# Is this function Schwartz?

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.

• For Schwartz, I also need polynomial boundedness in all derivatives. – Daniel Dec 15 '15 at 19:45
• Right, I was asleep. – Christian Remling Dec 15 '15 at 19:47
• Essentially, this is about controlling $\nabla\psi$ in terms of $\psi,\Delta\psi$ so en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/Riesz_transform should be useful. – Christian Remling Dec 15 '15 at 20:14
• @ChristianRemling Thank you for your suggestion! I actually tried some Sobolev-Type inequalities, but so far I didn't stumble over the Riez transform. The problem with the Sobolev-Type inequalities was usually, that I pay too much when passing to the sup. Riez transform looks useful in that regard. – Daniel Dec 15 '15 at 20:42

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.
• Why is $x_1|x|^{2p-2}v$ polynomially bounded? – Daniel Dec 16 '15 at 15:41
• It is bounded by $(|x|^{2p}+|x|^{2p-2})|v|$ which is in $L^2$. What is your doubt? – Piero D'Ancona Dec 16 '15 at 16:37
• Yeah, sadly what I actually meant seems to be right. I already tried something like this, the problem here is the "and so on"-part - in the induction step, I will get terms like $\partial_i|x|^{2p}D^\alpha v$ or similar (In my original attempt, I got the $|x|D^\alpha|x|^{2p} v$ from the question, as I came from the other side), and I have not figured out how to control them. – Daniel Dec 16 '15 at 20:59