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

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??

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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 $-\Delta v = f - v \in L^2(\Omega)$.

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In your last sentence, how can you rearrange the equation to give $-\Delta v \in L^2$? because the equation only holds in a weak sense (multiplied by a test function) and not pointwise? – C_Al Jul 29 '15 at 11:53
But the Laplacian is also defined in a weak sense, so it perfectly fits together. – gerw Jul 29 '15 at 18:51

$\Delta u\in L^2$ 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.

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