Let $\Omega \subset \mathbb{R}^n$ an open set, we define the sobolev space:
$$ W^{m,p}(\Omega) = \{u \in L^p(\Omega): D^\alpha u \in L^p(\Omega) \ \mbox{in the sense of distributions with} \ |\alpha| \le m\} $$
Considering $C^m(\bar{\Omega})$ with the norm: $$ ||\varphi||_{m,p} = \bigg(\sum_{|\alpha|\le m}\int_{\Omega}||D^\alpha \varphi||_{L^p(\Omega)}^p \bigg)^{1/p} $$ We define $H^{m,p}(\Omega)$ as the completion of $C^m(\bar{\Omega})$ according to this norm. My question is, $H^{m,p}(\Omega) = W^{m,p}(\Omega)$? If so, how to show it?
In general, the answer is no: Let $\Omega = (-1,0) \cup (0,1)$ and $m = 1$, $p = 2$. Then, $f = \chi_{(0,1)}$ belongs to $W^{1,2}(\Omega)$, but due to the jump at $t = 0$, it does not belong to $H^{1,2}(\Omega)$. (Note that convergence in $H^{1,2}(\Omega)$ implies uniform convergence due to $n = 1$. Hence, all functions in $H^{1,2}(\Omega)$ are continuous at $t = 0 $).
If you have some regularity of $\Omega$, the result should be true, but I do not have a reference.
Further, the result is unconditionally true if you replace $C^m(\bar\Omega)$ by $C^m(\Omega) \cap W^{m,p}(\Omega)$. This is the famous $H = W$ paper by Meyers and Serrin.