[proofplan] The construction extends $u \in W^{1,p}(\Omega)$ to all of $\mathbb{R}^n$ in three stages. First, a finite atlas of bi-Lipschitz boundary charts $\Phi_j: B_j \to B(0,1)$ flattens $\partial\Omega$ to $\{x_n = 0\}$ near each $\partial\Omega \cap B_j$, with $\Omega \cap B_j$ mapped onto the upper half-ball. Second, on each flattened chart, the [Half-Space Reflection](/theorems/3101) extends $u$ across the boundary; pulled back via $\Phi_j$, this gives an extension of a localised piece of $u$ across $\partial\Omega$ in $B_j$. Third, a smooth partition of unity $\{\zeta_0, \zeta_1, \dots, \zeta_k\}$ subordinate to $\{\Omega \setminus K\} \cup \{B_j\}_j$ (where $K$ is a compact set near $\partial\Omega$) glues the local extensions into a global one, with the support property (ii) coming from the cutoff structure. The norm bound (iii) accumulates from the bi-Lipschitz pull-back constants, the reflection constant $2^{1/p}$, the Leibniz rule applied to $\zeta_j u$, and finiteness of the cover. [/proofplan] [step:Set up the boundary atlas and flatten $\partial\Omega$ to a hyperplane in each chart] By definition of a Lipschitz domain, there exist finitely many open sets $B_1, \dots, B_k \subset \mathbb{R}^n$ covering $\partial\Omega$ and bi-Lipschitz maps \begin{align*} \Phi_j: B_j &\to B(0,1) \subset \mathbb{R}^n \\ x &\mapsto \Phi_j(x), \qquad j = 1, \dots, k, \end{align*} each satisfying $\Phi_j(B_j \cap \Omega) = B(0,1) \cap \mathbb{R}^n_+$ and $\Phi_j(B_j \cap \partial\Omega) = B(0,1) \cap \{x_n = 0\}$, where $\mathbb{R}^n_+ := \{(x', x_n) : x_n > 0\}$. Let $L_j \ge 1$ denote the maximum of the Lipschitz constants of $\Phi_j$ and $\Phi_j^{-1}$, and set $L := \max_j L_j$. Choose an open set $\Omega_0$ with $\overline{\Omega_0} \subset \Omega$ such that $\Omega \setminus \Omega_0 \subset \bigcup_{j=1}^k B_j$ — this is possible because $\partial\Omega \subset \bigcup_{j=1}^k B_j$ is open and $\partial\Omega$ is compact. Define the open set \begin{align*} V := \Omega_0 \cup \bigcup_{j=1}^k B_j, \end{align*} so that $V$ is a fixed open neighbourhood of $\overline{\Omega}$ in $\mathbb{R}^n$. We will arrange $\operatorname{supp}(Eu) \subset \overline{V}$ to fulfil property (ii). Fix a smooth partition of unity $\{\zeta_0, \zeta_1, \dots, \zeta_k\}$ on $\overline{\Omega}$ subordinate to the cover $\{\Omega_0, B_1, \dots, B_k\}$: \begin{align*} \zeta_j &\in C_c^\infty(\mathbb{R}^n), \qquad \operatorname{supp}\zeta_0 \subset \Omega_0, \quad \operatorname{supp}\zeta_j \subset B_j\ (j \ge 1), \\ 0 \le \zeta_j &\le 1, \qquad \sum_{j=0}^k \zeta_j = 1\ \text{on a neighbourhood of}\ \overline{\Omega}. \end{align*} Set $M := \max_{0 \le j \le k} \|\nabla\zeta_j\|_{L^\infty(\mathbb{R}^n)}$, a finite constant depending only on $\Omega$. [guided] The Lipschitz boundary hypothesis gives a finite atlas: there exist