Convolution with a mollifier is smooth, and derivatives pass under the integral sign

Fix $\epsilon >0$ and $x_0\in {\mathbb{R}}^n$. Choose $R>0$ with $\mathrm{supp}{\phi }_{\epsilon }\subseteq {\bar{B}}_R(0)$ and put $K={\bar{B}}_{R+1}(x_0)$. Every derivative of ${\phi }_{\epsilon }$ vanishes off its support. Thus, for $x\in B_1(x_0)$, all translated kernel derivatives vanish when $y\mathrm{\notin }K$. Each derivative ${\mathrm{\partial }}^{\alpha }{\phi }_{\epsilon }$ is bounded by a finite constant $C_{\alpha }$, since it is continuous with compact support.

For each multi-index $\alpha$ define $F_{\alpha }(x)=\int f(y){\mathrm{\partial }}^{\alpha }{\phi }_{\epsilon }(x-y)dy$. At fixed $x$, the integrand is Lebesgue measurable by [L4], because $f$ is measurable and the translated kernel derivative is continuous. On $B_1(x_0)$ its absolute value is bounded by $C_{\alpha }\mathrm{\mid }f\mathrm{\mid }{\mathbf{1}}_K$, which is integrable by local integrability. Consequently every $F_{\alpha }$ is well-defined. If $x_m\to x$ in $B_1(x_0)$, continuity of the kernel gives pointwise convergence of the integrands; [L5] with this same majorant gives $F_{\alpha }(x_m)\to F_{\alpha }(x)$. Thus every $F_{\alpha }$ is continuous locally, hence globally since $x_0$ was arbitrary.

Fix $x\in B_{1\mathrm{/}2}(x_0)$ and a coordinate $j$. For $\mathrm{\mid }t\mathrm{\mid }<1\mathrm{/}2$, the integrand $G(y,t)=f(y){\mathrm{\partial }}^{\alpha }{\phi }_{\epsilon }(x+t{e}_j-y)$ is integrable, differentiable in $t$, and has measurable derivative. The derivative is bounded by $C_{\alpha +e_j}\mathrm{\mid }f\mathrm{\mid }{\mathbf{1}}_K$ on this interval. Therefore [L2] yields ${\mathrm{\partial }}_j{F}_{\alpha }(x)=F_{\alpha +e_j}(x)$. This identity holds for every $\alpha ,j,x$.

Since $F_0=f\ast {\phi }_{\epsilon }$, repeated use of step 3.1 gives every ordered iterated derivative of $F_0$ as the integral of $f$ against the corresponding derivative of the smooth kernel. All these integrals are continuous by step 2.1. Hence $F_0\in C^{\mathrm{\infty }}$ by [L3], and in particular ${\mathrm{\partial }}^{\alpha }{F}_0=F_{\alpha }=f\ast ({\mathrm{\partial }}^{\alpha }{\phi }_{\epsilon })$ for every multi-index $\alpha$.