# Is $\lVert u \rVert_{L^2(M)} + \lVert \Delta u \rVert_{L^2(M)}$ equivalent to $\lVert u \rVert_{H^2(M)}$?

On a bounded Riemannian manifold without boundary, is it true that the norms $$\lVert u \rVert_{L^2(M)} + \lVert \Delta u \rVert_{L^2(M)}$$ is equivalent to the full $H^2$ norm $\lVert u \rVert_{H^2(M)}$?

Can i have a reference to this please?

Thanks

• I think that is true if $u\in H^2_0(M)$ from Sobolev inequality
– Lion
Feb 27, 2014 at 13:39

Yes, that's actually true. First, note that the inequality $$||u||_{L^2} + ||\Delta u ||_{L^2} \leq c ||u||_{H^2}$$ is always true (the constant may vary due to your definition though). Hence it remains to show that $$||u||_{H^2} \leq C \left( ||u||_{L^2} + ||\Delta u ||_{L^2} \right).$$ This follows from the fact that the Laplacian is elliptic and can e.g. be found in the book "Spin Geometry" by H. Blaine Lawson and Marie-Louise Michelsohn, see Theorem 5.2 (iii) in Chapter III, page 193.