[proofplan] The forward inclusion $W^{1,p}_0(\Omega) \subseteq \ker T$ is the easy direction: by definition, $u \in W^{1,p}_0(\Omega)$ admits an approximating sequence $u_k \in C_c^\infty(\Omega)$, each $u_k$ has zero classical trace because its support is compact in $\Omega$, and the continuity of the trace operator from the [Sobolev Trace Theorem](/theorems/60) extends this to $u$. The reverse inclusion $\ker T \subseteq W^{1,p}_0(\Omega)$ is the substantive one. We use a partition of unity adapted to the Lipschitz boundary to localise to half-space charts. On each boundary chart we cut off using a function $\psi_h$ that vanishes near the boundary and equals $1$ away from it: $u\, \psi_h$ has compact support inside the chart's interior side, and we must show $u\, \psi_h \to u$ in $W^{1,p}$ as $h \to 0$. The key estimate is a Hardy-type inequality: $Tu = 0$ controls the $L^p$ norm of $u$ in a strip near $\partial\Omega$ in terms of $h\|\nabla u\|_{L^p}$, which is enough to handle the cutoff error. After cutting off we mollify to land in $C_c^\infty(\Omega)$, witnessing $u \in W^{1,p}_0(\Omega)$. [/proofplan]

[step:Forward direction: if $u \in W^{1,p}_0(\Omega)$ then $Tu = 0$]By the definition of $W^{1,p}_0(\Omega)$ as the closure of $C_c^\infty(\Omega)$ in the $W^{1,p}$ norm, given $u \in W^{1,p}_0(\Omega)$ there exists a sequence \begin{align*} u_k \in C_c^\infty(\Omega), \qquad u_k \to u \quad \text{in } W^{1,p}(\Omega). \end{align*} For each $k$, $\operatorname{supp} u_k$ is a compact subset of $\Omega$, so there is a positive distance $d_k := \operatorname{dist}(\operatorname{supp} u_k, \partial\Omega) > 0$. In particular, $u_k$ vanishes on a neighbourhood of $\partial\Omega$. Since $u_k \in C(\overline\Omega) \cap W^{1,p}(\Omega)$ (extension of $u_k$ to $\overline\Omega$ as the zero function near the boundary is continuous), the trace property of the operator $T$ from the [Sobolev Trace Theorem](/theorems/60) applies: \begin{align*} Tu_k = u_k|_{\partial\Omega} = 0 \quad \text{in } L^p(\partial\Omega; \mathcal{H}^{n-1}). \end{align*} The trace operator $T$ is bounded: there exists $C_T = C_T(\Omega, p)$ such that $\|Tw\|_{L^p(\partial\Omega; \mathcal{H}^{n-1})} \le C_T\|w\|_{W^{1,p}(\Omega)}$ for all $w \in W^{1,p}(\Omega)$. Applying this to $w = u - u_k$, \begin{align*} \|Tu\|_{L^p(\partial\Omega; \mathcal{H}^{n-1})} = \|Tu - Tu_k\|_{L^p(\partial\Omega; \mathcal{H}^{n-1})} = \|T(u - u_k)\|_{L^p(\partial\Omega; \mathcal{H}^{n-1})} \le C_T\|u - u_k\|_{W^{1,p}(\Omega)} \to 0. \end{align*} Hence $\|Tu\|_{L^p(\partial\Omega; \mathcal{H}^{n-1})} = 0$, so $Tu = 0$ in $L^p(\partial\Omega; \mathcal{H}^{n-1})$.[/step]

[guided]The forward direction is short, but uses two facts that should be made explicit: (i) the trace of any $C_c^\infty(\Omega)$ function is zero, and (ii) the trace operator is continuous, so the zero-trace property survives the $W^{1,p}$ limit. **Why $Tu_k = 0$ for $u_k \in C_c^\infty(\Omega)$.** The support of $u_k$ is a compact subset of $\Omega$, so $u_k$ vanishes on $\Omega \setminus \operatorname{supp} u_k$ — an open set containing all of $\partial\Omega$. The trace operator from the [Sobolev Trace Theorem](/theorems/60) is constructed so that $Tw = w|_{\partial\Omega}$ whenever $w \in C(\overline\Omega) \cap W^{1,p}(\Omega)$. The function $u_k$, extended by zero to $\overline\Omega$, lies in $C(\overline\Omega) \cap W^{1,p}(\Omega)$ — extension by zero is continuous because $u_k$ is already zero near $\partial\Omega$. Hence $Tu_k = u_k|_{\partial\Omega} = 0$ in $L^p(\partial\Omega; \mathcal{H}^{n-1})$. **Why this passes to the limit.** Continuity of the trace operator from the [Sobolev Trace Theorem](/theorems/60): there exists $C_T = C_T(\Omega, p) > 0$ such that \begin{align*} \|Tw\|_{L^p(\partial\Omega; \mathcal{H}^{n-1})} \le C_T\|w\|_{W^{1,p}(\Omega)} \quad \text{for all } w \in W^{1,p}(\Omega). \end{align*} The hypotheses of this trace bound are: $\Omega$ bounded with Lipschitz boundary (in our hypothesis), $1 \le p < \infty$ (in our hypothesis). Apply the bound to $w = u - u_k$: \begin{align*} \|Tu - Tu_k\|_{L^p(\partial\Omega)} = \|T(u - u_k)\|_{L^p(\partial\Omega)} \le C_T\|u - u_k\|_{W^{1,p}(\Omega)} \to 0, \end{align*} where the linearity $T(u - u_k) = Tu - Tu_k$ and the convergence $u_k \to u$ in $W^{1,p}(\Omega)$ are used. With $Tu_k = 0$ for all $k$, $Tu = \lim Tu_k = 0$ in $L^p(\partial\Omega; \mathcal{H}^{n-1})$.[/guided]

[step:Reverse direction setup: localise via a boundary partition of unity] Let $u \in W^{1,p}(\Omega)$ with $Tu = 0$. We show $u \in W^{1,p}_0(\Omega)$. By the Lipschitz boundary hypothesis, $\partial\Omega$ admits a finite open cover $\{U_1, \dots, U_N\}$ together with bi-Lipschitz maps $\Phi_j: U_j \to B(0,1)$ that flatten the boundary in the sense that \begin{align*} \Phi_j(U_j \cap \Omega) = B(0,1) \cap \mathbb{R}^n_+, \qquad \Phi_j(U_j \cap \partial\Omega) = B(0,1) \cap \{x_n = 0\}. \end{align*} Choose an additional open set $U_0 \subset\subset \Omega$ such that $\{U_0, U_1, \dots, U_N\}$ covers $\overline\Omega$. Let $\{\zeta_j\}_{j=0}^N$ be a smooth partition of unity subordinate to this cover, with $\zeta_j \in C_c^\infty(\mathbb{R}^n)$, $\operatorname{supp}\zeta_j \subset U_j$, and $\sum_{j=0}^N \zeta_j \equiv 1$ on a neighbourhood of $\overline\Omega$. Decompose $u = \sum_{j=0}^N u_j$ with $u_j := \zeta_j u \in W^{1,p}(\Omega)$ (by the [Product Rule for Weak Derivatives](/theorems/3098), since $\zeta_j$ is smooth and bounded with bounded derivatives). The partition of unity has the linearity property \begin{align*} T u = T\!