# 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.

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 $$u \in W^{1,p}_0(\Omega)$$.