Nitsche methods for weak boundary enforcement

Let $F_K(\widehat{x})=b_K+A_K\widehat{x}$ be the reference map and let $\widehat{w}=w\circ F_K$. By [F2], $\widehat{w}\in W^{1,2}({\widehat{S}}_d)$ and its norm is bounded by a finite constant depending on $A_K$ times $\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,2}(K)}$. The surface Jacobian $J_F$ is positive and constant by [F3]. Define $T_{K,F}w$ by pullback of ${\gamma }_{\widehat{F}}\widehat{w}$; then $\mathrm{\parallel }{T}_{K,F}w{\mathrm{\parallel }}_{L^2(F)}^2=J_F\mathrm{\parallel }{\gamma }_{\widehat{F}}\widehat{w}{\mathrm{\parallel }}_{L^2(\widehat{F})}^2\le J_F{C}_{\widehat{F}}^2\mathrm{\parallel }\widehat{w}{\mathrm{\parallel }}_{W^{1,2}({\widehat{S}}_d)}^2\le C_{K,F}^2\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,2}(K)}^2\mathrm{.}$ Thus the trace is bounded, and it agrees with ordinary restriction for smooth $w$ by [F1].

Let $w\in H_0^1(\mathrm{\Omega };\mathbb{R})$. For each $m\ge 1$, the closure property [F4] and Countable Choice [F5] give ${\varphi }_m\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{R})$ with $\mathrm{\parallel }w-{\varphi }_m{\mathrm{\parallel }}_{W^{1,2}(\mathrm{\Omega })}<1\mathrm{/}m$. Restriction to $K$ cannot increase this norm, so step 1.1 gives $\mathrm{\parallel }{T}_{K,F}(w{\mathrm{\mid }}_K)-T_{K,F}({\varphi }_m{\mathrm{\mid }}_K){\mathrm{\parallel }}_{L^2(F)}\le C_{K,F}\mathrm{/}m\mathrm{.}$ The support of each ${\varphi }_m$ is compactly contained in $\mathrm{\Omega }$, hence ${\varphi }_m$ vanishes in a neighborhood of the boundary facet $F$ and its trace is zero by step 1.1. Letting $m\to \mathrm{\infty }$ proves $T_{K,F}(w{\mathrm{\mid }}_K)=0$.

If $p$ is polynomial on $K$, then it is smooth and step 1.1 identifies $T_{K,F}(p)$ with its classical restriction. Therefore for a global piecewise-polynomial $p\in H_0^1(\mathrm{\Omega })$, step 2.1 makes that restriction zero on each boundary facet. This proves the final assertion.

Apply [F1] to each component ${\mathrm{\partial }}_j{v}_h$, which has degree at most $k-1$. Summing the $d$ squared inequalities and using [F3] gives $\mathrm{\parallel }{\mathrm{\partial }}_{n_{K,F}}{v}_h{\mathrm{\parallel }}_{L^2(F)}^2\le \mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2(F)}^2\le C_0^2{h}_K^{-1}\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2(K)}^2$ for a finite $C_0$ depending only on $d,k$ and the shape bound.

Multiply step 1.1 by $h_F$ and use $h_F\mathrm{/}{h}_K\le C$ from [F2]. Choose $C_{\mathrm{t}\mathrm{r}}:=\max \{1,C_0\sqrt{C}\}$; this is positive, finite, uniform over the family, and proves the required bound.

If the family contains a cell, fix one of its facets and test the inequality with the affine polynomial $p(x):=n_{K,F}\cdot x$, which belongs to ${\mathbb{P}}_k(K)$ because $k\ge 1$. Then $\mathrm{\nabla }p=n_{K,F}$ and ${\mathrm{\partial }}_{n_{K,F}}p=1$, so $h_F\mathrm{\parallel }{\mathrm{\partial }}_{n_{K,F}}p{\mathrm{\parallel }}_F^2=(\mathrm{\mid }K\mathrm{\mid }\mathrm{/}\mathrm{\mid }F\mathrm{\mid })\mathrm{\mid }F\mathrm{\mid }=\mathrm{\mid }K\mathrm{\mid }=\mathrm{\parallel }\mathrm{\nabla }p{\mathrm{\parallel }}_K^2>0$. Thus, for a family containing a cell, every admissible positive constant satisfies $C_{\mathrm{t}\mathrm{r}}^2\ge 1$. For an empty family the bound is vacuous; in either case step 2.1 selects $C_{\mathrm{t}\mathrm{r}}\ge 1$.

