Is the inverse of a finite-propagation operator finite-propagation? A bounded operator $T:L^2(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n)$ is said to have finite propagation if there exists $r>0$ such that: if $\phi,\psi\in C_c(\mathbb{R}^n)$ have supports that are separated by a distance larger than $r$, then 
$$\psi T\phi=0\in\mathcal{B}(L^2(\mathbb{R}^n)).$$
Suppose $T$ is invertible and has finite propagation. Then does it follow that $T^{-1}$ has finite propagation?
Remark: I suspect the answer is no, but I'm looking for a concrete example.
 A: I think easy counterexamples can be produce using "discrete derivatives" whose inverse can be given in terms of primitives given by Riemann sums. The discrete derivatives have finite propagation, the primitives not so much. 
Edited solution: First, let me work in $L^2(\mathbb{T})$ instead of $L^2(\mathbb{R}^n)$. Let $\lambda_{\theta_0}$ be the left regular operator with respect to a number $\theta \in \mathbb{T}$ that is not a rational number times $2 \pi$. By standard Neumann series arguments
$$
  F = 1 - \frac12 \, \lambda_{\theta_0}
$$
is invertible and its propagation is $|\theta_0|$. Its inverse is given by the (absolutely) convergent series
$$
  F^{-1} = \sum_{j = 0}^\infty \frac1{2^j} \lambda_{j \, \theta_0}
$$
and since the orbit of $\theta_0$ is dense, for any nonzero $\psi \in C_c(\mathbb{T})_+$, $F^{-1}(\psi)$ would have total support. 
Similar arguments work in $L^2(\mathbb{R}^n)$. For instance the inverse above is still well defined and not of finite propagation (the support of $F^{-1}(\psi)$ for some $\psi \in C_c(\mathbb{R}^n)$ supported in a ball $B$ of radius $r < |\theta_0|$ would be the $$\bigcup_{k \geq 0} B + \theta_0 k$$.