\left(\sum_{j=0}^N u_j\right) = \sum_{j=0}^N T u_j, \end{align*} and additionally each $Tu_j$ is supported in $\overline{\Phi_j(U_j) \cap \{x_n = 0\}}$ (when transported to the half-space) or vanishes (for $j = 0$, since $\zeta_0$ is supported in $U_0 \subset\subset \Omega$). Because the $\zeta_j$ have disjoint supports near $\partial\Omega$ in the sense that $\sum_j \zeta_j(\Phi_j^{-1}(x', 0)) = 1$ on the boundary chart, the trace identities decouple: $Tu_j = \zeta_j|_{\partial\Omega}\, Tu = 0$ on $\partial\Omega$ for each $j \ge 1$, and $Tu_0 = 0$ by the same argument applied to the case where $\zeta_0|_{\partial\Omega} = 0$. If we can show that each $u_j \in W^{1,p}_0(\Omega)$, then $u = \sum_j u_j \in W^{1,p}_0(\Omega)$ by the linearity of the closure operation. So it suffices to handle each $u_j$ separately. **Interior piece $u_0$.** $u_0 = \zeta_0 u$ has $\operatorname{supp} u_0 \subset \overline{\operatorname{supp}\zeta_0} \subset U_0 \subset\subset \Omega$. Mollifying $u_0$ at scale $\varepsilon < \operatorname{dist}(\operatorname{supp} u_0, \partial\Omega)$ yields $\eta_\varepsilon * u_0 \in C_c^\infty(\Omega)$ approximating $u_0$ in $W^{1,p}(\Omega)$ as $\varepsilon \to 0$. So $u_0 \in W^{1,p}_0(\Omega)$. **Boundary pieces $u_j$, $j = 1, \dots, N$.** These require the substantial work in the next step. [/step]

[step:Reverse direction: cutoff against the boundary in a half-space chart]Fix $j \in \{1, \dots, N\}$. We work in the chart $\Phi_j: U_j \to B(0,1)$. Define \begin{align*} v: B(0,1) \cap \mathbb{R}^n_+ &\to \mathbb{R}, & v(x) := u_j(\Phi_j^{-1}(x)). \end{align*} Since $\Phi_j$ is bi-Lipschitz, $v \in W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)$ with $\|v\|_{W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)} \asymp \|u_j\|_{W^{1,p}(U_j \cap \Omega)}$, and $\operatorname{supp} v \subset\subset B(0,1)$ (because $\operatorname{supp} u_j \subset\subset U_j$). Furthermore, $Tv = 0$ on $B(0,1) \cap \{x_n = 0\}$ — this is the transport of $Tu_j = 0$ via the Lipschitz change of coordinates on $\partial\Omega$, which is itself bi-Lipschitz with respect to $\mathcal{H}^{n-1}$. We will show $v \in W^{1,p}_0(B(0,1) \cap \mathbb{R}^n_+)$, which transports back via $\Phi_j$ to $u_j \in W^{1,p}_0(U_j \cap \Omega) \subseteq W^{1,p}_0(\Omega)$ (extending $u_j$ by zero outside $U_j \cap \Omega$, using that $u_j$ already has compact support in $U_j$). For ease of notation, write $V := B(0,1) \cap \mathbb{R}^n_+$, with the boundary piece $\Sigma := B(0,1) \cap \{x_n = 0\}$. We have $v \in W^{1,p}(V)$, $\operatorname{supp} v \subset\subset B(0,1)$, $Tv|_\Sigma = 0$. **Step (a): Cutoff in the normal direction.** Pick a cutoff function $\eta \in C^\infty(\mathbb{R})$ with \begin{align*} \eta(s) = \begin{cases} 0, & s \le 1, \\ 1, & s \ge 2, \end{cases} \qquad 0 \le \eta \le 1, \qquad |\eta'| \le 2. \end{align*} For $h > 0$ define \begin{align*} \psi_h: \mathbb{R} \to \mathbb{R}, \qquad s \mapsto \eta(s/h), \end{align*} so $\psi_h(s) = 0$ for $s \le h$ and $\psi_h(s) = 1$ for $s \ge 2h$, with $|\psi_h'(s)| \le 2/h$. Define \begin{align*} v_h: V \to \mathbb{R}, \qquad v_h(x) := \psi_h(x_n) v(x). \end{align*} Then $v_h \in W^{1,p}(V)$ (by the [Product Rule for Weak Derivatives](/theorems/3098), with the smooth bounded factor $\psi_h(x_n)$), and $v_h$ vanishes on $\{x_n \le h\}$. In particular, $v_h$ has compact support in $V \cap \{x_n > h\}$, which is at strictly positive distance from $\Sigma$. **Step (b): $v_h \to v$ in $W^{1,p}(V)$ as $h \to 0$.** We must control two errors: \begin{align*} v - v_h &= (1 - \psi_h(x_n))v, \\ \nabla v - \nabla v_h &= (1 - \psi_h(x_n))\nabla v - \psi'_h(x_n) v \, e_n, \end{align*} where $e_n$ is the $n$-th coordinate vector and we used the product rule. For the $L^p$ part: $1 - \psi_h(x_n)$ is supported in $\{x_n \le 2h\}$, so \begin{align*} \|(1 - \psi_h(x_n))v\|_{L^p(V)}^p \le \int_{V \cap \{x_n \le 2h\}} |v|^p\, d\mathcal{L}^n. \end{align*} By the absolute continuity of the integral (since $|v|^p \in L^1(V)$), and $\mathcal{L}^n(V \cap \{x_n \le 2h\}) \to 0$ as $h \to 0$, this integral $\to 0$. For the gradient part, the first term \begin{align*} \|(1 - \psi_h(x_n))\nabla v\|_{L^p(V)}^p \le \int_{V \cap \{x_n \le 2h\}} |\nabla v|^p\, d\mathcal{L}^n \to 0 \end{align*} similarly. The harder term is $\psi_h'(x_n)\, v$: $|\psi_h'(x_n)| \le 2/h$ and $\psi_h'(x_n)$ is supported in $\{h \le x_n \le 2h\}$, so \begin{align*} \|\psi_h'(x_n)\, v\|_{L^p(V)}^p \le \frac{2^p}{h^p} \int_{V \cap \{h \le x_n \le 2h\}} |v|^p\, d\mathcal{L}^n. \end{align*} The naive bound $\int_{V \cap \{h \le x_n \le 2h\}}|v|^p \le h \cdot \sup_{0 \le x_n \le 2h} \int_{\Sigma}|v(x', x_n)|^p\, d\mathcal{H}^{n-1}(x')$ does not give what we need, because the prefactor $1/h^p$ would yield a divergent estimate. The decisive step is a strip-control inequality that exploits $Tv = 0$. **Step (c): Strip control for $W^{1,p}$ functions with zero trace on a half-space.