How to prove that $\lVert \Delta u \rVert_{L^2} + \lVert u \rVert_{L^2}$ and $\lVert u \rVert_{H^2}$ are equivalent norms? How to prove that $\lVert \Delta u \rVert_{L^2} + \lVert u \rVert_{L^2}$ and $\lVert u \rVert_{H^2}$ are equivalent norms on a bounded domain? I hear there is a way to do it by RRT but any other way is fine. Thanks.
 A: If you want an $H^2$ estimate up to the boundary for arbitrary bounded domains, I'm not sure it's even true. You can bound $\|u\|_{H^2}$ on compact subdomains (this is interior $H^2$ regularity for elliptic equations), or globally in domains with smooth boundary (this is boundary $H^2$ regularity). Both topics are covered in details in PDE by Evans (sections 6.3.1 and 6.3.2 of the 1st edition). It would be impractical to reproduce the proofs here, as they cover 4 and 5 pages, respectively. Besides, Evans' is a fine book to read for any pde_lover.
These lecture notes follow Evans pretty closely.
A: As user53153 wrote this is true for bounded smooth domains and can be directly obtained by the boundary regularity theory exposed in Evans.
BUT: consider the domain $\Omega=\{ (r \cos\phi,r \sin\phi); 0<r<1, 0<\phi<\omega\}$ for some $\pi<\omega<2\pi$. Then the function $u(r,\phi)=r^{\pi/\omega}\sin(\phi\pi/\omega)$ (in polar coordinates) satisfies $\Delta u=0$ in $\Omega$ and it is clearly bounded. On the other hand the second derivatives blow up as $r\rightarrow 0$, more specifically $u\not\in H^2$!
