[proofplan] The two inclusions are proved separately. For $\operatorname{Lip}(\Omega) \subseteq W^{1,\infty}(\Omega)$, a Lipschitz function is differentiable a.e. by Rademacher's theorem, and the classical derivative is bounded everywhere by the Lipschitz constant. We then check that this classical a.e.-derivative is the weak derivative by approximating with a smooth sequence (mollification of the Lipschitz extension to $\mathbb{R}^n$) and integrating by parts. For $W^{1,\infty}(\Omega) \subseteq \operatorname{Lip}(\Omega)$, we approximate $u \in W^{1,\infty}(\Omega)$ in $C^\infty(\overline\Omega) \cap W^{1,\infty}(\Omega)$ via the [Smooth Approximation Up to the Boundary](/theorems/3097) theorem (using the Lipschitz boundary hypothesis). Each smooth approximant is Lipschitz with constant controlled by $\|\nabla u_\varepsilon\|_{L^\infty}$, and these constants are uniformly bounded. Equicontinuity gives a Lipschitz limit. The norm equivalence then drops out of the explicit constants. The Lipschitz boundary hypothesis is used **only** in the second direction — it is what supplies the smooth approximation in $W^{1,\infty}(\Omega)$. [/proofplan] [step:Set up notation and recall the Lipschitz norm] For an open set $\Omega \subset \mathbb{R}^n$, the space of Lipschitz functions on $\Omega$ is \begin{align*} \operatorname{Lip}(\Omega) := \{u: \Omega \to \mathbb{R} : \exists L < \infty,\ |u(x) - u(y)| \le L|x - y| \text{ for all } x, y \in \Omega\}. \end{align*} The Lipschitz seminorm and norm are \begin{align*} [u]_{\operatorname{Lip}(\Omega)} := \sup_{\substack{x, y \in \Omega \\ x \ne y}} \frac{|u(x) - u(y)|}{|x - y|}, \qquad \|u\|_{\operatorname{Lip}(\Omega)} := \|u\|_{L^\infty(\Omega)} + [u]_{\operatorname{Lip}(\Omega)}. \end{align*} The Sobolev norm is \begin{align*} \|u\|_{W^{1,\infty}(\Omega)} := \|u\|_{L^\infty(\Omega)} + \|\nabla u\|_{L^\infty(\Omega)}, \end{align*} where $\|\nabla u\|_{L^\infty(\Omega)} := \max_{i=1, \dots, n}\|\partial_{x_i} u\|_{L^\infty(\Omega)}$ (we use the $\ell^\infty$ norm on $\mathbb{R}^n$ for the gradient; any other norm gives an equivalent $W^{1,\infty}$ norm). We will prove \begin{align*} \frac{1}{2}\|u\|_{\operatorname{Lip}(\Omega)} \le \|u\|_{W^{1,\infty}(\Omega)} \le C_\Omega \|u\|_{\operatorname{Lip}(\Omega)} \end{align*} for a constant $C_\Omega \ge 1$ depending only on $\Omega$, which is the asserted equivalence $\|u\|_{W^{1,\infty}} \asymp \|u\|_{\operatorname{Lip}}$. [/step] [step:Show $\operatorname{Lip}(\Omega) \subseteq W^{1,\infty}(\Omega)$ via Rademacher] Let $u \in \operatorname{Lip}(\Omega)$ with Lipschitz constant $L = [u]_{\operatorname{Lip}(\Omega)}$. By Rademacher's theorem (proven in GMT II as a corollary of the area formula or directly), $u$ is classically differentiable at $\mathcal{L}^n$-a.e. $x \in \Omega$. Denote the classical gradient (where it exists) by $\nabla_{\mathrm{cl}} u: \Omega \to \mathbb{R}^n$, extended to all of $\Omega$ by setting $\nabla_{\mathrm{cl}} u(x) := 0$ at the negligible set of non-differentiability points. **Step (a): $|\nabla_{\mathrm{cl}} u(x)| \le L$ a.e.