Let $\Omega$ be an open and bounded subset of $\mathbb{R}^n$ with a smooth boundary. Is $H_0^1(\Omega)\cap C^\infty(\Omega) \subseteq H^2(\Omega)$?
I suspect that that the answer is no, because continuity of the 2nd derivative of a function $f$ doesn't have to control its behaviour at the boundary. If all $f\in H_0^1(\Omega)\cap C^\infty(\Omega)$ were $0$ at the boundary the result could possibly follow, however that is not the case. I'm struggling to think of a counterexample though so I'd appreciate some help.
Following my train of thought, the next question is then would it be true that $H_0^1(\Omega)\cap C^\infty(\bar{\Omega}) \subseteq H^2(\Omega)$?
Consider the function $$u(x)=-x+x\log{x},\ x\in (0,1).$$
One can easily verify that $u\in H^1((0,1))\cap C^\infty((0,1))$ and $u\notin H^2((0,1))$.
Moreover, by using a smooth cutoff function, it can be redefined near the boundary value $x=1$ in such a way that it will belong to $H^1_0((0,1))\cap C^\infty((0,1))$ and still does not belong to $H^2((0,1))$.