Let $u \in L_{Loc}^1(\mathbb{R^n})$ $\alpha$-weak derivable. Let us denote by $u_{\epsilon}$ and $(D^{\alpha}u)_{\epsilon}$ respectively the mollified functions of $u$ and $D^{\alpha}u$, we can conclude that $D ^{\alpha}u_{\epsilon}(x)=(D^{\alpha}u)_{\epsilon}(x)$ for every $x \in \mathbb{R^n}$?
I tried this way:
Since $u_{\epsilon} \in C^\infty(\mathbb{R^n})$ then the weak $\alpha$-derivative coincides with the $\alpha$-derivative, so: $D^{\alpha}u_{\epsilon}(x)=\frac{1}{\epsilon^n} \int_{\mathbb{R^n}} u(y) D_x^{\alpha}\phi(\frac{y-x }{\epsilon}) dy$. Now observing that $D_x^{\alpha}\phi(\frac{y-x}{\epsilon})=D_y^{\alpha}\phi(\frac{y-x}{\epsilon})$ if $\alpha$ is even and $D_x^{\alpha}\phi(\frac{y-x}{\epsilon})=D_y^{\alpha}\phi(\frac{x-y}{\epsilon})$ if $\alpha$ is odd then we obtain:
$D^{\alpha}u_{\epsilon}(x)={(\frac{1}{\epsilon^n} \int_{\mathbb{R^n}} u(y) D_y^{\alpha} \phi(\frac{y-x}{\epsilon}) dy ,if \alpha text{ even}),(\frac{1}{\epsilon^n} \int_{\mathbb{R^n}} u(y ) D_y^{\alpha}\phi(\frac{x-y}{\epsilon}) dy ,if \alpha text{ odd}):}$
using that $\phi \in C_0^\infty(\mathbb{R^n})\subseteq C_0^{|\alpha|}(\mathbb{R^n})$ and that $u$ is $\alpha$-weak derivable, I get:
$D^{\alpha}u_{\epsilon}(x)={(\frac{1}{\epsilon^n} \int_{\mathbb{R^n}} D^{\alpha}u(y) \cdot \phi(\frac{y-x}{\epsilon}) dy ,if \alpha text{ even}),(\frac{1}{\epsilon^n} \int_{\mathbb{R^n}} D^ {\alpha}u(y) v \phi(\frac{x-y}{\epsilon}) dy ,if \alpha text{ odd}):}$
Finally, since $\phi$ is even we obtain:
$D^{\alpha}u_{\epsilon}(x)=\frac{1}{\epsilon^n} \int_{\mathbb{R^n}} D^{\alpha}u(y) \cdot \phi(\frac{y-x}{\epsilon}) dy=(D^{\alpha}u)_{\epsilon}(x)$
(I think i also need to suppose that $D^{\alpha}u \in L_{Loc}^1(\mathbb{R^n})$)
I wanted to know if everything was right or if there was something wrong, thanks.