Reading through Chapter 8 of Brezis. We see that $C_{c}^{\infty}(\mathbb{R})$ is dense in $W^{1,\,p}(I)$. Later on in the chapter we define $W^{1,\,p}_{0}(I)$ to be the closure of $C_{c}^{1}(I)$ in $W^{1,\,p}(I)$. This $W^{1,\,p}_{0}(I)$ space has really confused me.
What happens when look at $C_{c}^{\infty}(I)$ which causes us to lose density in $W^{1,\,p}(I)$?
This question is linked to the first question. Suppose $(u_{n})\in C_{c}^{1}(I)$, then $(u_{n}')\in C_{c}(I)$. Both $C_{c}^{1}(I)$ and $C_{c}(I)$ are dense in $L^{p}(I)$, so $u_{n}\rightarrow u$ and $u_{n}'\rightarrow g$ in $L^{p}(I)$. Hence why cannot we not say that $u_{n}\rightarrow u$ in $W^{1,\,p}$ by taking $u'=g$?
By definition $\overline{C_{c}^{1}(I)}=W^{1,\,p}_{0}(I)$ in the $W^{1,\,p}$ norm. So how does one show density of $C_{c}^{\infty}(I)$ in $W^{1,\,p}_{0}(I)$? Is it sufficient to show that $C_{c}^{\infty}(I)$ is dense in $C_{c}^{1}(I)$ with respect to the supremum norm?
It's not true that $\mathcal C_c^\infty (I)$ is dense in $W^{1,p}(I)$ when $I\neq \mathbb R$. What is true is $\mathcal C_c^\infty (\mathbb R)$ is dense in $W^{1,p}(\mathbb R)$. And indeed, if $I\neq \mathbb R$ is an interval, we define $W_0^{1,p}(I)$ as the closure of $\mathcal C_c^\infty (I)$ in $W^{1,p}(I)$.