Let $\phi\in\mathcal S(\mathbb R^n)$ be a Schwartz function vanishing at $0$. Then the FTC implies $$ \phi(x) = x\cdot \int_0^1 \nabla \phi (tx)\ dt.$$ Let this integral term be $\psi(x) := \int_0^1 \nabla \phi(tx)dt$. Is it true that $\psi$ is (component-wise) a Schwartz function?
It is at least $C^\infty$. I tried for a while to come up with counterexamples and failed. I have proved it in dimension $1$, but I need a replacement for division in higher dimensions?
The result is false for arbitrary functions $p$ that solve $\phi(x)= x\cdot p(x)$, even if $\phi$ is Schwartz ($0 = \binom x y \cdot \binom {-y}x$). Also, $\psi$ is not Schwartz if the condition $\phi(0)=0$ is dropped, as then (in 1D) $\psi(x)= \frac{\phi(x)-\phi(0)}x\sim\frac{\phi(0)}x$ for large $x$ which is not enough decay. So a positive proof would need to use something about Schwartz functions and something about vanishing at zero (the two assumptions).
I can provide a detailed attempt if wanted; in fact I have deleted many "answers" after noticing errors. In particular I have tried merely using the Schwartz seminorms to bound the integral, but it seems that either many factors of $t^{-1}$ appear which make the integral diverge, or (trying to treat $t\ll 1$ separately) I am unable to achieve large decay. Of course, as previously mentioned, the issue at $t=0$ should be somehow resolved with the vanishing at $x=0$.
This comes out of trying to understand what the correct fix for a typo in M Zworski's Semiclassical Analysis p.32 (Screenshot,Google books link), in a proof that the Fourier Transform is an automorphism of $\mathcal S$. The online errata (screenshot copy) says to set $\phi(0)=0$, but I'm unsure if the preceeding paragraph is correct with this correction (and in any case, it is stated that $\psi$ may not be in $\mathcal S$ without proof.)
Thoughts appreciated!
Zworski, Maciej, Semiclassical analysis, Graduate Studies in Mathematics 138. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-8320-4/hbk). xii, 431 p. (2012). ZBL1252.58001.
Now I have come to the conclusion that it is not true:
Take $\phi(x,y) = -y e^{-(x^2+y^2)/2},$ which is in $\mathcal{S}(\mathbb{R}^2)$ and satisfies $\phi(0,0)=0.$ Then $\partial_x\phi(x,y) = xy e^{-(x^2+y^2)/2}$ so the $x$-component of $\psi$ is $$ \psi_x(x,y) = \int_0^1 t^2xy e^{-t^2(x^2+y^2)/2} \, dt = \{ s:= t\sqrt{x^2+y^2} \} \\ = xy\int_0^{\sqrt{x^2+y^2}} \frac{s^2}{x^2+y^2} e^{-s^2/2} \, \frac{ds}{\sqrt{x^2+y^2}} \\ = \frac{xy}{(x^2+y^2)^{3/2}}\int_0^{\sqrt{x^2+y^2}} s^2 e^{-s^2/2} \, ds . $$ Now study $\psi_x(x,y)$ along $(x,y)=(r,r),$ i.e. $$ \psi_x(r,r) = \frac{2^{-3/2}}{r} \int_0^{\sqrt{2}r} s^2 e^{-s^2/2} \, ds . $$ As $r\to\infty$ this will have the asymptotic behavior $$ \psi_x(r,r) \sim \frac{2^{-3/2}}{r} \int_0^{\infty} s^2 e^{-s^2/2} \, ds = \frac{2^{-3/2}}{r} \sqrt{\frac{\pi}{2}} $$ which does not tend to $0$ fast enough, so $\psi_x \not\in \mathcal{S}(\mathbb{R}^2).$