Why the existence of such an operator implies that $\mathcal C^1_c(\mathbb R^d)$ is dense in $W^{1,p}(\Omega )$?

Question

We have the following theorem : suppose $\Omega \mathbb R^n$ is of class $\mathcal C^1$ with a bounded boundary. There exists an extension operator $$P:W^{1,p}(\Omega )\longrightarrow W^{1,p}(\mathbb R^d)$$ such that for all $u\in W^{1,p}(\Omega )$, $$Pu|_\Omega =u,\quad \|Pu\|_{L^p(\Omega )}\leq C\|u\|_{L^p(\Omega )},\quad \text{and}\quad \|Pu\|_{W^{1,p}(\Omega )}\leq C\|u\|_{W^{1,p}(\Omega )}.$$

1) First of all, is $P$ continuous?

2) Why does it imply that $\mathcal C^1_c(\mathbb R^d)$ is dense in $W^{1,p}(\Omega )$ ?

I recall that $\mathcal C^1_c(\mathbb R^d)$ mean functions that are $\mathcal C^1(\mathbb R^d)$ which have compact support.

Answer 1

First of all, to my knowledge, the operator $P$ you wrote there is called the extension operator.

The existence of extension operator depends on the geometric property of domain. The most common requirement could be "the boundary of $\Omega$ is Lipschitz".

For 1, note that $P$ is a linear operator, so it is continuous.

For 2, I guess the argument here is since $C_c^\infty(\mathbb R^d)$ is dense in $W^{1,p}(\mathbb R^d)$, and hence by using extension operator, you can prove that the density result in $\Omega$.

You may look for some Sobolev space book about more information of extension operator. For example, this book is good.