** For any unit vector $e \in \mathbb{R}^n$ and any point $x$ where $u$ is differentiable, \begin{align*} |\partial_e u(x)| = \lim_{h \to 0}\left|\frac{u(x + he) - u(x)}{h}\right| \le \lim_{h \to 0}\frac{L|h|}{|h|} = L. \end{align*} Taking $e = e_i$ for each coordinate, $|\partial_{x_i} u(x)| \le L$ a.e. (in the $\ell^\infty$ norm); in particular $|\nabla_{\mathrm{cl}} u(x)| \le L$ in the $\ell^\infty$ sense. Hence the classical gradient is in $L^\infty(\Omega; \mathbb{R}^n)$ with $\|\nabla_{\mathrm{cl}} u\|_{L^\infty(\Omega)} \le L$. **Step (b): The classical gradient is the weak gradient.** We must verify that for every $\varphi \in C_c^\infty(\Omega)$ and every $i = 1, \dots, n$, \begin{align*} \int_\Omega u\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega \partial_{x_i, \mathrm{cl}} u\, \varphi\, d\mathcal{L}^n. \end{align*} Fix such $\varphi$ and let $K := \operatorname{supp}\varphi \subset\subset \Omega$. Choose $\delta_0 \in (0, \operatorname{dist}(K, \partial\Omega))$. For $\delta \in (0, \delta_0)$, let $\eta_\delta$ be a standard mollifier and set \begin{align*} u_\delta: \{x \in \Omega : \operatorname{dist}(x, \partial\Omega) > \delta\} &\to \mathbb{R}, & u_\delta(x) := (\eta_\delta * u)(x). \end{align*} Then $u_\delta \in C^\infty$ on its domain, $u_\delta \to u$ uniformly on $K$ (since $u$ is uniformly continuous on $K$, as a Lipschitz function), and the classical chain rule gives $\partial_{x_i} u_\delta = \eta_\delta * \partial_{x_i, \mathrm{cl}} u$ on the domain of $u_\delta$ (using that the distributional derivative of $u$ on $\Omega$ — once we admit it exists — coincides with the convolution). Let us prove this last identity differently, without circularity. The formula $\partial_{x_i} u_\delta = (\partial_{x_i}\eta_\delta) * u$ is the standard derivative-of-convolution formula, which requires only $u \in L^1_{\mathrm{loc}}(\Omega)$. By Rademacher and the classical chain rule applied to $u(y) = u(y - z) + \int_0^1 \nabla_{\mathrm{cl}} u(y - z + tz) \cdot z\, dt$ — the fundamental theorem of calculus along a line segment, valid for the Lipschitz $u$ — and Fubini's theorem, \begin{align*} \partial_{x_i} u_\delta(x) = \int_{\mathbb{R}^n} \partial_{x_i}\eta_\delta(x - y)\, u(y)\, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n}\eta_\delta(x - y)\, \partial_{y_i, \mathrm{cl}} u(y)\, d\mathcal{L}^n(y), \end{align*} where the last equality is by the FTC + Fubini argument (or, equivalently, integration by parts with the mollifier, which is justified for $u$ Lipschitz). Thus $\partial_{x_i} u_\delta = \eta_\delta * \partial_{x_i, \mathrm{cl}} u$ on its domain. Now integrate by parts on $K$, where $\varphi$ has compact support: for $\delta < \delta_0$, \begin{align*} \int_\Omega u_\delta\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega \partial_{x_i} u_\delta\, \varphi\, d\mathcal{L}^n = -\int_\Omega (\eta_\delta * \partial_{x_i, \mathrm{cl}} u)\, \varphi\, d\mathcal{L}^n. \end{align*} Letting $\delta \to 0$: the LHS converges to $\int_\Omega u\, \partial_{x_i}\varphi\, d\mathcal{L}^n$ by uniform convergence on $K$. The RHS converges to $-\int_\Omega \partial_{x_i, \mathrm{cl}} u\, \varphi\, d\mathcal{L}^n$ by the standard $L^p$ convergence of mollification: $\|\eta_\delta * \partial_{x_i, \mathrm{cl}} u - \partial_{x_i, \mathrm{cl}} u\|_{L^1(K)} \to 0$ (using $\partial_{x_i, \mathrm{cl}} u \in L^\infty \subset L^1_{\mathrm{loc}}$ and continuity of translation in $L^1$). Hence $\partial_{x_i, \mathrm{cl}} u$ is the weak partial derivative of $u$ in the $x_i$ direction, and $\partial_{x_i} u = \partial_{x_i, \mathrm{cl}} u \in L^\infty(\Omega)$. **Step (c): Norm bound.** From Step (a) and Step (b), $u \in W^{1,\infty}(\Omega)$ and $\|\nabla u\|_{L^\infty(\Omega)} \le [u]_{\operatorname{Lip}(\Omega)}$. Combined with the immediate inequality $\|u\|_{L^\infty(\Omega)} \le \|u\|_{L^\infty(\Omega)}$ (i.e., $\|u\|_{L^\infty} \le \|u\|_{\operatorname{Lip}} = \|u\|_{L^\infty} + [u]_{\operatorname{Lip}}$), \begin{align*} \|u\|_{W^{1,\infty}(\Omega)} = \|u\|_{L^\infty(\Omega)} + \|\nabla u\|_{L^\infty(\Omega)} \le \|u\|_{L^\infty(\Omega)} + [u]_{\operatorname{Lip}(\Omega)} = \|u\|_{\operatorname{Lip}(\Omega)}. \end{align*} This is the inclusion $\operatorname{Lip}(\Omega) \subseteq W^{1,\infty}(\Omega)$ with $\|u\|_{W^{1,\infty}} \le \|u\|_{\operatorname{Lip}}$. The Lipschitz boundary hypothesis is **not** used in this direction. [guided] The forward inclusion has two ingredients: Rademacher's theorem (a Lipschitz function is differentiable a.e.) gives a candidate for the gradient, and we then verify that this classical a.e.-gradient also serves as the weak gradient. The verification is needed because $W^{1,\infty}$ is defined via the distributional derivative, not the classical pointwise one — although for Lipschitz functions the two coincide, this is a theorem and not a definition. **Rademacher gives the classical a.e.-derivative.** A Lipschitz function $u: \Omega \to \mathbb{R}$ is differentiable $\mathcal{L}^n$-a.e. on $\Omega$ — this is Rademacher's theorem. Defined a.e. and extended by zero on the null set of bad points, the classical gradient $\nabla_{\mathrm{cl}} u$ has $|\nabla_{\mathrm{cl}} u(x)| \le L = [u]_{\operatorname{Lip}}$ everywhere by the difference-quotient bound: for any unit vector $e$ and any differentiability point $x$, $|\partial_e u(x)| = \lim_{h \to 0} |u(x + he) - u(x)|/|h| \le L$. So the classical gradient lies in $L^\infty(\Omega; \mathbb{R}^n)$. **Why we still must verify the weak-derivative identity.** Membership in $W^{1,\infty}(\Omega)$ requires the **distributional** derivative to be in $L^\infty$. A priori, the classical gradient (an a.e.-defined function) need not coincide with the distributional gradient (a distribution). For Lipschitz functions, the two do coincide, but this is the substantive content. We prove it by mollification. **Mollification setup.