** This is the key estimate. [claim:Strip control near the boundary] Let $w \in W^{1,p}(V)$ with $Tw|_\Sigma = 0$, and suppose $\operatorname{supp} w \subset\subset B(0,1)$. Then for every $h \in (0, 1/2)$, \begin{align*} \frac{1}{h^p}\int_{V \cap \{h \le x_n \le 2h\}} |w|^p\, d\mathcal{L}^n \le C(p) \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w|^p\, d\mathcal{L}^n, \end{align*} where $C(p) := 2^p (p/(p-1))^p$ for $p > 1$ and $C(1) := 2$. In particular, the right-hand side $\to 0$ as $h \to 0$ by absolute continuity of the integral. [/claim] [proof] Density of $C_c^\infty(\Omega)$ in $W^{1,p}_0(\Omega)$ is the definition of $W^{1,p}_0$, but we cannot use that here for $w$ on the half-space — we have not yet established that $w$ is in any such closure. Instead we use density of $C^\infty(\overline V) \cap W^{1,p}(V)$ in $W^{1,p}(V)$ on the Lipschitz set $V$ (this is the [Smooth Approximation Up to the Boundary](/theorems/3097) theorem), combined with continuity of the trace map on $V$ (which is the [Sobolev Trace Theorem](/theorems/60), applied at the prior level $V$, not at the level being proved): both ingredients are independent of the trace-kernel characterisation we are presently establishing on $\Omega$. Pick a sequence $w_k \in C^\infty(\overline V) \cap W^{1,p}(V)$ with $w_k \to w$ in $W^{1,p}(V)$. Continuity of the trace on $V$ yields $w_k|_\Sigma = Tw_k \to Tw = 0$ in $L^p(\Sigma; \mathcal{H}^{n-1})$. We bound the strip integral for $w_k$ first and then pass to the limit. For $w_k \in C^\infty(\overline V)$, the fundamental theorem of calculus along the segment $\{(x', t) : 0 \le t \le x_n\}$ gives, for $(x', x_n) \in V$ with $0 < x_n \le 2h$, \begin{align*} w_k(x', x_n) = w_k(x', 0) + \int_0^{x_n} \partial_{x_n} w_k(x', t)\, d\mathcal{L}^1(t). \end{align*} **Case $p = 1$.** Take absolute values and integrate over the strip $V \cap \{h \le x_n \le 2h\}$. Using $\int_0^{x_n}|\partial_{x_n}w_k(x',t)|\,dt \le \int_0^{2h}|\partial_{x_n}w_k(x',t)|\,dt$ for $x_n \le 2h$: \begin{align*} \int_{V \cap \{h \le x_n \le 2h\}} |w_k|\, d\mathcal{L}^n &\le \int_{V \cap \{h \le x_n \le 2h\}} |w_k(x', 0)|\, d\mathcal{L}^n + \int_{V \cap \{h \le x_n \le 2h\}} \int_0^{2h} |\partial_{x_n} w_k(x', t)|\, dt\, d\mathcal{L}^n \\ &\le h \int_\Sigma |w_k(x', 0)|\, d\mathcal{H}^{n-1} + h \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w_k(x', t)|\, d\mathcal{L}^n. \end{align*} The first inequality bounds $|w_k|$ by the trace value plus the integrated derivative; the second bounds the $x_n$-integration on the strip by its width $h$ (and uses Fubini for the swapped order of integration in the second term, with $x'$ ranging over $\Sigma$ and the $dt$ integration absorbing the strip thickness). Dividing by $h$ on both sides: \begin{align*} \frac{1}{h}\int_{V \cap \{h \le x_n \le 2h\}} |w_k|\, d\mathcal{L}^n \le \int_\Sigma |w_k(x', 0)|\, d\mathcal{H}^{n-1} + \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w_k|\, d\mathcal{L}^n. \end{align*} Letting $k \to \infty$: the left side converges to $h^{-1}\int_{V \cap \{h \le x_n \le 2h\}}|w|\, d\mathcal{L}^n$ (by $L^1$ convergence of $w_k \to w$), the trace term converges to $\int_\Sigma |Tw|\, d\mathcal{H}^{n-1} = 0$ (by trace continuity and $Tw = 0$), and the gradient term converges to $\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w|\, d\mathcal{L}^n$. This gives the $p = 1$ bound with $C(1) = 1 \le 2$. **Case $p > 1$.** Trace zero means $w_k(x', 0) \to 0$ in $L^p(\Sigma)$, so by selecting a further subsequence we may modify each $w_k$ (subtracting a thin boundary layer of size $\to 0$ in $W^{1,p}$) to assume $w_k(x', 0) \equiv 0$ on $\Sigma$ in what follows. Concretely: replace $w_k$ by $w_k - \chi_k(x_n) w_k(x', 0)$ where $\chi_k \in C_c^\infty([0, 1))$ is a smooth cutoff with $\chi_k(0) = 1$, $\chi_k \equiv 0$ on $[1/k, 1)$, $|\chi_k'| \le 2k$. The modification has $W^{1,p}(V)$ norm controlled by $\|w_k(\cdot,0)\|_{L^p(\Sigma)}(1 + 2k\cdot k^{-1/p}\cdot |\Sigma|^{1/p})$, which $\to 0$ along a subsequence by trace continuity (choosing $k$ large enough relative to $\|w_k(\cdot,0)\|_{L^p(\Sigma)}$). Replace $w_k$ by the modified sequence; we may now assume $w_k(x',0) = 0$. For $w_k$ with $w_k(x', 0) = 0$, the fundamental theorem reduces to $w_k(x', x_n) = \int_0^{x_n} \partial_{x_n} w_k(x', t)\, dt$. Apply Hölder's inequality with exponents $p$ and $p' = p/(p-1)$: \begin{align*} |w_k(x', x_n)|^p \le \left(\int_0^{x_n} |\partial_{x_n} w_k(x', t)|\, dt\right)^p \le x_n^{p-1} \int_0^{x_n} |\partial_{x_n} w_k(x', t)|^p\, dt. \end{align*} Integrate over $V \cap \{h \le x_n \le 2h\}$ (so $x_n^{p-1} \le (2h)^{p-1}$) and over $\Sigma \subset \mathbb{R}^{n-1}$ via Fubini: \begin{align*} \int_{V \cap \{h \le x_n \le 2h\}} |w_k|^p\, d\mathcal{L}^n &\le (2h)^{p-1} \int_\Sigma \int_h^{2h} \int_0^{x_n} |\partial_{x_n} w_k(x', t)|^p\, dt\, dx_n\, d\mathcal{H}^{n-1}(x') \\ &\le (2h)^{p-1} \cdot h \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w_k|^p\, d\mathcal{L}^n, \end{align*} where the last inequality uses $\int_h^{2h} \int_0^{x_n} g(t)\, dt\, dx_n \le h \int_0^{2h} g(t)\, dt$ for $g \ge 0$. Dividing by $h^p$: \begin{align*} \frac{1}{h^p}\int_{V \cap \{h \le x_n \le 2h\}} |w_k|^p\, d\mathcal{L}^n \le 2^{p-1} \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w_k|^p\, d\mathcal{L}^n \le 2^p \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w_k|^p\, d\mathcal{L}^n. \end{align*} Letting $k \to \infty$, both sides pass to the limit by $L^p$ convergence, yielding the bound for $w$ with constant $2^{p-1} \le 2^p$. In both cases, $\int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} w|^p\, d\mathcal{L}^n \to 0$ as $h \to 0$ by absolute continuity of the integral, since $|\partial_{x_n} w|^p \in L^1(V)$. [/proof] **Step (d): Apply the strip bound to control the cutoff error.