open sets $B_1, \dots, B_k$ in $\mathbb{R}^n$ that together cover $\partial\Omega$, and for each $B_j$ a bi-Lipschitz homeomorphism $\Phi_j: B_j \to B(0,1)$ that **straightens** the boundary inside $B_j$. Concretely, $\Phi_j$ sends the portion of $\Omega$ inside $B_j$ to the upper half of the unit ball, and the portion of $\partial\Omega$ inside $B_j$ to the equatorial disk. Each $\Phi_j$ and its inverse have Lipschitz constants bounded by some $L_j \ge 1$, and we set $L := \max_j L_j$ to denote the worst case. We need a fixed open neighbourhood $V$ of $\overline\Omega$ that contains the supports of all our extensions; eventually property (ii) will read $\operatorname{supp}(Eu) \subset \overline V$. To build $V$, pick an open set $\Omega_0$ deep inside $\Omega$ — that is, $\overline{\Omega_0} \subset \Omega$ — with the property that everything outside $\Omega_0$ in $\Omega$ is already covered by the boundary charts: \begin{align*} \Omega \setminus \Omega_0 \subset \bigcup_{j=1}^k B_j. \end{align*} This is possible because $\partial\Omega$ is compact (being a closed subset of $\overline\Omega$, which is compact since $\Omega$ is bounded) and the union $\bigcup_j B_j$ is an open cover of $\partial\Omega$, so there is room to thicken $\partial\Omega$ slightly inwards while staying inside the cover. Then set \begin{align*} V := \Omega_0 \cup \bigcup_{j=1}^k B_j, \end{align*} which is an open neighbourhood of $\overline\Omega$. We now choose a partition of unity $\{\zeta_0, \zeta_1, \dots, \zeta_k\}$ subordinate to the cover $\{\Omega_0, B_1, \dots, B_k\}$ of $\overline\Omega$. Standard PoU theorems give us smooth $\zeta_j \in C_c^\infty(\mathbb{R}^n)$ with $\operatorname{supp}\zeta_0 \subset \Omega_0$, $\operatorname{supp}\zeta_j \subset B_j$ for $j \ge 1$, $0 \le \zeta_j \le 1$, and $\sum_{j=0}^k \zeta_j \equiv 1$ on a neighbourhood of $\overline\Omega$. Each $\nabla\zeta_j$ is bounded; we record $M := \max_j \|\nabla\zeta_j\|_{L^\infty}$ for later use in the Leibniz rule. The interior cutoff $\zeta_0 u$ is supported in $\overline{\Omega_0} \subset \Omega$, which has positive distance from $\partial\Omega$ since $\overline{\Omega_0}$ is compact and $\overline{\Omega_0} \cap \partial\Omega = \emptyset$; therefore the zero extension of $\zeta_0 u$ to $\mathbb{R}^n$ lies in $W^{1,p}(\mathbb{R}^n)$ with the same norm, by the standard zero-extension lemma for $W^{1,p}$ functions whose support is compactly contained in an open set. The boundary cutoffs $\zeta_j u$ for $j \ge 1$ are supported in $B_j$ but generally **not** away from $\partial\Omega$ — extending them across $\partial\Omega$ is precisely the role of Stage 2 below. [/guided] [/step] [step:Pull back each boundary piece $\zeta_j u$ via $\Phi_j$ to a $W^{1,p}$ function on the upper half-ball] Fix $j \in \{1, \dots, k\}$. The function $\zeta_j u$ belongs