I have a question about Dirichlet form.
Let $\Omega$ be an Euclidean domain of $\mathbb{R}^{N}$ and $X=\bar{\Omega}$. The measure $m$ on the Borel $\sigma$ algebra $\mathcal{B}(X)$ is given by $m(A)=\lambda(A \cap \Omega)$ for all $A \in \mathcal{B}(X)$ with $\lambda $ the Lebesgue measure. It follows that $L^{2}(\Omega)=L^{2}(X,\mathcal{B}(X),m)$. We define a Dirichlet form on $L^{2}(\Omega)$ by \begin{equation*} \mathcal{E}(f,g)=\int_{\Omega}\left(\nabla f,\nabla g \right)\,dx,\quad f,g \in \widetilde{H}^{1}(\Omega), \end{equation*} where $\widetilde{H}^{1}(\Omega)=\text{closure of }H^{1}(\Omega)\cap C_{c}(\bar{\Omega}) \text{ in } H^{1}(\Omega)$. $C_{c}(\bar{\Omega})$ denotes all continuous reak valued function on $\bar{\Omega}$ with support and $H^{1}(\Omega) \cap C_{c}(\bar{\Omega})=\left\{ f \left| \right._{\Omega} \in H^{1}(\Omega) : f \in C_{c}(\bar{\Omega}) \right\}$.
Question
I want to check the following assertion:
\begin{align*} &(1) \quad \widetilde{H}^{1}(\Omega) \cap C_{c}(\bar{\Omega}) \text{ is dense in } C_{c}(\bar{\Omega}) \text{ w.r.t. sup norm}. \end{align*}
My attempt
(1): It is enough to show that for all $ f \in C_{c}(\bar{\Omega})$, $\epsilon>0$, there exists $g \in H_{1}(\Omega) \cap C_{c}(\bar{\Omega})$ such that $\|f-g\|<\epsilon$, where $\|\cdot\|$ is sup norm.
Take $f \in C_{c}(\bar{\Omega})$. By Tietze extension theorem, we can find $F \in C_{c}(\mathbb{R}^{n})$ such that $F=f$ on $\bar{\Omega}$. Define $F_{\delta}=\int_{\mathbb{R}^{n}}j_{\delta}(x-y)F(y)\,dy$, where $j_{\delta}$ is standard mollifier. Then $F_{\delta } \to f$ uniformly on $\text{supp} [f]$ and $F_{\delta} \in C_{c}^{\infty}(\mathbb{R}^{n})$. But I don't know how to prove $F_{\delta} \left|_{\Omega} \right. \in H_{1}(\Omega) \cap C_{c}(\bar{\Omega})$.
Please tell me how to prove (2). Thank you in advance.
To check $F_\delta = j_\delta * F$ restricted to $\Omega$ is in $H^1(\Omega)$, i.e., $D F_\delta |_\Omega \in L^2(\Omega)$, we have from properties of convolution that \begin{align*} D F_\delta = D(j_\delta * F) = (D j_\delta) * F. \end{align*} Now, $(D j_\delta) * F$ has compact support $K$ (contained in the compact support of $F$) and is smooth (inheriting smoothness of $D j_\delta$ -- this is another property of convolution), and hence belongs to $C_c^\infty(K)$. It follows that it has finite $L^2$ norm since $C_c^\infty(K) \subset C_c(K) \subset L^2(K)$.