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

Daniel Doz
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Rosa
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1 Answers1

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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 $\frac{\partial^2 u}{\partial x \partial y} \equiv 1$.

stgatilov
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    One should also mention that such an estimate is possible, if one can control the boundary values of $u$ and $\Omega$ is sufficiently regular. In particular, for $\Omega$ being the (open) unit ball in $\mathbb{R}^n$, one has $|u|{H^2(\Omega)} \le C , |\Delta u|{L^2(\Omega)}$ for all $u \in H_0^1(\Omega)$ with $\Delta u \in L^2(\Omega)$. – gerw Jan 19 '16 at 11:42