@NancyBoy I guess that this new $F$ is not analytic, but that the proof is not as simple as for $\chi\ast\chi$ in my answer. In fact the product of the Fourier transforms of your functions $\chi_j$ cannot be exponentially decreasing. Maybe my post "Fourier transform of Analytic Functions" (see the linked posts above on the right) may help understand that business further.

The solution of $(\sharp)$ given above is a smooth function which is not analytic in any neighborhood of the origin. Since $0$ is the only singular point (at a finite distance) of that equation, any solution will be analytic on $\mathbb R^*$ (consequence of the P.S.), so in that sense, this counterexample is optimal.

Sharp and natural: a great example. Thx.

Would it be easier to consider Gevrey functions of order $s>1$? hen we would have for $W$ neighborhood of $0$, $\forall k\in \mathbb N, \quad \sup_{x\in W}\vert f^{(k)}(x)\vert\le C_W^{1+k} k^{sk}. $ In particular the linear form $$ f\mapsto \sum_{k\ge 0} \frac{f^{(k)}(0)}{k!}k^{-(s-1)k}\rho^k=T_\rho(f), $$ would be well defined for $\vert\rho\vert$ small enough.

Following your assumptions, the commutator $[\hat P, \hat Q]$ should be skew-adjoint, so cannot be the identity.