** Fix $\varphi \in C_c^\infty(\Omega)$ with $K := \operatorname{supp}\varphi \subset\subset \Omega$, and choose $\delta_0 \in (0, \operatorname{dist}(K, \partial\Omega))$. Pick a standard mollifier $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\operatorname{supp}\eta \subset B(0,1)$, $\int\eta\, d\mathcal{L}^n = 1$, and define $\eta_\delta(x) := \delta^{-n}\eta(x/\delta)$. Set \begin{align*} u_\delta: \Omega_\delta &\to \mathbb{R}, & u_\delta(x) := (\eta_\delta * u)(x), \quad \Omega_\delta := \{x \in \Omega : \operatorname{dist}(x, \partial\Omega) > \delta\}. \end{align*} For $\delta < \delta_0$, $K \subset \Omega_\delta$, and $u_\delta$ is well-defined and $C^\infty$ on $K$. **Convergence on $K$.** $u$ is uniformly continuous on the compact $K' := \{x : \operatorname{dist}(x, K) \le \delta_0\} \subset\subset \Omega$ (by Lipschitz, with modulus $L\delta$). Standard mollifier theory then gives $u_\delta \to u$ uniformly on $K$ as $\delta \to 0$. **Identifying $\partial_{x_i} u_\delta = \eta_\delta * \partial_{x_i, \mathrm{cl}} u$.** Two routes: Route A (via the formula $(D\eta_\delta) * u = \eta_\delta * Du$ when $u$ is AC along lines): differentiate under the integral sign, \begin{align*} \partial_{x_i} u_\delta(x) = \int_{\mathbb{R}^n} \partial_{x_i}\eta_\delta(x - y)\, u(y)\, d\mathcal{L}^n(y). \end{align*} Then integrate by parts in $y_i$ — legitimate because $u$ is Lipschitz hence absolutely continuous along lines (a consequence of Rademacher and Fubini), so the FTC applies along almost every line parallel to $e_i$: \begin{align*} \int_{\mathbb{R}^n}\partial_{x_i}\eta_\delta(x - y)\, u(y)\, d\mathcal{L}^n(y) = -\int_{\mathbb{R}^n}\partial_{y_i}\eta_\delta(x - y)\, u(y)\, d\mathcal{L}^n(y) \stackrel{\text{IBP in } y_i}{=} \int_{\mathbb{R}^n}\eta_\delta(x - y)\, \partial_{y_i, \mathrm{cl}} u(y)\, d\mathcal{L}^n(y). \end{align*} The $-$ sign is from the chain rule for $\eta_\delta(x - y)$, and the integration by parts has no boundary term because $\eta_\delta(x - \cdot)$ has compact support strictly inside $\Omega$ for $\delta < \delta_0$ and $x \in K$. Route B: simply note that mollification of an $L^1_{\mathrm{loc}}$ function commutes with weak derivatives; this is a standard fact, but verifying it for $u$ a priori only Lipschitz uses essentially the same argument as Route A. **The integration-by-parts identity.** With $u_\delta \in C^\infty(K)$ and $\varphi \in C_c^\infty(K)$, classical integration by parts gives \begin{align*} \int_\Omega u_\delta\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega \partial_{x_i} u_\delta\, \varphi\, d\mathcal{L}^n = -\int_\Omega (\eta_\delta * \partial_{x_i, \mathrm{cl}} u)\, \varphi\, d\mathcal{L}^n. \end{align*} **Limit as $\delta \to 0$.** Uniform convergence $u_\delta \to u$ on $K$ and $\partial_{x_i}\varphi \in L^\infty(\Omega)$ with compact support in $K$ give \begin{align*} \int_\Omega u_\delta\, \partial_{x_i}\varphi\, d\mathcal{L}^n \to \int_\Omega u\, \partial_{x_i}\varphi\, d\mathcal{L}^n. \end{align*} For the RHS: $\partial_{x_i, \mathrm{cl}} u \in L^\infty(\Omega) \subset L^1(K)$, and continuity of translation in $L^1$ gives $\eta_\delta * \partial_{x_i, \mathrm{cl}} u \to \partial_{x_i, \mathrm{cl}} u$ in $L^1(K)$. Combined with $\varphi \in L^\infty$, \begin{align*} -\int_\Omega(\eta_\delta * \partial_{x_i, \mathrm{cl}} u)\varphi\, d\mathcal{L}^n \to -\int_\Omega \partial_{x_i, \mathrm{cl}} u\, \varphi\, d\mathcal{L}^n. \end{align*} The limiting identity is $\int_\Omega u\, \partial_{x_i}\varphi\, d\mathcal{L}^n = -\int_\Omega \partial_{x_i, \mathrm{cl}} u\, \varphi\, d\mathcal{L}^n$, which is the definition of the weak derivative. With $\partial_{x_i, \mathrm{cl}} u \in L^\infty(\Omega)$, $u \in W^{1,\infty}(\Omega)$ with $\partial_{x_i} u = \partial_{x_i, \mathrm{cl}} u$. **Norm estimate.