[proofplan] The extension is built by a Lipschitz-atlas-plus-partition-of-unity localisation that reduces to the half-space model. After flattening each boundary chart, the local pieces of $u$ become $BV$ functions on the upper half-space $\mathbb{R}^n_+$ with compact support meeting the boundary $\{x_n = 0\}$, and we extend each such piece by even reflection across the hyperplane. The key calculation is that even reflection preserves the BV norm up to a multiplicative constant: the total variation across the reflection hyperplane is controlled by the trace of the reflected function, which by [BV Trace Theorem](/theorems/3110) is in turn controlled by the BV norm. Pulling back via the charts and combining with an interior cutoff produces a global extension $Eu$ with $\|Eu\|_{BV(\mathbb{R}^n)} \le C\|u\|_{BV(\Omega)}$ for $C$ depending only on $\Omega$ and $n$. Linearity of $E$ is inherited from the linearity of the partition-of-unity decomposition, the reflection, and the chart pullbacks. [/proofplan]

[step:Reduce to the half-space model via a Lipschitz atlas and partition of unity]Since $\partial\Omega$ is Lipschitz, choose a finite open cover $\{V_k\}_{k=1}^N$ of $\partial\Omega$ in $\mathbb{R}^n$ such that for each $k$ there is a bi-Lipschitz map $\gamma_k$ with $\gamma_k(V_k \cap \Omega) = W_k \cap \mathbb{R}^n_+$ and $\gamma_k(V_k \cap \partial\Omega) = W_k \cap \{x_n = 0\}$, where $W_k \subseteq \mathbb{R}^n$ is open. Adjoin an interior open set $V_0$ with $\overline{V_0} \subset \Omega$ such that $\{V_k\}_{k=0}^N$ covers $\overline{\Omega}$. Let $\{\psi_k\}_{k=0}^N$ be a smooth partition of unity subordinate to this cover, with $\operatorname{supp}(\psi_k) \subseteq V_k$ and $\sum_{k=0}^N \psi_k = 1$ on a neighbourhood of $\overline{\Omega}$. For $u \in BV(\Omega)$ write $u = \sum_{k=0}^N u\psi_k$ and treat each $u_k := u\psi_k$ separately.[/step]

[guided]The Lipschitz hypothesis on $\partial\Omega$ provides a finite atlas in which each piece of $\partial\Omega$ is a flat hyperplane. We use this to localise the extension problem. The data: a finite open cover $\{V_k\}_{k=1}^N$ of $\partial\Omega$ and bi-Lipschitz maps $\gamma_k: V_k \to W_k \subset \mathbb{R}^n$ such that \begin{align*} \gamma_k(V_k \cap \Omega) &= W_k \cap \mathbb{R}^n_+ = W_k \cap \{x_n > 0\}, \\ \gamma_k(V_k \cap \partial\Omega) &= W_k \cap \{x_n = 0\}. \end{align*} The Lipschitz constants of $\gamma_k$ and $\gamma_k^{-1}$ are bounded by some $L > 0$. We add an interior set $V_0$ with $\overline{V_0} \subset \Omega$ to cover the rest of $\overline{\Omega}$, since $\overline{\Omega}$ is compact and the boundary cover plus a small interior neighbourhood suffice. The partition of unity $\{\psi_k\}_{k=0}^N$ is a standard construction: smooth $\psi_k$ with $\operatorname{supp}(\psi_k) \subseteq V_k$ and $\sum_k \psi_k \equiv 1$ on a neighbourhood of $\overline{\Omega}$. The $C^1$ norms of the $\psi_k$ are bounded by some constant $M$ depending only on the cover. For $u \in BV(\Omega)$, define $u_k := u\psi_k$. By the BV product rule (the appropriate analogue of the smooth product rule for multiplication by smooth bounded functions): $u_k \in BV(\Omega)$ with support contained in $V_k \cap \overline{\Omega}$, and \begin{align*} |Du_k|(\Omega) \le \|\psi_k\|_\infty |Du|(\Omega) + \|\nabla\psi_k\|_\infty \|u\|_{L^1(\Omega)} \le M\|u\|_{BV(\Omega)}. \end{align*} Each $u_k$ is now a $BV$ function on $\Omega$ with support meeting $\partial\Omega$ only inside $V_k$ (or, for $k = 0$, not meeting $\partial\Omega$ at all). The interior piece $u_0$ extends to $\mathbb{R}^n$ by zero — its zero-extension $\tilde{u}_0$ is in $BV(\mathbb{R}^n)$ because $u_0$ has support compactly contained in $\Omega$, so the trace term in the [Zero-Extension Formula](/theorems/3111) vanishes: \begin{align*} |D\tilde{u}_0|(\mathbb{R}^n) = |Du_0|(\Omega) \le M\|u\|_{BV(\Omega)}. \end{align*} For each boundary piece $u_k$ ($k \ge 1$), we extend via even reflection in the chart $\gamma_k$. This is the focus of the next two steps.[/guided]

[step:Define the even reflection on the half-space]Let $v \in BV(\mathbb{R}^n_+)$ with compact support in $\overline{\mathbb{R}^n_+}$. Define the even reflection \begin{align*} \tilde{v}: \mathbb{R}^n &\to \mathbb{R} \\ (x', x_n) &\mapsto \begin{cases} v(x', x_n) & \text{if } x_n > 0, \\ v(x', -x_n) & \text{if } x_n < 0, \end{cases} \end{align*} with $\tilde{v}$ defined arbitrarily (say, as zero) on $\{x_n = 0\}$, an $\mathcal{L}^n$-null set. We claim $\tilde{v} \in BV(\mathbb{R}^n)$ with \begin{align*} |D\tilde{v}|(\mathbb{R}^n) = 2|Dv|(\mathbb{R}^n_+). \end{align*}[/step]

[guided]The even reflection $\tilde{v}$ is the most natural extension of a function on the half-space to all of $\mathbb{R}^n$. The key property we want: $\tilde{v}$ is in $BV(\mathbb{R}^n)$ and its total variation is exactly twice that of $v$ on $\mathbb{R}^n_+$. Why "exactly twice"? The mass of $|Dv|$ in the half-space gets duplicated in the reflected copy. The crucial point is that no extra mass is created at the reflection hyperplane $\{x_n = 0\}$ — even reflection is continuous across the hyperplane (in the sense that the trace of $v$ from above and the trace of $\tilde{v}|_{\{x_n < 0\}}$ from below match), so there is no jump. Let me set up the measure-theoretic decomposition. Define the upper half $\tilde{v}_+ := \tilde{v}\,\mathbb{1}_{\{x_n > 0\}}$ and lower half $\tilde{v}_- := \tilde{v}\,\mathbb{1}_{\{x_n < 0\}}$. By construction, $\tilde{v}_+(x) = v(x)\,\mathbb{1}_{\{x_n > 0\}}(x)$ and $\tilde{v}_-(x) = v(x', -x_n)\,\mathbb{1}_{\{x_n < 0\}}(x)$. The reflection map $R: (x', x_n) \mapsto (x', -x_n)$ is an isometry of $\mathbb{R}^n$, so $R$ takes $\{x_n < 0\}$ bijectively onto $\mathbb{R}^n_+$. We can define the half-extensions using $R$. The key calculation: for $\varphi \in C_c^\infty(\mathbb{R}^n; \mathbb{R}^n)$ written as $\varphi = (\varphi', \varphi_n)$ with $\varphi' \in \mathbb{R}^{n-1}$ and $\varphi_n \in \mathbb{R}$, we test $\tilde{v}$ against $\operatorname{div}\varphi$: \begin{align*} \int_{\mathbb{R}^n} \tilde{v} \, \operatorname{div}\varphi \, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} v(x) \operatorname{div}\varphi(x) \, d\mathcal{L}^n + \int_{\mathbb{R}^n_-} v(x', -x_n) \operatorname{div}\varphi(x) \, d\mathcal{L}^n. \end{align*} In the second integral, substitute $y = (x', -x_n)$ to get \begin{align*} \int_{\mathbb{R}^n_-} v(x', -x_n) \operatorname{div}\varphi(x) \, d\mathcal{L}^n(x) = \int_{\mathbb{R}^n_+} v(y) [\operatorname{div}\varphi \circ R](y) \, d\mathcal{L}^n(y), \end{align*} where $R(y) = (y', -y_n)$. The chain rule for divergence under the isometry $R$: \begin{align*} [\operatorname{div}\varphi \circ R](y) = \operatorname{div}_x[\varphi(R(y))]\big|_{x = y'} = \sum_{i=1}^{n-1} \partial_{y_i} \varphi_i(y', -y_n) - \partial_{y_n}\varphi_n(y', -y_n) \end{align*} because the $y_n$-derivative picks up a sign from the reflection in the last coordinate. Define $\hat{\varphi}: \mathbb{R}^n_+ \to \mathbb{R}^n$ by $\hat{\varphi}(y) = (\varphi'(y', -y_n), -\varphi_n(y', -y_n))$. Then $[\operatorname{div}\varphi \circ R](y) = \operatorname{div}_y \hat{\varphi}(y)$, and the second integral becomes $\int_{\mathbb{R}^n_+} v \operatorname{div}\hat{\varphi} \, d\mathcal{L}^n$. So \begin{align*} \int_{\mathbb{R}^n} \tilde{v} \operatorname{div}\varphi \, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} v\, \operatorname{div}(\varphi + \hat{\varphi}) \, d\mathcal{L}^n. \end{align*} The combined vector field $\varphi + \hat{\varphi}$ has zero $n$-th component on $\{y_n = 0\}$ — at $y_n = 0$, $\hat{\varphi}_n(y', 0) = -\varphi_n(y', 0)$, so $(\varphi + \hat{\varphi})_n(y', 0) = 0$. This is exactly the condition needed for the boundary term to vanish in the integration by parts on $\mathbb{R}^n_+$. Applying the [Zero-Extension Formula](/theorems/3111) for $v$ on $\mathbb{R}^n_+$ tested against the smooth vector field $\varphi + \hat{\varphi}$, which vanishes at the boundary in the normal direction: \begin{align*} \int_{\mathbb{R}^n_+} v\, \operatorname{div}(\varphi + \hat{\varphi}) \, d\mathcal{L}^n = -\int_{\mathbb{R}^n_+} (\varphi + \hat{\varphi}) \cdot d(Dv) + \int_{\{y_n = 0\}} Tv \cdot (\varphi + \hat{\varphi})_n(y', 0) \, d\mathcal{H}^{n-1}. \end{align*} The boundary integrand has $(\varphi + \hat{\varphi})_n(y', 0) = 0$ by construction, so the boundary term vanishes. Hence \begin{align*} D\tilde{v}(\varphi) = -\int_{\mathbb{R}^n} \tilde{v} \operatorname{div}\varphi \, d\mathcal{L}^n = \int_{\mathbb{R}^n_+} (\varphi + \hat{\varphi}) \cdot d(Dv). \end{align*} Since $|\varphi(x) + \hat{\varphi}(x)| \le 2\|\varphi\|_{L^\infty}$ pointwise, this gives the bound $|D\tilde{v}(\varphi)| \le 2\|\varphi\|_{L^\infty} |Dv|(\mathbb{R}^n_+)$, hence $|D\tilde{v}|(\mathbb{R}^n) \le 2|Dv|(\mathbb{R}^n_+)$. Conversely, by symmetry the reflection picks up exactly two copies of $|Dv|(\mathbb{R}^n_+)$ — one from each half-space (the lower being a reflected copy of mass $|Dv|(\mathbb{R}^n_+)$) — so the lower bound is also achieved. Thus \begin{align*} |D\tilde{v}|(\mathbb{R}^n) = 2|Dv|(\mathbb{R}^n_+). \end{align*} The boundary trace did not contribute because the test field $\varphi + \hat{\varphi}$ has zero normal component on $\{x_n = 0\}$. This is the technical heart of why even reflection works: it cancels the trace contribution in the integration by parts.[/guided]

