Is $H^2(\Omega) \cap H^1_0$ the same space as $H^2_0(\Omega)$ when $\Omega$ is a bounded open subset of $\mathbb{R}^d$?

Question

I suppose yes since by inclusion of $L^p$ spaces associated to finite measures, H^2 should be in H^1. But in my lecture notes the teacher writes $H^2(\Omega) \cap H^1_0$ instead of $H^2_0(\Omega)$

Answer 1

To me, $H^2(\Omega) \cap H^1_0(\Omega)$ is the space of functions in $H^1_0(\Omega)$ (ie: admitting a weak derivative and and vanishing on the boundary of $\Omega$) for which the weak derivative admits a weak derivative, and $H^2_0(\Omega)$ is the space of functions in $H^1_0(\Omega)$ admitting a weak derivative which vanishes on the boundary.