TITLE: Is $C^\infty_{0}(\bar{\Omega})$ dense in the hilbert space $W^{2,2}(\Omega)\cap W^{1,2}_{0}(\Omega)$
QUESTION [7 upvotes]: Assume $\Omega$ is open bounded domain in $\mathbb R^n$ 

Is $C^\infty_{0}(\bar{\Omega})$ dense in the hilbert space $W^{2,2}(\Omega)\cap W^{1,2}_{0}(\Omega)$ with inner product $$(u,v)=\int_{\Omega} \Delta u \Delta v $$

I ask this question because I have proved an equality for $C^\infty_{0}(\bar{\Omega})$ functions and I want to prove it for the hilbert space mentioned with density argument

REPLY [2 votes]: There's an issue with your notation. I assume that $C_0^\infty$ means functions with compact support, but then $C^\infty_{0}(\bar{\Omega}) = C^\infty(\bar{\Omega})$. You probably meant to write $C^\infty_{0}(\Omega)$. 
The answer is no. Functions in $C_0^\infty$ clearly have zero trace of $\nabla u$ on $\partial \Omega$. Since this trace is continuous in $W^{2,2}$, the same property holds for functions in the closure of $C_0^\infty$ (which is denoted by $W_0^{2,2}$). 
On the other hand, $W^{2,2} \cap W_0^{1,2}$ consists exactly of functions in $W^{2,2}$ having zero trace. For example, $1-|x|^2$ lies in $W^{2,2} \cap W_0^{1,2}(B(0,1))$ (since it has zero trace), but its derivatives have non-zero traces. 
If you're not familiar with the characterization of $W_0^{1,2}$ in terms of traces, you can show by hand that $1-|x|^2$ can be approximated by functions with smooth support.