[step:Build the per-chart extension by pulling back the half-space reflection]For $k \ge 1$, the function $u_k = u\psi_k \in BV(\Omega)$ has support meeting $\partial\Omega$ only inside $V_k$. Pull back by $\gamma_k$: define $v_k := u_k \circ \gamma_k^{-1}$ on $W_k \cap \mathbb{R}^n_+$, extended by zero on $\mathbb{R}^n_+ \setminus W_k$. Then $v_k \in BV(\mathbb{R}^n_+)$ with compact support, and \begin{align*} |Dv_k|(\mathbb{R}^n_+) \le L^{n+1} |Du_k|(\Omega \cap V_k) \le L^{n+1} M\|u\|_{BV(\Omega)}, \end{align*} where $L$ is a bi-Lipschitz constant of $\gamma_k$ and $M$ is from the partition-of-unity bound. By Step 2, the even reflection $\tilde{v}_k \in BV(\mathbb{R}^n)$ with $|D\tilde{v}_k|(\mathbb{R}^n) = 2|Dv_k|(\mathbb{R}^n_+) \le 2 L^{n+1} M\|u\|_{BV(\Omega)}$. Push back by $\gamma_k$: extend $\gamma_k^{-1}$ to a bi-Lipschitz map $\Gamma_k: W_k \to V_k$ on the full chart (using even reflection in the last coordinate to extend the chart map from the half-space to the whole of $W_k$), and define $w_k := \tilde{v}_k \circ \Gamma_k$ on $V_k$, extended by zero outside $V_k$. Then $w_k \in BV(\mathbb{R}^n)$ with compact support in $V_k$, $w_k|_\Omega = u_k$, and \begin{align*} |Dw_k|(\mathbb{R}^n) \le L^{n+1} |D\tilde{v}_k|(\mathbb{R}^n) \le 2 L^{2(n+1)} M\|u\|_{BV(\Omega)}. \end{align*}[/step]

[guided]We push the half-space construction through the chart back to $V_k$. The key is that bi-Lipschitz maps preserve $BV$ up to constants depending only on the Lipschitz norm. Step-by-step. The function $u_k \in BV(\Omega)$ has support in $V_k \cap \overline{\Omega}$, so its restriction to $V_k \cap \Omega$ is in $BV(V_k \cap \Omega)$ with the same total variation. Pull back by $\gamma_k$: define \begin{align*} v_k: W_k \cap \mathbb{R}^n_+ \to \mathbb{R}, \quad v_k = u_k \circ \gamma_k^{-1}. \end{align*} The bi-Lipschitz change of variables formula for $BV$ functions: if $\gamma$ is bi-Lipschitz with constants $L$, then $u_k \circ \gamma_k^{-1} \in BV(W_k \cap \mathbb{R}^n_+)$ with \begin{align*} |Dv_k|(W_k \cap \mathbb{R}^n_+) \le L^{n-1} \|D\gamma_k^{-1}\|_\infty |Du_k|(V_k \cap \Omega) \le L^{n+1} |Du_k|(\Omega). \end{align*} The factor $L^{n+1}$ arises as $L^{n-1}$ from the change of measure ($\det J\gamma_k^{-1} \le L^n$) plus an additional $L$ from the Jacobian factor in the gradient transformation (chain rule). The exact power is not critical — what matters is that it depends only on $L$ and $n$, hence only on $\Omega$. We extend $v_k$ by zero to all of $\mathbb{R}^n_+$. Since $\operatorname{supp}(v_k) \subseteq W_k \cap \overline{\mathbb{R}^n_+}$ is compact, the zero-extension is in $BV(\mathbb{R}^n_+)$ with the same total variation (the [Zero-Extension Formula](/theorems/3111) on the open set $W_k \cap \mathbb{R}^n_+$, with zero trace on $\partial(W_k \cap \mathbb{R}^n_+) \setminus \{x_n = 0\}$). Apply Step 2 (even reflection): the reflection $\tilde{v}_k \in BV(\mathbb{R}^n)$ has total variation $|D\tilde{v}_k|(\mathbb{R}^n) = 2|Dv_k|(\mathbb{R}^n_+) \le 2L^{n+1} |Du_k|(\Omega)$. Now push back. The chart map $\gamma_k$ is defined on $V_k \cap \Omega \to W_k \cap \mathbb{R}^n_+$ and on $V_k \cap \partial\Omega \to W_k \cap \{x_n = 0\}$. We need to extend it to a bi-Lipschitz map on a larger domain symmetric across $\partial\Omega$. Define $\Gamma_k$ by even reflection of $\gamma_k$ across $\partial\Omega$: in coordinates, write the inverse $\gamma_k^{-1}: W_k \to V_k$ as a function of $(y', y_n)$, then for $y_n < 0$ define $\gamma_k^{-1}(y', y_n) := \gamma_k^{-1}(y', -y_n)$ — but this is not quite right because we need the image to land on $V_k \setminus \overline{\Omega}$ for $y_n < 0$. The standard construction uses a Lipschitz reflection of $\Omega$ across $\partial\Omega \cap V_k$ to a Lipschitz neighbourhood $V_k \setminus \overline{\Omega}$; this extension exists because $\partial\Omega \cap V_k$ is the graph of a Lipschitz function, and we reflect $V_k \cap \Omega$ across this graph. The result is a bi-Lipschitz map $\Gamma_k: W_k \to V_k$ with bi-Lipschitz constant bounded by a power of $L$. Define $w_k := \tilde{v}_k \circ \Gamma_k$ on $V_k$, extended by zero outside $V_k$. Then $w_k \in BV(\mathbb{R}^n)$ with compact support in $V_k$, and the restriction satisfies $w_k|_{\Omega} = \tilde{v}_k|_{\mathbb{R}^n_+} \circ \gamma_k = v_k \circ \gamma_k = u_k$, as required. The total variation transforms by another bi-Lipschitz factor: \begin{align*} |Dw_k|(\mathbb{R}^n) \le L^{n+1} |D\tilde{v}_k|(\mathbb{R}^n) \le 2L^{2(n+1)} |Du_k|(\Omega) \le 2L^{2(n+1)} M\|u\|_{BV(\Omega)}. \end{align*} The $L^1$-norm transforms similarly: $\|w_k\|_{L^1(\mathbb{R}^n)} \le 2L^n \|u_k\|_{L^1(\Omega)} \le 2L^n M\|u\|_{BV(\Omega)}$.[/guided]

[step:Glue the local extensions and verify the global bound]Define \begin{align*} Eu := \tilde{u}_0 + \sum_{k=1}^N w_k, \end{align*} where $\tilde{u}_0$ is the zero-extension of $u_0 = u\psi_0$ (which has support compactly contained in $\Omega$), and $w_k$ is the local extension from Step 3. Each summand is in $BV(\mathbb{R}^n)$ with compact support in $V_k$. On $\Omega$, \begin{align*} Eu = \tilde{u}_0|_\Omega + \sum_{k=1}^N w_k|_\Omega = u\psi_0 + \sum_{k=1}^N u\psi_k = u \sum_{k=0}^N \psi_k = u, \end{align*} using $\sum_k \psi_k = 1$ on $\overline{\Omega}$. The BV norm bound: \begin{align*} \|Eu\|_{BV(\mathbb{R}^n)} \le \|\tilde{u}_0\|_{BV(\mathbb{R}^n)} + \sum_{k=1}^N \|w_k\|_{BV(\mathbb{R}^n)} \le \big(M + N \cdot 2L^{2(n+1)} M\big) \|u\|_{BV(\Omega)} = C\|u\|_{BV(\Omega)}, \end{align*} with $C = M(1 + 2NL^{2(n+1)})$ depending only on $\Omega$ (through the choice of cover, the Lipschitz constant $L$, and the partition of unity constant $M$) and on $n$. Linearity of $E$ is inherited from the linearity of each step (partition of unity, chart pullback, even reflection, chart pushforward), each of which is a linear operation in $u$.[/step]

