Proof of Brezis Theorem 9.17. Theorem 9.17 of Brezis, "Functional Analysis, Sobolev Spaces, and Partial Differential Equations," states the following claim.

Suppose that $\Omega$ is of class $C^1$. Let
$u \in W^{1,p}(\Omega) \cap C(\overline{\Omega})$ with $1 \leq p < \infty$.
Then the following properties equivalent:
(i) $u = 0$ on $\Gamma = \partial \Omega$.
(ii) $u \in W^{1,p}_0 (\Omega)$.

I need the proof of (i) $\Rightarrow$ (ii).
The Brezis' book just prove this when $\textrm{supp} \, u$ is a compact subset of $\Omega$.
And, the Brezis' book says "in the general case in which $\textrm{supp} \, u$ is not bounded,
consider the sequence $(\zeta_n u)$ (where $(\zeta_n)$ is a sequence of cut-off functions)."
However, I cannot prove the general case. I want to show that $\|\nabla u - \nabla u_n\|_{L^p} \to 0$, but if I try to do it, $\nabla \zeta_n$ appear and I cannot handle it.
 A: You have a lot of freedom when you choose cut-off functions, so much so you can usually impose conditions on their gradients. Here's what I mean: let $\zeta \in C^\infty(\mathbb{R})$ be such that $$
\begin{cases}
\zeta(t) = 1 \text{ if } t \leqslant 1, \, \zeta(t) = 0 \text{ if } t \geqslant 2 \\
0 \leqslant \zeta \leqslant 1  \\
0\leqslant \zeta' \leqslant 2
\end{cases}
$$ and define for $k\in \mathbb{N}$, $\zeta_k(x) = \zeta (\vert x \vert /k)$. Then $\zeta_k$ satisfy
$$
\begin{cases}
\zeta_k(t) = 1 \text{ in } B_k, \, \zeta_k(t) = 0 \text{ in } \mathbb{R}^n \backslash B_{k+1} \\
0 \leqslant \zeta_k \leqslant 1  \\
0\leqslant \vert D \zeta_k \vert \leqslant 2/k . 
\end{cases}$$
Since $\zeta_k u $ satisfy the assumptions of the previous lemma in Brezis, it follows $\zeta_k u \in W^{1,p}_0(\Omega)$. Then since $\vert\zeta_k u - u \vert^p \leqslant \vert u \vert^p$, it follows from dominated convergence that
\begin{align*}
\| \zeta_k u - u \|_{L^{p}(\Omega)}^p &= \int_\Omega  \vert\zeta_k u - u \vert^p d x  \to 0 \qquad \text{ as } k \to \infty. 
\end{align*} Moreover, \begin{align*}
\| \nabla(\zeta_k u)  - \nabla u \|_{L^{p}(\Omega)} &=  \bigg (  \int_\Omega \vert \nabla(\zeta_k u) - \nabla u\vert^p d x \bigg) ^{1/p} \\
&\leqslant \bigg (  \int_\Omega \vert u\nabla\zeta_k \vert^p d x \bigg) ^{1/p} + \bigg (  \int_\Omega \vert \zeta_k \nabla u - \nabla u\vert^p d x \bigg) ^{1/p} \\
&\leqslant\frac 2 k \bigg (  \int_\Omega \vert u \vert^p d x \bigg) ^{1/p} + \bigg (  \int_\Omega \vert \zeta_k \nabla u - \nabla u\vert^p d x \bigg) ^{1/p}\\
&\to 0 \qquad \text{ as } k \to \infty.
\end{align*} (Dominated convergence was used again for the second integral on the right hand side). Since $W^{1,p}_0(\Omega)$ is closed in $W^{1,p}(\Omega)$, it follows $u \in W^{1,p}_0(\Omega)$.