** $\|\nabla u\|_{L^\infty} = \|\nabla_{\mathrm{cl}} u\|_{L^\infty} \le L = [u]_{\operatorname{Lip}}$. Adding $\|u\|_{L^\infty}$ to both sides gives $\|u\|_{W^{1,\infty}} \le \|u\|_{\operatorname{Lip}}$. **Lipschitz boundary irrelevant here.** The argument used only $\Omega$ being open. The Lipschitz boundary plays no role in this direction. [/guided] [/step] [step:Show $W^{1,\infty}(\Omega) \subseteq \operatorname{Lip}(\Omega)$ via smooth approximation] Let $u \in W^{1,\infty}(\Omega)$. We show $u$ has a representative that is Lipschitz on $\Omega$ with $[u]_{\operatorname{Lip}(\Omega)} \le C_\Omega \|\nabla u\|_{L^\infty(\Omega)}$, for a constant $C_\Omega$ depending only on $\Omega$. **Step (a): Smooth approximation up to $\partial\Omega$.** Since $\Omega$ has Lipschitz boundary, the [Smooth Approximation Up to the Boundary](/theorems/3097) theorem provides a sequence \begin{align*} u_\varepsilon \in C^\infty(\overline\Omega) \cap W^{1,\infty}(\Omega), \qquad \|u_\varepsilon - u\|_{W^{1,p}(\Omega)} \to 0 \quad \text{as } \varepsilon \to 0 \end{align*} for any $p < \infty$. The original Meyers-Serrin / Smooth-Approximation-Up-to-the-Boundary statement is for $1 \le p < \infty$, but the standard construction (truncate, mollify on each chart, glue with a partition of unity) preserves the $L^\infty$ bounds: \begin{align*} \|u_\varepsilon\|_{L^\infty(\Omega)} \le \|u\|_{L^\infty(\Omega)}, \qquad \|\nabla u_\varepsilon\|_{L^\infty(\Omega)} \le C_\Omega \|\nabla u\|_{L^\infty(\Omega)}, \end{align*} where $C_\Omega \ge 1$ is a geometric constant depending on the partition of unity, the Lipschitz constants of the boundary charts, and the cutoff used in the construction (it does not depend on $u$ or $\varepsilon$). The inflation $C_\Omega$ over $1$ is unavoidable: localising via $\zeta_j u$ produces $\nabla\zeta_j \cdot u$ terms whose $L^\infty$ norm is bounded by $\|\nabla\zeta_j\|_{L^\infty}\|u\|_{L^\infty}$, and similarly the bi-Lipschitz changes of variable inflate norms by their bi-Lipschitz constant. In the special case $\Omega = \mathbb{R}^n_+$ or $\Omega$ convex, $C_\Omega = 1$ suffices. We also have, from $W^{1,p}$ convergence and a subsequence argument, $u_\varepsilon \to u$ a.e. on $\Omega$. **Step (b): Each $u_\varepsilon$ is Lipschitz with controlled constant.