First consider the hyperplane ${\mathbb{R}}^{d-1}\times \{0\}$ and $U\in \mathcal{S}({\mathbb{R}}^d)$. Write $G(x^{\mathrm{\prime }})=U(x^{\mathrm{\prime }},0)$. Fourier inversion in the last coordinate and Fubini on the absolutely integrable Schwartz integrands give $\widehat{G}({\xi }^{\mathrm{\prime }})={\int }_{\mathbb{R}}\widehat{U}({\xi }^{\mathrm{\prime }},\eta )d\eta$.

Apply Cauchy–Schwarz in $\eta$ with weights $⟨({\xi }^{\mathrm{\prime }},\eta ){⟩}^s$ and its reciprocal; this follows by expanding $0\le \mathrm{\parallel }f-\lambda g{\mathrm{\parallel }}_2^2$ and minimizing over the complex scalar $\lambda$. Both weighted factors lie in $L^2(\mathbb{R})$ for fixed ${\xi }^{\mathrm{\prime }}$, the reciprocal factor by [A1] and the weighted Fourier transform because $U$ is Schwartz. The reciprocal-square integral is ${\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^{-s}d\eta =C_s(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2-s}\le C_s,$ by the substitution $\eta =(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}t$ and [A1, F5]. Therefore $\mathrm{\mid }\widehat{G}({\xi }^{\mathrm{\prime }}){\mathrm{\mid }}^2\le C_s{\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^s\mathrm{\mid }\widehat{U}({\xi }^{\mathrm{\prime }},\eta ){\mathrm{\mid }}^{2}d\eta$. Tonelli [F4] and Plancherel [F1] give $\mathrm{\parallel }G{\mathrm{\parallel }}_{L^2({\mathbb{R}}^{d-1})}\le C_s\mathrm{\parallel }U{\mathrm{\parallel }}_{H^s({\mathbb{R}}^d)}$.

Since Schwartz space is dense in the completion [F2], the estimate of step 2.1 makes the restrictions of every Cauchy representative a Cauchy sequence in $L^2$; completeness [F3] defines a bounded trace on the hyperplane. The same estimate applied to the difference of two representatives proves independence of the representative and linearity.