to $W^{1,p}(\Omega)$ by the [Product Rule for Weak Derivatives](/theorems/3098): since $\zeta_j \in C_c^\infty(\mathbb{R}^n) \subset L^\infty$ and $u \in W^{1,p}(\Omega) \cap L^\infty_{\mathrm{loc}}$ when $p < \infty$ (and $u \in L^\infty(\Omega)$ by definition when $p = \infty$), we have \begin{align*} \nabla(\zeta_j u) = \zeta_j\, \nabla u + u\, \nabla\zeta_j \quad \text{a.e. on } \Omega, \end{align*} and consequently \begin{align*} \|\zeta_j u\|_{W^{1,p}(\Omega)} &\le \|\zeta_j u\|_{L^p(\Omega)} + \|\zeta_j \nabla u\|_{L^p(\Omega)} + \|u \nabla\zeta_j\|_{L^p(\Omega)} \\ &\le (1 + M)\|u\|_{W^{1,p}(\Omega)}, \end{align*} where we used $0 \le \zeta_j \le 1$ and $|\nabla\zeta_j| \le M$. (For $p = \infty$ replace the chain by the obvious $L^\infty$ estimate; the bound $(1 + M)$ is identical.) Here we have applied the product rule when $u \in L^\infty$; in general for $u \in W^{1,p}$ with $\zeta_j$ smooth and compactly supported the same identity holds, since smoothness and boundedness of $\zeta_j$ ensure $\zeta_j u \in W^{1,p}$ regardless of whether $u \in L^\infty$. The function $\zeta_j u$ extends by zero to $W^{1,p}(B_j \cap \Omega)$, since $\zeta_j$ vanishes near $\partial B_j$. Define the pulled-back function \begin{align*} v_j: B(0,1) \cap \mathbb{R}^n_+ &\to \mathbb{R} \\ y &\mapsto (\zeta_j u)(\Phi_j^{-1}(y)). \end{align*} The change-of-variables theorem for Sobolev functions under bi-Lipschitz maps yields $v_j \in W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)$, with the chain rule \begin{align*} \nabla v_j(y) = \nabla(\zeta_j u)(\Phi_j^{-1}(y))\, J\Phi_j^{-1}(y) \quad \text{a.e. } y \in B(0,1) \cap \mathbb{R}^n_+, \end{align*} where $J\Phi_j^{-1}(y)$ is the Jacobian matrix of $\Phi_j^{-1}$ at $y$ (which exists $\mathcal{L}^n$-a.e. by Rademacher's theorem applied to the Lipschitz map $\Phi_j^{-1}$). The $L^p$ bounds are \begin{align*} \|v_j\|_{L^p(B(0,1) \cap \mathbb{R}^n_+)}^p &= \int_{B_j \cap \Omega} |\zeta_j u(x)|^p\, |\det J\Phi_j(x)|\, d\mathcal{L}^n(x) \le L^n \|\zeta_j u\|_{L^p(\Omega)}^p, \\ \|\nabla v_j\|_{L^p(B(0,1) \cap \mathbb{R}^n_+)}^p &\le L^{p+n} \|\nabla(\zeta_j u)\|_{L^p(\Omega)}^p, \end{align*} where $|\det J\Phi_j| \le L^n$ holds $\mathcal{L}^n$-a.e. by the Lipschitz bound, and $|J\Phi_j^{-1}| \le L$ similarly. Combining, \begin{align*} \|v_j\|_{W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)} \le L^{(p+n)/p}\|\zeta_j u\|_{W^{1,p}(\Omega)} \le L^{(p+n)/p}(1 + M)\|u\|_{W^{1,p}(\Omega)}. \end{align*} Since $\zeta_j$ vanishes near $\partial B_j$, the function $v_j$ vanishes near $\partial B(0,1)$ — that is, $v_j \in W^{1,p}_{\text{loc, side-supp}}$ has support compactly contained in $B(0,1)$ in the side and top directions, but generically reaches the equator $\{y_n = 0\}$ from above. Extending $v_j$ by zero in the side and top directions gives a function $\tilde v_j \in W^{1,p}(\mathbb{R}^n_+)$ with $\operatorname{supp}\tilde v_j \subset \overline{B(0,1)} \cap \overline{\mathbb{R}^n_+}$. Strict containment of the support away from $\partial B(0,1)$ in the lateral directions ensures $\tilde v_j \in W^{1,p}(\mathbb{R}^n_+)$ via the standard zero-extension lemma for $W^{1,p}$ functions vanishing on a Lipschitz portion of the boundary; the $W^{1,p}$ norm is preserved. [guided] The aim of this step is to take the localised piece $\zeta_j u$, supported in $B_j \cap \overline\Omega$, and transport it via $\Phi_j$ to a $W^{1,p}$ function on the upper half of the unit ball. Two things have to be checked: that $\zeta_j u$ itself is in $W^{1,p}$ with a controlled norm, and that the pullback under the bi-Lipschitz map $\Phi_j$ preserves the Sobolev structure with a constant depending only on $L$ and $n$. **$\zeta_j u \in W^{1,p}(\Omega)$ with explicit norm bound.** Since $\zeta_j \in C_c^\infty(\mathbb{R}^n)$, in particular $\zeta_j \in L^\infty(\Omega)$ with $\|\zeta_j\|_{L^\infty} \le 1$ and $|\nabla\zeta_j| \le M$. The [Product Rule for Weak Derivatives](/theorems/3098) gives that $\zeta_j u \in W^{1,p}(\Omega)$ with \begin{align*} \nabla(\zeta_j u) = \zeta_j \nabla u + u \nabla\zeta_j \quad \mathcal{L}^n\text{-a.e. on } \Omega. \end{align*} Taking $L^p$ norms and applying the triangle inequality, \begin{align*} \|\nabla(\zeta_j u)\|_{L^p(\Omega)} \le \|\zeta_j \nabla u\|_{L^p(\Omega)} + \|u \nabla\zeta_j\|_{L^p(\Omega)} \le \|\nabla u\|_{L^p(\Omega)} + M\|u\|_{L^p(\Omega)}, \end{align*} using $0 \le \zeta_j \le 1$ and $|\nabla\zeta_j| \le M$. Adding $\|\zeta_j u\|_{L^p(\Omega)} \le \|u\|_{L^p(\Omega)}$, \begin{align*} \|\zeta_j u\|_{W^{1,p}(\Omega)} \le (1 + M)\|u\|_{W^{1,p}(\Omega)}. \end{align*} **Pulling back via $\Phi_j$.** Define the pullback as the map \begin{align*} v_j: B(0,1) \cap \mathbb{R}^n_+ &\to \mathbb{R}, & y &\mapsto (\zeta_j u)(\Phi_j^{-1}(y)). \end{align*} The change of variables for $W^{1,p}$ under bi-Lipschitz maps is a standard result: if $\Phi: U \to V$ is bi-Lipschitz between open sets in $\mathbb{R}^n$ and $w \in W^{1,p}(V)$, then $w \circ \Phi \in W^{1,p}(U)$ with \begin{align*} \nabla(w \circ \Phi)(x) = (\nabla w)(\Phi(x))\, J\Phi(x) \quad \mathcal{L}^n\text{-a.e. } x \in U. \end{align*} Here $J\Phi(x)$ is the Jacobian matrix of $\Phi$, which exists a.e. by Rademacher's theorem ([3069](/theorems/3069)). Applied to $w = \zeta_j u \in W^{1,p}(B_j \cap \Omega)$ and $\Phi = \Phi_j^{-1}: B(0,1) \cap \mathbb{R}^n_+ \to B_j \cap \Omega$, this yields the claimed identity for $\nabla v_j$. **$L^p$ bound on $v_j$.