I have read through the Meyers-Serrin theorem, and would like to understand why a simpler argument would not work. The theorem states that $C^{\infty}(\Omega)$ is dense in $W^{k,p}(\Omega), 1 \le p < +\infty.$
In the following we assume $k = 1$ and $\rho_{\epsilon} $a sequence of mollifiers.
For $u \in W^{1,p}(\Omega),$ we consider $u, \nabla u \in L^p(\mathbb{R}^n),$ through natural extension through zero. Then we know:
$u*\rho_{\epsilon} \rightarrow u$
$\nabla u*\rho_{\epsilon} \rightarrow \nabla u$
where both convergences are in $L^p(\mathbb{R}^n)$. From here we find:
$u*\rho_{\epsilon} |_{\Omega} \rightarrow u|_{\Omega} $
$\nabla u*\rho_{\epsilon}|_{\Omega} \rightarrow \nabla u|_{\Omega} $
here both convergences are in $L^p(\Omega).$ So we have almoast proven convergence in $W^{1,p}(\Omega),$ as soon as we know that $\nabla u*\rho_{\epsilon}|_{\Omega} = \nabla (u*\rho_{\epsilon}|_{\Omega})$ as a distribution on $\Omega.$ Is this last statement false? Because it seems true to me, and it should follow from the general results on derivatives of a convolution with a distribution.
Thanks for the help in making my trhoughts clearer! :)
The very first step already fails: The extension by zero will in general not produce a function $u$ that has a weak derivative on $\mathbb R^n$. Think of $u\equiv 1$ on $\Omega$.
This implies that one cannot mollify in a way which uses values of $u$ outside of $\Omega$. Hence, the proof uses translations to stay away from the boundary of $\Omega$.