[proofplan] We show that [mollification](/page/Standard%20Mollifier) commutes with weak differentiation and then invoke the $L^p_{\mathrm{loc}}$ convergence of mollifiers to conclude Sobolev convergence. The key identity $D^\alpha(u_\varepsilon) = (D^\alpha u)_\varepsilon$ on $U_\varepsilon$ reduces the $W^{k,p}$ approximation problem to finitely many instances of $L^p$ mollifier convergence, one for each multi-index $|\alpha| \le k$. [/proofplan] [step:Establish smoothness of $u_\varepsilon$ on $U_\varepsilon$ and define the interior subdomain] Define $U_\varepsilon := \{x \in U : \operatorname{dist}(x, \partial U) > \varepsilon\}$. For any $x \in U_\varepsilon$, the map $y \mapsto \eta_\varepsilon(x - y)\, u(y)$ has compact support $\overline{B(x,\varepsilon)} \subset U$. Since $\eta_\varepsilon \in C_c^\infty(B(0,\varepsilon))$, the function $x \mapsto \eta_\varepsilon(x - y)$ is smooth in $x$ for each fixed $y$, and all $x$-derivatives $D^\beta_x[\eta_\varepsilon(x - y)]$ are bounded and compactly supported. By iterated differentiation under the integral sign (justified by dominated convergence, since $u \in L^1_{\mathrm{loc}}(U)$ and the derivatives of $\eta_\varepsilon$ are bounded with support uniformly contained in $U$ for $x$ in any compact subset of $U_\varepsilon$), the function $u_\varepsilon$ has continuous partial derivatives of all orders on $U_\varepsilon$, i.e., $u_\varepsilon \in C^\infty(U_\varepsilon)$. [/step] [step:Show that mollification commutes with weak differentiation on $U_\varepsilon$] Fix $x \in U_\varepsilon$ and a multi-index $\alpha$ with $|\alpha| \le k$. Since $\eta_\varepsilon \in C_c^\infty(B(0,\varepsilon))$ and $\operatorname{dist}(x, \partial U) > \varepsilon$ (by definition of $U_\varepsilon$), the integrand $y \mapsto \eta_\varepsilon(x - y) u(y)$ has compact support in $U$ as a function of $y$. Differentiating under the integral (justified by the smoothness and compact support of $\eta_\varepsilon$): \begin{align*} D^\alpha u_\varepsilon(x) &= D^\alpha_x \int_U \eta_\varepsilon(x - y)\, u(y) \, d\mathcal{L}^n(y) \\ &= \int_U D^\alpha_x[\eta_\varepsilon(x - y)]\, u(y) \, d\mathcal{L}^n(y). \end{align*} The sign-flip identity $D^\alpha_x[\eta_\varepsilon(x - y)] = (-1)^{|\alpha|} D^\alpha_y[\eta_\varepsilon(x - y)]$ holds because each $\partial_{x_j}$ applied to $\eta_\varepsilon(x - y)$ produces a factor $-1$ relative to $\partial_{y_j}$. Define $\phi: U \to \mathbb{R}$ by $\phi(y) := \eta_\varepsilon(x - y)$. Since $x \in U_\varepsilon$, we have $\operatorname{supp}(\phi) = \overline{B(x,\varepsilon)} \subset U$, so $\phi \in C_c^\infty(U)$ is a valid test function. By the definition of the [weak derivative](/page/Weak%20Derivative) $D^\alpha u$: \begin{align*} \int_U u(y)\, (-1)^{|\alpha|} D^\alpha_y \phi(y) \, d\mathcal{L}^n(y) &= \int_U D^\alpha u(y)\, \phi(y) \, d\mathcal{L}^n(y). \end{align*} Substituting $\phi(y) = \eta_\varepsilon(x - y)$: \begin{align*} D^\alpha u_\varepsilon(x) &= \int_U \eta_\varepsilon(x - y)\, D^\alpha u(y) \, d\mathcal{L}^n(y) = (\eta_\varepsilon * D^\alpha u)(x) = (D^\alpha u)_\varepsilon(x). \end{align*} [guided] We want to show that $D^\alpha(u_\varepsilon) = (D^\alpha u)_\varepsilon$ on $U_\varepsilon$, meaning the mollification of the weak derivative equals the classical derivative of the mollification. Why does this matter? Because it reduces convergence in the full Sobolev norm to convergence of mollifiers in $L^p$, which is already known. Fix $x \in U_\varepsilon$ (recall $U_\varepsilon = \{x \in U : \operatorname{dist}(x, \partial U) > \varepsilon\}$, defined in the previous step) and a multi-index $\alpha$ with $|\alpha| \le k$. The mollification is defined by \begin{align*} u_\varepsilon(x) &= \int_U \eta_\varepsilon(x - y)\, u(y) \, d\mathcal{L}^n(y). \end{align*} Since $\eta_\varepsilon \in C_c^\infty(B(0,\varepsilon))$ and $\operatorname{dist}(x, \partial U) > \varepsilon$, the support of $y \mapsto \eta_\varepsilon(x - y)$ is the closed ball $\overline{B(x,\varepsilon)}$, which lies entirely inside $U$. Therefore the integrand is compactly supported in $U$, and we may differentiate under the integral sign with respect to $x$: \begin{align*} D^\alpha u_\varepsilon(x) &= \int_U D^\alpha_x[\eta_\varepsilon(x - y)]\, u(y) \, d\mathcal{L}^n(y). \end{align*} Now we use a sign-flip identity: $D^\alpha_x[\eta_\varepsilon(x - y)] = (-1)^{|\alpha|} D^\alpha_y[\eta_\varepsilon(x - y)]$. This holds because $\eta_\varepsilon(x - y)$ depends on $x - y$, so each partial derivative $\partial_{x_j}$ produces the same result as $-\partial_{y_j}$, and $|\alpha|$ such sign flips yield the factor $(-1)^{|\alpha|}$. The trick is to recognise the resulting integral as a weak-derivative pairing. Define $\phi: U \to \mathbb{R}$ by $\phi(y) := \eta_\varepsilon(x - y)$. Since $x \in U_\varepsilon$, the support $\operatorname{supp}(\phi) = \overline{B(x, \varepsilon)}$ is a compact subset of $U$, so $\phi \in C_c^\infty(U)$ qualifies as a test function. The definition of the weak derivative $D^\alpha u$ states that for all $\phi \in C_c^\infty(U)$: \begin{align*} \int_U u(y)\, (-1)^{|\alpha|} D^\alpha_y \phi(y) \, d\mathcal{L}^n(y) &= \int_U D^\alpha u(y)\, \phi(y) \, d\mathcal{L}^n(y). \end{align*} Applying this with $\phi(y) = \eta_\varepsilon(x - y)$ transfers all derivatives from $u$ onto $\eta_\varepsilon$: \begin{align*} D^\alpha u_\varepsilon(x) &= \int_U (-1)^{|\alpha|} D^\alpha_y[\eta_\varepsilon(x - y)]\, u(y) \, d\mathcal{L}^n(y) = \int_U \eta_\varepsilon(x - y)\, D^\alpha u(y) \, d\mathcal{L}^n(y). \end{align*} The right-hand side is precisely $(D^\alpha u)_\varepsilon(x)$, the mollification of the weak derivative. This completes the commutativity identity $D^\alpha(u_\varepsilon) = (D^\alpha u)_\varepsilon$ on $U_\varepsilon$. [/guided] [/step] [step:Expand the Sobolev norm and apply $L^p_{\mathrm{loc}}$ mollifier convergence to each term] Let $V \subset\subset U$ be a compact subset. By the definition of the [Sobolev space](/page/Sobolev%20Space) norm: \begin{align*} \|u_\varepsilon - u\|_{W^{k,p}(V)}^p &= \sum_{|\alpha| \le k} \|D^\alpha u_\varepsilon - D^\alpha u\|_{L^p(V)}^p. \end{align*} The commutativity identity from the previous step gives $D^\alpha u_\varepsilon = (D^\alpha u)_\varepsilon$ on $U_\varepsilon$ (and hence on $V$ for $\varepsilon$ sufficiently small), so: \begin{align*} \|u_\varepsilon - u\|_{W^{k,p}(V)}^p &= \sum_{|\alpha| \le k} \|(D^\alpha u)_\varepsilon - D^\alpha u\|_{L^p(V)}^p. \end{align*} Since $u \in W^{k,p}(U)$, each weak derivative $D^\alpha u$ belongs to $L^p(U) \subset L^p_{\mathrm{loc}}(U)$. By [Theorem 48](/theorems/48) (convergence of mollifiers in $L^p_{\mathrm{loc}}$), for each multi-index $\alpha$ with $|\alpha| \le k$: \begin{align*} \lim_{\varepsilon \to 0} \|(D^\alpha u)_\varepsilon - D^\alpha u\|_{L^p(V)} &= 0. \end{align*} Since the sum over $\{|\alpha| \le k\}$ is finite, every term tends to zero, and therefore $\|u_\varepsilon - u\|_{W^{k,p}(V)} \to 0$ as $\varepsilon \to 0$. [/step]