Relation between seminorm of Sobolev space and $L^2$ norm

Question

Let we have the seminorm of second derivative of $u$ in $H^2(\Omega)$ i.e.

$|u|_{H^2(\Omega)}=\int_{\Omega} \sum_{|\alpha|=2} D^{\alpha}u $.

Can we derive that $|u|_{H^2(\Omega)}\leq C||\Delta u||_{L^2(\Omega)}$?

Answer 1

No, you can't.

Consider $\Omega$ being a unit circle on $\mathbb{R}^2$. Then:

$$ \int\limits_{\Omega} \sum_{|\alpha|=2} | D^{\alpha}u | = \int\limits_{x^2 + y^2 < 1} \left( \left| \frac{\partial^2 u}{\partial x^2} \right| + \left| \frac{\partial^2 u}{\partial x \partial y} \right| + \left| \frac{\partial^2 u}{\partial y^2} \right| \right) dx \, dy $$

Obviously, mixed derivative is included here, but it is not included in the Laplace operator.

For example, take $u(x, y) = xy$. Clearly, $\Delta u \equiv 0$, but your seminorm has nonzero value, because $\frac{\partial^2 u}{\partial x \partial y} \equiv 1$.