Zero-boundary conforming polynomials lie in W0 1p and satisfy Poincare

If $\mathrm{\Omega }=\mathrm{\varnothing }$, then $v=0$ and the conclusion is immediate. Assume henceforth $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$. Extend $v$ by zero to $\tilde{v}$ on ${\mathbb{R}}^d$ and define $g_i$ to be its cellwise derivative on mesh cells and zero outside $\bar{\mathrm{\Omega }}$. On each cell apply the simplex Green formula to $v\phi e_i$ for an arbitrary $\phi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$. Fact [F1] gives the cell identity. Interior facet terms cancel by [F2]; boundary facet terms vanish by the zero-boundary clause of [F2]. Summing the finitely many cell identities gives ${\int }_{{\mathbb{R}}^d}\tilde{v}{\mathrm{\partial }}_i\phi =-{\int }_{{\mathbb{R}}^d}{g}_i\phi \mathrm{.}$ Both functions are in $L^p$, so [F7] gives $\tilde{v}\in W^{1,p}({\mathbb{R}}^d)$ with gradient $g$.

Put $d_{\mathrm{\Omega }}(x)=\mathrm{dist}(x,{\mathbb{R}}^d\setminus \mathrm{\Omega })$, which is $1$-Lipschitz. Choose a smooth scalar step $\theta$ with $\theta (s)=0$ for $s\le 1$, $\theta (s)=1$ for $s\ge 2$, and $0\le \theta \le 1$. For $ϵ>0$ set ${\zeta }_{ϵ}(x)=\theta (d_{\mathrm{\Omega }}(x)\mathrm{/}ϵ)$ and convolve it with a nonnegative smooth mollifier of integral one supported in the ball of radius $ϵ\mathrm{/}4$, obtaining ${\eta }_{ϵ}$. Then ${\eta }_{ϵ}$ is smooth, supported in $\mathrm{\Omega }$, equals zero where $d_{\mathrm{\Omega }}\le 3ϵ\mathrm{/}4$, equals one where $d_{\mathrm{\Omega }}\ge 9ϵ\mathrm{/}4$, and $\mathrm{\parallel }\mathrm{\nabla }{\eta }_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }{\theta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}ϵ$. Since $\mathrm{\Omega }$ is bounded, its support is compact in $\mathrm{\Omega }$.

Let $M$ be the maximum of the cellwise suprema of $\mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }$. Across every interior facet the polynomial traces agree; on every boundary facet they agree with zero. For a point $x$ in the mesh skeleton, its minimal mesh face is the unique mesh face whose relative interior contains $x$; the local star of that face means the union of the cells containing it. If the minimal face is a facet, there are exactly two incident cells at an interior facet and exactly one at a boundary facet. In the interior case their values at $x$ agree by the matching polynomial traces; in the boundary case the single cell value agrees with exterior zero by the zero boundary trace. For the remaining faces, let $c\ge 2$ be their codimension. The incident simplicial cones in their local star cut a small sphere in the normal space of the face, of dimension $c-1$, into a finite face-to-face triangulation of the local directions in $\mathrm{\Omega }$. Within each connected component, top-dimensional pieces are connected by facet adjacency: perturb a path in that open component to avoid the finite union of codimension-two faces. At an interior point the direction set is the whole sphere, which is connected because $c\ge 2$, so matching traces make all incident polynomial values agree. At a boundary point every component has a boundary direction; following facet adjacencies reaches a boundary facet, where the polynomial value is zero, so all cell limits in that component are zero. Thus in all cases the cellwise continuous representatives and the exterior zero fit continuously on the whole finite polyhedral partition. Restrict this continuous function to any line segment. The segment is divided by the finitely many cell facets into finitely many intervals; on each cell interval its derivative along the segment has magnitude at most $M$, and on exterior intervals it is zero. Continuity at the interval endpoints then gives $\mathrm{\mid }\tilde{v}(x)-\tilde{v}(y)\mathrm{\mid }\le M\mathrm{\mid }x-y\mathrm{\mid }$ for all $x,y$. Thus $\mathrm{\mid }v(x)\mathrm{\mid }\le M\mathrm{dist}(x,{\mathbb{R}}^d\setminus \mathrm{\Omega })$.

The difference $v-{\eta }_{ϵ}v$ and its weak gradient are supported in $E_{ϵ}=\{x\in \mathrm{\Omega }:d_{\mathrm{\Omega }}(x)<9ϵ\mathrm{/}4\}$. On this layer, step 2.1 gives $\mathrm{\mid }v\mathrm{\mid }\le (9\mathrm{/}4)Mϵ$, and step 1.2 gives $\mathrm{\mid }v\mathrm{\nabla }{\eta }_{ϵ}\mathrm{\mid }\le (9\mathrm{/}4)M\mathrm{\parallel }{\theta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Also $\mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }\le M$. For $n\in {\mathbb{N}}_0$, the sets $E_{1\mathrm{/}(n+1)}$ decrease to the empty set and $\mathrm{\mid }{E}_1\mathrm{\mid }\le \mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }<\mathrm{\infty }$, so [F6] gives $\mathrm{\mid }{E}_{1\mathrm{/}(n+1)}\mathrm{\mid }\to 0$. These bounds show $\mathrm{\parallel }v-{\eta }_{1\mathrm{/}(n+1)}v{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}\to 0$ for finite $p$.

Each ${\eta }_{1\mathrm{/}(n+1)}v$ has compact support in $\mathrm{\Omega }$. By [F5] extend it by zero to $W^{1,p}({\mathbb{R}}^d)$ and approximate there by ${\varphi }_j\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$. Choose a smooth cutoff $\chi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one near the compact support of ${\eta }_{1\mathrm{/}(n+1)}v$. Then $\chi {\varphi }_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ and, by the weak product rule, $\chi {\varphi }_j\to {\eta }_{1\mathrm{/}(n+1)}v$ in $W^{1,p}(\mathrm{\Omega })$. A diagonal choice with step 3.1 proves $v\in W_0^{1,p}(\mathrm{\Omega })$ by [F3].

A bounded set lies in a slab of finite width $b-a$. Apply [F4] to the membership from step 4.1, obtaining the displayed inequality with $C_{\mathrm{\Omega }}=C(p)(b-a)$ for the fixed exponent $p$.