This trace is local. For Schwartz $W$, let $W_t(x^{\mathrm{\prime }}):=W(x^{\mathrm{\prime }},t)$. Partial Plancherel in the tangential variables, Fubini [F4], and the weighted Cauchy--Schwarz estimate of step 2.1 give $\mathrm{\parallel }{W}_t-W_0{\mathrm{\parallel }}_{L^2({\mathbb{R}}^{d-1})}^2\le {\int }_{{\mathbb{R}}^{d-1}}{J}_t({\xi }^{\mathrm{\prime }})G({\xi }^{\mathrm{\prime }})d{\xi }^{\mathrm{\prime }},$ where $J_t({\xi }^{\mathrm{\prime }}):={\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^{-s}\mathrm{\mid }{e}^{2\pi i\eta t}-1{\mathrm{\mid }}^{2}d\eta$ and $G({\xi }^{\mathrm{\prime }}):={\int }_{\mathbb{R}}(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2+{\eta }^2{)}^s\mathrm{\mid }\widehat{W}({\xi }^{\mathrm{\prime }},\eta ){\mathrm{\mid }}^{2}d\eta$. Choose $0<\delta <\min \{1,2s-1\}$. Since $\mathrm{\mid }{e}^{i\alpha }-1{\mathrm{\mid }}^2\le C_{\delta }\mathrm{\mid }\alpha {\mathrm{\mid }}^{\delta }$, the substitution $\eta =(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}r$ using [F5] gives $J_t({\xi }^{\mathrm{\prime }})\le C_{s,\delta }\mathrm{\mid }t{\mathrm{\mid }}^{\delta }(1+\mathrm{\mid }{\xi }^{\mathrm{\prime }}{\mathrm{\mid }}^2{)}^{(1-2s+\delta )\mathrm{/}2}\le C_{s,\delta }\mathrm{\mid }t{\mathrm{\mid }}^{\delta }$. Also $\int G({\xi }^{\mathrm{\prime }})d{\xi }^{\mathrm{\prime }}=\mathrm{\parallel }W{\mathrm{\parallel }}_{H^s({\mathbb{R}}^d)}^2$ by [F1, F4]. Hence $\mathrm{\parallel }{W}_t-W_0{\mathrm{\parallel }}_2^2\le C_{s,\delta }\mathrm{\mid }t{\mathrm{\mid }}^{\delta }\mathrm{\parallel }W{\mathrm{\parallel }}_{H^s}^2$. Applying this estimate to differences of two slice parameters, density of Schwartz space and completeness [F3] give every $W\in H^s({\mathbb{R}}^d)$ a continuous $L^2({\mathbb{R}}^{d-1})$ slice representative whose value at $t=0$ is the trace from step 3.1. It agrees with the ordinary sections for almost every $t$: choose Schwartz approximants with summable squared $L^2({\mathbb{R}}^d)$ errors and apply Fubini [F4] to their squared slice errors. If two extensions agree on $K$, their difference vanishes on an interior prism beside every relatively compact patch $E$ of $F$. Fubini [F4] makes almost every interior slice zero on $E$; continuity then makes the trace zero there. A countable exhaustion covers $F$ up to its lower-dimensional boundary, which has surface measure zero, so the trace is independent of the extension. For a translation $b$ and orthogonal map $R$, $\widehat{W(R^{-1}(\cdot -b))}(\xi )=e^{-2\pi ib\cdot \xi }\widehat{W}(R^{-1}\xi )$ by change of variables [F5]; since $\mathrm{\mid }\det R\mathrm{\mid }=1$ and $⟨R\xi ⟩=⟨\xi ⟩$, the whole-space norm is invariant. Thus the hyperplane estimate from step 2.1 applies to the supporting plane of $F$ and then restricts to $F$.

For $U\in \mathcal{S}({\mathbb{R}}^d)$ and $0<s<1$, the map $(x,y)\mapsto (x,z=y-x)$ is an invertible linear $C^1$ diffeomorphism of ${\mathbb{R}}^{2d}$ with determinant of absolute value $1$. Apply the change-of-variables formula in [F6] to the nonnegative integrand, then Tonelli and Plancherel in the $x$ variable, to obtain ${\int }_{{\mathbb{R}}^d}{\int }_{{\mathbb{R}}^d}\frac{\mathrm{\mid }U(x)-U(y){\mathrm{\mid }}^2}{\mathrm{\mid }x-y{\mathrm{\mid }}^{d+2s}}dxdy={\int }_{{\mathbb{R}}^d}\mathrm{\mid }\widehat{U}(\xi ){\mathrm{\mid }}^2\left({\int }_{{\mathbb{R}}^d}\frac{\mathrm{\mid }{e}^{2\pi i\xi \cdot z}-1{\mathrm{\mid }}^2}{\mathrm{\mid }z{\mathrm{\mid }}^{d+2s}}dz\right)d\xi =c_{d,s}{\int }_{{\mathbb{R}}^d}\mathrm{\mid }\xi {\mathrm{\mid }}^{2s}\mathrm{\mid }\widehat{U}(\xi ){\mathrm{\mid }}^{2}d\xi \mathrm{.}$ The inner integral is finite because its numerator is at most $C\mathrm{\mid }z{\mathrm{\mid }}^2$ near zero and at most $4$ at infinity, giving radial integrals ${\int }_0^1{r}^{1-2s}dr$ and ${\int }_1^{\mathrm{\infty }}{r}^{-1-2s}dr$; it is positive for $\xi \mathrm{\ne }0$. Rotation and dilation, each covered by [F6], give the factor $c_{d,s}\mathrm{\mid }\xi {\mathrm{\mid }}^{2s}$. The vector-valued version follows by applying this identity to each component and summing. Since $1+\mathrm{\mid }\xi {\mathrm{\mid }}^{2s}$ is comparable to $⟨\xi {⟩}^{2s}$, this identifies the whole-space intrinsic norm with the Bessel norm on Schwartz functions.

