How to prove that $\mathcal{D}(\Omega)$ is dense in $\mathcal{D}'(\Omega)$? How to prove that $\mathcal{D}(\Omega)$ is dense in $\mathcal{D}'(\Omega)$? Where $\mathcal{D}(\Omega)$ is the space of test functions with support compact and  $\mathcal{D}'(\Omega)$ is the distribution space.
Thanks in advance.
 A: The first result you need is the following:
(i) If $u \in \mathcal{D}'(\mathbb{R}^n)$ and $\varphi \in \mathcal{D}(\mathbb{R}^n)$, then $u \ast \varphi \in \mathcal{E}(\mathbb{R}^n)$ and $(u \ast \varphi) \ast \psi = u \ast (\varphi \ast \psi)$ $\forall \psi \in C^{\infty}_c(\mathbb{R}^n)$ 
We show that $\mathcal{D}(\mathbb{R}^n)$ is dense in $\mathcal{D}'(\mathbb{R}^n)$.This means that if $u \in \mathcal{D}'(\mathbb{R}^n)$ is fixed, and $\lbrace \psi_\epsilon \rbrace_{\epsilon > 0}$ is standard mollifier, then $u \ast \psi_\epsilon \rightarrow u$ in $\mathcal{D}'(\mathbb{R}^n)$ , i.e.
\begin{align*}
\displaystyle \lim_{\epsilon \rightarrow 0^+} \int_{\mathbb{R}^n} (u \ast \psi_\epsilon)(x) \varphi(x) dx = \langle \varphi , u \rangle , \forall \varphi \in \mathcal{D}(\mathbb{R}^n)
\end{align*}
Proof. It is known that $\forall \varphi \in \mathcal{D}(\mathbb{R}^n)$ fixed, we have $\psi_\epsilon \ast \varphi \rightarrow \varphi$ and $D^\alpha \psi_\epsilon \ast \varphi = D^\alpha \psi_\epsilon \ast \varphi \rightarrow D^\alpha \varphi$ uniformly $\forall K \subset \Omega$ and $\forall \alpha \in \mathbb{N}^n$, where $K$ is a compact.
Then, or the characterization of the convergence in $\mathcal{D}(\mathbb{R}^n)$, follows that $\psi_\epsilon \ast \varphi \rightarrow \varphi$ with respect vectorial topology  $\mathcal{T}$ of $\mathcal{D}(\mathbb{R}^n)$.
Consider $\widetilde{\varphi}(x):=\varphi(-x)$, $u \in \mathcal{D}'(\mathbb{R}^n)$, $\varphi \in \mathcal{D}(\mathbb{R}^n)$ then it is well-defined convolution\begin{align*}
\displaystyle (u \ast \varphi)(x)=\langle \varphi(x-y),u(y) \rangle = \langle \varphi(x -\cdot), u(\cdot) \rangle = \langle \widetilde{\varphi}(x) , u \rangle = (u \ast \varphi)(0)
\end{align*}
Therefore, by (i) and since $u \in \mathcal{D}'(\mathbb{R}^n)$ is continuous, follows that
\begin{align*}
\langle \varphi , u \rangle &= (u \ast \varphi)(0) \\
&= \langle \widetilde{\varphi}(x) , u \rangle \\
& = \lim_{\epsilon \rightarrow 0^+} \langle (\psi_\epsilon \ast \widetilde{\varphi})(0) , u \rangle \\
&= \lim_{\epsilon \rightarrow 0^+} (u \ast (\psi_\epsilon \ast \widetilde{\varphi}))(0) \\
& = \lim_{\epsilon \rightarrow 0^+} ((u \ast \psi_\epsilon) \ast \widetilde{\varphi})(0) \\
&=\lim_{\epsilon \rightarrow 0^+} \int_{\mathbb{R}^n} (u \ast \psi_\epsilon)(y) \widetilde{\varphi}(\cdot - y) dy  \\
&= \lim_{\epsilon \rightarrow 0^+} \int_{\mathbb{R}^n} (u \ast \psi_\epsilon)(y) \varphi(y) dy.
\end{align*}
We show that $\mathcal{D}(\Omega)$ is dense in $\mathcal{D}'(\Omega)$. 
Proof.
Consider  $\Omega_k = \lbrace x \in \Omega : \mathrm{dist}(x, \partial \Omega) > 1/k \rbrace \cap B(0,k)$ , $\forall k \in \mathbb{N}$. Then, $\Omega_k \subset \Omega_{k+1}$ and $\Omega=\bigcup_{k \in \mathbb{N}}\Omega_k$, and by regular version of Urysohn lemma we have 
\begin{align*}
\exists \psi_k \in \mathcal{D}(\mathbb{R}^n) : \mathrm{supp}(\psi_k) \subset \Omega_{k+1}, \psi_k(x)=1 , \forall x \in \overline{\Omega}_k
\end{align*}
In particular, $\mathrm{supp}(\psi_k u) \subset \mathrm{supp}(\psi_k) \cap \mathrm{supp}(u) \subset \mathrm{supp}(\psi_k)$, and $\psi_k u \rightarrow u$ in $\mathcal{D}'(\Omega)$, and $\psi_k u \in \mathcal{E}'(\mathbb{R}^n)$ is a distribution with compact support.
Consider a standard mollifier $\lbrace \eta_\epsilon \rbrace_{\epsilon >0}$ such that $\mathrm{supp}(\eta_\epsilon) \subset B(0,\epsilon)$ and assuming $0 < \epsilon_k < \mathrm{dist}(\mathrm{supp}(\psi_k), \partial \Omega_{k+1})$ such that $\epsilon_k \rightarrow 0$ decrease.
By properties of the support convolution's, we have
\begin{align*}
\mathrm{supp}(\eta_{\epsilon_k} \ast (\psi_k u)) \subset B(0,\epsilon_k) + \mathrm{supp}(\psi_k) \subset \Omega_{k+1}.
\end{align*} 
is a compact since it's closed and bounded, and then $\eta_{\epsilon_k} \ast (\psi_k u) \in \mathcal{D}(\Omega_{k+1}) \subset \mathcal{D}(\Omega)$, and for the preceding result  $\eta_{\epsilon_k} \ast (\psi_k u) \rightarrow u$ in $\mathcal{D}'(\mathbb{R}^n)$  and then also in $\mathcal{D}'(\Omega)$.