** The claim above gives directly \begin{align*} \|\psi_h'(x_n)\, v\|_{L^p(V)}^p \le \frac{2^p}{h^p}\int_{V \cap \{h \le x_n \le 2h\}} |v|^p\, d\mathcal{L}^n \le 2^p \cdot C(p) \int_{V \cap \{0 < x_n \le 2h\}} |\partial_{x_n} v|^p\, d\mathcal{L}^n, \end{align*} and the right side $\to 0$ as $h \to 0$ by absolute continuity of $\int_V |\partial_{x_n} v|^p\, d\mathcal{L}^n < \infty$. Combining with the $L^p$ convergence of $v_h \to v$ from Step (b) and the convergence of the easier gradient term $(1 - \psi_h(x_n))\nabla v$, we obtain $v_h \to v$ in $W^{1,p}(V)$.[/step]

[guided]This step does the geometric work of the reverse direction. It is the central technical ingredient: every other step is essentially bookkeeping around the cutoff-and-mollify scheme. The narrative below walks through the chart setup, the transported trace-zero condition, the choice of cutoff, the four error pieces in the $W^{1,p}$ norm, the strip-control inequality split by $p = 1$ versus $p > 1$, and the final assembly into a single $W^{1,p}(V)$-approximation of $v$ by functions vanishing on a fixed normal layer near $\Sigma$. **Chart setup: transport to the half-space.** Fix a boundary index $j \in \{1, \dots, N\}$. From the partition-of-unity setup of the previous step, the chart $\Phi_j: U_j \to B(0,1)$ is bi-Lipschitz and flattens the boundary in the precise sense \begin{align*} \Phi_j(U_j \cap \Omega) = B(0,1) \cap \mathbb{R}^n_+, \qquad \Phi_j(U_j \cap \partial\Omega) = B(0,1) \cap \{x_n = 0\}. \end{align*} Define the transported function \begin{align*} v: B(0,1) \cap \mathbb{R}^n_+ \to \mathbb{R}, \qquad v(x) := u_j(\Phi_j^{-1}(x)). \end{align*} The bi-Lipschitz invariance of $W^{1,p}$ norms — composition with a bi-Lipschitz map preserves $W^{1,p}$ membership, with norm equivalence governed by the bi-Lipschitz constant $L_j$ of $\Phi_j$ and its inverse — yields $v \in W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)$ together with the two-sided bound \begin{align*} L_j^{-(n+1)/p}\, \|u_j\|_{W^{1,p}(U_j \cap \Omega)} \le \|v\|_{W^{1,p}(B(0,1) \cap \mathbb{R}^n_+)} \le L_j^{(n+1)/p}\, \|u_j\|_{W^{1,p}(U_j \cap \Omega)}, \end{align*} written compactly as $\|v\|_{W^{1,p}} \asymp \|u_j\|_{W^{1,p}(U_j \cap \Omega)}$. Since $\operatorname{supp} u_j \subset \overline{\operatorname{supp}\zeta_j} \subset\subset U_j$, the transported support $\operatorname{supp} v \subset\subset B(0,1)$ is bounded away from the lateral boundary $\partial B(0,1)$ — only the flat face $\Sigma := B(0,1) \cap \{x_n = 0\}$ remains as a relevant boundary. **Transported trace-zero condition.** The trace-zero hypothesis $Tu_j = 0$ on $\partial\Omega$ transports to $Tv|_\Sigma = 0$ on $\Sigma$. The chain of identifications: the boundary trace operator on $V$ is defined via the [Sobolev Trace Theorem](/theorems/60) at the prior level (on the half-space chart $V$, which is itself a bounded Lipschitz domain), the change of coordinates $\Phi_j$ restricted to $\partial\Omega \cap U_j$ is bi-Lipschitz with respect to the surface measure $\mathcal{H}^{n-1}$ (Lipschitz maps push forward $\mathcal{H}^{n-1}$ to a measure absolutely continuous with respect to $\mathcal{H}^{n-1}$, with Radon-Nikodym density bounded between $L_j^{-(n-1)}$ and $L_j^{n-1}$), and the trace map on $V$ pulled back through $\Phi_j$ agrees on smooth functions with the trace map on $\Omega$ — by density, the agreement passes to all of $W^{1,p}$. Therefore the equality $Tv = 0$ on $\Sigma$ is the precise translation of $Tu_j = 0$ on $\partial\Omega$. **Goal in the chart.** With the data $v \in W^{1,p}(V)$, $\operatorname{supp} v \subset\subset B(0,1)$, $Tv|_\Sigma = 0$, the goal of this step is to produce a $W^{1,p}(V)$-approximation of $v$ by a one-parameter family $\{v_h\}_{h > 0}$ where each $v_h$ vanishes on a fixed normal layer $\{x_n \le h\}$. The next step's mollification then lands $v_h$ in $C_c^\infty$, and after pulling back through the bi-Lipschitz chart with a second mollification, we obtain $u_j$ as a $W^{1,p}(\Omega)$-limit of $C_c^\infty(\Omega)$ functions, witnessing $u_j \in W^{1,p}_0(\Omega)$. **Strategy.** Pick a cutoff $\psi_h(x_n)$ depending only on the normal coordinate $x_n$, vanishing for $x_n \le h$ and equal to $1$ for $x_n \ge 2h$, with $|\psi_h'| \le 2/h$. The product $v_h := \psi_h(x_n)\, v$ vanishes on $\{x_n \le h\}$, so it has compact support strictly inside $V$ at distance $\ge h$ from $\Sigma$. The $L^p$ part of $v - v_h$ is supported in the thin strip $\{x_n \le 2h\}$ and vanishes by absolute continuity of the integral. The gradient part splits into a tame term $(1 - \psi_h(x_n))\nabla v$ and a dangerous term $\psi_h'(x_n)\, v\, e_n$ with prefactor $2/h$ on a strip of width $h$. Without the trace-zero hypothesis the dangerous term would scale as $h^{1-p}\, \|v|_\Sigma\|_{L^p(\Sigma)}^p$ — divergent as $h \to 0$ for $p > 1$ (and bounded but non-vanishing for $p = 1$). The trace-zero hypothesis $Tv = 0$ supplies the missing decay: $v$ vanishes (in trace sense) along $\Sigma$, so $|v(x', x_n)|$ scales like $\int_0^{x_n}|\partial_{x_n} v|$, picking up a factor of $x_n$ that exactly cancels the $1/h$ blowup. **Step (a): construct the cutoff.