** $u_\varepsilon \in C^\infty(\overline\Omega)$ has a continuous classical gradient $\nabla u_\varepsilon$ on $\overline\Omega$, and its $L^\infty$ norm equals its sup norm. For any $x, y \in \Omega$, if the segment $[x, y] := \{x + t(y - x) : t \in [0, 1]\}$ is contained in $\Omega$, the fundamental theorem of calculus along the segment gives \begin{align*} |u_\varepsilon(x) - u_\varepsilon(y)| = \left|\int_0^1 \nabla u_\varepsilon(x + t(y - x)) \cdot (y - x)\, d\mathcal{L}^1(t)\right| \le \|\nabla u_\varepsilon\|_{L^\infty(\Omega)}|x - y|, \end{align*} where we used Cauchy-Schwarz $|\nabla u_\varepsilon \cdot (y - x)| \le |\nabla u_\varepsilon|\, |y - x|$ and the $\ell^2$ norm bound $|\nabla u_\varepsilon| \le \sqrt{n}\|\nabla u_\varepsilon\|_{L^\infty}$ in the $\ell^\infty$ sense (a constant $\sqrt n$ that we absorb into $C_\Omega$). For points $x, y \in \Omega$ where the segment $[x,y]$ leaves $\Omega$, the bound above does not directly apply. However, since $\Omega$ has Lipschitz boundary, $\Omega$ is **quasi-convex**: there exists a constant $K_\Omega \ge 1$ such that any two points $x, y \in \Omega$ can be joined by a curve $\gamma_{x,y}: [0, 1] \to \overline\Omega$ of length $\ell(\gamma_{x,y}) \le K_\Omega |x - y|$. (This is a standard property of Lipschitz domains; for convex $\Omega$, $K_\Omega = 1$.) Applying the FTC along $\gamma_{x,y}$, \begin{align*} |u_\varepsilon(x) - u_\varepsilon(y)| \le \|\nabla u_\varepsilon\|_{L^\infty(\overline\Omega)}\, \ell(\gamma_{x,y}) \le K_\Omega\, \|\nabla u_\varepsilon\|_{L^\infty(\overline\Omega)}\, |x - y|. \end{align*} Thus $u_\varepsilon$ is Lipschitz on $\Omega$ with constant \begin{align*} [u_\varepsilon]_{\operatorname{Lip}(\Omega)} \le K_\Omega \|\nabla u_\varepsilon\|_{L^\infty(\Omega)} \le K_\Omega C_\Omega \|\nabla u\|_{L^\infty(\Omega)} =: C'_\Omega \|\nabla u\|_{L^\infty(\Omega)}. \end{align*} The constant $C'_\Omega := K_\Omega C_\Omega$ is finite, depends only on $\Omega$, and is independent of $\varepsilon$. **Step (c): Pass to the pointwise limit.** From Step (a), $u_\varepsilon \to u$ a.e. on $\Omega$. Pick any two points $x, y \in \Omega$ at which $u_\varepsilon \to u$ — this is a.e. in $\Omega \times \Omega$ for the product measure. Then \begin{align*} |u(x) - u(y)| = \lim_{\varepsilon \to 0} |u_\varepsilon(x) - u_\varepsilon(y)| \le C'_\Omega \|\nabla u\|_{L^\infty(\Omega)} |x - y|. \end{align*} This holds for $\mathcal{L}^n$-a.e. $x, y \in \Omega$. By redefining $u$ on a $\mathcal{L}^n$-null set, $u$ has a Lipschitz representative — call it $u$ also — on a set of full measure, which extends uniquely to a Lipschitz function on $\Omega$ by continuity (since Lipschitz functions on a dense subset of an open connected set extend to Lipschitz functions on the whole set; for a general open Lipschitz domain $\Omega$, decompose into connected components, each of which is itself a Lipschitz domain by definition, and apply the argument componentwise). Thus $u \in \operatorname{Lip}(\Omega)$ with $[u]_{\operatorname{Lip}(\Omega)} \le C'_\Omega \|\nabla u\|_{L^\infty(\Omega)}$ and $\|u\|_{L^\infty(\Omega)}$ unchanged from the $W^{1,\infty}$ definition. Adding, \begin{align*} \|u\|_{\operatorname{Lip}(\Omega)} = \|u\|_{L^\infty(\Omega)} + [u]_{\operatorname{Lip}(\Omega)} \le \|u\|_{L^\infty(\Omega)} + C'_\Omega \|\nabla u\|_{L^\infty(\Omega)} \le C'_\Omega \|u\|_{W^{1,\infty}(\Omega)}. \end{align*} This is the inclusion $W^{1,\infty}(\Omega) \subseteq \operatorname{Lip}(\Omega)$ with $\|u\|_{\operatorname{Lip}} \le C'_\Omega \|u\|_{W^{1,\infty}}$. [guided] The reverse inclusion is the substantive part, and the Lipschitz boundary hypothesis enters here in two distinct places: (i) supplying smooth approximations up to the boundary, and (ii) supplying quasi-convexity of $\Omega$ so that two points can be joined by paths of comparable length to the chord. **Smooth approximation up to the boundary.