I want to mollify a function $f\in L^1_{loc}(\Omega)$, $\Omega$ open in $\mathbb R^n$. So, we take a sequence of standard mollifiers $\{\phi_\epsilon\}$, for example, and do the convolution $f*\phi_\epsilon$.
I've read that, if $\Omega$ is not the whole $\mathbb R^n$, $f*\phi_\epsilon(x)$ is well defined for all $x\in \Omega$ such that $0<\epsilon<d(x,\partial \Omega)$: does this mean that $u_\epsilon$ is well defined only for that $x$'s? If it were so, I wouldn't understand why. I see that if $x\in \Omega$ and if $d(x,\partial \Omega)<\epsilon$, then $\text{supp }\phi_{\epsilon}(\,x-\cdot\,)$ is not entirely contained in $\Omega$. Is this a problem? When I have to write down the integral I get $$f*\phi_\epsilon(x):=\int_\Omega \phi_\epsilon(x-y)f(y)dy=\int_{\Omega\cap \overline{B(x,\epsilon)}} \phi_\epsilon(x-y)f(y),$$
so, even if $x\in \Omega$ but $d(x,\partial \Omega)<\epsilon$, $u_\epsilon(x)$ seems well defined to me. What's wrong? Where's the problem?
No, it is not a problem. What you are doing is to extend $f$ by setting it equal to $0$ outside of $\Omega$ and then doing the convolution with the extension.