Let us consider a self-adjoint elliptic pseudodifferential operator $P \in OPS^2$ on a compact manifold $M$ such that $spec(P) \subset (0, \infty)$. Is the norm $(Pu, u)^{1/2}$ on $H^1(M)$ equivalent to $||u||_{H^1(M)}$, the usual Sobolev norm? A reference would be greatly appreciated.
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Yes. Take the functional square root $Q:=\sqrt{P}$. It is a 1st order, elliptic, selfadjoint positive $\psi$do on $M$ according to the results of Seeley. Then $(Pu,u)^{1/2}=\Vert Qu\Vert_{L^2}$, $\forall u\in C^\infty (M)$. The operator $Q$ induces a continuous bijective linear map
$$Q:H^1(M)\to L^2(M). $$
Using the open mapping theorem we deduce that there exist $0<c<C$ such that
$$c \Vert u\Vert_{H^1}\leq \Vert Qu\Vert_{L^2}\leq C \Vert u\Vert_{H^1},\;\;\forall u\in H^1(M). $$
answered Apr 4, 2015 at 9:44
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$\begingroup$ If $M$ were a non-compact manifold, would the same argument work? $\endgroup$ Jun 3, 2015 at 0:58
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$\begingroup$ It would work once you show that $P :H^2(M)\to L^2(M)$ is onto. It is not clear to me why, when $M$ is noncompact, the operator $P$, viewed as an unbounded operator $L^2(M)\to L^2(M)$ with domain $H^2(M)$, is self-adjoint. $\endgroup$ Jun 3, 2015 at 14:06