Closure of $C_0^{\infty}$ in $W^{k,p}(\Omega)$

Question

Why is it that in the definition of $W_0^{k,p}(\Omega)$ for $\Omega$ with boundary smooth enough, we only have $D^{\alpha}u$ for all $0\leq|\alpha|\leq k-1$ vanishing at the boundary and not $D^{\alpha}u$ with $|\alpha|=k$?

Thanks a lot!

Answer 1

Think about $L^p(\Omega)$ first. The closure of $C_0^\infty(\Omega)$ in $L^p(\Omega)$ is all of $L^p(\Omega)$. That is, the vanishing at $\partial \Omega$ of functions in $C_0^\infty(\Omega)$ in no way affects the boundary behavior of functions in the closure.

Now look at $W^{1,p}(\Omega)$. The closure of $C_0^\infty(\Omega)$ in $W^{1,p}(\Omega)$ is $W_0^{1,p}(\Omega)$. If $u \in W^{1,p}(\Omega)$, $\{u_k\} \subset C_0^\infty(\Omega)$, and $u_k \to u$ in the $W^{1,p}$-norm, the convergence in $L^p$ of $\nabla u_k$ to $\nabla u$ is not enough to imply that $\nabla u$ vanishes on $\partial \Omega$ (as in the preceding paragraph) but it, along with $u_k \to u$ in $L^p(\Omega)$, is enough to imply that $u$ vanishes (in an appropriate weak sense) on $\partial \Omega$.