TITLE: Is this a pseudodifferential operator?
QUESTION [5 upvotes]: 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 $$\langle s,t\rangle_{H^1}=\langle s,t\rangle_{L^2}+\langle Ds,Dt\rangle_{L^2}.$$

REPLY [2 votes]: 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.