We construct finitely many boundary charts. At each boundary point choose an inward direction in the interior cone of the simplex; in a sufficiently small rotated box the active facet inequalities solve for the vertical coordinate and give $t>g(x^{\mathrm{\prime }})$, where $g$ is the maximum of finitely many affine functions. Inactive facets stay away after shrinking the box. If the affine functions are ${\mathrm{\ell }}_j$, then $\mathrm{\mid }g(x^{\mathrm{\prime }})-g(y^{\mathrm{\prime }})\mathrm{\mid }\le L\mathrm{\mid }{x}^{\mathrm{\prime }}-y^{\mathrm{\prime }}\mathrm{\mid }$ with $L:={\max }_j\mathrm{\mid }\mathrm{\nabla }{\mathrm{\ell }}_j\mathrm{\mid }$, so $g$ is Lipschitz in this explicit sense. Compactness gives a finite cover of the boundary by smaller such boxes and one interior open set covering the remaining compact subset. Choose nonnegative smooth cutoffs ${\eta }_j$ supported in the corresponding larger chart sets and positive on the smaller cover. Their sum is positive on a neighborhood of the closed simplex, so ${\psi }_j:={\eta }_j\mathrm{/}({\sum }_i{\eta }_i)$ there form a smooth finite partition of unity. Multiplication by each cutoff is bounded in the intrinsic norm: expand $\chi (x)w(x)-\chi (y)w(y)$ into $\chi (x)(w(x)-w(y))+w(y)(\chi (x)-\chi (y))$; the second term is bounded by $C\mathrm{\parallel }w{\mathrm{\parallel }}_2^2$ because $\mathrm{\mid }\chi (x)-\chi (y)\mathrm{\mid }\le \min (C\mathrm{\mid }x-y\mathrm{\mid },2\mathrm{\parallel }\chi {\mathrm{\parallel }}_{\mathrm{\infty }})$ and the resulting radial kernel is integrable for $0<s<1$.

The identity extends to the completion, and that completion is exactly the $L^2$ functions with finite displayed intrinsic seminorm. For the converse density, choose smooth cutoffs ${\chi }_R$ equal to one on $B_R$, supported in $B_{2R}$, with $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$. For any such function $W$, the difference quotient of $(1-{\chi }_R)W$ is the sum of $(1-{\chi }_R(x))(W(x)-W(y))$ and $W(y)({\chi }_R(y)-{\chi }_R(x))$. Dominated convergence sends the first term to zero in the weighted pair $L^2$ norm; for the second, integration in $x-y$ gives at most $C{R}^{-2s}\mathrm{\parallel }W{\mathrm{\parallel }}_2^2$. Thus ${\chi }_{R}W\to W$ in the intrinsic norm. For compactly supported $W$, translation is continuous in $L^2$ both for $W$ and for $G_W(x,y):=(W(x)-W(y))\mathrm{/}\mathrm{\mid }x-y{\mathrm{\mid }}^{d\mathrm{/}2+s}$ under the simultaneous translation $(x,y)\mapsto (x-a,y-a)$. Choose a nonnegative smooth compactly supported unit-mass mollifier $\rho$. For a scalar function $Z$ and probability density ${\rho }_{\epsilon }$, scalar Jensen gives $\mathrm{\mid }\int {\rho }_{\epsilon }(a)Z(a)da{\mathrm{\mid }}^2\le \int {\rho }_{\epsilon }(a)\mathrm{\mid }Z(a){\mathrm{\mid }}^{2}da$. For a finite-dimensional Euclidean-valued $Z=(Z_1,\dots ,Z_m)$, apply this to each real component and sum: $\mathrm{\mid }\int {\rho }_{\epsilon }Z{\mathrm{\mid }}^2={\sum }_{j=1}^m\mathrm{\mid }\int {\rho }_{\epsilon }{Z}_j{\mathrm{\mid }}^2\le {\sum }_{j=1}^m\int {\rho }_{\epsilon }\mathrm{\mid }{Z}_j{\mathrm{\mid }}^2=\int {\rho }_{\epsilon }\mathrm{\mid }Z{\mathrm{\mid }}^2$. Applying this inequality to the scalar or vector differences for $W$ and $G_W$, then Tonelli, gives $\mathrm{\parallel }{\rho }_{\epsilon }\ast W-W{\mathrm{\parallel }}_2^2+[{\rho }_{\epsilon }\ast W-W{]}_{s,{\mathbb{R}}^d}^2\le \int {\rho }_{\epsilon }(a)(\mathrm{\parallel }W(\cdot -a)-W{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{G}_W(\cdot -a,\cdot -a)-G_W{\mathrm{\parallel }}_{L^2({\mathbb{R}}^{2d})}^2)da\mathrm{.}$ The right side tends to zero because both translation differences tend to zero as $a\to 0$ and the mollifier supports shrink to zero. The convolution is smooth by [F5], so $C_c^{\mathrm{\infty }}$ is dense. Conversely, an intrinsic-Cauchy smooth sequence has an $L^2$ limit and a weighted-pair-$L^2$ limit; an almost-everywhere convergent subsequence identifies the latter with the difference quotient of the former. Completeness and [F1] therefore identify the completion with the intrinsic space.

