Let $\Omega\subset\mathbb{R}^n$ be open, $1\le p <\infty$, $k\in\mathbb{N}$. Meyers-Serrin tells us: $$ C^\infty(\Omega)\cap W^{k,p}(\Omega)\stackrel{dense}{\subset} W^{k,p}(\Omega)$$ Can one deduce the follwing?
- $$C^k(\Omega)\cap W^{k,p}(\Omega)\stackrel{dense}{\subset} W^{k,p}(\Omega)$$
- $$C^k(\Omega)\stackrel{dense}{\subset} W^{k,p}(\Omega) \qquad \text{for $\Omega$ bounded}$$
And what about $C^k(\overline{\Omega})$ ?
(1) Since $C^\infty(\Omega)\subset C^k(\Omega)$ it follows that $C^\infty(\Omega) \cap W^{k,p}(\Omega) \subset C^k(\Omega)\cap W^{k,p}(\Omega) \subset W^{k,p}(\Omega)$. Now the left-hand set is dense, hence the the set in the middle as well.
(2) This is not true as $C^k(\Omega)$ is not a subset of $W^{k,p}(\Omega)$. Take for instance $\Omega=(0,1)$. Then the functions $u(x)=x^{-s}$ are in $C^k(\Omega)$ but not in $W^{k,p}(\Omega)$ for large $s$.
(3) If we take $C^k(\bar\Omega)$ instead then $C^k(\bar\Omega)\subset W^{k,p}(\Omega)$ for bounded sets $\Omega$. However, density can be proven only under some assumptions on $\Omega$.
Here, $C^k(\bar\Omega)$ denotes the space of $k$-times continuously differentiable functions on $\Omega$ with bounded and uniformly continuous derivatives up to order $k$. (I am used to this definition, it is used, e.g. in Adams' book on Sobolev spaces)