What is the dense subset in $H_0^1(\Omega)\cap H^2(\Omega)$ I came across this problem when I try to prove that for space $H(\Omega):=H_0^1(\Omega)\cap H^2(\Omega)$, where $\Omega$ is open bounded with nice boundary, then the norm $\|u\|_1:=\|\Delta u\|_{L^2}$ is equivalent to the norm $\|u\|_{H^2}$ in the usual sense. 
This problem can be proved by using open mapping theorem and showing that $\|u\|_1$ is actually make $H$ as a Banach space.
However, I was thinking by using the fact that the norm $\|u\|_2:=\|u\|_{L^2}+\|D^2u\|_{L^2}$ is already an equivalent norm of $\|u\|_{H^2}$ and the term $\|u\|_{L^2}$ will be taken care by Poincare. Then I only need to show that $\|D^2u\|_{L^2(\Omega)}\leq C\|\Delta u\|_{L^2(\Omega)}$ for some constant $C$.
I remembered when I deal with the weak solution of Biharmonic functions, I proved that 
$$ \int_\Omega \partial_i\partial_j u\cdot \partial_i\partial_j u\,dx=\int_\Omega \partial_i\partial_i u\cdot \partial_j\partial_j u\,dx \tag 1$$
by using density argument. 
So here I was going to use the same approach. But I can not see what function is dense in $H$ under $H^2$ norm. I don't think $C_c^\infty(\Omega)$ will work here because it will give us $H_0^2$ but not $H$. But we can not use $C^\infty(\bar \Omega)$ neither because we will lose $H_0^1$. So, what function should I use as the approximate function here?
i.e., can I somehow use smooth function to approx $u\in H$ under $H^2$ norm?

Update
This question has been solved but I want to update that we can generalize this to $H^{2k}(\Omega)\cap H_0^k(\Omega)$. That is, the space $H^{2k}(\Omega)\cap H_0^k(\Omega)$ has equivalent norm $\| \Delta^k u\|_{L^2(\Omega)}$ and it can be proved by using regularity of Laplace equation. (The equivalent of $H^2\cap H_0^1$ can be proved by regularity in only TWO steps!)
 A: We have that (see Evans in regularity part) for each $f\in L^2(\Omega)$, the problem 
$$
\left\{ \begin{array}{cc}
 -\Delta u =f&\mbox{ in $\Omega$,} \\
  u=0 &\mbox{ on $\partial\Omega$,}
       \end{array} \right.
$$
has an unique solution $u\in H$ satisfying $\|u\|_{H^2}\le C\|f\|_2$. 
Fix $u\in H$ and write $-\Delta u=f$. Let $f_n\in C^\infty(\overline{\Omega})$ be a sequence satisfying $f_n\to f$ in $L^2(\Omega)$. Let $u_n\in C^2(\overline{\Omega})\cap C_0(\overline{\Omega})$ be the unique solution of 
$$
\left\{ \begin{array}{cc}
 -\Delta u_n =f_n&\mbox{ in $\Omega$,} \\
  u_n=0 &\mbox{ on $\partial\Omega$,}
       \end{array} \right.
$$
therefore
$$
\left\{ \begin{array}{cc}
 -\Delta(u- u_n) =f-f_n&\mbox{ in $\Omega$,} \\
 u- u_n=0 &\mbox{ on $\partial\Omega$.}
       \end{array} \right.
$$
Thus, $\|u-u_n\|_{H^2(\Omega)}\le C\|f-f_n\|_2$, which implies that $u_n\to u$ in the $H^2$ norm.
Remark: Although the previous construction shows that $C^2(\overline{\Omega})\cap C_0(\overline{\Omega})$ is dense in $H$, it would be nice to do it directly, i.e. by constructing an explicitly sequence $u_n$, using a mollifier sequence.
Update: I will give an answer for the question proposed in the remark. In Burenkov's book chapter 2, there is a method to regularize a function in $W^{k,p}(\Omega)$, which preserves boundary values.
By applying this method here, we can prove that $C^2(\overline{\Omega})\cap C_0(\overline{\Omega})$ is dense in $H$.