For a boundary-localized piece, reflect across its graph by $R(x^{\mathrm{\prime }},t)=(x^{\mathrm{\prime }},2g(x^{\mathrm{\prime }})-t)$ and set the extension below the graph to $w(R(x^{\mathrm{\prime }},t))$. The horizontal coordinates are unchanged and the vertical difference is bounded by $\mathrm{\mid }t-u\mathrm{\mid }+2L\mathrm{\mid }{x}^{\mathrm{\prime }}-y^{\mathrm{\prime }}\mathrm{\mid }$, so $\mathrm{\mid }Rx-Ry\mathrm{\mid }\le C\mathrm{\mid }x-y\mathrm{\mid }$; since $R^2$ is the identity, the reverse bound follows by applying the same estimate to $Rx,Ry$. Thus $R$ is bi-Lipschitz. Because $g$ is a finite maximum of affine functions, partition each integration region into the finitely many open affine pieces; their boundaries lie in finitely many hyperplanes and are null. On each piece $R$ is an affine $C^1$ diffeomorphism with determinant of absolute value $1$. Apply the nonnegative change-of-variables formula in [F6] on each such piece (and in the $y$ variable for the mixed term); Tonelli combines the resulting estimates. Consequently same-side double integrals are bounded by a constant multiple of the upper-side integral. For $x$ above and $y$ below the graph, write their heights above the graph as $a,b\ge 0$. Then $a+b\le \mathrm{\mid }{x}_d-y_d\mathrm{\mid }+L\mathrm{\mid }{x}^{\mathrm{\prime }}-y^{\mathrm{\prime }}\mathrm{\mid }$, hence $\mathrm{\mid }x-Ry\mathrm{\mid }\le C\mathrm{\mid }x-y\mathrm{\mid }$; the same piecewise affine changes of variables $y\mapsto Ry$ bound the mixed integral by the upper-side seminorm. Choose chart supports with positive margin to the artificial box edges, so cross pairs for extension by zero contribute at most $C\mathrm{\parallel }w{\mathrm{\parallel }}_2^2$; the interior piece has positive distance from the simplex boundary and obeys the same bound. Sum the finitely many extensions of ${\psi }_{j}w$. The result equals $w$ on the reference simplex and is bounded in intrinsic norm.

Restricting any whole-space extension decreases both the $L^2$ norm and the nonnegative pair integral. Conversely, step 2.2 supplies a bounded extension of every intrinsic reference-simplex function; together with step 2.1 this proves equivalence of the intrinsic and restriction $H^s$ norms for $0<s<1$. Apply the whole-space Fourier hyperplane trace bound [F2] to that extension; restriction to the reference facet gives its trace, and uniqueness in [F2] identifies it with the existing trace. At $s=1$, [F4] gives on the reference simplex $\mathrm{\parallel }Tw{\mathrm{\parallel }}_F^2\le C(\mathrm{\parallel }w{\mathrm{\parallel }}_K^2+\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_K^2)$ for weak $H^1$ functions; density identifies it with the restriction trace whenever the input has a whole-space extension.

