I would like to prove that the following definitions of $H^1_0(\Omega)$ are equivalent. In what follows $\Omega$ is supposed to be a bounded domain with smooth boundary.
$$1) \qquad H^1_0(\Omega) = \overline{C_0^{\infty}}^{\|\cdot\|_{H^1}}$$ $$2) \qquad H_0^1(\Omega) = \{u \in H^1(\Omega)\ :\ u_{|_{\partial \Omega}} = 0 \}$$ In the second definition $u_{|_{\partial \Omega}} = 0$ means that $Tu = 0$, where $T$ is the trace operator. Can anyone help me?
Let $$ A = \overline{C_0^{\infty}}^{\|\cdot\|_{H^1}}$$ and $$B = \{u \in H^1(\Omega)\ :\ u_{|_{\partial \Omega}} = 0 \}.$$ We also know $H^1(\Omega) = \overline{C^\infty}^{\|\cdot\|_{H^1}}$. Then, $A \subset B$ is apparent, since $\varphi|_{\partial\Omega} = 0$ for all $\varphi \in C_0^\infty(\Omega)$ and since the trace operator is continuous.
For $B \subset A$ let $v \in B$ be given. Then, there is a $\varphi \in C^\infty(\Omega)$ with $\|v - \varphi\|_{H^1} \le \varepsilon$. Finally, you can take a smooth cut-off function $\psi \in C_0^\infty(\Omega)$, such that $\| v - \varphi\,\psi\|_{H^1} \le 2 \, \varepsilon$.