** By the change of variables formula for Lebesgue integrals under bi-Lipschitz maps, \begin{align*} \int_{B(0,1) \cap \mathbb{R}^n_+} |v_j(y)|^p\, d\mathcal{L}^n(y) = \int_{B_j \cap \Omega} |\zeta_j u(x)|^p\, |\det J\Phi_j(x)|\, d\mathcal{L}^n(x). \end{align*} Since $\Phi_j$ has Lipschitz constant $\le L$, each entry of $J\Phi_j$ is bounded by $L$, hence $|\det J\Phi_j| \le L^n$ a.e. (using the Hadamard inequality or simply that an $n \times n$ matrix with entries of absolute value $\le L$ has determinant of absolute value $\le n^{n/2} L^n$, but more sharply $\le L^n$ when the matrix is the Jacobian of a Lipschitz map: each row has Euclidean norm $\le L$, and the determinant is bounded by the product of row norms by the Hadamard inequality, giving $L^n$). Therefore \begin{align*} \|v_j\|_{L^p(B(0,1) \cap \mathbb{R}^n_+)}^p \le L^n \|\zeta_j u\|_{L^p(\Omega)}^p. \end{align*} **$L^p$ bound on $\nabla v_j$.** From $\nabla v_j(y) = (\nabla(\zeta_j u))(\Phi_j^{-1}(y))\, J\Phi_j^{-1}(y)$ and $|J\Phi_j^{-1}| \le L$ a.e., \begin{align*} |\nabla v_j(y)| \le L\, |\nabla(\zeta_j u)(\Phi_j^{-1}(y))| \quad \mathcal{L}^n\text{-a.e. } y. \end{align*} Raising to the $p$-th power and applying the change-of-variables formula again, \begin{align*} \|\nabla v_j\|_{L^p}^p \le L^p \int_{B(0,1) \cap \mathbb{R}^n_+} |\nabla(\zeta_j u)(\Phi_j^{-1}(y))|^p\, d\mathcal{L}^n(y) \le L^{p+n}\|\nabla(\zeta_j u)\|_{L^p(\Omega)}^p. \end{align*} **Combined Sobolev bound.** Adding the $L^p$ and $\nabla$ contributions, \begin{align*} \|v_j\|_{W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)} \le L^{(p+n)/p}\|\zeta_j u\|_{W^{1,p}(\Omega)} \le L^{(p+n)/p}(1 + M)\|u\|_{W^{1,p}(\Omega)}. \end{align*} **Extension by zero in lateral directions.** $\zeta_j$ vanishes near the boundary of $B_j$ (it has compact support strictly inside $B_j$), so $v_j$ vanishes near the lateral and top portion of $\partial B(0,1)$, i.e. $\partial B(0,1) \cap \overline{\mathbb{R}^n_+}$. The only portion of the boundary where $v_j$ may not vanish is the equator $\{y_n = 0\} \cap B(0,1)$ — and that is exactly the boundary across which we will reflect in Stage 2. Extending by zero in all directions other than the equator gives $\tilde v_j \in W^{1,p}(\mathbb{R}^n_+)$ with $\operatorname{supp}\tilde v_j \subset \overline{B(0,1)} \cap \overline{\mathbb{R}^n_+}$, and the same $W^{1,p}$ norm. [/guided] [/step] [step:Reflect $\tilde v_j$ across $\{y_n = 0\}$ to obtain a $W^{1,p}$ extension on all of $\mathbb{R}^n$] Apply the [Half-Space Reflection](/theorems/3101) to $\tilde v_j \in W^{1,p}(\mathbb{R}^n_+)$: \begin{align*} R: W^{1,p}(\mathbb{R}^n_+) &\to W^{1,p}(\mathbb{R}^n) \\ \tilde v_j &\mapsto R\tilde v_j, \qquad (R\tilde v_j)(y', y_n) := \begin{cases}\tilde v_j(y', y_n), & y_n > 0, \\ \tilde v_j(y', -y_n), & y_n < 0.