** Pick a profile $\eta \in C^\infty(\mathbb{R})$ with \begin{align*} \eta(s) = \begin{cases} 0, & s \le 1, \\ 1, & s \ge 2, \end{cases} \qquad 0 \le \eta \le 1, \qquad |\eta'| \le 2. \end{align*} Such a profile exists: take $\eta$ to be a $C^\infty$ regularization of the piecewise-linear function that equals $0$ on $(-\infty, 1]$, equals $s - 1$ on $[1, 2]$, equals $1$ on $[2, \infty)$ — this is Lipschitz with constant $1$, and a small mollification gives a smooth function with derivative bound $\le 2$. For each $h > 0$ define \begin{align*} \psi_h: \mathbb{R} \to \mathbb{R}, \qquad s \mapsto \eta(s/h), \end{align*} so that $\psi_h(s) = 0$ for $s \le h$, $\psi_h(s) = 1$ for $s \ge 2h$, and by the chain rule $\psi_h'(s) = h^{-1}\eta'(s/h)$ with $|\psi_h'| \le 2/h$ and $\operatorname{supp}\psi_h' \subset [h, 2h]$. Then define the truncated function \begin{align*} v_h: V \to \mathbb{R}, \qquad v_h(x) := \psi_h(x_n)\, v(x). \end{align*} The product lies in $W^{1,p}(V)$ by the [Product Rule for Weak Derivatives](/theorems/3098), since $\psi_h(x_n)$ is smooth on $\overline V$ with $\psi_h(x_n)$ and $\partial_{x_n}\psi_h(x_n)$ both bounded. The weak gradient is \begin{align*} \nabla v_h(x) = \psi_h(x_n)\nabla v(x) + \psi_h'(x_n)\, v(x)\, e_n, \end{align*} where $e_n$ is the $n$-th coordinate unit vector. The support inclusion $\operatorname{supp} v_h \subset V \cap \{x_n \ge h\}$ follows from $\psi_h \equiv 0$ on $\{x_n \le h\}$. **Step (b): the $L^p$ error vanishes.** The pointwise difference is \begin{align*} v(x) - v_h(x) = (1 - \psi_h(x_n))\, v(x), \end{align*} and $1 - \psi_h(x_n)$ is supported in $\{x_n \le 2h\}$ and bounded by $1$ in absolute value. Therefore \begin{align*} \|v - v_h\|_{L^p(V)}^p \le \int_{V \cap \{x_n \le 2h\}} |v|^p\, d\mathcal{L}^n. \end{align*} The right-hand side $\to 0$ as $h \to 0$ by absolute continuity of the integral: $|v|^p \in L^1(V)$, the strips $V \cap \{x_n \le 2h\}$ have $\mathcal{L}^n$-measure $\to 0$ (bounded by $2h \cdot \mathcal{L}^{n-1}(\Sigma)$), so $\int_{V \cap \{x_n \le 2h\}}|v|^p\, d\mathcal{L}^n \to 0$. Equivalently, by dominated convergence applied to the integrand $(1 - \psi_h(x_n))^p |v(x)|^p$, which is bounded by $|v|^p \in L^1(V)$ and converges to $0$ a.e. on $V$ (since $\psi_h(x_n) \to 1$ for every $x_n > 0$). **Step (c): the tame gradient term.** From the product rule and the gradient formula above, \begin{align*} \nabla v(x) - \nabla v_h(x) = (1 - \psi_h(x_n))\nabla v(x) - \psi_h'(x_n)\, v(x)\, e_n. \end{align*} The first piece is supported in $\{x_n \le 2h\}$ and bounded pointwise by $|\nabla v|$, so the same absolute-continuity argument gives \begin{align*} \|(1 - \psi_h(x_n))\nabla v\|_{L^p(V)}^p \le \int_{V \cap \{x_n \le 2h\}} |\nabla v|^p\, d\mathcal{L}^n \to 0 \quad \text{as } h \to 0, \end{align*} because $|\nabla v|^p \in L^1(V)$ and the strip measure $\to 0$. **Step (d): the dangerous gradient term.** What remains is \begin{align*} \|\psi_h'(x_n)\, v\|_{L^p(V)}^p \le \left(\frac{2}{h}\right)^p \int_{V \cap \{h \le x_n \le 2h\}} |v|^p\, d\mathcal{L}^n = \frac{2^p}{h^p}\int_{V \cap \{h \le x_n \le 2h\}}|v|^p\, d\mathcal{L}^n, \end{align*} using $|\psi_h'| \le 2/h$ and $\operatorname{supp}\psi_h' \subset [h, 2h]$. Naively, the strip $V \cap \{h \le x_n \le 2h\}$ has $\mathcal{L}^n$-measure of order $h$, so $\int_{V \cap \{h \le x_n \le 2h\}}|v|^p$ scales like $h$ (assuming $v$ is bounded), and $h^{-p} \cdot h = h^{1-p}$ — divergent for $p > 1$. The trace-zero hypothesis is what saves us: it forces $v$ to vanish at $\Sigma$, so the strip integral picks up an additional factor of $h^p$ from a Hardy-type estimate. **Step (e): the strip-control inequality.** The key bound, proved in detail in the [exact] block, is \begin{align*} \frac{1}{h^p}\int_{V \cap \{h \le x_n \le 2h\}}|v|^p\, d\mathcal{L}^n \le C(p) \int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} v|^p\, d\mathcal{L}^n, \end{align*} valid for every $v \in W^{1,p}(V)$ with $Tv|_\Sigma = 0$ and $\operatorname{supp} v \subset\subset B(0,1)$, with constant $C(p) = 2^p$ for $p > 1$ and $C(1) = 2$. The strategy of the proof: approximate $v$ by smooth functions $w_k \in C^\infty(\overline V) \cap W^{1,p}(V)$ via [Smooth Approximation Up to the Boundary](/theorems/3097) (a result that holds at the prior level $V$ and is independent of the trace-kernel characterisation we are establishing), use trace continuity from the [Sobolev Trace Theorem](/theorems/60) to transfer $Tv = 0$ to $w_k|_\Sigma \to 0$ in $L^p(\Sigma)$, prove the strip bound for smooth $w_k$ via the fundamental theorem of calculus along the normal segment, then pass to the limit. **Step (e1): the $p = 1$ subcase.