Poincaré inequality for $H^2(\Omega)$-functions

Question

I am wondering if given a bounded open subset of $\mathbb{R}^2$ with Lipschitz boundary, then there exists $C>0$ such that, for every $u\in H^2(\Omega)$, $$ \|u\|_{H^2(\Omega)}\le C \left(\|D^2u\|_{L^2(\Omega)}+\|\frac{\partial u}{\partial n}\|_{L^2(\partial\Omega)}+\|u\|_{L^2(\partial\Omega)}\right). $$

Answer 1

Only the inequality $$ \|u\|_{H^1} \le C (\|D^2u\|_{L^2} + \|u\|_{L^2}) \ \forall u\in H^2 $$ has to be proven, as the first-order terms are the only terms of the $H^2$ norm that are missing on the right-hand side.

You can argue by contradiction: assume there is no such $C$, then there is a sequence $(u_k)$ such that $$ \|u_k\|_{H^1} > k (\|D^2u_k\|_{L^2} + \|u_k\|_{L^2}). $$ Obviously, we can choose $u_k$ such that $\|u_k\|_{H^1}=1$. Then the inequality above implies that $(u_k)$ is bounded in $H^2$. Then after extracting a subsequence, $u_k\rightharpoonup u$ in $H^2$ and $u_k\to u$ in $H^1$ (by compact embeddings).

Since $\|u_k\|_{L^2}\to 0$ by the construction of $u_k$, it follows $u=0$. This is a contradiction to $u_k\to u$ in $H^1$ and $\|u_k\|_{H^1}=1$.