functions in $H_0^1(\Omega) \cap C(\overline{\Omega})$ are zero on the boundary

Question

I want to solve the following problem

Let $\Omega \subseteq \mathbb{R}^n$ be a bounded open set with $C^\infty$-smooth boundary. Show that for any $u \in H_0^1(\Omega) \cap C(\overline{\Omega})$ we have $u \big|_{\partial \Omega}=0$.

Hint: look first at the case $\Omega=(0,1)^n \subset \mathbb{R}^n$.

My attempt:

My understanding is that $H_0^1(\Omega)$ is the closure of $C_c^\infty({\Omega})$ with the $H^1$ norm. Since all functions in the latter are zero at the boundary, isn't it trivial that functions in the former (i.e. the closure) are zero at the boundary as well?

Thank you!

Answer 1

The statement can be proven using the properties of the trace mapping $\tau$, where $\tau:H^1(\Omega) \to L^2(\partial \Omega)$ is continuous, and $\tau v = v|_{\partial \Omega}$ for continuous $v\in C(\bar \Omega)$.

Due to the density of $C_c^\infty(\Omega)$ in $H^1_0(\Omega)$, there are functions $v_k \in C_c^\infty(\Omega)$ converging to $u$ in $H^1(\Omega)$.

Then $\tau(v_k)=0$ for all $k$, and $\tau(v_k)\to \tau(u)$ shows that $\tau(u)=0$. Now $u$ is in $C(\bar \Omega)$, thus $\tau(u)=u|_{\partial\Omega}=0$.