Why is the mollification $\frac{1}{r^n}\int_\Omega\varrho\left(\frac{|x_0-x|}{r}\right)f(x)\;dx$ of an integrable $f$ infinitely differentiable? Let $\Omega\subseteq\mathbb{R}^n$ be a bounded domain and $\overline{B}_r(x_0)\subseteq\Omega$ be the closed ball around $x_0\in\Omega$ with radius $r>0$. Let $\lambda_n$ be the Lebesgue measure on the Borel-$\sigma$-algebra on $\mathbb{R}^n$. Given $f\in\mathcal{L}^1\left(\left.\lambda_n \right|_\Omega\right)$, $$f_r(x_0):=\frac{1}{r^n}\int_\Omega\varrho\left(\frac{|x_0-x|}{r}\right)f(x)\;d\lambda_n(x)$$ with $$\varrho(t):=\begin{cases}\alpha\exp\left(\frac 1{t^2-1}\right)&\text{, if }t\in [0,1]\\\ 0&\text{, otherwise}\end{cases}$$ and $\alpha\in\mathbb{R}$ chosen such that $$\int_{\mathbb{R}^n}\varrho\left(|x|\right)d\lambda_n(x)=1 ,$$ is called _mollification_ of $f$. How can we prove, that $f_r$ is infinitely differentiable with respect to $x_0$?

The mollification is the convolution product of a $C^{\infty}_c$ function $\varrho$ and a function $f$ integrable:

$$f_r = \varrho \star f$$

And it suffice to show that for $g$ differentiable with compact support and $f$ integrable, you get

$$(f\star g )' = f\star g'$$

And this is not hard to show as :

$$(f\star g )'(t) = \left( \int_{\mathbb{R}} f(x) g(t-x) dx \right)' = \int_{\mathbb{R}} f(x) g'(t-x) dx = (f\star g')(t)$$

By derivation under the integral theorem

Then you get that $f_r$ is infinitely differentiable by a quick reccurence