** For $w_k \in C^\infty(\overline V)$ approximating $v$ in $W^{1,1}(V)$, the fundamental theorem of calculus along the normal segment from $(x', 0)$ to $(x', x_n)$ yields \begin{align*} w_k(x', x_n) = w_k(x', 0) + \int_0^{x_n}\partial_{x_n} w_k(x', t)\, d\mathcal{L}^1(t). \end{align*} Take absolute values, integrate over the strip $V \cap \{h \le x_n \le 2h\}$, and use $\int_0^{x_n}|\partial_{x_n} w_k(x', t)|\, dt \le \int_0^{2h}|\partial_{x_n} w_k(x', t)|\, dt$ for $x_n \le 2h$: \begin{align*} \int_{V \cap \{h \le x_n \le 2h\}} |w_k|\, d\mathcal{L}^n &\le \int_{V \cap \{h \le x_n \le 2h\}} |w_k(x', 0)|\, d\mathcal{L}^n \\ &\quad + \int_{V \cap \{h \le x_n \le 2h\}} \int_0^{2h} |\partial_{x_n} w_k(x', t)|\, dt\, d\mathcal{L}^n. \end{align*} The first term: $|w_k(x', 0)|$ depends only on $x'$, so the inner $x_n$-integration over $[h, 2h]$ produces a factor $h$, giving $h \int_\Sigma |w_k(\cdot, 0)|\, d\mathcal{H}^{n-1}$. The second term: by Fubini, swap the $dt$ and $dx_n$ integrations, the $x_n$-integration over $[h, 2h]$ produces a factor $h$, giving $h \int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w_k|\, d\mathcal{L}^n$. Dividing by $h$: \begin{align*} \frac{1}{h}\int_{V \cap \{h \le x_n \le 2h\}} |w_k|\, d\mathcal{L}^n \le \int_\Sigma |w_k(\cdot, 0)|\, d\mathcal{H}^{n-1} + \int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w_k|\, d\mathcal{L}^n. \end{align*} Pass to the limit $k \to \infty$. The left side: by $L^1$ convergence $w_k \to v$, the strip integral $\int_{V \cap \{h \le x_n \le 2h\}}|w_k|\, d\mathcal{L}^n \to \int_{V \cap \{h \le x_n \le 2h\}}|v|\, d\mathcal{L}^n$. The first term on the right: by trace continuity, $w_k|_\Sigma = Tw_k \to Tv = 0$ in $L^1(\Sigma)$, so $\int_\Sigma|w_k(\cdot, 0)|\, d\mathcal{H}^{n-1} \to 0$. The second term on the right: by $L^1$ convergence of the gradient, $\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w_k|\, d\mathcal{L}^n \to \int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} v|\, d\mathcal{L}^n$. The limit is \begin{align*} \frac{1}{h}\int_{V \cap \{h \le x_n \le 2h\}}|v|\, d\mathcal{L}^n \le \int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} v|\, d\mathcal{L}^n, \end{align*} which is the $p = 1$ strip bound with $C(1) = 1 \le 2$. **Step (e2): the $p > 1$ subcase.** Trace continuity gives $w_k|_\Sigma \to 0$ in $L^p(\Sigma)$. Modify the smooth approximating sequence so the boundary trace is identically zero: replace $w_k$ by $\widetilde w_k := w_k - \chi_k(x_n)\, w_k(x', 0)$ where $\chi_k \in C_c^\infty([0, 1))$ is a cutoff with $\chi_k(0) = 1$, $\chi_k \equiv 0$ on $[1/k, 1)$, and $|\chi_k'| \le 2k$. The modification has $W^{1,p}(V)$ norm controlled by $\|w_k(\cdot, 0)\|_{L^p(\Sigma)}$ multiplied by a factor depending on $k$: explicitly, \begin{align*} \|\chi_k(x_n)\, w_k(x', 0)\|_{L^p(V)}^p &\le \int_0^{1/k}|\chi_k(x_n)|^p\, dx_n \cdot \int_\Sigma |w_k(\cdot, 0)|^p\, d\mathcal{H}^{n-1} \le k^{-1}\, \|w_k(\cdot, 0)\|_{L^p(\Sigma)}^p, \\ \|\partial_{x_n}[\chi_k(x_n)\, w_k(x', 0)]\|_{L^p(V)}^p &\le (2k)^p \cdot k^{-1}\, \|w_k(\cdot, 0)\|_{L^p(\Sigma)}^p = (2k)^p\, k^{-1}\, \|w_k(\cdot, 0)\|_{L^p(\Sigma)}^p. \end{align*} The gradient term is the worrying one: it has factor $k^{p-1}$ multiplying $\|w_k(\cdot, 0)\|_{L^p(\Sigma)}^p$. Choose the index $k$ along a subsequence so that $\|w_k(\cdot, 0)\|_{L^p(\Sigma)}^p$ decays faster than $k^{-(p-1)}$ — this is possible because $w_k|_\Sigma \to 0$ in $L^p(\Sigma)$, so passing to a subsequence with rapid decay is fine. With this subsequence choice the modification satisfies $\widetilde w_k \to v$ in $W^{1,p}(V)$ and $\widetilde w_k(x', 0) \equiv 0$. Rename $\widetilde w_k$ back to $w_k$. For the modified $w_k$ with $w_k(x', 0) = 0$, the fundamental theorem reduces to \begin{align*} w_k(x', x_n) = \int_0^{x_n}\partial_{x_n} w_k(x', t)\, dt. \end{align*} Apply Hölder's inequality with conjugate exponents $p$ and $p' = p/(p-1)$ to the inner integral, treating $1$ on $[0, x_n]$ as the second factor: \begin{align*} \left|\int_0^{x_n}\partial_{x_n} w_k(x', t)\, dt\right| \le \left(\int_0^{x_n} 1\, dt\right)^{1/p'} \left(\int_0^{x_n}|\partial_{x_n} w_k(x', t)|^p\, dt\right)^{1/p} = x_n^{(p-1)/p}\left(\int_0^{x_n}|\partial_{x_n} w_k|^p\, dt\right)^{1/p}. \end{align*} Raising to the $p$-th power: \begin{align*} |w_k(x', x_n)|^p \le x_n^{p-1}\int_0^{x_n}|\partial_{x_n} w_k(x', t)|^p\, dt. \end{align*} Integrate over the strip $V \cap \{h \le x_n \le 2h\}$. Use $x_n^{p-1} \le (2h)^{p-1}$ on the strip, and $\int_0^{x_n}(\cdot)\, dt \le \int_0^{2h}(\cdot)\, dt$ for $x_n \le 2h$: \begin{align*} \int_{V \cap \{h \le x_n \le 2h\}} |w_k|^p\, d\mathcal{L}^n &\le (2h)^{p-1}\int_\Sigma\int_h^{2h}\int_0^{x_n}|\partial_{x_n} w_k(x', t)|^p\, dt\, dx_n\, d\mathcal{H}^{n-1}(x') \\ &\le (2h)^{p-1}\int_\Sigma\int_h^{2h}\int_0^{2h}|\partial_{x_n} w_k(x', t)|^p\, dt\, dx_n\, d\mathcal{H}^{n-1}(x'). \end{align*} The inner two integrals over $t \in [0, 2h]$ and $x_n \in [h, 2h]$ are independent of one another (the $t$-integration absorbs the strip thickness), so by Fubini the $x_n$-integration produces a factor of $h$: \begin{align*} \int_{V \cap \{h \le x_n \le 2h\}} |w_k|^p\, d\mathcal{L}^n \le (2h)^{p-1} \cdot h \int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w_k|^p\, d\mathcal{L}^n. \end{align*} Divide by $h^p$: \begin{align*} \frac{1}{h^p}\int_{V \cap \{h \le x_n \le 2h\}}|w_k|^p\, d\mathcal{L}^n \le 2^{p-1}\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w_k|^p\, d\mathcal{L}^n \le 2^p\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} w_k|^p\, d\mathcal{L}^n. \end{align*} Pass to the limit $k \to \infty$: both sides converge by $L^p$ convergence of $w_k$ and $\partial_{x_n} w_k$, giving the $p > 1$ strip bound for $v$ with $C(p) = 2^p$. **Step (e3): assemble the strip control.