[proofplan] We compute $\partial_{x_i} u_\varepsilon$ in two ways and equate them. First, differentiation under the integral sign — justified by [Leibniz Integral Rule](/theorems/???) using the smoothness and compact support of $\eta_\varepsilon$ together with $u \in L^p_{\mathrm{loc}}$ — gives $\partial_{x_i} u_\varepsilon(x) = \int u(y)\, \partial_{x_i}\eta_\varepsilon(x-y)\, d\mathcal{L}^n(y)$. Next, the elementary identity $\partial_{x_i}\eta_\varepsilon(x-y) = -\partial_{y_i}\eta_\varepsilon(x-y)$ converts the derivative on $x$ into a derivative on $y$, allowing us to invoke the definition of the weak derivative of $u$ with test function $\eta_\varepsilon(x - \cdot)$. The smallness condition $\varepsilon < \operatorname{dist}(V, \partial\Omega)$ guarantees the test function is supported inside $\Omega$, which is the precise hypothesis needed to apply the weak derivative identity. The $W^{1,p}(V)$ convergence then follows from $L^p$ convergence of mollifications, applied to $u$ and to each weak derivative. [/proofplan] [step:Differentiate $u_\varepsilon$ under the integral sign] Let $V \subset\subset \Omega$ be open with $\overline{V} \subset \Omega$, fix $\varepsilon < \operatorname{dist}(V, \partial\Omega)$, and let $\eta_\varepsilon \in C_c^\infty(\mathbb{R}^n)$ be the standard mollifier ($\operatorname{supp} \eta_\varepsilon \subseteq \overline{B}(0, \varepsilon)$, $\int_{\mathbb{R}^n} \eta_\varepsilon\, d\mathcal{L}^n = 1$). Extend $u$ by zero outside $\Omega$ if needed; for $x \in V$ and $y \in \overline{B}(x, \varepsilon)$, the assumption $\varepsilon < \operatorname{dist}(V, \partial\Omega)$ ensures $y \in \Omega$, so the convolution \begin{align*} u_\varepsilon: V &\to \mathbb{R} \\ x &\mapsto \int_{\mathbb{R}^n} u(y)\, \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y) = \int_{B(x, \varepsilon)} u(y)\, \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y) \end{align*} involves only the values of $u$ inside $\Omega$ and is well-defined. To compute $\partial_{x_i} u_\varepsilon$, we apply the [Leibniz Integral Rule](/theorems/???). For each $x \in V$, choose a closed ball $\overline{B}(x, r) \subset V$ small enough that the union $\bigcup_{x' \in \overline{B}(x, r)} B(x', \varepsilon)$ is contained in a fixed compact set $K \subset \Omega$ (take $K = \overline{B}(x, r + \varepsilon) \cap \overline{\Omega}$; by hypothesis on $\varepsilon$, $K \subset \Omega$). Then for $x' \in \overline{B}(x, r)$, \begin{align*} u_\varepsilon(x') = \int_K u(y)\, \eta_\varepsilon(x' - y)\, d\mathcal{L}^n(y). \end{align*} The integrand $(x', y) \mapsto u(y)\, \eta_\varepsilon(x' - y)$ has the following properties on $\overline{B}(x, r) \times K$: (i) for each fixed $x'$ it is measurable in $y$; (ii) for $\mathcal{L}^n$-a.e. $y \in K$ it is $C^\infty$ in $x'$ since $\eta_\varepsilon$ is $C^\infty$; (iii) the partial derivative $\partial_{x'_i}[u(y)\, \eta_\varepsilon(x' - y)] = u(y)\, (\partial_{x_i}\eta_\varepsilon)(x' - y)$ is dominated by $|u(y)| \cdot \|\partial_{x_i}\eta_\varepsilon\|_{L^\infty}$, which lies in $L^1(K)$ because $u \in L^p(K) \subseteq L^1(K)$ on the compact set $K \subset \Omega$. Hence the [Leibniz Integral Rule](/theorems/???) applies and \begin{align*} \partial_{x_i} u_\varepsilon(x) = \int_{\mathbb{R}^n} u(y)\, (\partial_{x_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y). \end{align*} The same argument applied iteratively to $\partial_{x_i}\eta_\varepsilon$ in place of $\eta_\varepsilon$ shows that $u_\varepsilon \in