The inverse of a $\Psi$DO is a $\Psi$DO

Question

The following question looks quite simple, but unfortunately I was not able to find an answer in the literature so far.

Let $A \in OPS^m(X)$, $m \in \mathbb R$, be a pseudodifferential operator on a compact manifold $X$. If $A$ is invertible, is it true that the inverse $A^{-1}$ is actually a pseudo-differential operator $A^{-1} \in OPS^{-m}(X)$?

By invertble I mean that $A^{-1}$ is defined on $C^\infty(X)$ and in this space $AA^{-1} = A^{-1}A = \mathrm{Id}$.

For example, a similar statement is used in the beginning of p.293 of [M.Taylor, Pseudodifferential Operators, 1981]:

If $\in OPS^m$ is elliptic, positive self-adjoint operator on a compact manifold $X$, or order $m>0$, then $(I+P)^{-1} \in OPS^{-m}$ is compact.

Since it is not explained, I think it must be quite obvious.

Answer 1

This is a partial answer, which is too large for a comment. I've put a bounty on the question, as I am curious for a general solution.

The inverse $A^{-1}$ is a pseudodifferential operator in (at least) the following two situations:

Situation A: Suppose that additionally we have:

1. $A$ is elliptic and formally self-adjoint. (Ellipticity possibly follows from invertibility?)
2. The inverse is defined over all distributions $\mathscr{D}'(M)$

In that case we have $BA-\mathrm{id}=K$ for a paramatrix $B$ ($\psi$do of order $-m$) and and a smoothing operator $K$. By our assumptions we have $A^{-1}f = Bf-KA^{-1}f$ for all $f\in \mathscr{D}'(M)$. Now $KA^{-1}$ maps $\mathscr{D'}$(M) into $C^\infty(M)$ (as $K$ is smoothing) and due to self-adjointness of $A$ (and consequently $A^{-1}$), also $(KA^{-1})^*$ has this mapping property. This implies that $KA^{-1}$ is a smoothing operator. Hence $A^{-1}$ is the sum of an order $-m$ $\psi$do and a smoothing operator, which makes it into a $\psi$do itself.

Situation B: For this we don't need a priori knowledge of invertibility of $A$, but rather assume:

1. $A$ is classical and has a positive order $m>0$
2. There exists a sector $S=\{z\in \mathbb{C}:\vert z \vert <\epsilon \text{ or } \vert \arg(z)-\theta\vert<\delta\}\subset \mathbb{C}$ (for $\epsilon,\delta>0,\theta\in (-\pi,\pi]$) which is avoided by the principal symbol $\sigma_A$. By this I mean that $\sigma_A(x,\xi)-z\neq 0$ for all $(x,\xi)\in T^*M\backslash 0$, $z\in S$.

In this case one can construct complex powers $A^z$ ($z\in\mathbb{C}$), which are $\psi$do's of order $m\mathrm{Re}z$ and satisfy the expected algebraic identities. In particular $A^{-1}$ exists and is a $\psi$do.

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Edit (May 2022): Shubin's book on pseudodifferential operators contains the `inverse operator theorem' as Theorem 8.2 and answers the question in a more general context than Situation A above.

Assume that $A$ is a pseudodifferential operator of order $m>0$ on a closed manifold $M$ and let $A_0$ be the closure of $A\vert_{C^\infty(M)}$ in $L^2(M)$.

1. $A$ is elliptic (more generally, it is enough for $A$ to lie in the hypoelliptic class $HL^{m,m}_{\rho,\delta}$ for $1-\rho \le \delta <\rho$)

2. $0\notin \sigma(A_0)$, the $L^2$-spectrum of $A_0$

Then $A_0^{-1}$ is the $L^2$-bounded extension of a pseudodifferential operator of order $-m$.