Pull back by the shape-regular affine map. The map from the reference open simplex to the cell interior and its product map on the double integral are invertible affine $C^1$ diffeomorphisms. Apply the nonnegative formula in [F6] to the $L^2$ and pair-integral terms; the determinant and singular-value bounds in [F3] show that for $0<s<1$ the $L^2$ term on the reference simplex is bounded by $C{h}_K^{-d}\mathrm{\parallel }w{\mathrm{\parallel }}_K^2$, and its pair seminorm by $C{h}_K^{2s-d}\mathrm{\mid }w{\mathrm{\mid }}_{s,K}^2$: the two volume Jacobians contribute $h_K^{-2d}$ and the inverse-map distance contributes at most $C{h}_K^{d+2s}$. Multiplying the reference trace bound by the facet Jacobian $C{h}_K^{d-1}$ yields the asserted terms $h_K^{-1}\mathrm{\parallel }w{\mathrm{\parallel }}_K^2$ and $h_K^{2s-1}\mathrm{\mid }w{\mathrm{\mid }}_{s,K}^2$. At $s=1$, write $\widehat{w}=w\circ F_K$, where $F_K(\widehat{x})=b_K+A_K\widehat{x}$. The affine weak-derivative pullback [F8] gives ${\mathrm{\nabla }}_{\widehat{x}}\widehat{w}=A_K^T(\mathrm{\nabla }w)\circ F_K$. Together with [F6], its full volume-scaled norm bound is $\mathrm{\parallel }{\mathrm{\nabla }}_{\widehat{x}}\widehat{w}{\mathrm{\parallel }}_{L^2(\widehat{K})}\le \mathrm{\parallel }{A}_K\mathrm{\parallel }\mathrm{\mid }\det A_K{\mathrm{\mid }}^{-1\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{1-d\mathrm{/}2}\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2(K)}$, while $\mathrm{\parallel }\widehat{w}{\mathrm{\parallel }}_{L^2(\widehat{K})}^2=\mathrm{\mid }\det A_K{\mathrm{\mid }}^{-1}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(K)}^2$. Multiply the squared reference trace inequality of step 3.1 by the facet surface Jacobian $J_F\le C{h}_K^{d-1}$ from [F3]. This gives $\mathrm{\parallel }{T}_{F}w{\mathrm{\parallel }}_{L^2(F)}^2\le C{J}_F\mathrm{\mid }\det A_K{\mathrm{\mid }}^{-1}(\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(K)}^2+\mathrm{\parallel }{A}_K{\mathrm{\parallel }}^2\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2(K)}^2)\le C(h_K^{-1}\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(K)}^2+h_K\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2(K)}^2)$. The same calculation applies componentwise to vector fields and the squared component norms add.

For $0<s<1$, the identity $\mathrm{\parallel }w-w_K{\mathrm{\parallel }}_K^2=(2\mathrm{\mid }K\mathrm{\mid }{)}^{-1}{\int }_K{\int }_K\mathrm{\mid }w(x)-w(y){\mathrm{\mid }}^{2}dxdy$ and $\mathrm{\mid }x-y\mathrm{\mid }\le h_K$ give $\mathrm{\parallel }w-w_K{\mathrm{\parallel }}_K^2\le (2\mathrm{\mid }K\mathrm{\mid }{)}^{-1}{h}_K^{d+2s}\mathrm{\mid }w{\mathrm{\mid }}_{s,K}^2\le C{h}_K^{2s}\mathrm{\mid }w{\mathrm{\mid }}_{s,K}^2$ by [F3]. At $s=1$, apply the reference mean-zero Poincaré inequality [F4] and the affine weak-derivative pullback [F8] with the geometric bounds [F3]. This proves the stated fractional Poincaré bound and the endpoint convention.

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 $C_{\mathrm{\partial }}=\sqrt{N_{\mathrm{\partial }}}{C}_{\mathrm{t}\mathrm{r}}\ge 1$.