^2$. Taking square roots proves the first inequality.

Choose a point $x_0$ in the interior of a mesh cell. The finite union of cells is compact and bounded, so the ray $x_0+tv$ has a last point in that union for every unit direction $v$. Choose $v$ outside the finitely many measure-zero sets of directions whose ray meets a cell face of dimension at most $d-2$ or is parallel to a facet plane; the excluded directions are measure zero because each such face has dimension at most $d-2$ and radial projection from $x_0$ has dimension at most $d-2$ on the $(d-1)$-sphere. The last point therefore lies in the relative interior of a facet and the ray crosses it transversally. If two cells meet at a point in the relative interior of a common facet, their interiors lie on opposite sides: if both lay on the same side of its affine hyperplane, a sufficiently small point displaced to that side near the facet point would lie in the interior of both nondegenerate simplices, contradicting their disjoint interiors in [F3]. Thus such a shared-facet point is interior to the union, contrary to its being the last point on the ray. The last point is consequently on a one-cell boundary facet. Since the local result gives $C_{\mathrm{t}\mathrm{r}}\ge 1$, we obtain $1\le N_{\mathrm{\partial }}$ and $C_{\mathrm{\partial }}=\sqrt{N_{\mathrm{\partial }}}{C}_{\mathrm{t}\mathrm{r}}\ge 1$.

Apply [F4] on each facet and then finite-sum Cauchy–Schwarz to obtain $\mid {\sum }_F{\int }_F({\mathrm{\partial }}_n{v}_h)v_h\mid \le {\sum }_F(h_F^{1\mathrm{/}2}\mathrm{\parallel }{\mathrm{\partial }}_n{v}_h{\mathrm{\parallel }}_F)(h_F^{-1\mathrm{/}2}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_F)\le {\left({\sum }_F{h}_F\mathrm{\parallel }{\mathrm{\partial }}_n{v}_h{\mathrm{\parallel }}_F^2\right)}^{1\mathrm{/}2}{\left({\sum }_F{h}_F^{-1}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_F^2\right)}^{1\mathrm{/}2}\mathrm{.}$ Substitute the first inequality from step 1.1 to get the stated bound.