C^\infty(V)$. [guided] We start with the formula $u_\varepsilon(x) = \int u(y)\, \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y)$ and want to compute $\partial_{x_i} u_\varepsilon$. The natural move is to push the derivative inside the integral. The issue is that "differentiating under the integral sign" requires a justification — typically domination of the partial derivative by an integrable function. Let us first arrange the geometry. The convolution at $x \in V$ involves only $y \in B(x, \varepsilon)$ (because $\eta_\varepsilon$ is supported in $\overline{B}(0, \varepsilon)$). For $x \in V$ and $\varepsilon < \operatorname{dist}(V, \partial\Omega)$, every such $y$ lies in $\Omega$, so the integrand is well-defined and equals zero for $y$ outside the union of these balls. To get a uniform integrable dominator, restrict to a compact neighbourhood of any $x \in V$: choose a small ball $\overline{B}(x, r) \subset V$ and let $K = \overline{B}(x, r + \varepsilon) \cap \overline{\Omega}$. We have $K \subset \Omega$ because $r + \varepsilon$ can be made smaller than $\operatorname{dist}(\overline{B}(x, r) \cup V, \partial\Omega)$ by choosing $r$ small. Now apply the [Leibniz Integral Rule](/theorems/???). Its hypotheses are: 1. The integrand $(x', y) \mapsto u(y)\, \eta_\varepsilon(x' - y)$ is measurable in $y$ for each fixed $x'$. Yes — $u$ is measurable, $\eta_\varepsilon$ is smooth. 2. The integrand is differentiable in $x'$ for a.e. $y$. Yes — $\eta_\varepsilon$ is $C^\infty$. 3. The partial derivative $\partial_{x'_i}[u(y)\, \eta_\varepsilon(x' - y)] = u(y)\, (\partial_{x_i}\eta_\varepsilon)(x' - y)$ is dominated, uniformly in $x' \in \overline{B}(x, r)$, by an integrable function of $y$. Take $g(y) := |u(y)| \cdot \|\partial_{x_i}\eta_\varepsilon\|_{L^\infty(\mathbb{R}^n)} \cdot \mathbb{1}_K(y)$. The supremum norm is finite since $\partial_{x_i}\eta_\varepsilon \in C_c^\infty$. The remaining factor $|u| \cdot \mathbb{1}_K$ is integrable because $u \in L^p_{\mathrm{loc}}(\Omega)$ and $K \subset \Omega$ is compact: by [Inclusion of $L^p$ Spaces on Finite Measure Spaces](/theorems/???), $u \in L^p(K) \subseteq L^1(K)$. All hypotheses verified, the rule gives \begin{align*} \partial_{x_i} u_\varepsilon(x) = \int_{\mathbb{R}^n} u(y)\, (\partial_{x_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y). \end{align*} To upgrade this to $u_\varepsilon \in C^\infty(V)$, repeat the argument inductively: each derivative $\partial_{x_i} u_\varepsilon$ is itself a convolution with a smooth compactly supported function $\partial_{x_i}\eta_\varepsilon$, and the same hypotheses hold, so we can differentiate again, and so on. Each iteration costs at most a factor of $\|D^\alpha \eta_\varepsilon\|_{L^\infty}$ in the dominator, so all derivatives exist and are continuous on $V$. [/guided] [/step] [step:Convert the $x$-derivative on $\eta_\varepsilon$ into a $y$-derivative via the sign flip] For any $x, y \in \mathbb{R}^n$ and $i \in \{1, \dots, n\}$, by the chain rule applied to $\eta_\varepsilon$ at the point $x - y$, \begin{align*} (\partial_{x_i}\eta_\varepsilon)(x - y) = -(\partial_{y_i}\eta_\varepsilon)(x - y), \end{align*} where $\partial_{x_i}$ denotes the $i$-th partial derivative with respect to the variable indicated. Substituting into the formula for $\partial_{x_i} u_\varepsilon$ from Step 1, \begin{align*} \partial_{x_i} u_\varepsilon(x) = -\int_{\mathbb{R}^n} u(y)\, (\partial_{y_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y). \end{align*} [/step] [step:Apply