** In both subcases, \begin{align*} \frac{1}{h^p}\int_{V \cap \{h \le x_n \le 2h\}}|v|^p\, d\mathcal{L}^n \le C(p)\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} v|^p\, d\mathcal{L}^n, \end{align*} with $C(1) = 2$ and $C(p) = 2^p$ for $p > 1$. The right-hand side $\to 0$ as $h \to 0$ by absolute continuity of the integral, since $|\partial_{x_n} v|^p \in L^1(V)$ and the strip $V \cap \{0 < x_n \le 2h\}$ has $\mathcal{L}^n$-measure $\to 0$. Therefore the dangerous gradient term satisfies \begin{align*} \|\psi_h'(x_n)\, v\|_{L^p(V)}^p \le \frac{2^p}{h^p}\int_{V \cap \{h \le x_n \le 2h\}}|v|^p\, d\mathcal{L}^n \le 2^p\, C(p)\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} v|^p\, d\mathcal{L}^n \to 0 \end{align*} as $h \to 0$. **Step (f): final assembly of the $W^{1,p}$ approximation.** Combining steps (b), (c), and (e3): \begin{align*} \|v - v_h\|_{W^{1,p}(V)}^p &= \|v - v_h\|_{L^p(V)}^p + \|\nabla v - \nabla v_h\|_{L^p(V)}^p \\ &\le \int_{V \cap \{x_n \le 2h\}}|v|^p\, d\mathcal{L}^n + 2^{p-1}\left(\int_{V \cap \{x_n \le 2h\}}|\nabla v|^p\, d\mathcal{L}^n + 2^p\, C(p)\int_{V \cap \{0 < x_n \le 2h\}}|\partial_{x_n} v|^p\, d\mathcal{L}^n\right), \end{align*} where the factor $2^{p-1}$ comes from $|a + b|^p \le 2^{p-1}(|a|^p + |b|^p)$ applied to $\nabla v - \nabla v_h = (1 - \psi_h(x_n))\nabla v - \psi_h'(x_n)\, v\, e_n$. Each term on the right $\to 0$ as $h \to 0$, so $\|v - v_h\|_{W^{1,p}(V)} \to 0$. **Conclusion of this step.** The cutoff $v_h := \psi_h(x_n)\, v \in W^{1,p}(V)$ has compact support in $V \cap \{x_n \ge h\}$ — at distance $\ge h$ from $\Sigma$ and at distance $\ge \operatorname{dist}(\operatorname{supp} v, \partial B(0,1))$ from the lateral boundary $\partial B(0,1)$. As $h \to 0$, $v_h \to v$ in $W^{1,p}(V)$. The next step mollifies $v_h$ to land in $C_c^\infty(V)$, then transports back through the bi-Lipschitz chart $\Phi_j^{-1}$ with a second mollification to recover smoothness lost under the Lipschitz pullback.[/guided]

[step:Reverse direction: mollify the cutoff to land in $C_c^\infty$]The function $v_h \in W^{1,p}(V)$ has compact support in $V \cap \{x_n > h\}$. Pick a small parameter $\rho > 0$ and let $\eta_\rho$ be a standard mollifier with $\operatorname{supp}\eta_\rho \subset B(0,\rho)$. Define \begin{align*} v_{h,\rho}: \mathbb{R}^n \to \mathbb{R}, \qquad v_{h,\rho} := \eta_\rho * v_h, \end{align*} where $v_h$ is extended by zero outside $V \cap \{x_n > h\}$. (The zero extension lies in $W^{1,p}(\mathbb{R}^n)$: since $v_h$ has compact support strictly inside $V$, the zero extension across $\Sigma$ has weak gradient equal to the zero extension of $\nabla v_h$, with no singular boundary contribution.) For $\rho < h/2$, $v_{h,\rho}$ has support in $V \cap \{x_n > h/2\} \subset V \cap \{x_n > 0\} \subset V$, so $v_{h,\rho} \in C_c^\infty(V)$. By [Differentiation Through Convolution](/theorems/3096), $v_{h,\rho} \to v_h$ in $W^{1,p}(V)$ as $\rho \to 0$. Combining with $v_h \to v$ in $W^{1,p}(V)$ from the previous step, by a diagonal argument we extract a sequence $v_{h_k, \rho_k} \in C_c^\infty(V)$ with $v_{h_k, \rho_k} \to v$ in $W^{1,p}(V)$. **Pull back through the bi-Lipschitz chart, then mollify again.** Since $\Phi_j$ is only bi-Lipschitz (not smooth), the pullback $\tilde v_k := v_{h_k, \rho_k} \circ \Phi_j$ on $U_j \cap \Omega$ is Lipschitz with compact support inside $U_j \cap \Omega$, but is not in $C^\infty$ in general. We perform a second mollification to recover smoothness. Extend each $\tilde v_k$ by zero to $\mathbb{R}^n$ (the support of $\tilde v_k$ is contained in $\Phi_j^{-1}(\overline{V \cap \{x_n > h_k/2\}}) \subset U_j \cap \Omega$, which is at strictly positive distance $\delta_k > 0$ from $\partial U_j \cup \partial\Omega$, since $\Phi_j$ is bi-Lipschitz and $\Phi_j(\partial(U_j\cap\Omega)) \supset \partial V$ is at positive distance from $\overline{V \cap \{x_n > h_k/2\}}$). The zero extension is Lipschitz on $\mathbb{R}^n$, hence in $W^{1,p}(\mathbb{R}^n)$. Pick a second mollifier $\eta_{\sigma_k}$ with $\sigma_k < \delta_k / 2$ and define \begin{align*} \tilde v_{k, \sigma_k} := \eta_{\sigma_k} * \tilde v_k \in C^\infty(\mathbb{R}^n). \end{align*} The convolution is smooth (smoothness of the convolution of any $L^1_{\text{loc}}$ function with a $C^\infty$ kernel of compact support — see [Differentiation Through Convolution](/theorems/3096)). The support is contained in $\operatorname{supp}\tilde v_k + B(0, \sigma_k) \subset U_j \cap \Omega$ (by the choice of $\sigma_k$), so $\tilde v_{k, \sigma_k} \in C_c^\infty(U_j \cap \Omega)$. By [Differentiation Through Convolution](/theorems/3096) and the $W^{1,p}(\mathbb{R}^n)$ membership of $\tilde v_k$, $\tilde v_{k, \sigma_k} \to \tilde v_k$ in $W^{1,p}(\mathbb{R}^n)$ as $\sigma_k \to 0$. By bi-Lipschitz invariance of the $W^{1,p}$ norm under $\Phi_j$, $\tilde v_k \to v \circ \Phi_j = u_j$ in $W^{1,p}(U_j \cap \Omega)$. Choosing $\sigma_k \to 0$ fast enough relative to the convergence rate of $\tilde v_k \to u_j$, the diagonal sequence $\tilde v_{k, \sigma_k} \in C_c^\infty(U_j \cap \Omega) \subset C_c^\infty(\Omega)$ converges to $u_j$ in $W^{1,p}(\Omega)$. Hence $u_j \in W^{1,p}_0(\Omega)$. Combining with $u_0 \in W^{1,p}_0(\Omega)$ from the setup step, \begin{align*} u = \sum_{j=0}^N u_j \in W^{1,p}_0(\Omega). \end{align*} This completes the reverse inclusion $\ker T \subseteq W^{1,p}_0(\Omega)$.[/step]

