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.

  • $\begingroup$ What is RRT short for? $\endgroup$ – Sam Jan 9 '13 at 16:06
  • $\begingroup$ @SamL. It's Riesz Representation Theorem $\endgroup$ – pde_lover Jan 9 '13 at 16:13
  • $\begingroup$ Why the downvote? I didn't put any working done because I have no idea how to start, and I have searched to find a proof but no luck. $\endgroup$ – pde_lover Jan 9 '13 at 16:35
  • 2
    $\begingroup$ Wasn't my down-vote btw... $\endgroup$ – Sam Jan 9 '13 at 18:48

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.


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 $u\not\in H^2$!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.