the definition of weak derivative with test function $\eta_\varepsilon(x - \cdot)$] Fix $x \in V$. Define the test function \begin{align*} \phi_x: \Omega &\to \mathbb{R} \\ y &\mapsto \eta_\varepsilon(x - y). \end{align*} Since $\eta_\varepsilon \in C_c^\infty(\mathbb{R}^n)$ with $\operatorname{supp} \eta_\varepsilon \subseteq \overline{B}(0, \varepsilon)$, we have $\phi_x \in C^\infty(\mathbb{R}^n)$ with $\operatorname{supp} \phi_x \subseteq \overline{B}(x, \varepsilon)$. The hypothesis $\varepsilon < \operatorname{dist}(V, \partial\Omega)$ together with $x \in V$ gives $\overline{B}(x, \varepsilon) \subset \Omega$, so $\operatorname{supp} \phi_x$ is a compact subset of $\Omega$. Hence $\phi_x \in C_c^\infty(\Omega)$, qualifying as a valid test function for the weak derivative. Applying the [Definition of Weak Derivative](/page/Weak%20Derivative) for $u \in W^{1,p}(\Omega)$ with $\phi = \phi_x$, \begin{align*} \int_\Omega u(y)\, (\partial_{y_i}\phi_x)(y)\, d\mathcal{L}^n(y) = -\int_\Omega (\partial_{x_i} u)(y)\, \phi_x(y)\, d\mathcal{L}^n(y). \end{align*} We compute $(\partial_{y_i}\phi_x)(y) = (\partial_{y_i}\eta_\varepsilon)(x - y)$ from the definition of $\phi_x$ (the chain rule gives this without a sign flip — we are differentiating $\eta_\varepsilon(x - y)$ as a function of $y$). Substituting, \begin{align*} \int_\Omega u(y)\, (\partial_{y_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y) = -\int_\Omega (\partial_{x_i} u)(y)\, \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y). \end{align*} The right-hand side is exactly $-(\partial_{x_i} u) * \eta_\varepsilon(x)$. Since the integrand on the left vanishes outside $\Omega$ (because $\phi_x$ is supported in $\Omega$) and $u$ has been extended by zero, we may replace the domain of integration on the left by $\mathbb{R}^n$: \begin{align*} \int_{\mathbb{R}^n} u(y)\, (\partial_{y_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y) = -\int_{\mathbb{R}^n} (\partial_{x_i} u)(y)\, \eta_\varepsilon(x - y)\, d\mathcal{L}^n(y) = -((\partial_{x_i} u) * \eta_\varepsilon)(x). \end{align*} [guided] We have $\partial_{x_i} u_\varepsilon(x) = -\int u(y)\, (\partial_{y_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y)$. The integrand contains $u(y)$ multiplied by a $y$-derivative of a smooth compactly supported function — this is exactly the right side of the integration-by-parts identity that defines the weak derivative of $u$. So we want to interpret the function $y \mapsto \eta_\varepsilon(x - y)$ as a test function and apply the weak derivative formula. Define \begin{align*} \phi_x: \Omega &\to \mathbb{R} \\ y &\mapsto \eta_\varepsilon(x - y). \end{align*} For $\phi_x$ to qualify as a [Test Function](/page/Test%20Function) on $\Omega$, two conditions must hold: $\phi_x \in C^\infty(\Omega)$ and $\operatorname{supp} \phi_x$ is compact in $\Omega$. - Smoothness: $\eta_\varepsilon \in C^\infty(\mathbb{R}^n)$, so $\phi_x \in C^\infty(\Omega)$. - Compact support: $\operatorname{supp} \phi_x = x - \operatorname{supp} \eta_\varepsilon \subseteq x - \overline{B}(0, \varepsilon) = \overline{B}(x, \varepsilon)$. We need $\overline{B}(x, \varepsilon) \subset \Omega$. This is exactly where the hypothesis $\varepsilon < \operatorname{dist}(V, \partial\Omega)$ is used: for any $x \in V$, $\operatorname{dist}(x, \partial\Omega) \geq \operatorname{dist}(V, \partial\Omega) > \varepsilon$, so $\overline{B}(x, \varepsilon) \subset \Omega$. Both conditions met, $\phi_x \in C_c^\infty(\Omega)$. The [Definition of Weak Derivative](/page/Weak%20Derivative) for $u \in W^{1,p}(\Omega)$ with test function $\phi_x$ reads \begin{align*} \int_\Omega u(y)\, (\partial_{y_i}\phi_x)(y)\, d\mathcal{L}^n(y) = -\int_\Omega (\partial_{x_i} u)(y)\, \phi_x(y)\, d\mathcal{L}^n(y). \end{align*} Compute $(\partial_{y_i}\phi_x)(y) = \frac{\partial}{\partial y_i}\eta_\varepsilon(x - y) = (\partial_{y_i}\eta_\varepsilon)(x - y)$, where the chain rule for the inner map $y \mapsto x - y$ has Jacobian $-I$ — but we are looking at the derivative *as a function of $y$*, with no implicit sign flip. (Compare: in Step 2 we had $(\partial_{x_i}\eta_\varepsilon)(x-y)$ and asked how it relates to $(\partial_{y_i}\eta_\varepsilon)(x-y)$ — *those* are derivatives at the same argument with respect to *different* variables, and they differ by a sign.) Substituting, the left side becomes $\int_\Omega u(y)\, (\partial_{y_i}\eta_\varepsilon)(x-y)\, d\mathcal{L}^n(y)$ and the right side becomes $-((\partial_{x_i} u) * \eta_\varepsilon)(x)$. Because $\phi_x$ is supported in $\Omega$ and we have extended $u$ by zero outside $\Omega$, the integrals over $\Omega$ and over $\mathbb{R}^n$ agree: \begin{align*} \int_{\mathbb{R}^n} u(y)\, (\partial_{y_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y) = -((\partial_{x_i} u) * \eta_\varepsilon)(x). \end{align*} [/guided] [/step] [step:Combine to identify $\partial_{x_i} u_\varepsilon = (\partial_{x_i} u) * \eta_\varepsilon$ on $V$] Substituting the result of Step 3 into the formula from Step 2, \begin{align*} \partial_{x_i} u_\varepsilon(x) = -\int_{\mathbb{R}^n} u(y)\, (\partial_{y_i}\eta_\varepsilon)(x - y)\, d\mathcal{L}^n(y) = -\bigl(-((\partial_{x_i} u) * \eta_\varepsilon)(x)\bigr) = ((\partial_{x_i} u) * \eta_\varepsilon)(x). \end{align*} This holds for every $x \in V$, establishing the identity \begin{align*} \partial_{x_i} u_\varepsilon = (\partial_{x_i} u) * \eta_\varepsilon \quad \text{on } V. \end{align*} [/step] [step:Deduce $W^{1,p}(V)$ convergence as $\varepsilon \to 0$] We must show $u_\varepsilon \to u$ in $L^p(V)$ and $\partial_{x_i} u_\varepsilon \to \partial_{x_i} u$ in $L^p(V)$ for each $i = 1, \dots, n$. For the function values, the [Convergence of Mollifications in $L^p$](/theorems/???) theorem states: if $f \in L^p(\Omega)$ for some $1 \leq p < \infty$ and $V \subset\subset \Omega$ is open, then $f_\varepsilon = f * \eta_\varepsilon \to f$ in $L^p(V)$ as $\varepsilon \to 0$. Applying this with $f = u \in L^p(\Omega)$ gives $u_\varepsilon \to u$ in $L^p(V)$. For the derivatives, apply the same theorem with $f = \partial_{x_i} u \in L^p(\Omega)$ (the weak derivative, an $L^p$ function): $(\partial_{x_i} u) * \eta_\varepsilon \to \partial_{x_i} u$ in $L^p(V)$. By Step 4, $(\partial_{x_i} u) * \eta_\varepsilon = \partial_{x_i} u_\varepsilon$ on $V$, so \begin{align*} \partial_{x_i} u_\varepsilon \to \partial_{x_i} u \quad \text{in } L^p(V), \quad i = 1, \dots, n. \end{align*} Combining, \begin{align*} \|u_\varepsilon - u\|_{W^{1,p}(V)}^p = \|u_\varepsilon - u\|_{L^p(V)}^p + \sum_{i=1}^n \|\partial_{x_i} u_\varepsilon - \partial_{x_i} u\|_{L^p(V)}^p \xrightarrow{\varepsilon \to 0} 0, \end{align*} which is the claimed $W^{1,p}(V)$ convergence. The hypothesis $p < \infty$ enters precisely here: the [Convergence of Mollifications in $L^p$](/theorems/???) theorem fails for $p = \infty$ (indeed, mollifications converge only in $L^\infty_{\mathrm{loc}}$ when the function is continuous, and otherwise weakly-$*$ but not in norm). [/step]