I am trying to prove that mollified functions (as defined on p.31 of Showalter's book Hilbert Space Methods in PDEs) are smooth ($C^\infty$). I am not sure how to conclude the proof (in particular, I don't see how to go from my last line to concluding that $f_\epsilon \in C^\infty (\mathbb R^n$)) , and I also would appreciate any critique of my work-in-progress proof- something feels wrong here. I summarize the needed definitions and the proposition:
For $\epsilon > 0$, suppose $\varphi_\epsilon$ is $C^\infty$ on $\mathbb{R}^n$ and has compact support, along with the following properties: $\varphi_\epsilon \geq 0$, $\text{supp}(\varphi_\epsilon) \subset \{x \in \mathbb{R}^n:|x| \leq \epsilon\},$ and $\int \varphi_\epsilon = 1$. Let $f \in L^1 (G)$ where $G$ is open in $\mathbb{R}^n$ and define the mollified function $$f_\epsilon (x) = \int_{\mathbb{R}^n} f(x-y) \varphi_\epsilon (y) dy, x \in \mathbb{R}^n.$$ Then for each $\epsilon > 0, f_\epsilon \in C^\infty (\mathbb{R}^n)$.
Here is my attempt at a proof:
Consider some arbitrary partial derivative $$ D \equiv \frac{\partial^\alpha}{\partial_{x_1}^{\alpha_1} \cdots \partial_{x_n}^{\alpha_n}}$$ of order $\alpha = \sum_{i=1}^n \alpha_i$. We evaluate $$Df_\epsilon (x) = D \int\color{red}{f(x-y)} \varphi_\epsilon (y) dy$$ by defining $s = x-y$ (note that $\frac{\partial s}{\partial x} = 1$) and rewriting as $$Df_\epsilon (x) = D \int f(s) \varphi_\epsilon (x-s) ds = \int f(s) D \varphi_\epsilon (x-s) ds,$$ where the second equality follows from $\varphi_\epsilon$ being smooth. But then we may rewrite this once more as $$\int f(x) D\varphi_\epsilon (x-y) dy = f(x) \int D\varphi_\epsilon (x-y) dy$$.