\end{cases} \end{align*} By Theorem 3101, $R\tilde v_j \in W^{1,p}(\mathbb{R}^n)$ with \begin{align*} \|R\tilde v_j\|_{W^{1,p}(\mathbb{R}^n)} \le 2^{1/p}\|\tilde v_j\|_{W^{1,p}(\mathbb{R}^n_+)}. \end{align*} Moreover $\operatorname{supp}(R\tilde v_j) \subset \overline{B(0,1)}$, since the reflection maps the support $\subset \overline{B(0,1)} \cap \overline{\mathbb{R}^n_+}$ into $\overline{B(0,1)} \cap \overline{\mathbb{R}^n_-}$, and $\overline{B(0,1)}$ is invariant under $(y', y_n) \mapsto (y', -y_n)$. Now push back to $\mathbb{R}^n$ via $\Phi_j$. The map $\Phi_j: B_j \to B(0,1)$ does not directly extend to a bi-Lipschitz map $\mathbb{R}^n \to \mathbb{R}^n$, but we only need to define the extension on $B_j$ and zero outside. Define \begin{align*} w_j: \mathbb{R}^n &\to \mathbb{R} \\ x &\mapsto \begin{cases}(R\tilde v_j)(\Phi_j(x)), & x \in B_j, \\ 0, & x \notin B_j.\end{cases} \end{align*} Since $\operatorname{supp}(R\tilde v_j) \subset \overline{B(0,1)}$ and $\Phi_j$ is a bi-Lipschitz homeomorphism $B_j \to B(0,1)$, the function $w_j$ has $\operatorname{supp} w_j \subset \Phi_j^{-1}(\overline{B(0,1)}) = \overline{B_j} \cap B_j$ extended to its closure inside $B_j$; in particular $w_j$ vanishes near $\partial B_j$, so the zero extension produces $w_j \in W^{1,p}(\mathbb{R}^n)$. Moreover, by the same change-of-variables argument as in the previous step (now in the reverse direction), \begin{align*} \|w_j\|_{W^{1,p}(\mathbb{R}^n)} \le L^{(p+n)/p}\|R\tilde v_j\|_{W^{1,p}(\mathbb{R}^n)} \le L^{(p+n)/p} \cdot 2^{1/p}\|\tilde v_j\|_{W^{1,p}(\mathbb{R}^n_+)}. \end{align*} Combining with the bound from the previous step, \begin{align*} \|w_j\|_{W^{1,p}(\mathbb{R}^n)} \le 2^{1/p}\, L^{2(p+n)/p}(1 + M)\|u\|_{W^{1,p}(\Omega)} =: C_j\|u\|_{W^{1,p}(\Omega)}, \end{align*} with $C_j = C_j(\Omega, p, n)$ depending only on $\Omega$, $p$, and $n$ (since $L = L(\Omega)$ and $M = M(\Omega)$). The function $w_j$ has the restriction property $w_j|_\Omega \cdot \mathbb{1}_{B_j} = (\zeta_j u)|_\Omega \cdot \mathbb{1}_{B_j}$ in $\Omega$: indeed, for $x \in B_j \cap \Omega$, $\Phi_j(x) \in B(0,1) \cap \mathbb{R}^n_+$, where $R\tilde v_j$ coincides with $\tilde v_j$, and $\tilde v_j(\Phi_j(x)) = (\zeta_j u)(x)$ by the definition of $v_j$. Outside $B_j$, both $w_j$ and $\zeta_j u$ are zero (since $\operatorname{supp}\zeta_j \subset B_j$). [guided] **Reflecting on the flat half-ball.** Theorem 3101 — [Half-Space Reflection](/theorems/3101) — provides a bounded linear extension operator $R: W^{1,p}(\mathbb{R}^n_+) \to W^{1,p}(\mathbb{R}^n)$ with operator norm $2^{1/p}$ (for $1 \le p < \infty$; norm $1$ for $p = \infty$). Applied to $\tilde v_j$, the reflected function $R\tilde v_j$ is in $W^{1,p}(\mathbb{R}^n)$, satisfies $R\tilde v_j|_{\mathbb{R}^n_+} = \tilde v_j$, and obeys \begin{align*} \|R\tilde v_j\|_{W^{1,p}(\mathbb{R}^n)} \le 2^{1/p}\|\tilde v_j\|_{W^{1,p}(\mathbb{R}^n_+)}. \end{align*} The hypothesis of Theorem 3101 is exactly that $\tilde v_j \in W^{1,p}(\mathbb{R}^n_+)$, which we established in the previous step. **Why the support stays inside the unit ball.