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

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!

• No it is not trivial. Note that for example $C_c^\infty (0,1)$ is a dense subspace of $L^2(0,1)$. The former space always has vanishing boundary terms whereas the function $x\mapsto 1$ is a $L^2(0,1)$ function. – Peter Jun 10 '15 at 12:27
• @Peter: I have rejected your edit even though at the fundamental level it was correct: if you spot a mathematical mistake (as opposed to formatting issues) in a post, it is better to signal it to the author in a comment. Thus, the author becomes aware of the mistake; even more, what if there is no mistake after all? It is better to have a dialogue with the author in these cases, rather than to silently make changes. – Alex M. Jun 10 '15 at 12:49

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 $\tau(u)=u|_{\partial\Omega}=0$.
• @daw Thank you , why $\tau(v_k)\to \tau(u)$ is true ? I know that $C_c^\infty(\Omega)$ is dense in $H^1_0(\Omega)$ but why the convergense is true in $L^2$?? – Eli Elizirov Jun 17 '15 at 2:22
• This is continuity of the trace operator $\tau$. – daw Jun 17 '15 at 6:03