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. 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?" 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. Jun 9 '18 at 3:55

Yes, it is a classical pseudo-differential operator of order $$-1$$ with principal symbol $$\vert p_D(x,\xi)\vert^{-1}$$ where $$p_D$$ is the principal symbol of $$D$$; it is also possible to prove that you have an asymptotic expansion for the (total) symbol $$q$$ of $$1/\sqrt{1+D^2}$$, providing $$q-\sum_{0\le j
To get this you can use the Richard Beals characterization of pseudo-differential operators in the paper [MR0435933] or its refinements by Jean-Michel Bony in the article [MR1482829]. Basically, given an operator $$A$$ produced by an operator-theoretic construction such as your $$1/\sqrt{1+D^2}$$, you have to check the $$L^2$$ boundedness of commutators of $$A$$ with simple operators which are quantization of coordinates, e.g. $$x_j, D_{x_j}$$ in the $$\mathbb R^n$$ case.