** $\operatorname{supp}\tilde v_j \subset \overline{B(0,1)} \cap \overline{\mathbb{R}^n_+}$ by construction. The reflection $(y', y_n) \mapsto (y', -y_n)$ preserves the unit ball $\overline{B(0,1)}$ (since $|y'|^2 + y_n^2 = |y'|^2 + (-y_n)^2$), hence sends the upper-half support into a subset of $\overline{B(0,1)} \cap \overline{\mathbb{R}^n_-}$. The full support of $R\tilde v_j$ is therefore contained in $\overline{B(0,1)}$. **Pushing back to $B_j \subset \mathbb{R}^n$.** We want to transfer $R\tilde v_j$ from $B(0,1)$ back to the original geometry inside $B_j$. Define \begin{align*} w_j: \mathbb{R}^n &\to \mathbb{R}, & x &\mapsto \begin{cases}(R\tilde v_j)(\Phi_j(x)), & x \in B_j, \\ 0, & x \notin B_j.\end{cases} \end{align*} This is well-defined because $\Phi_j$ is a bi-Lipschitz homeomorphism $B_j \to B(0,1)$. To check that $w_j \in W^{1,p}(\mathbb{R}^n)$ — and not merely $W^{1,p}(B_j)$ — we need the gluing across $\partial B_j$ to be benign. The key is that $\operatorname{supp}(R\tilde v_j) \subset \overline{B(0,1)}$, hence after pulling back, $\operatorname{supp}(w_j|_{B_j}) \subset \Phi_j^{-1}(\overline{B(0,1)})$. Now $\Phi_j^{-1}$ is continuous on $\overline{B(0,1)}$ and maps into $\overline{B_j}$, so $\Phi_j^{-1}(\overline{B(0,1)}) \subset \overline{B_j}$. **Crucially**, we also need the support of $w_j$ in $B_j$ to be **away** from $\partial B_j$, i.e. there must be a positive gap. This is guaranteed because $\zeta_j$ has compact support strictly inside $B_j$ — let $K_j := \operatorname{supp}\zeta_j \subset B_j$ be compact. Then $v_j$ is supported inside $\Phi_j(K_j)$, which is a compact subset of $B(0,1)$. After reflection, $R\tilde v_j$ is supported in $\Phi_j(K_j) \cup \{(y', -y_n) : (y', y_n) \in \Phi_j(K_j)\}$, still a compact subset of $B(0,1)$. Pulling back, $w_j|_{B_j}$ is supported in a compact subset of $B_j$, with positive distance to $\partial B_j$. Therefore the zero extension across $\partial B_j$ is a $W^{1,p}$ function — the standard fact that a $W^{1,p}$ function with compact support inside an open set extends by zero to all of $\mathbb{R}^n$ in $W^{1,p}$. **Norm bound for $w_j$.** The change-of-variables computation in the reverse direction works exactly as before: \begin{align*} \|w_j\|_{L^p(\mathbb{R}^n)}^p = \int_{B_j} |R\tilde v_j(\Phi_j(x))|^p\, d\mathcal{L}^n(x) = \int_{B(0,1)}|R\tilde v_j(y)|^p\, |\det J\Phi_j^{-1}(y)|\, d\mathcal{L}^n(y) \le L^n\|R\tilde v_j\|_{L^p(B(0,1))}^p, \end{align*} and similarly $\|\nabla w_j\|_{L^p}^p \le L^{p+n}\|\nabla(R\tilde v_j)\|_{L^p}^p$, leading to \begin{align*} \|w_j\|_{W^{1,p}(\mathbb{R}^n)} \le L^{(p+n)/p}\|R\tilde v_j\|_{W^{1,p}(\mathbb{R}^n)} \le L^{(p+n)/p}\cdot 2^{1/p}\|\tilde v_j\|_{W^{1,p}(\mathbb{R}^n_+)}. \end{align*} Chaining this with the bound $\|\tilde v_j\|_{W^{1,p}(\mathbb{R}^n_+)} \le L^{(p+n)/p}(1+M)\|u\|_{W^{1,p}(\Omega)}$ from the previous step, \begin{align*} \|w_j\|_{W^{1,p}(\mathbb{R}^n)} \le 2^{1/p}\, L^{2(p+n)/p}(1+M)\|u\|_{W^{1,p}(\Omega)} =: C_j\|u\|_{W^{1,p}(\Omega)}. \end{align*} The constant $C_j$ depends only on $\Omega$ (through $L$ and $M$), $p$, and $n$ — not on $u$. **Restriction property.