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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.

cmk
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Appliqué
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  • I don't know anything about this, but I'll try nonetheless. A pseudo differential operator is associated to a function that they call the symbol, correct? Might it be the case that the pseudo diff.op. is invertible iff the symbol never vanishes? In that case, maybe the reciprocal of the symbol is the symbol of the inverse. – Giuseppe Negro Nov 30 '17 at 13:02
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    @GiuseppeNegro Actually, there is a formula for the symbol of the parametrix of an elliptic PsDO (inverse modulo infinitely smoothing PsDO), and I would rather expect that this formula holds for the symbol of $A^{-1}$. But it only requires non-vanishing of the symbol $a(x,\xi)$ for large $|\xi|$ – Appliqué Nov 30 '17 at 13:06
  • I see. Anyway, to fix ideas: what if $a(x, \xi)=|\xi|^2$? The corresponding operator is the Laplacian, I think. Do you consider this operator as invertible or not? – Giuseppe Negro Nov 30 '17 at 13:11
  • @GiuseppeNegro Maybe $\Delta - 1$ would be better? $\Delta$ has an eigenvalue zero on $X$, and so is not invertible – Appliqué Nov 30 '17 at 13:23
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    I am asking because I suspect that the answer to your question is negative. – Giuseppe Negro Nov 30 '17 at 15:31

1 Answers1

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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.

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 $-m$.

Jan Bohr
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