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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 $$\int f(x) D\varphi_\epsilon (x-y) dy = f(x) \int D\varphi_\epsilon (x-y) dy$$.

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The substitution in the integral is wrong. You must formally set $y=x-s$ and the resulting integral is in $ds$. After that you prove what you want with the argument you were proposing.

The last equality you write is wrong. It is the same mistake that you made in the part in red now. But you are done. You don't need that last equality. You can differentiate under the integral sign as you did, and that quantity is continuous by dominated convergence.

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