Let $\Omega \subset \mathbb R^n$ be a bounded set. Taking the Hilbert space $$W^2=\{v\in \mathcal S'(\Omega):\ v\in L^2,\ |\nabla v|\in L^2,\ \|D^2v\|\in L^2\ \}$$ in order to prove an analogue of the Poincaré inequality $$\exists C:\ \forall v\in ?\subset W^2(\Omega)\qquad \|v\|_{W^2}\le C\|\Delta v\|_{L^2}$$ we have to restrict to a subspace where:
- the functions take value zero at the boundary?
- the functions and their normal derivatives take value zero at the boundary?
I would expect the second option, nevertheless, even for $\Omega=(-1,1)$ I cannot find an example of function $v\in W^2(\Omega)$ where the boundary value is zero and $$\|v\|_{W^2}>>\|v''\|_{L^2}\qquad v(-1)=v(1)=0$$ so I wonder if we can have a control of the norm even in case 1.
Assume $u \in H^{1}_{0}(\Omega)\cap H^{2}(\Omega)$ and let $f=-\Delta{u}$. Then $u$ is the unique weak solution of $$-\Delta{u}=f $$ on $\Omega$ with zero-boundary conditions on $\partial{\Omega}$ and we can therefore use standard regularity estimates (see Evans Chapter 6 for example) to conclude that $$||u||_{H^{2}} \le C \cdot ||\Delta{u}||_{L^{2}}$$ for some constant $C>0$ that depends only on $\Omega$.
So if $\Omega$ is sufficiently regular, all you need is zero trace!