** Since $\Omega$ has Lipschitz boundary, [Smooth Approximation Up to the Boundary](/theorems/3097) provides $u_\varepsilon \in C^\infty(\overline\Omega) \cap W^{1,p}(\Omega)$ with $\|u_\varepsilon - u\|_{W^{1,p}} \to 0$ for any fixed $p < \infty$. The standard construction — partition of unity over chart neighbourhoods, translate-then-mollify on each boundary chart, mollify directly on the interior chart, sum — preserves $L^\infty$ bounds with a multiplicative constant: each step (multiplication by a $C^\infty$ cutoff, bi-Lipschitz change of variables to flatten the boundary chart, translation, mollification) maps $L^\infty$ to $L^\infty$ with a controlled increase. The result is \begin{align*} \|u_\varepsilon\|_{L^\infty(\Omega)} \le \|u\|_{L^\infty(\Omega)}, \qquad \|\nabla u_\varepsilon\|_{L^\infty(\Omega)} \le C_\Omega \|\nabla u\|_{L^\infty(\Omega)}, \end{align*} with $C_\Omega \ge 1$ a constant depending only on the boundary chart bi-Lipschitz constants and the cutoff functions in the partition of unity. The first inequality is the easier one (mollification with a non-negative kernel of unit mass cannot increase the $L^\infty$ norm), but the second has the inflation $C_\Omega$ from the chart constants — for example, if $\Phi: B_j \to \mathbb{R}^n$ is the bi-Lipschitz chart with constants $L^{-1} \le |\det J\Phi| \le L$ and $|\nabla\Phi|, |\nabla\Phi^{-1}| \le L$, then the chain rule gives $|\nabla(u \circ \Phi^{-1})|_\infty \le L|\nabla u|_\infty$, and likewise for the inverse change of variables. **Why $C_\Omega \ge 1$ rather than $1$.** For convex $\Omega$, the smooth approximation can be done by a simple radial scaling and mollification, both of which preserve the gradient $L^\infty$ norm exactly, and $C_\Omega = 1$. For general Lipschitz $\Omega$ the chart geometry forces a finite $C_\Omega$ depending on the boundary roughness. **Subsequence with a.e. convergence.** From $u_\varepsilon \to u$ in $L^p(\Omega)$ for any $p < \infty$ — note $u \in L^\infty \subset L^p$ on bounded $\Omega$ — we extract a subsequence with $u_{\varepsilon_k} \to u$ a.e. on $\Omega$. **Lipschitz constant of $u_\varepsilon$.** The smooth $u_\varepsilon$ has continuous gradient on $\overline\Omega$ and $\|\nabla u_\varepsilon\|_{L^\infty(\Omega)} = \sup_{\overline\Omega}|\nabla u_\varepsilon|$. For two points $x, y \in \Omega$, we want to bound $|u_\varepsilon(x) - u_\varepsilon(y)|$ by $|x - y|$ times the gradient norm. The fundamental theorem of calculus along a curve $\gamma: [0, 1] \to \overline\Omega$ joining $x$ and $y$ gives \begin{align*} |u_\varepsilon(x) - u_\varepsilon(y)| = \left|\int_0^1 \nabla u_\varepsilon(\gamma(t)) \cdot \gamma'(t)\, d\mathcal{L}^1(t)\right| \le \|\nabla u_\varepsilon\|_{L^\infty(\overline\Omega)}\, \ell(\gamma). \end{align*} We need to choose $\gamma$ with $\ell(\gamma) \le K_\Omega |x - y|$. For convex $\Omega$, the straight segment works with $K_\Omega = 1$. **Quasi-convexity of Lipschitz domains.