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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 $$\langle s,t\rangle_{H^1}=\langle s,t\rangle_{L^2}+\langle Ds,Dt\rangle_{L^2}.$$

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    $\begingroup$ Yes. All of the operations used to define $A$ in terms of $D$ are well defined in the space of pseudodifferential operators. $\endgroup$
    – Deane Yang
    Commented Jun 8, 2018 at 18:50
  • $\begingroup$ 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?" $\endgroup$
    – geometricK
    Commented Jun 9, 2018 at 3:24
  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Jun 9, 2018 at 3:55

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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 $$ q-\sum_{0\le j<N} q_j\quad\text{ is a symbol of order $-1-N$, $q_0=\vert p_D\vert^{-1}$, $q_j$ of order $-1-j$.} $$

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.

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