I already asked this question here on MSE, didn't get an answer, and I'm still stuck with it.

Suppose I have a smooth function $\psi$ from $\mathbb{R}^n$ to $\mathbb{C}$, for which I know that $$ \sup_{x\in\mathbb{R}^n}\left||x|^k\Delta^{p}\psi(x)\right|<\infty $$ for all $k,p\in\mathbb{N}_0$. Is this already sufficient for $\psi$ to be a Schwartz function?

I have tried fooling around with the Fourier transform, the problem seems to be however, that I am not able to control terms of the form $$ |x|D^p(|x|^{2k}\hat{\psi}) $$In particular, it doesn't seem to be possible to adapt the proof, that the Fourier transform is an isomorphism of Schwartz spaces, or I'm just too dumb to see it. I also tried to apply a vast number of different inequalities, but I didn't find anything that fits my problem. I didn't find a counterexample either.

Background: I have a Schrödinger resolvent, i.e. a solution $\psi$ of the equation $$ (-\Delta+V-\lambda)\psi=\varphi $$ where $-\Delta+V$ is e.s.a. on an appropiate kernel, $\varphi$ is Schwartz and $V$ is $C^\infty$ and slowly growing. Using semigroup estimates, I managed to prove that $\psi$ is polynomially bounded, rearranging the equation gives me that $\Delta\psi$ is polynomially bounded.

Any help in form of proofs, proof ideas or possible counterexamples would be greatly appreciated.