** For $x \in B_j \cap \Omega$, $\Phi_j(x) \in B(0,1) \cap \mathbb{R}^n_+$, so $R\tilde v_j(\Phi_j(x)) = \tilde v_j(\Phi_j(x)) = v_j(\Phi_j(x)) = \zeta_j u(\Phi_j^{-1}(\Phi_j(x))) = \zeta_j u(x)$. Outside $B_j$, both sides vanish. So $w_j = \zeta_j u$ on $\Omega$. [/guided] [/step] [step:Glue the local extensions via the partition of unity to define $E$] Define the extension operator \begin{align*} E: W^{1,p}(\Omega) &\to W^{1,p}(\mathbb{R}^n) \\ u &\mapsto Eu := \widetilde{\zeta_0 u} + \sum_{j=1}^k w_j, \end{align*} where $\widetilde{\zeta_0 u}$ denotes the extension of $\zeta_0 u \in W^{1,p}(\Omega)$ by zero to $\mathbb{R}^n$. Since $\operatorname{supp}\zeta_0 \subset \Omega_0$ with $\overline{\Omega_0} \subset \Omega$, the function $\zeta_0 u$ has support compactly contained in $\Omega$ and the zero extension is a $W^{1,p}$ function with $\|\widetilde{\zeta_0 u}\|_{W^{1,p}(\mathbb{R}^n)} = \|\zeta_0 u\|_{W^{1,p}(\Omega)} \le (1 + M)\|u\|_{W^{1,p}(\Omega)}$. Linearity of $E$ follows because each step (multiplication by $\zeta_j$, pullback, reflection $R$, push-back) is linear, and the sum of linear operators is linear. **Verification of (i) — restriction.** For $x \in \Omega$, $\sum_{j=0}^k \zeta_j(x) = 1$, and: - $\widetilde{\zeta_0 u}(x) = \zeta_0(x) u(x)$; - $w_j(x) = \zeta_j(x) u(x)$ for each $j \ge 1$ (established at the end of the previous step). Summing, \begin{align*} Eu(x) = \zeta_0(x) u(x) + \sum_{j=1}^k \zeta_j(x) u(x) = u(x) \sum_{j=0}^k \zeta_j(x) = u(x) \quad \text{for } x \in \Omega. \end{align*} **Verification of (ii) — support.** Each $w_j$ satisfies $\operatorname{supp} w_j \subset \overline{B_j} \subset \overline{V}$, and $\operatorname{supp}\widetilde{\zeta_0 u} \subset \overline{\Omega_0} \subset \overline{V}$. Hence $\operatorname{supp}(Eu) \subset \overline{V}$. Since $V$ is fixed and $V \supset\supset \Omega$ (a strict superset with $\overline\Omega \subset V$), property (ii) holds with this $V$. **Verification of (iii) — norm bound.** By the triangle inequality, \begin{align*} \|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le \|\widetilde{\zeta_0 u}\|_{W^{1,p}(\mathbb{R}^n)} + \sum_{j=1}^k \|w_j\|_{W^{1,p}(\mathbb{R}^n)} \le \left[(1 + M) + \sum_{j=1}^k C_j\right]\|u\|_{W^{1,p}(\Omega)}. \end{align*} Setting $C(\Omega, p) := (1 + M) + \sum_{j=1}^k C_j$, which depends only on $\Omega$ (through $L$, $M$, $k$), on $p$, and on $n$, gives the desired estimate $\|Eu\|_{W^{1,p}(\mathbb{R}^n)} \le C(\Omega, p)\|u\|_{W^{1,p}(\Omega)}$. This completes the construction and verification of all three properties. [/step]