I already asked this question [here](https://math.stackexchange.com/questions/1397339/another-equivalent-characterization-of-schwartz-function "Another equivalent characterization of Schwartz function?") on MSE, didn't get an answer, and I'm still stuck with it. 




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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?




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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.



Is this function Schwartz?
I think the bound is sufficient to prove that $u$ is a Schwartz function.


Your assumption implies that $x^\alpha \Delta^p u$ is in $L^2$ for all $\alpha,k$. Thus for the Fourier transform $v$ of $u$ we have that $D^\alpha(|x|^{2p}v)$ is in $L^2$ for all $\alpha,p$. Take $\alpha=0$: we get that $|x|^{2p}v$ is in $L^2$ for all $p$. Now take $\alpha=(1,0,...,0)$: we have $|x|^{2p}\partial\_1v+2p|x|^{2p-2}x\_1v$ is in $L^2$ for all $p$, which by the previous step implies $|x|^{2p}\partial\_1v$ is in $L^2$ for all $p$. And so on. Finally you get that $|x|^{2p}D^\alpha v$ is in $L^2$ for all $\alpha,p$ thus $v$ is Schwartz.


