The space $\{ u \in H^1(\Omega) \mid \Delta u \in L^2(\Omega)\}$

Question

For a open Lipschitz domain $\Omega$, consider the space $$A =\{ u \in H^1(\Omega) \mid \Delta u \in L^2(\Omega)\}.$$

Now I heard somewhere that all the second derivatives of a function $u$ are controlled by its Laplacian (because $|u|_{L^2} + |\Delta u|_{L^2}$ is equivalent to $|u|_{H^2}$) so then is not $A=H^2(\Omega)$?

Why is this space used??

Answer 1

If $\Omega$ is convex or smooth, one knows that the solution $v$ of \begin{align*}-\Delta v + v &= f \quad\text{in }\Omega\\\frac{\partial}{\partial n}v + v &= 0 \quad\text{on }\Gamma\end{align*} Belongs to $H^2(\Omega)$ for all $f \in L^2(\Omega)$. In this regular case, you get indeed $A = H^2(\Omega)$.

However, this is not the case, e.g., for the L-shaped domain. Then, the solution with $f = 1$ belongs to $H^1(\Omega)$ but not to $H^2(\Omega)$. However, you still have $-\Delta v = f - v \in L^2(\Omega)$.

Edit: In the case without boundary conditions, this is not true. Consider $g \in H^{1/2}(\partial\Omega) \setminus H^{3/2}(\partial\Omega)$ and let $u$ be the harmonic extension of $g$. Then, you have $u \in H^1(\Omega)$ and $\Delta u \in L^2(\Omega)$. However, $u \not\in H^2(\Omega)$, since its trace $g$ does not belong to $H^{3/2}(\partial\Omega)$.

Answer 2

$\Delta u\in L^2$ is equivalent to the $H^2$ seminorm in the whole space (you can use Riesz transforms), or in the periodic case (same reason).

However, if you try to obtain the same result in a bounded domain, integrating by parts you get some "extra" terms coming from the boundary. If you impose some boundary conditions (for instance $u,Du=0$) for your functions then you can handle this new terms.

In your set $A$ there is no conditions on the value of the function of its derivatives in the boundary.