3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$$c \Vert u\Vert_{H^1}\leq \Vert Qu\Vert_{L^2}\leq C \Vert u\Vert_{H^1},\;\;\forall u\in H^1(M). $$

$\endgroup$
  • $\begingroup$ If $M$ were a non-compact manifold, would the same argument work? $\endgroup$ – Peter Halburt Jun 3 '15 at 0:58
  • $\begingroup$ 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. $\endgroup$ – Liviu Nicolaescu Jun 3 '15 at 14:06

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.