[guided]With $v_h$ already vanishing in a strip near $\Sigma$, the cutoff has done the geometric work and what remains is to smooth $v_h$ to land in $C_c^\infty$. This is a standard mollification: **Mollifier setup.** Pick a non-negative $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\operatorname{supp}\eta \subset B(0,1)$ and $\int\eta\, d\mathcal{L}^n = 1$. Set $\eta_\rho(x) := \rho^{-n}\eta(x/\rho)$. **Zero extension.** $v_h$ has support in $V \cap \{x_n > h\}$, and we extend it by zero outside $V$ to obtain a function on $\mathbb{R}^n$. The zero extension lies in $W^{1,p}(\mathbb{R}^n)$ because $\operatorname{supp} v_h$ has positive distance from $\partial V$ in directions not on $\Sigma$, and $v_h$ vanishes near $\Sigma$ — so the zero extension across all of $\partial V$ is regular. Specifically: the only boundary of $V$ across which $v_h$ might fail to extend regularly is $\Sigma$, but $v_h$ already vanishes in the strip $\{x_n \le h\}$, making the zero extension across $\Sigma$ effectively a zero-extension from a region away from $\Sigma$. So the weak gradient of the zero extension on $\mathbb{R}^n$ equals the zero extension of $\nabla v_h$. **Mollification.** Define $v_{h, \rho} := \eta_\rho * v_h$ on $\mathbb{R}^n$. By [Differentiation Through Convolution](/theorems/3096), $v_{h,\rho} \in C^\infty(\mathbb{R}^n)$ with $\partial_{x_i}(v_{h,\rho}) = \eta_\rho * \partial_{x_i} v_h$, and $v_{h,\rho} \to v_h$ in $W^{1,p}(\mathbb{R}^n)$ as $\rho \to 0$. **Support.** $\operatorname{supp} v_{h,\rho} \subset \operatorname{supp} v_h + B(0, \rho)$. Choosing $\rho < h/2$, we have $\operatorname{supp} v_{h,\rho} \subset V \cap \{x_n > h - \rho\} \subset V \cap \{x_n > h/2\} \subset\subset V$. Hence $v_{h,\rho} \in C_c^\infty(V)$. **Diagonal extraction.** From $v_h \to v$ as $h \to 0$ and $v_{h,\rho} \to v_h$ as $\rho \to 0$, by a standard diagonal argument we pick sequences $h_k \to 0$ and $\rho_k \to 0$ with $\rho_k < h_k/2$ such that $v_{h_k, \rho_k} \to v$ in $W^{1,p}(V)$. **Pull-back to $\Omega$ — the double-mollification.** The chart $\Phi_j: U_j \to B(0,1)$ is bi-Lipschitz, not smooth, so naively pulling back $v_{h_k, \rho_k} \in C_c^\infty(V)$ via $\Phi_j$ produces a function that is only Lipschitz on $U_j \cap \Omega$. We need a second mollification to recover smoothness. Define $\tilde v_k := v_{h_k, \rho_k} \circ \Phi_j$ on $U_j \cap \Omega$. As composition of a $C_c^\infty$ function with a bi-Lipschitz map, $\tilde v_k$ is Lipschitz with compact support in $U_j \cap \Omega$. Compactness of support transfers because $\Phi_j$ is a homeomorphism: $\operatorname{supp}\tilde v_k = \Phi_j^{-1}(\operatorname{supp} v_{h_k,\rho_k})$ is the bi-Lipschitz image of a compact set in $V \cap \{x_n > h_k/2\}$, hence compact in $U_j \cap \Omega$. Set $\delta_k := \operatorname{dist}(\operatorname{supp}\tilde v_k, \partial(U_j \cap \Omega)) > 0$. Extend $\tilde v_k$ by zero to $\mathbb{R}^n$. The extension is Lipschitz on $\mathbb{R}^n$ (Lipschitz functions with compact support extend by zero to Lipschitz functions on $\mathbb{R}^n$) and lies in $W^{1,p}(\mathbb{R}^n)$ with $\|\nabla \tilde v_k\|_{L^\infty} < \infty$. Mollify with a second mollifier: pick $\sigma_k < \delta_k / 2$, set $\tilde v_{k, \sigma_k} := \eta_{\sigma_k} * \tilde v_k$. Three properties: (i) **Smoothness.** Convolution of any locally integrable function with $\eta_{\sigma_k} \in C_c^\infty$ is $C^\infty$, by [Differentiation Through Convolution](/theorems/3096). (ii) **Compact support inside $U_j \cap \Omega$.** $\operatorname{supp}\tilde v_{k,\sigma_k} \subset \operatorname{supp}\tilde v_k + B(0, \sigma_k)$, and the right side is contained in $U_j \cap \Omega$ because $\sigma_k < \delta_k/2$ and $\operatorname{supp}\tilde v_k$ is at distance $\delta_k$ from $\partial(U_j \cap \Omega)$. Hence $\tilde v_{k, \sigma_k} \in C_c^\infty(U_j \cap \Omega)$. (iii) **$W^{1,p}$ approximation.** $\tilde v_{k, \sigma_k} \to \tilde v_k$ in $W^{1,p}(\mathbb{R}^n)$ as $\sigma_k \to 0$, again by [Differentiation Through Convolution](/theorems/3096) (the standard mollifier approximation theorem applies because $\tilde v_k$ and $\nabla \tilde v_k$ are in $L^p(\mathbb{R}^n)$). **Diagonal extraction in the chart.** From $\tilde v_k \to u_j$ in $W^{1,p}(U_j \cap \Omega)$ (by bi-Lipschitz invariance of $W^{1,p}$ norms applied to $v_{h_k, \rho_k} \to v$ in $W^{1,p}(V)$) and $\tilde v_{k, \sigma_k} \to \tilde v_k$ in $W^{1,p}$ as $\sigma_k \to 0$, choosing $\sigma_k \to 0$ rapidly relative to the rate of $\tilde v_k \to u_j$ gives $\tilde v_{k, \sigma_k} \to u_j$ in $W^{1,p}(\Omega)$ along the diagonal. Since each $\tilde v_{k, \sigma_k} \in C_c^\infty(U_j \cap \Omega) \subset C_c^\infty(\Omega)$, this exhibits $u_j$ as a $W^{1,p}$ limit of $C_c^\infty(\Omega)$ functions. Hence $u_j \in W^{1,p}_0(\Omega)$. **Combining the pieces.** $u = \sum_{j=0}^N u_j$ with each $u_j \in W^{1,p}_0(\Omega)$. The space $W^{1,p}_0(\Omega)$ is a closed subspace of $W^{1,p}(\Omega)$, and is in particular closed under finite sums. Hence $u \in W^{1,p}_0(\Omega)$.[/guided]

[step:Combine the two inclusions] The first step established $W^{1,p}_0(\Omega) \subseteq \ker T$. The second through fourth steps established $\ker T \subseteq W^{1,p}_0(\Omega)$. Together, \begin{align*} W^{1,p}_0(\Omega) = \ker T = \{u \in W^{1,p}(\Omega) : Tu = 0 \text{ in } L^p(\partial\Omega; \mathcal{H}^{n-1})\}, \end{align*} which is the asserted characterisation. [/step]