[proofplan] We prove the three properties of mollification separately. Smoothness follows from differentiation under the integral sign (justified by dominated convergence). The support property follows from a contrapositive argument using the support of $\rho_\varepsilon$. The $L^p$ convergence uses Minkowski's integral inequality to reduce to the $L^p$-continuity of translation, then applies dominated convergence. [/proofplan] [step:Prove smoothness by differentiating under the integral sign] Write $(f * \rho_\varepsilon)(x) = \int_{\mathbb{R}^n} f(y) \, \rho_\varepsilon(x - y) \, d\mathcal{L}^n(y)$. For each fixed $y$, the map $x \mapsto \rho_\varepsilon(x - y)$ is $C^\infty$, and \begin{align*} |\partial_x^\alpha \rho_\varepsilon(x - y)| &\leq \varepsilon^{-n - |\alpha|} \|\partial^\alpha \rho\|_{L^\infty}, \end{align*} with support in $\{x : |x - y| \leq \varepsilon\}$. For $f \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$, the function $y \mapsto f(y) \, \partial_x^\alpha \rho_\varepsilon(x - y)$ is integrable for each $x$ (since $\partial^\alpha \rho_\varepsilon$ is bounded and compactly supported). The [dominated convergence theorem](/theorems/4) justifies passing $\partial_x^\alpha$ under the integral: the difference quotient is dominated by $|f(y)| \cdot \varepsilon^{-n-|\alpha|-1} \|\partial^{\alpha + e_j} \rho\|_{L^\infty} \cdot \mathbb{1}_{|x-y| \leq \varepsilon}$, which is integrable against $f$ on the compact ball $\overline{B}(x, \varepsilon)$. This gives \begin{align*} \partial^\alpha(f * \rho_\varepsilon)(x) &= \int_{\mathbb{R}^n} f(y) \, (\partial^\alpha \rho_\varepsilon)(x - y) \, d\mathcal{L}^n(y) = (f * \partial^\alpha \rho_\varepsilon)(x). \end{align*} The right-hand side is continuous in $x$ by another application of dominated convergence. Since this holds for all multi-indices $\alpha$, $f * \rho_\varepsilon \in C^\infty(\mathbb{R}^n)$. [/step] [step:Prove the support property by contrapositive] Suppose $x \notin \mathrm{supp}(f) + \overline{B}(0, \varepsilon)$. Then for every $y \in \mathrm{supp}(f)$, $|x - y| > \varepsilon$, so $\rho_\varepsilon(x - y) = 0$ (since $\mathrm{supp}(\rho_\varepsilon) = \overline{B}(0, \varepsilon)$). The integrand $f(y) \rho_\varepsilon(x - y)$ therefore vanishes for $\mathcal{L}^n$-a.e. $y$: when $y \notin \mathrm{supp}(f)$, $f(y) = 0$ a.e.; when $y \in \mathrm{supp}(f)$, $\rho_\varepsilon(x - y) = 0$. Hence $(f * \rho_\varepsilon)(x) = 0$, proving $\mathrm{supp}(f * \rho_\varepsilon) \subseteq \mathrm{supp}(f) + \overline{B}(0, \varepsilon)$. [/step] [step:Prove $L^p$ convergence via Minkowski's integral inequality and continuity of translation] Using $\int_{\mathbb{R}^n} \rho_\varepsilon \, d\mathcal{L}^n = 1$, write \begin{align*} (f * \rho_\varepsilon)(x) - f(x) &= \int_{\mathbb{R}^n} \rho_\varepsilon(y) \bigl(f(x - y) - f(x)\bigr) \, d\mathcal{L}^n(y). \end{align*} Substituting $z = y/\varepsilon$: \begin{align*} (f * \rho_\varepsilon)(x) - f(x) &= \int_{\mathbb{R}^n} \rho(z) \bigl(f(x - \varepsilon z) - f(x)\bigr) \, d\mathcal{L}^n(z). \end{align*} By [Minkowski's integral inequality](/theorems/464) (interchanging the $L^p_x$ norm and the $d\mathcal{L}^n(z)$ integral): \begin{align*} \|f * \rho_\varepsilon - f\|_{L^p} &\leq \int_{\mathbb{R}^n} \rho(z) \, \|f(\cdot - \varepsilon z) - f(\cdot)\|_{L^p} \, d\mathcal{L}^n(z). \end{align*} The integrand is bounded by $2\rho(z) \|f\|_{L^p}$ (using the triangle inequality and translation-invariance of the $L^p$ norm), which is integrable since $\int \rho \, d\mathcal{L}^n = 1$. For each fixed $z$ with $|z| \leq 1$, the $L^p$-continuity of translation gives \begin{align*} \|f(\cdot - \varepsilon z) - f(\cdot)\|_{L^p} &\to 0 \quad \text{as } \varepsilon \to 0. \end{align*} By the [dominated convergence theorem](/theorems/4) applied to the outer $d\mathcal{L}^n(z)$ integral, $\|f * \rho_\varepsilon - f\|_{L^p} \to 0$. [guided] The strategy is to write $f * \rho_\varepsilon - f$ as an integral of translations of $f$ weighted by the mollifier, then pull the $L^p$ norm inside the integral. The key step is Minkowski's integral inequality, which is the continuous analogue of the triangle inequality: it says $\|x \mapsto \int K(y)f(x-y)\,d\mathcal{L}^n(y)\|_{L^p_x} \leq \int |K(y)| \|f(\cdot - y)\|_{L^p}\,d\mathcal{L}^n(y)$. After applying this, the problem reduces to showing that $\|f(\cdot - \varepsilon z) - f(\cdot)\|_{L^p} \to 0$ for each fixed $z$. This is the $L^p$-continuity of translation, a standard property of $L^p$ spaces for $1 \leq p < \infty$. The dominated convergence theorem then allows us to pass the limit inside the outer integral over $z$, since the integrand $\rho(z)\|f(\cdot - \varepsilon z) - f(\cdot)\|_{L^p}$ is bounded by $2\rho(z)\|f\|_{L^p}$, which is integrable. Note that the result fails for $p = \infty$ because $L^\infty$-translation is not continuous in general. [/guided] [/step]