I am reading the proof the following statement provided in Functional Analysis, Sobolev Spaces and Partial Differential Equations, by haim Brezis:
If $u \in W_0^{1,p}(I)$, then $u=0$ on $\partial I$
Where $I \subset \mathbb{R}$ is an open interval. I do not understand the proof given, which is as follows:
If $u \in W_0^{1,p}$ then there exists a sequence $(u_n) \subset C^1_c(I)$ such that $u_n \to u$ in $W^{1,p}(I)$.
Then $u_n \to u$ uniformly on $\bar{I}$ and as a consequence $u=0$ on $\partial{I}$.
I understand the first part, $W_0^{1,p}$ is defined as the closure of $ C^1_c(I)$ in $W^{1,p}(I)$.
But how is the uniform convergence proved from the $W^{1,p}$-convergence? And why is the uniform convergence needed? Isn't it enough to have pointwise convergence? All of the functions $u_n$ are equal to zero on $\partial(I)$, so if they converge pointwise to a function $u$, then $u=0$ on $\partial(I)$, doesn't it?
As far as I know, $W^{1,p}$-convergence implies $L^p$ convergence of both $u_n$ and its derivative, but I don't know how to show uniform convergence.