Inclusions between $H^1_0(\Omega) \cap H^2(\Omega)$ and $H^2_0(\Omega)$

Question

Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. I would like to understand something about the relationships between these two spaces: $$H^1_0(\Omega) \cap H^2(\Omega)\quad \text{and}\quad H^2_0(\Omega).$$

Is there any inclusion? If there is not, could you exhibit a counterexample? Thank you!

Answer 1

Using the definitions, you should be able to prove that $H^2_0(\Omega) \subset H_0^1(\Omega) \cap H^2(\Omega)$. The reverse inclusion does not hold; as a counterexample, try picking a really nice $\Omega$ (such as a disc) and writing down a well-behaved function that vanishes at the boundary, but whose first derivatives do not. Then such a function will be in $H_0^1(\Omega) \cap H^2(\Omega)$ but not $H_0^2(\Omega)$.

Answer 2

Just to continue Chris Wong's argument.

Let $\Omega = \{x\in \mathbb{R}^n: \|x\|_{\infty}= \max|x_i|<1 \}$ be an $n$-cube with side length $2$, then it can be verified that: $$ u = \prod_{i=1}^n \sin(2\pi x_i) $$ is in $H^1_0(\Omega)\cap H^2(\Omega)$, since on boundary of the $n$-cube, there is some $x_i = \pm 1$. However for $\partial u/\partial x_i$ will give you a cosine which does not vanish on the boundary when dot product with normal to the boundary, hence $\partial u/\partial \nu$ is not zero (though the tangentially it does vanish).

To visualize just consider $n=2$, the square $(-1,1)^2$, and $u = \sin(2\pi x)\sin(2\pi y)$.