# Is this a pseudodifferential operator?

Let $M$ be a non-compact manifold and $D$ a first-order self-adjoint elliptic differential operator on $M$. Then is the bounded operator

$$A:=\sqrt{(D^2+1)^{-1}}:L^2(M)\rightarrow H^1(M)$$

a pseudodifferential operator? More precisely, is there a pseudodifferential operator $P:C_c^\infty(M)\rightarrow C_c^\infty(M)$ of order $-1$ such that $P$ extends to $A$?

Here I am defining the inner product on $H^1(M)$ by $$\langle s,t\rangle_{H^1}=\langle s,t\rangle_{L^2}+\langle Ds,Dt\rangle_{L^2}.$$

• Yes. All of the operations used to define $A$ in terms of $D$ are well defined in the space of pseudodifferential operators. – Deane Yang Jun 8 '18 at 18:50
• But, for instance, it is a non-trivial result of Seeley in the setting of a compact manifold that powers of elliptic operators are still pseudodifferential. I guess I'm asking a version of the question "does Seeley's result work in the non-compact setting?" – ougoah Jun 9 '18 at 3:24
• For an arbitrary power, It might be a nontrivial theorem but proving that the square of a pseudodifferential and operator and the square root of a positive pseudodifferential operator are pseudodifferential is straightforward using the symbol calculus. – Deane Yang Jun 9 '18 at 3:55