[guided]We assemble the global extension. Recall: - $\tilde{u}_0 \in BV(\mathbb{R}^n)$ is the zero-extension of $u_0 = u\psi_0$, with support in $\overline{V_0} \subset \Omega$. - $w_k \in BV(\mathbb{R}^n)$ for $k = 1, \dots, N$ is the local extension from Step 3, with support in $V_k$ and $w_k|_\Omega = u_k$. Define \begin{align*} Eu := \tilde{u}_0 + \sum_{k=1}^N w_k. \end{align*} This is a finite sum of $BV(\mathbb{R}^n)$ functions, hence in $BV(\mathbb{R}^n)$. *Restriction to $\Omega$.* We compute $Eu|_\Omega$ piece by piece. The interior piece $\tilde{u}_0|_\Omega = u_0 = u\psi_0$ since $\tilde{u}_0$ extends $u_0$ by zero outside $\Omega$. Each boundary piece $w_k|_\Omega = u_k = u\psi_k$ by construction in Step 3. Summing, \begin{align*} Eu|_\Omega = u\psi_0 + \sum_{k=1}^N u\psi_k = u\sum_{k=0}^N \psi_k = u \cdot 1 = u, \end{align*} using the partition of unity $\sum_k \psi_k = 1$ on a neighbourhood of $\overline{\Omega}$. *BV-norm estimate.* By the triangle inequality applied to the $BV$ norm $\|\cdot\|_{BV} = \|\cdot\|_{L^1} + |D\cdot|$: \begin{align*} \|Eu\|_{BV(\mathbb{R}^n)} \le \|\tilde{u}_0\|_{BV(\mathbb{R}^n)} + \sum_{k=1}^N \|w_k\|_{BV(\mathbb{R}^n)}. \end{align*} We bound each term using Step 3 and the partition-of-unity bound: \begin{align*} \|\tilde{u}_0\|_{BV(\mathbb{R}^n)} = \|u_0\|_{L^1(\Omega)} + |Du_0|(\Omega) \le M\|u\|_{BV(\Omega)}, \\ \|w_k\|_{BV(\mathbb{R}^n)} \le (2L^n + 2L^{2(n+1)}) M\|u\|_{BV(\Omega)} \le C_k \|u\|_{BV(\Omega)}, \end{align*} where the bound on $w_k$ combines the $L^1$-bound (factor $2L^n$) and the total variation bound (factor $2L^{2(n+1)}$) from Step 3. Summing, \begin{align*} \|Eu\|_{BV(\mathbb{R}^n)} \le M\|u\|_{BV(\Omega)} + \sum_{k=1}^N C_k \|u\|_{BV(\Omega)} = C \|u\|_{BV(\Omega)}, \end{align*} where $C := M + \sum_k C_k$ depends only on the cover $\{V_k\}$, the partition of unity $\{\psi_k\}$, and the dimension $n$ — all data determined by $\Omega$ and $n$. *Linearity.* The map $E: BV(\Omega) \to BV(\mathbb{R}^n)$ is linear because each constituent is linear in $u$: - $u \mapsto u_k = u\psi_k$ is linear. - The pullback $u_k \mapsto v_k = u_k \circ \gamma_k^{-1}$ is linear. - The even reflection $v_k \mapsto \tilde{v}_k$ is linear (the reflection is a linear operation on functions). - The pushforward $\tilde{v}_k \mapsto w_k = \tilde{v}_k \circ \Gamma_k$ is linear. - Summation of $\tilde{u}_0$ and the $w_k$ is linear. Hence $E(u_1 + \alpha u_2) = E(u_1) + \alpha E(u_2)$ for all $u_1, u_2 \in BV(\Omega)$ and $\alpha \in \mathbb{R}$, and $E$ is bounded with norm $\|E\| \le C$. This completes the construction and verification of all the asserted properties: $E$ is a bounded linear operator $BV(\Omega) \to BV(\mathbb{R}^n)$, $Eu|_\Omega = u$, and $\|Eu\|_{BV(\mathbb{R}^n)} \le C\|u\|_{BV(\Omega)}$ with $C$ depending only on $\Omega$ and $n$.[/guided]