I have some difficulties in proving the following theorem:
Let $f\in C_{c}^{k}(\mathbb{R}^n)$. Let $\rho\in C_{c}^{\infty}(\mathbb{R}^n)$ be such that $$\rho\geq0,\,\ \mbox{supp}\, \rho\subset\{|x|\leq 1\},\, \int\rho\, dx=1.$$ Let $\varepsilon>0$ and put $$f_{\varepsilon}(x)=\varepsilon^{-n}\int f(y)\rho (\frac{x-y}{\varepsilon})dy.$$ Then $f_{\varepsilon}\in C_{c}^{\infty}(\mathbb{R}^n)$ and the support is of $f_{\varepsilon}$ is contained in the $\varepsilon$ - neighbourhood of the support $f$, moreover, if $|\alpha|\leq k$, then $\partial^{\alpha}f_{\varepsilon}$ converge uniformly to $\partial^{\alpha}f$ as $\varepsilon \to 0$.
The first statement is obvious, i.e., $f_{\varepsilon}\in C_{c}^{\infty}(\mathbb{R}^n)$. Why the support of $f_{\varepsilon}$ is contained in the $\varepsilon$ - neighbourhood of the support $f$? In order to prove the convergance, I consider $k=0$ at first. Let $z=(x-y)/\varepsilon$. Then $$y=x-\varepsilon\cdot x,\,\, dy=-\varepsilon dz.$$ So $$f_{\varepsilon}(x)=-\varepsilon^{-n+1}\int f(x-\varepsilon z)\rho(z)dz$$ and $$|f_{\varepsilon}(x)-f(x)|=|\varepsilon^{-n+1}|\cdot |\int(\frac{f(x)}{\varepsilon^{-n+1}}-f_{\varepsilon}(x-\varepsilon z))\rho (z)dz|.$$ What should be done next?