Nitsche consistency and energy quasi-optimality

The coercivity threshold in the Statement and [F3] give $a_h^{\theta }(v_h,v_h)\ge {\lambda }_{\theta }\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{E,h}^2$ for all $v_h\in V_h$. If $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{E,h}=0$, then the weak gradient of $v_h$ is zero. For any ball $B$ with $\bar{B}\subset \mathrm{\Omega }$, choose a slightly larger ball $B^{\mathrm{\prime }}$ still compactly contained in $\mathrm{\Omega }$ and a smooth cutoff $\chi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one on a neighborhood of $\overline{B^{\mathrm{\prime }}}$, using [F5]. The product $\chi v_h$ lies in $H^1(\mathrm{\Omega })$ with compact support by [F5]; let $V$ be its zero extension to ${\mathbb{R}}^d$. On a neighborhood of $\overline{B^{\mathrm{\prime }}}$, $\mathrm{\nabla }V=\mathrm{\nabla }{v}_h=0$. For sufficiently small $\epsilon$, [F5] says $V_{\epsilon }={\rho }_{\epsilon }\ast V$ is smooth and has gradient ${\rho }_{\epsilon }\ast \mathrm{\nabla }V=0$ on $B$, so it is constant there. The rescaled kernels are an $L^1$ approximate identity, and [F5] gives $V_{\epsilon }\to V=v_h$ in $L^2(B)$; because $B$ has positive measure, the constants $V_{\epsilon }{\mathrm{\mid }}_B$ are Cauchy and $v_h$ is a.e. constant on $B$. The points polygonally reachable from a fixed point form a nonempty open and relatively closed subset of connected $\mathrm{\Omega }$: if a point of $\mathrm{\Omega }$ is in the relative closure, a small ball in $\mathrm{\Omega }$ meets the reachable set and its line segment reaches the point. Hence any two points are joined by a polygonal path. A compact polygonal path is covered by a finite chain of overlapping interior balls, so the local constants agree. Thus every point has a neighborhood on which $v_h$ equals the same constant $c$ almost everywhere; second countability gives a countable subcover of $\mathrm{\Omega }$, and the union of its null exceptional sets is null. On each cell its polynomial agrees a.e. with $c$ on the cell interior and therefore equals $c$ everywhere on the cell; its boundary trace is $c$. The global flux-bound supplier [F2] guarantees a boundary facet; that facet has positive measure and $h_F>0$, so the zero boundary term in $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{E,h}$ forces $c=0$. Thus the energy seminorm is a norm on $V_h$.

The global identity [F4] gives $a_h^0(u,w_h)={\mathrm{\ell }}_h^0(w_h)$: its volume and first boundary-flux terms equal $(f,w_h{)}_{\mathrm{\Omega }}$, and the penalty uses $g=u{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$. For $\theta =1$, the additional term $-{\sum }_F{\int }_{F}u{\mathrm{\partial }}_n{w}_h$ equals the added load term $-{\sum }_F{\int }_{F}g{\mathrm{\partial }}_n{w}_h$, so $a_h^1(u,w_h)={\mathrm{\ell }}_h^1(w_h)$ as well. Subtracting the discrete equation yields $a_h^{\theta }(u-u_h^{\theta },w_h)=0$ for all $w_h\in V_h$.

For any comparison function $z\in {\mathcal{X}}_h$ and any $w_h\in V_h$, weighted Cauchy–Schwarz bounds the four form terms by $\mathrm{\mid }{a}_h^{\theta }(z,w_h)\mathrm{\mid }\le (2+\theta C_{\mathrm{\partial }}+{\gamma }_0)\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\#},h}\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_{\mathrm{\#},h}\mathrm{.}$ Here the first flux term uses the flux component of $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\#},h}$ and the boundary-value component of $\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_{\mathrm{\#},h}$; the transposed flux uses [F2] for $w_h$; the penalty uses the two boundary-value components. Also [F2] gives $\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_{\mathrm{\#},h}^2\le (1+C_{\mathrm{\partial }}^2)\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_{E,h}^2$, so [F3] implies $a_h^{\theta }(w_h,w_h)\ge {\alpha }_{\theta }\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_{\mathrm{\#},h}^2$ with ${\alpha }_{\theta }:={\lambda }_{\theta }\mathrm{/}(1+C_{\mathrm{\partial }}^2)$.

Define the linear map $A_h:V_h\to V_h^{\ast }$ by $(A_h{v}_h)(w_h)=a_h^{\theta }(v_h,w_h)$. If $A_h{v}_h=0$, testing with $v_h$ and using [F3] gives ${\lambda }_{\theta }\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_{E,h}^2\le 0$, hence $v_h=0$ by step 1.1. Therefore $A_h$ is injective; since $V_h$ and its dual are finite-dimensional with equal dimension, $A_h$ is bijective. The load ${\mathrm{\ell }}_h^{\theta }$ is a finite linear functional on $V_h$, so the discrete equation has a unique solution.

Fix $v_h\in V_h$ and set $w_h:=u_h^{\theta }-v_h$. Step 1.2 gives $a_h^{\theta }(w_h,w_h)=a_h^{\theta }(u-v_h,w_h)$. By step 1.3, if $w_h\mathrm{\ne }0$ then $\mathrm{\parallel }{w}_h{\mathrm{\parallel }}_{\mathrm{\#},h}\le \frac{(2+\theta C_{\mathrm{\partial }}+{\gamma }_0)(1+C_{\mathrm{\partial }}^2)}{{\lambda }_{\theta }}\mathrm{\parallel }u-v_h{\mathrm{\parallel }}_{\mathrm{\#},h};$ if $w_h=0$ the same inequality is immediate. The triangle inequality and infimum over $v_h$ prove the displayed quasi-optimal estimate. If [A2] holds, substituting its right-hand side proves the stated conditional rate.