A single collar-reflection and free-tail exterior extension

If $C=\mathrm{\varnothing }$, take $E=I$; henceforth let $C\mathrm{\ne }\mathrm{\varnothing }$. Choose $\eta \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$ equal to $1$ near $C$. Both $\mathrm{\partial }\mathrm{\Omega }\subset C$ and $K=\mathrm{supp}\eta \cap \bar{\mathrm{\Omega }}$ are closed and bounded, hence compact, and $\mathrm{\partial }\mathrm{\Omega }\subset K$. At each boundary point use the stated graph data and put $\mathrm{\Phi }(x)=(y,z-h(y))$. This is a smooth coordinate map with inverse ${\mathrm{\Phi }}^{-1}(y,t)=R^{-1}(y,h(y)+t)$, taking $\mathrm{\Omega }\cap W$ exactly to $\mathrm{\Phi }(W)\cap \{t>0\}$. Since $\mathrm{\Phi }(W)$ is open and contains $\mathrm{\Phi }(a)=(y_a,0)$, choose a bounded ball $B_a$ centered at $y_a$ and $b_a>0$ so that ${\bar{B}}_a\times [-b_a,b_a]\subset \mathrm{\Phi }(W)$, with ${\bar{B}}_a\subset B$. Choose a concentric $B_a^{\mathrm{\prime }}$ with $\overline{B_a^{\mathrm{\prime }}}\subset B_a$ and $0<b_a^{\mathrm{\prime }}<b_a$. The smaller chart $U_a^{\mathrm{\prime }}={\mathrm{\Phi }}^{-1}(B_a^{\mathrm{\prime }}\times (-b_a^{\mathrm{\prime }},b_a^{\mathrm{\prime }}))$ has compact closure inside $U_a={\mathrm{\Phi }}^{-1}(B_a\times (-b_a,b_a))$. Reflection $(y,t)\mapsto (y,-t)$ preserves the smaller closed cylinder, giving the required reflected-support margin. The first derivatives of these chart maps and their inverses are bounded on the larger closed cylinder because the derivatives of $h$ are continuous on ${\bar{B}}_a$; their Jacobians have absolute determinant one. For $n=1$, use $\mathrm{\Phi }(x)=\epsilon (x-a)$ and two nested symmetric intervals, which have the same properties.

The sets $U_a^{\mathrm{\prime }}$ cover the compact boundary, so retain finitely many, say $U_1^{\mathrm{\prime }},\dots ,U_m^{\mathrm{\prime }}$. The set $K\setminus {\bigcup }_i{U}_i^{\mathrm{\prime }}$ is closed and bounded, hence compact, and lies in $\mathrm{\Omega }$: every point of $\bar{\mathrm{\Omega }}$ outside $\mathrm{\Omega }$ is a boundary point and was covered. Cover this remainder by finitely many balls $V_j$ with ${\bar{V}}_j\subset \mathrm{\Omega }$. Apply the finite ambient partition supplier to the resulting finite open cover of $K$, obtaining ${\chi }_i\in C_c^{\mathrm{\infty }}(U_i^{\mathrm{\prime }})$ and ${\chi }_{0j}\in C_c^{\mathrm{\infty }}(V_j)$ with total sum one near $K$. Put ${\rho }_i=\eta {\chi }_i$ and ${\rho }_0=\eta {\sum }_j{\chi }_{0j}$. Then ${\rho }_0$ has compact support in $\mathrm{\Omega }$ and ${\rho }_0+{\sum }_i{\rho }_i=\eta$ on $\mathrm{\Omega }$: wherever $\eta$ is nonzero in $\mathrm{\Omega }$ the point lies in $K$, and elsewhere both sides are zero. Each ${\rho }_i$ has compact support in the smaller symmetric chart from step 1.1, so its reflected support stays compactly inside $U_i$. Choose a final smooth cutoff compactly supported in $U_i$ and equal to one near both these supports. These are precisely the chart and cutoff data needed by the localization supplier. All this data is fixed independently of $q$.

For $u\in L^q(\mathrm{\Omega })$ define $T_{\mathrm{\infty }}u$ as the zero extension of $(1-\eta )u$. It is $L^q$ bounded. If $u\in W^{1,q}$, fix $\varphi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^n)$; then $(1-\eta )\varphi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, so the global weak derivative of $T_{\mathrm{\infty }}u$ is the zero extension of $(1-\eta ){\mathrm{\partial }}_{j}u-({\mathrm{\partial }}_j\eta )u$. Thus $\mathrm{\parallel }{T}_{\mathrm{\infty }}u{\mathrm{\parallel }}_{W^{1,q}}\le C_{\eta }\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,q}}$. This test argument does not require bounded support of the tail. The zero extension $T_{0}u$ of ${\rho }_{0}u$ has both endpoint bounds by its compact interior support.

Define $Eu=T_{\mathrm{\infty }}u+T_{0}u+{\sum }_{i=1}^m{T}_{i}u$, using the chart operators of the localization supplier with the data in step 2.1. This is one formula for every $L^q$ input; for a $W^{1,q}$ input it is exactly the same formula. The finite sum of the $L^q$ bounds gives $\mathrm{\parallel }Eu{\mathrm{\parallel }}_q\le C_q\mathrm{\parallel }u{\mathrm{\parallel }}_q$, and the finite sum of the first-derivative bounds gives $\mathrm{\parallel }Eu{\mathrm{\parallel }}_{W^{1,q}}\le {\tilde{C}}_q\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,q}}$. On $\mathrm{\Omega }$ the sum equals $((1-\eta )+{\rho }_0+{\sum }_i{\rho }_i)u=u$. This proves $PE=I$, the norm bounds and the common-operator assertion. All choices depend only on the geometry and cutoffs; reflection is the same even formula for every $q$, which proves the final claim.