$W^{m,p}_{0}(\mathbb{R}^d) = W^{m,p}(\mathbb{R}^d)$ and continuously imbedded

Question

Can anyone help me solve the following results? If $1\leq p < \infty$, then

(a) $W^{m,p}_{0}(\mathbb{R}^d) = W^{m,p}(\mathbb{R}^d)$,

(b) $W^{m,p}_{0}(\Omega) \hookrightarrow W_{m,p}(\Omega)$, (continuously imbedded),

where $W^{m,p}_{0}(\mathbb{R}^d)$ the closure in $ W^{m,p}(\mathbb{R}^d)$ of $C_{c}^{\infty}(\Omega)(test function)$ and

$W^{m,p}(\mathbb{R}^d)$ is space sobolev.

Answer 1

Hint: $C^{\infty}(\mathbb{R}^{n}) \cap W^{m,p}(\mathbb{R}^{n})$ is dense in $W^{m,p}(\mathbb{R}^{n})$ with respect to the Sobolev norm. Now show that $C_{0}^{\infty}(\mathbb{R}^{n})$ is dense in $C^{\infty}(\mathbb{R}^{n}) \cap W^{m,p}(\mathbb{R}^{n})$ with respect to the Sobolev norm. To this end, note that if $u \in L^{p}(\mathbb{R}^{n})$, we can approximate it - in $L^{p}$ - by a function with compact support.