** For non-convex $\Omega$, the segment $[x, y]$ may leave $\Omega$. The relevant geometric fact is **quasi-convexity** of Lipschitz domains: there exists $K_\Omega < \infty$ such that any two $x, y \in \Omega$ are joined by a path in $\overline\Omega$ of length at most $K_\Omega|x - y|$. The proof of quasi-convexity for Lipschitz domains uses the chart structure: locally, the boundary is a Lipschitz graph, and within each chart one can connect points by paths whose length is bounded by a multiple of the Euclidean distance. Globally, transitions between charts cost only finitely many short detours. **Pointwise Lipschitz bound for $u_\varepsilon$.** With $\gamma$ of length $\le K_\Omega|x - y|$, \begin{align*} |u_\varepsilon(x) - u_\varepsilon(y)| \le K_\Omega \|\nabla u_\varepsilon\|_{L^\infty(\Omega)}|x - y| \le K_\Omega C_\Omega \|\nabla u\|_{L^\infty(\Omega)}|x - y| =: C'_\Omega \|\nabla u\|_{L^\infty(\Omega)}|x - y|. \end{align*} **Pass to the limit.** At any pair $(x, y)$ where $u_{\varepsilon_k}(x) \to u(x)$ and $u_{\varepsilon_k}(y) \to u(y)$ — i.e., for a.e. $(x, y) \in \Omega \times \Omega$ — taking $k \to \infty$: \begin{align*} |u(x) - u(y)| = \lim |u_{\varepsilon_k}(x) - u_{\varepsilon_k}(y)| \le C'_\Omega \|\nabla u\|_{L^\infty(\Omega)}|x - y|. \end{align*} This Lipschitz bound holds on the full-measure subset $E \subset \Omega$ where $u_{\varepsilon_k} \to u$. Since $u$ is uniformly continuous (in fact Lipschitz) on $E$ and $E$ is dense in $\Omega$, $u$ extends uniquely to a Lipschitz function on $\Omega$ with the same constant. Equivalently, the equivalence class $u \in W^{1,\infty}(\Omega)$ contains a Lipschitz representative. **Norm bound.** $\|u\|_{\operatorname{Lip}} = \|u\|_{L^\infty} + [u]_{\operatorname{Lip}} \le \|u\|_{L^\infty} + C'_\Omega \|\nabla u\|_{L^\infty} \le C'_\Omega(\|u\|_{L^\infty} + \|\nabla u\|_{L^\infty}) = C'_\Omega \|u\|_{W^{1,\infty}}$, since $C'_\Omega \ge 1$. [/guided] [/step] [step:Combine to obtain norm equivalence] The forward inclusion (step "Show $\operatorname{Lip}(\Omega) \subseteq W^{1,\infty}(\Omega)$") gives $\|u\|_{W^{1,\infty}} \le \|u\|_{\operatorname{Lip}}$, and the reverse inclusion (step "Show $W^{1,\infty}(\Omega) \subseteq \operatorname{Lip}(\Omega)$") gives $\|u\|_{\operatorname{Lip}} \le C'_\Omega \|u\|_{W^{1,\infty}}$. Together, \begin{align*} \|u\|_{W^{1,\infty}(\Omega)} \le \|u\|_{\operatorname{Lip}(\Omega)} \le C'_\Omega \|u\|_{W^{1,\infty}(\Omega)}, \end{align*} which is $\|u\|_{W^{1,\infty}(\Omega)} \asymp \|u\|_{\operatorname{Lip}(\Omega)}$. The set equality $W^{1,\infty}(\Omega) = \operatorname{Lip}(\Omega)$ follows: every element of one space lies in the other, and the equivalence classes (mod a.e. equality on the Sobolev side) coincide because Lipschitz representatives are unique up to no-sets. [/step]