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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.

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What is RRT short for? – Sam Jan 9 '13 at 16:06
@SamL. It's Riesz Representation Theorem – pde_lover Jan 9 '13 at 16:13
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. – pde_lover Jan 9 '13 at 16:35
Wasn't my down-vote btw... – Sam Jan 9 '13 at 18:48
up vote 6 down vote accepted

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.

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Thanks a lot!!! – pde_lover Jan 9 '13 at 22:19

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 $u\not\in H^2$!

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